THE ULTIMATE REVISION VIDEO - HIGHER GCSE | GCSE 2025

Summary

Highlights

• A giant 106-question paper is used to revise almost every topic in the GCSE Higher course 📄
• The video mixes worked solutions with revision advice, not just answers 🧠
• Lots of classic exam areas show up, including factorising, simultaneous equations, and trigonometry 📐
• There are plenty of graph questions, from straight lines to quadratics and transformations 📈
• The video repeatedly reminds you that good revision means doing questions yourself first, not just watching solutions 👀

Key Takeaways

• Practice first, then check solutions so you can target the topics you actually struggle with 🎯
• Higher GCSE maths covers a massive range of skills, from algebra to geometry to statistics 📚
• Knowing key formulas and calculator methods can save loads of time in the exam ⏱️
• You’ll often need to move between representations like graphs, equations, tables, and diagrams 🔄
• Always watch out for little exam details like whether a question is non-calculator or needs exact values ✍️

Overview

Chapters

• 00:00 - 40:00: Algebra Foundations: Simplifying, Factorising, Indices and Fractions The video introduces a large GCSE higher maths past-paper solution set and explains how to use it effectively: try questions first, then mark and revise weaknesses. It then begins the algebra foundations section with expanding and simplifying an expression, factorising a term by taking out common factors, and writing 92 as a product of prime factors using a factor tree.
• 40:00 - 80:00: Number Skills: Percentages, Fractions, Standard Form and Estimation Worked with vectors: multiplying and subtracting vectors, drawing the negative of a vector, and identifying a perpendicular vector on a grid.
• 80:00 - 120:00: Graphs, Coordinates and Linear Relationships Calculated the area of a trapezium using 1/2(a+b)h, expanded and rearranged the resulting equation, factorized to find x values, and rejected the invalid negative-length solution to conclude x = 6.
• 120:00 - 160:00: Geometry and Mensuration: Constructions, Area, Volume and Transformations The chapter covered a wide range of GCSE-style geometry and mensuration topics, beginning with comparing approximations and ordering surds/powers, then moving into counting possibilities with multiplication principles and sketching the cosine graph over 0° to 360° using key values at 0°, 90°, 180°, 270° and 360°.
• 160:00 - 200:00: Trigonometry, Vectors and Proportional Reasoning The chapter begins with reading and interpreting a histogram to estimate frequencies from frequency density, then uses the total frequency to locate quartiles and calculate the interquartile range, demonstrating how to convert histogram areas into counts and how to estimate quartiles by cumulative frequency positions.
• 200:00 - 220:00: Advanced Algebra, Proof and Statistics Problem Solving Solved coordinate-pair and graph transformation questions by identifying solution points, reflecting a function in the x-axis, and applying a translation vector to a quadratic. Expanded the transformed function and simplified it to obtain the final equation.

THE ULTIMATE REVISION VIDEO - HIGHER GCSE | GCSE 2025 Transcription

• Segment 1: 00:00 - 02:30 if you're preparing for GCSE higher maths then this is the video that's going to make the biggest difference to your grades I've written a huge past paper with 106 different questions and it covers every single topic in the course and in this video I'm going to go through the solutions to all of those questions you can find a link to this paper in the video's description the best way to approach this is to try each of the questions yourself first and then come back to the video to mark them you'll find that you get some of them right and some of them you get wrong and it's those that you can revise in more detail I've even put a checklist of all of the topics at the front of the paper so you can keep track of your progress to complete the paper fully you're going to need all of these bits of equipment a pencil a ruler a pen a pair of compasses and a scientific calculator when you go through the paper you'll see this symbol if you do see the symbol by a question then this is a paper one Topic in which case you shouldn't use a calculator for that question and before we begin a brief disclaimer just watching this video is not enough to guarantee you're going to pass your exams there are many videos on on YouTube with people claiming you can pass your exams with a couple of hours of preparation but this is just not the case if this were the case we just shut down schools and everybody could learn everything from YouTube this is of course an incredible resource but you need to practice the topics that you struggle in in more detail after in addition whilst all of the topics are featured in this video not all of the variations are most of the topics in the GCSE Ma's course can be asked in a variety of different ways I've only included one example of each Topic in this paper otherwise it would be ridiculously long if you want to
• Segment 2: 00:00 - 02:30 practice any of the topics in more detail afterwards I'd recommend checking out the other videos on my channel and if you've done all of these the next best way to revise is doing past papers if you head over to my website FirstClass math.com you'll be able to find all of the past papers on there for you now that we've got all of that out the way it's time to go through the answers to the questions whilst watching remember that these are video Solutions I'm going to explain what I'm doing as I'm going along but in some cases if you've never seen the topic before it may seem a bit too fast if that's the case then definitely check out the other videos on my channel to learn those Topics in more detail in this first question we've been asked to expand and simplify so we're going to start with the expanding part and let's start with the first bracket here to expand this we need to multiply 7 by X which is 7 x and then 7 by -3 which is -21 Now we move on to the second bracket so here we've got -2 lots of x - 10 so we do -2 * X which is -2X and then -2 * -10 which is POS 20 so that's the expanding part of the question done now we need to simplify
• Segment 3: 02:30 - 05:00 you can see we have two terms in X here 7 x and - 2x 7 x take 2x is 5X and then we have a -21 and + 20 - 21 + 20 is -1 so the answer to this first question is 5x - 1 the second part of the question we need to factorize to do this we're going to put brackets back in So if we take the question but first of all look at the coefficients we're looking for a number that goes into 8 and 10 the largest number that we could pick here is two so we're going to factorize out a two then we look for each of the letters here we have X2 and x and we want to take out as many of those as possible but since there's only one X in the second term we're just going to take out X and then we've got Y and Y cubed once again we want to factorize out as many of these y's as possible but since there's only one in the first one we can only take out one y then we write our brackets and we need to work out what goes inside to work out what goes inside the brackets we need to think about what we would multiply 2x y by to get 8 x^2 Y and then also - 10 X Y Cub so let's start with the first term we want to get to an 8 but we've already factorized out of two so we must need to multiply by four then we're looking to get X squ but we've already got one of the X's so we need one more and then we're looking to get y but we've already factorized out the Y so we don't need anything else so 2x y * 4X will give you 8x2 y now we'll move on to the second term we need a negative and to get from 2 to 10 you need need to multiply by five then we're looking to get X but we've already factorized out an X so if we move on to the Y's we need y to the^ 3 and we've already got one y there so we need y^2 so this is the answer to the question in this question we need to write 92 as a product of its prime factors to do this we take the number in the question 92 and we think of two
• Segment 4: 02:30 - 05:00 numbers that multiply to make it I'm going to go for 46 * 2 if any of these numbers are prime numbers we Circle those two is a prime number so we can Circle that one but 46 is not this means we can split 46 once again I'm going to go for 23 * 2 now this time they're both prime numbers two is prime but also 23 is prime so we Circle both of those now we just take all of the numbers in the circles and multiply them together so we've got 2 * 2 * 23 since we got 2 * 2 here we can write that in an index form as 2^ 2 so the answer to this question would be 2^ s *
• Segment 5: 05:00 - 07:30 23 in this question we have Chloe who buys a phone for £120 she then sells it for £138 and we need to work out the percentage profit to do this we work out the profit first of all you can work out the profit by doing 138 take away 120 which gives you £ 18 so Chloe has made £18 in profit but the question wants us to find the percentage profit to do this we can use the following formula we do the change divide by the original multipli by 100 so the change is referring to the profit in the question that was 18 divide by the original price that's £120 and then multiplied by 100 you can type this into a calculator and you'll get the answer 15 so the answer to the question is Chloe made 15% profit for the first part of this question we're going to convert the mixed numbers into improper fractions to do this we multip multiply the four by the 2 and then add the 1 4 * 2 is 8 + 1 is 9 so this could be written as 9 / 2 we keep the denominator the same now we can do the same with the second fraction so this time is 1 * 4 then + 3 1 * 4 is 4 + 3 is 7 so this fraction is the same as 7 over 4 so the question could be Rewritten as 9/2 IDE 7 over4 when you divide fractions you can just multiply by the re reciprocal of the second fraction so if we keep the 9/2 the same change the Divide to a multiply but take the reciprocal of the second fraction that's a bit like flipping it upside down so 4 over 7 instead of 7 over 4 now we can just multiply these we do 9 * 4 that's 36 and 2 * 7 that's 14 you could simplify this fraction by dividing them both by 2 36 / 2 is 18 and 14 / 2 is 7 finally we need to convert this to a mixed number like it asks us to do in the question so we need to work out how many sevens go into 18 that's two and then how many would be left over or two sevens are 14 so there must be four left over to get to 18 so it's four and we
• Segment 6: 05:00 - 07:30 keep the denominator the same still at seven so the answer to this one is 2 and 47th for Part B of this question we're also going to convert them into improper fractions first so we do 3 * 3 + 2 which is 11 so this is the same as 11 over 3 and for the second one 1 * 5 + 2 which
• Segment 7: 07:30 - 10:00 is 7 so that's the same as 7 over 5 so the question could be Rewritten as 11 over3 subtract 7 over 5 now to subtract fractions you need a common denominator so we're going to look for a common denominator for three and five the lowest common multiple of those is 15 so we're going to rewrite both of these as something over 15 now we need to work out the numerators for the first fraction the three has turned into a 15 so we've multiplied it by five so we need to also so multiply the 11 on the top by 5 11 * 5 is 55 for the second fraction the five on the bottom has become a 15 so we've multiplied by three so on the top we need to multiply by three as well 7 * 3 is 21 now we can subtract these fractions we do 55 take away 21 which is 34 and it's over 15 still once again we need to give this as a mixed number so how many 15s go into 34 well that's two and there are four remainder so four over 15 in this question we've been asked to make B the subject of the formula if we look at the formula and its current state R is the subject because we can see that R is on its own on the left hand side if we look at the right hand side there are two other things with the B we've got the N9 and we've got the minus P so what we're going to do is first of all add P to both sides if we add P to the right hand side that will cancel the negative P that's there so we'll just be left with 9B on the left hand side of if we add P we have R and then we add P to that that's just r+ P now on the right hand side we have 9B this means 9 multiplied by B so we're going to divide by 9 on both sides if we divide by 9 on the right hand side the nines will cancel and which just left with b and on the left hand side we've got r+ P but we're going to divide that by 9 we would just write that as r+ P
• Segment 8: 07:30 - 10:00 divide by 9 we can now see that b is the subject because it's on its own but we would normally write the subject on the left hand side though so if we just flip this round we end up with b = r + P / 9 for the first part of this question we need to solve this inequality to solve an inequality you do exactly as you would for solving an equation but instead of writing an equal symbol you write the inequality symbol so let's begin by adding 2x to both sides of this if we add 2x on the left hand side we have 3x + 2x which is 5X and then we also have this + 10
• Segment 9: 10:00 - 12:30 if we add 2x to the right hand side that will cancel thetive - 2x that's already there so we're just left with five so we write less than or equal to 5 now we can subtract 10 from both sides if we do this on the left the 10 will cancel so we have 5x and on the right five take away 10 is -5 finally we can divide both sides by 5 5x / 5 is just 1 x and then -5 / 5 is -1 so the solution to this one X is less than or equal to -1 for the second part of the question we need to represent the inequality we just found on the number line shown so that's this inequality here to do this we're first of all going to draw a circle at the number given in the inequality so -1 then we need to decide which side the arrow is going to go the numbers we're interested in are less than or equal to netive 1 so we draw an arrow going off to the left make sure it goes all the way to the end of the number line finally we need to decide if we're going to shade in the circle or not If the symbol in the inequality is a less than or equal to or a greater than or equal to you need to shade in the circle since L1 is less than or equal to we would shade in the circle if however the inequality symbol was a regular less than without the equal to part or a regular greater than without the equal to part you would leave the circle open for the final part of this question we need to write down all of the possible values of P for this inequality here think of this inequality as having three sections to it we got the left hand side that's 9 we got the right hand side that's -3 and then in the middle we've got 2 pus 5 now whatever we do here we're going to do it to all three of those sections we're going to start by adding five so if we write the inequality symbols as they are on the left we've got 9 add five that's -4 in the middle if we add five the neg five will cancel so it's just 2 p and on the
• Segment 10: 10:00 - 12:30 right -3 add 5 is 2 now we can divide by two on all three of those sections so if we write the inequality symbols once again and on the left -4 / 2 is -2 2 p / 2 is just 1 p and then 2/ 2 is 1 so we end up with this inequality here which is that P is in between -2 and 1 but we've been told in the question that P is an integer which is a whole number so all of the whole numbers between -2 and 1 are2 -1 0 and 1 however this inequality symbol here doesn't say less than or equal to
• Segment 11: 12:30 - 15:00 it just says less than this means that P can't actually be equal to one so we need to remove one from the list so the final answer to this question is -2 -1 and 0 for the first part of this question we need to find the reciprocal of 1.25 to find the reciprocal of any number you can just do one over that number so we need to do 1 over 1.25 this is just the same as 1 divide by 1.25 which you can type into your calculator it will give you the answer 0.8 for the second part of the question we need to write down an error interval to do this I like to draw out a number line I always place the number from the question so 6.4 in the center of that number line now we've been told that n has been rounded to one decimal place this means we need to find the number that's one decimal place below and above 6.4 one decimal place below is 6.3 and one decimal place above is 6.5 now we can find the error interval by looking at this region here the bounds of which are halfway in between 6.3 and 6.4 and also halfway in between 6.4 and 6.5 so halfway between 6.3 and 6.4 that's this one here 6.35 and halfway in between 6.4 and 6.5 is 6.45 we can now use those two numbers to complete the answer to the question it's 6.35 and 6.45 for the first part of this question we need to write down the modal class the modal class is the one with the highest frequency you can see from the table the highest frequency is 28 so the class that corresponds to 28 is 10 to 20 for the next part of the question we need to write down the interval that contains the median we're told in the question there are 60 students so if you do 60 + 1 which is 61 and then divide this by two so 61 divide by 2 you get 30.5 this tells you the person that's in the middle is the 30.5 person now
• Segment 12: 12:30 - 15:00 obviously you can't have half a person but for the purposes of this question it won't matter so we need to look down the table and count up all the frequencies until we go past 30.5 so in the first group we've got 28 then if we include the next group that's another 13 people that takes us to 41 you can see we must have now gone past the 30.5 person so the interval is the second one 20 to 30
• Segment 13: 15:00 - 17:30 for the final part of the question we need to write down an estimate for the meantime spent revising to do this we're going to do another copy of the table first of all we write down all of the midpoints of all of the groups so the first group from 10 to 20 the midpoint there is 15 the next group 20 to 30 is 25 and then 35 and 45 you can now multiply all of these midpoints by the frequencies so 15 * 28 that's 420 25 * 13 that's 3 25 then we get 4555 and 270 if we now add up this final column here we end up with 1,470 this represents an estimate for the total time spent revising for all of the students but we want the mean time spent revising so we need to divide it by how many students there are which is the sum of this column which is 60 so if we do 1,470 divide by 60 we get the answer to the mean which is 24.5 in this question we need to type this into our calculator we're going to start by pressing the fraction button then it's 1.8 then we need to do cubed you can find that near the top of this calculator and then we need to get to the bottom of the fraction so I'm going to press the down arrow Here and Now on the bottom I need square root which is this button over here and then 17 and then I need to go outside of the square root so I press the right arrow and then subtract two then we just press equals and we end up with this number here and the question does say to write down the whole calculator display so we're going to write down the whole display which is this number here for the second part of the question we need to round the answer to the previous part to three significant figures so if we take the number here the two at the front of the number is the first significant figure this is the second and this is the third so this is how far we need to go when we round this number so if we draw a line after that four we
• Segment 14: 15:00 - 17:30 need to check this six to tell us if we round up or down and since six is above five we're going to round this one up so the four is going to round up to a five so the answer would be 2.75 in this question we need to use a ruler encompasses to construct the line BP that bisects the angle ABC this is asking us to do an angle bis sector to do an angle bisect you're going to need a pair of compasses you take the pair of compasses and you place the sharp point right at the point of the angle on B like this you then draw yourself an arc that goes across both of the lines of the angle so like
• Segment 15: 17:30 - 20:00 this this Arc will cross the lines in two places here and here we're now going to draw one Arc at each of these crosses making sure to keep the pair of compasses the same size when we switch so first of all place the pair of compasses at this cross and draw yourself an arc like this and then making sure they stay the same size place them here on this cross and draw an ark like this where these two arcs intersect will be the point P you can now connect up B to P to form the line BP which is the completed angle bis sector in this question we've been asked to draw the graph of y = 3x + 2 to do this we're going to draw a table of values so we have X and Y now the X values are actually given to us we're told in the question it needs to go from -2 to 3 so let's put those into the table -2 -1 0 1 2 3 now we need to work out the Y values but we're going to use this formula here to get any of the Y values we're going to do three lots of the x value plus 2 now I like to start with x equal 0 because that's an easy one so when x equal 0 y will be three lots of 0 + 2 which is just two so when xal 0 y equal 2 we can add a two into the table now let's move on to when X is 1 we do 3 lots of 1 + 2 which is five then when x = 2 we do 3 lots of 2+ + 2 which is 8 and then for x = 3 3 of 3 + 2 which is 11 then we move on to the negative 1es we need to do 3 lots of - 1 + 2 which is -1 and finally 3 lots of -2 + 2 is -4 now some people have some difficulty with the negative ones here and since this is a straight line graph you may notice that we're just subtracting three each time so once you've done the positive numbers here you could just do two take away three which gives you NE -1 and then1 takeway 3 is -4 now we need to plot each of the coordinates so the first one is at -2 -4 that goes here the next one is it negative 1 negative 1
• Segment 16: 17:30 - 20:00 that goes here then we have 02 that goes here then 15 that's here 28 and 311 once You' plotted all of those you can join them up with a nice straight line it would look something like this for the second part of the question we have to decide if the point with coordinates 25 77 lies on the line so if
• Segment 17: 20:00 - 22:30 we take the coordinates 2577 the 25 is the x coordinate and the 77 is the y-coordinate so we just need to check that if we substitute in these numbers we end up with something that makes sense so let's replace x with 25 and see if we get 77 so remember it's 3x + 2 so three lots of 25 + 2 325 so 75 + 2 which is 77 which is the y-coordinate since this works the point is on the line so we could say yes in this question we need to work out an estimate for this calculation this means we don't need to do it exactly we just need to know roughly and we can do this by rounding each of the numbers to one significant figure let's start with the 699 to one significant figure that's 700 then we move on to 32 to one significant figure that's 30 and finally the 0.21 on the bottom term would round to 0.2 so we just need to work this out to work this out we can start with the top 700 * 30 well we can do 7 * 3 that's 21 and then we need three more zeros and on the bottom we just have 0.2 some people find this a little bit tricky to divide by 0.2 you're better off multiplying both the top and bottom of this by 10 if we multiply the top by 10 we get 210,000 and if we multiply the Bottom by 10 0.2 * 10 is just 2 this calculation gives the same answer so we can do 210,000 IDE 2 which is 105,000 and the answer to the question in this question we're told the circumference of the circle is 20% greater than the perimeter of the triangle we're also told that the triangle is an equilateral triangle which means all of the sides are the same length so we know that this is eight and this is eight as well so the perimeter of the triangle is not too hard to find we just add up all of those sides 8 + 8 + 8 is 24 now if we've been told the circumference is 20% greater G
• Segment 18: 20:00 - 22:30 than this we just need to increase that number by 20% so if you do 100% plus 20% you get 120% and then take 120 and divide it by 100 you get 1.2 this is a multiplier for increasing by 20% so we can multiply the 24 by 1.2 which gives us 28.8 so this is the circumference of the circle but we've been asked to find the area of the circle to find the area of a circle you need to know its radius so the formula for the circumference of a circle is pi *
• Segment 19: 22:30 - 25:00 diameter now that we know the circumference we can use this formula to work out the diameter if the circumference is equal to Pi * the diameter we can divide the circumference by pi to get the diameter so if we take the circumference which we know is 28.8 and divide it by pi we get this number here which must be the diameter of the circle the diameter is always twice the radius so if we divide this by two we end up with the radius which is this number here now that we have the radius we can work out the area using the area of a circle formula to work out the area of a circle use area = < * R 2 so we just need to do PI * the radius squar that we found 4.58 and so on and if you type this into your calculator you get this number here the question says to round it to three significant figures so if we round it we'll end up with 66.0 for this question we need to find the number of triangular tiles in pattern n this just means we need to find an M term of a sequence where the sequence is just the number of triangles in each of the patterns so let's write down that sequence for pattern number one there are six triangles in the second pattern there are eight and the third pattern if you count them all you get 10 we just need to find the MTH term of this sequence to get from 6 to 8 you add two and also to get from 8 to 10 you add two so it's a linear sequence of the form 2 N since it goes up by two each time but 2 N just describes the two times table so if we just write the two times table above the sequence 2 4 6 and so on we need to work out how to get from the blue numbers to the red numbers now in all cases to get from 2 to 6 4 to 8 6 to 10 and so on you're going to add four this means the MTH term of this sequence is 2 N which is the blue sequence plus 4 which gives you the red sequence in the second part of the question we're told that rich is going to make one of the patterns from the
• Segment 20: 22:30 - 25:00 sequence but he's going to use 88 total tiles this means those tiles could be square or triangular we need to work out how many square tiles were it's used so let's now write out a sequence which isn't the number of triangular tiles but the number of total tiles in pattern number one you've got six triangles and two squares so that's eight tiles in pattern number two you have four squares eight triangles that's 12 tiles total and if you count up all of those tiles in pattern number three it's 16 this is once again a linear sequence this goes up by four each time so it must be a four end sequence if we write the four
• Segment 21: 25:00 - 27:30 times table above the sequence 4 8 12 and so on we can see that to get from the blue sequence to the red sequence you must add four again so this one is 4 n + 4 now we know that rich is going to use 88 total tiles so if we say 4 n + 4 equals 88 we can find out which pattern number he's making this is just an equation to solve if you subtract four from both sides you get 4N = 84 then divide both sides by four you find that n is equal to 21 this means rich is making pattern number 21 but the question said how many square tiles is he using well if you count up the number of square tiles in each pattern and compare it to the pattern number you'll see it's just double the first pattern pattern number one has two pattern number two has four pattern number three has six and so on so if we've got pattern number 21 we just need to double 21 21 * 2 is 42 in this question we're told the number of people visiting a cinema on Saturday was 20% more than it was on Friday so from Friday if we add 20% we get the number for Saturday then we're told the number of people who visited on a Saturday that's 9,840 so let's write that underneath Saturday and we need to work out the number that visited on Friday so let's call that number X since we don't know what it is yet so if we take X then add 20% we need to get to 9,840 to increase a number by 20% you multiply it by its multiplier 1.2 so we end up with x * by 1.2 = 9840 to solve this you can divide both sides by 1.2 on the left this will give you X and on the right 9,840 IDE by 1.2 you can use your calculator for this and it will give you the answer 8,200 which is the answer to the question for the first part of this question we need to simplify a fraction let's start
• Segment 22: 25:00 - 27:30 with the numbers so we can do 15 / 3 quite easily that's five then if we move on to the letters we got X to 6 / X when you divide like this you're going to subtract the powers the power of X on the bottom is 1 so we do six take away 1 which is five so x 6 / X is x ^ 5 then if we look at the Y's we've got y 8 / Y 2 again we'll subtract the powers 8 take away 2 is 6 so it's y ^
• Segment 23: 27:30 - 30:00 6 for the second part of the question we have 2 y cubed all in a bracket to the^ five if you raise something to the power five you times it by itself five times so we've got what's inside the bracket 2 y cubed times by itself five times so 2 y Cub * 2 y Cub * 2 y cub and again and again you can see we've got that five times so we just need to times all of these together we'll start start by multiplying all of the twos 2 * 2 * 2 * 2 * 2 is 32 then we look at the Y we've got y 3 * by itself five times here we can just add up all of those threes 3 add 3 add 3 add 3 add three or 3 * 5 which is 15 so it's y ^ 15 for the first part of this question we need to expand and simplify a double bracket to do this we multiply everything in the first bracket by everything in the second bracket so let's start with x * X X which is x^2 then x * 8 which is-8 X then -5 * X which is -5x and finally -5 * 8 is a postive 40 we can then simplify these middle two x terms here 8X subtract 5x is -3x so we end up with the answer x^2 subtract 13x + 40 in the second part of the question we've been asked to factorize this means we need to put it back into its bracketed form this is a a particular type of factorization called the difference of two squares because 4x^2 is a square and 9 is a square and it's the difference because we've got a subtraction you should notice that 2x * 2x is 4x^2 and 3 * 3 is 9 this means our brackets must be made up of a 2X and a three and when it's a difference of two squares one of them is positive and one of them is negative so we end up with 2x + 3 and 2x - 3 for the final part of this question we need to solve a quadratic equation this one we can do by factorizing we're going to look at the final term here8 we need to think of two numbers that would multiply to give ne8 you could have 1 * 8 or 2 * -4 but you could also flip the signs around here
• Segment 24: 27:30 - 30:00 and have 8 * 1 and 4 * -2 all of those pairs of numbers will multiply to give 8 but only one of those is going to add to give two we need to get 2x here so if we add all of these together the first one 1 at8 is is -7 2 add4 is -2 8 add -1 is
• Segment 25: 30:00 - 32:30 7 and 4 add -2 is 2 so the one we're interested in is the bottom one here you can see it was 4 and -2 which means the brackets will be x + 4 x subtract 2 now the question didn't say factorize otherwise we'd be finished it said solve and you can see it says equal Z over here so this one also equals zero now what we have here is two brackets multiplied together to give zero if you're going to multiply two things together to get zero then at least one of them must be zero so either the first bracket zero or the second bracket zero this just gives us two equations to solve for the first one you subtract four from both sides and you'll get x = -4 and the second one you add two to both sides and you get xal 2 in this question we've been told that we have a floor that has an area of 10 squ m and it's going to be tiled by three workers in 8 hours but we need to tile a bigger floor it's going to be 25 M squ and we're also going to have more workers so four workers for the moment let's stick with the 10 s m floor we've been told that if it's 10 square m we can tile it with three workers in 8 hours if you multiply 3 * 8 this tells you it takes 24 worker hours to tile that floor now let's think about how long it would take to tile the same floor of the same size with an extra worker so we're four workers how many hours would that take well we know it takes 24 worker hours to complete the tiling of the floor but we have four workers so they're going to share out that time equally so if we do 24 divide by 4 we end up with six so it would take 6 hours for four workers to do the same size floor and it makes sense that it would take 6 hours since you have more workers it would take less time but the question didn't want us to do an area of 10 squ m still it's a bigger area of 25 squ M but how many times bigger is is that well if you do 25 divide by 10 you find out this floor
• Segment 26: 30:00 - 32:30 is 2.5 times bigger if the floor is 2.5 times bigger it will take 2.5 times as long so we found out earlier it was going to take four workers 6 hours to do the smaller floor so we just need to multiply this 6 by 2.5 to see how long it would take to do the larger floor 6 * 2.5 is 15 which is the answer to the question 15 hours in this question we have a lry that takes a d direct 60 M route from A to B traveling at an average speed of 50
• Segment 27: 32:30 - 35:00 mph so it's going along this route here and its speed is 50 Mi per hour we've then got a car that's going from A to B as well but it's going through Point C along the Route so the car will start at a go to C and then go from C to B we're also given information about the speed of the car between a and C so the blue part of the route for that part it's going at 60 M hour as its average speed then we're told that both vehicles to part A at the same time and arrive at B at the same time and we need to work out the average speed of the car between C and B now if you look at the red section and the blue section of this diagram you can see we have both of the distances and both of the speeds so we can use the formula time equals Distance divid by speed to work out the time for each of those sections let's do the red one first so the distance for the red one is 60 Mi and we're going to divide it by its average speed which is 50 miles hour 60 divide by 50 is 1.2 hours then we'll do the same from a to c the blue section so this time the distance is 45 and the speed is 60 45 divid 60 is 0.75 hours now since we know the car and the Lorry take the same amount of total time to complete their Journeys we can subtract these two to find out how long the purple section is from C to B so if we do 1.2 hours takeway 0.75 hours we get 0.45 hours that means the car must take 0.45 5 hours to go from C to B now that we know the time it takes and we know its distance we can work out the speed we use the formula speed equals distance ID by time the distance in this one is 18 and the time we just worked that out is 0.45 hours 18 / 0.45 is 40 mph and that's the answer to the question in the first part of this question we're looking for the point that's the outlier the outlier is the point that doesn't fit with the trend for the rest of the points so that's this one here the coordinates of this
• Segment 28: 32:30 - 35:00 outlier are 32 and 620 for all of the other points we need to write down the type of correlation since this data has an upward Trend we'd say it has a positive correlation for the final part of the question we have a person who is aged 55 and we need to use the graph to estimate their reaction time to do this you need to draw a line of best fit a line of best fit is a line that that goes through the general trend of the points we've been given so it would look
• Segment 29: 35:00 - 37:30 something like this now everyone would draw a slightly different line of best fit but there are a few key features to look out for lots of people tend to start their line of best fit right in the bottom left corner but in fact in most cases it probably won't do secondly a good role of f is to try and get the same amount of crossair above and below the line you can see on mine they're roughly equally spread above and below the line finally you need to make sure it's following the general trend of the point remember we said this was a positive correlation you can see the points go in an upward Trend so our line must go upwards too now we can use this line to find out the reaction time of that person who was age 55 so if we go to 55 on the bottom axis where it's the age and go up to the line of best fit and then read off the corresponding reaction time you can see that comes in at around 560 milliseconds now you may get a slightly different answer depending on your line of best fit for the first part of this question we need to describe the single transformation that Maps triangle a onto triangle B now there are four different types of transformation you need to know they are reflection rotation translation and enlargement now we can see that shape a to shape B here has just been moved and we call that a translation so you would write down translation to get the first Mark when describing a translation you need to give a vector to describe how it's been moved so if we take the shape a and see how do we move it to get to shape B well we go one square to the right two squares to the right and three squares to the right so we could say it's a translation by the vector and then we draw a vector and since it went three to the right we put a three at the top and since it didn't go up or down at all we put a zero at the bottom for the second part of this
• Segment 30: 35:00 - 37:30 question we need to do a rotation it's going to be 90° anticlockwise about the 0.10 first of all take a piece of tracing paper and Trace the shape then you're going to Mark a cross at the point given so 1 Z that's here then place your pen on that cross and turn the tracing paper 90° anticlockwise that's one4 of a turn to the left so like this you can then draw the shape in this place and remove the tracing paper it says to label at Triangle C so we'll do that as well for the final part of this question we've been asked to do a reflection of triangle a in the line yal 2 the line yal 2 is a horizontal line that cuts the y axis at the point two so this one here we can now reflect triangle a in that line it will just flip down below that line since it's touching the line and look something like this and the
• Segment 31: 37:30 - 40:00 question asks us to label this triangle D in this question we've been asked to work out the perimeter of the kite to do its perimeter we need to know the length of all of its sides you can see we know two of them but we don't know the other two so let's start by trying to work out the length of this one here from D to C if we just look at this triangle here we can see we have a right angle in it so it's a right angle triangle this means we can use Pythagoras to find the missing length also notice in the question though it says to give our answer in centim we've got one of the sides is 78 CM but the other one's in meters so let's convert the 1.3 M into 130 CM now we're ready to use Pythagoras so if we Square both of those sides 130 squar is this and 78 squ is this obviously use your calculator for those then we need to decide if we're going to add or subtract those sides since this side here the one we're finding is not the hypotenuse we're going to subtract them so if we subtract those numbers we end up with this now to find the ANW for the length of DC we need to square root that and we end up with the answer 104 so from D to C is 104 but since it's a kite that's also the distance from B to C now we know all of the sides of the kite we can just add those together we've got 78 add 78 add 104 add 104 and this gives you the answer 364 cm in this question we're told that Craig spends 30 minutes longer revising chemistry than he spends revising physics if we look at the pie chart you can see the angle for biology is 90° that's a right angle so we can work out the angle for physics as well since all three of the angles need to add up to 360 so if we do 90 + 160 we get 250 and if we subtract that from 360 we get 110 so the angle for physics must be 110 now if Craig is going to spend 30 more minutes on chemistry than he is on
• Segment 32: 37:30 - 40:00 physics that 30 minutes must be represented by the difference in their angles if you subtract the angles 160 take 110 you get 50° this means in our pie chart 50° represents 30 minutes but we need to work out how long Craig is going to spend revising biology and in the pie chart that's 90° so we need to work out how we can get from 50° to 90° I'm going to work out how we get to 10° first to get from 50° to 10° I would divide by five so if I divide 30 minutes by 5 I get 6 minutes this means every 10 10° of the pie chart represents 6 minutes now how can I get from 10° to
• Segment 33: 40:00 - 42:30 90° well that's just Times by 9 so 6 * 9 is 54 minutes so the answer to this question is 54 minutes in this question we need to work out the vector 2 a subtract B using the vectors A and B we've been given so we need two lots of the vector a which is -65 and we're going to subtract the vector B which is 2 -2 to work this out we're first of all going to work out what two lots of vector a is to multiply a vector by two you just double the numbers in the vector so instead of -6 we've got -12 and instead of 5 we've got 10 so it's -12 10 subtract Vector B which is 2 -2 now we just have two calculations to do along the top we've got -12 subtract 2 if you do -12 subtract 2 you get -14 on the bottom we've got 10 subtract -2 When you subtract a negative number that's just the same as adding the positive version so it's the same as 10 add 2 which is 12 so the answer to this question is -42 for Part B of the question we need to draw and label the vector Nega a the vector negative a is the same as Vector a apart from we switched the signs so instead of being -65 it's 65 now when we draw the picture of this we need to consider what direction that Vector is telling us to go the top number tells you to move left or right and since it's positive it's right so it's six to the right the bottom number is for up and down and since it's negative we're going to go down so five down so we just pick a point on this grid and then go six right and then five down and this is the final point we can draw a line that connects these two together and we put an arrow on that line to indicate the direction the vector goes for the final part of the question we need to draw another Vector but it needs to be perpendicular to Vector B vector B is this one here that means to move two right and two down so if we
• Segment 34: 40:00 - 42:30 were to draw Vector B it would look like this two to the right two down and an arrow to indicate the direction but we don't need to draw Vector B we need to draw one that's perpendicular to it a vector that's perpendicular to this will cross it at 90° so we need to draw any Vector that goes in this direction instead it doesn't actually matter how long that Vector is as long as it goes in this direction it could be this long this long it doesn't matter in this question we've been asked to
• Segment 35: 42:30 - 45:00 work out the maximum number of Pancakes that can be made with these ingredients we've been told the recipe for 12 pancakes in the question if we look at the first ingredient flour for 12 pancakes you'll need 300 G of flour but we have more flour than this available we have 1,500 G so we can definitely make more than 12 pancakes but the question is how many well if we do 1,500 divide by 300 we end up with five this means we could repeat the recipe for 12 pancakes five times with all that flour that we have how would this work for milk well for 12 pancakes we need 400 milliliters but again we have more 1,800 milliliters so if we do 1,800 divide by 400 we end up with 4.5 so we can repeat this recipe for 12 pancakes 4 and 1 half times if we do the same for eggs we need two eggs we've got 11 eggs if you do 11 divide by 2 you get 5.5 this means we have enough eggs to repeat this recipe 5 and 1 half times now the most important of these threee numbers is the smallest one 4.5 since even though we have enough flour to repeat this recipe five times and eggs to do it five and a half times that doesn't matter because we would have run out of milk so the lowest one is the most important one and that's the one we're going to need to use so if we remember we're trying to work out the maximum number of Pancakes and we know we can repeat this recipe for 12 pancakes 4 and 1/2 times so if we do 12 * 4.5 we end up with 54 which which is the maximum number of Pancakes in this question we've been asked to write some numbers in order of size and we need to start with the smallest one to compare these numbers we're going to convert them all to decimals so let's start with the first one here 8.6 * 10 ^ 4 timesing a number by 10 ^ 4 is timesing it by 10 four times so let's take 8.6 and times it by 10 once and a second time a third time and a fourth time so this number is 86,000 the next one here 0.086 is
• Segment 36: 42:30 - 45:00 already a decimal and if we move on to the next one we have 86 * 10^ 2 that's timesing by 10 twice so if we Times by 10 once and twice we end up with 8,600 and for the final number it's 8,600 * 10 ^-4 this is like dividing by 10 four times so if we start with 8,600 and divide by 10 second time a third time and a fourth time we find this is equal to 0.86 6 now it's quite easy to see which one's the smallest that's
• Segment 37: 45:00 - 47:30 0.086 so let's put that onto the answer line the next one is 0.86 but remember in the question that was 8,600 * 10 ^-4 the next one is 8,600 that's 86 * 10^ 2 and the final one 86,000 which is 8.6 * 10 ^ 4 for this question we need to draw a frequency polygon whenever we do a frequency polygon we need the midpoints of all of the groups so for the first group from 60 to 70 the midpoint is 65 the next group from 70 to 80 that's 75 and we get 85 95 and 105 now what we need to do is plot at each of these midpoints the associated frequency so the first midpoint is 65 and the frequency is three so we go across to 65 on the bottom and go up to three we do the same with 75 and 12 which goes here and 85 and 15 that goes here 95 and 44 four that's all the way up here 105 and 26 that goes here you then connect up each of these points in order from left to right like this with straight lines and there's your completed frequency polygon in this question we've been given the universal set of numbers from 1 to 10 this means the only numbers we need to place inside this diagram are the numbers from 1 to 10 we've been told that set a is the multiples of three so this is the three times table but again we're only interested in the numbers that are from 1 to 10 10 so the multiples of three that are from 1 to 10 are 3 6 and 9 then we have set B which says the factors of 12 so we want to know any factors of 12 that are also from the numbers 1 to 10 so that's 1 2 3 4 and 6 notice we don't put the number 12 now we're going to go through each of these numbers and see which sets they're in to help us place them in the ven diagram let's start with the number one the number one is not in the set a it's not a multiple of three but it is in the set B the factors of 12 so it goes in inside the B Circle only this is exactly the same for the number two but then the number three is
• Segment 38: 45:00 - 47:30 in set a and set B so it goes in the overlap in the middle here then we move on to four that's only in set B again and then we hit five now five is not in set a or set B so it goes outside of the circles but still inside the rectangle then six that's in both of them so into the middle again then seven that's not in a or b so that goes in the outside then on to eight that's the same name is seven so in the outside again nine is in set a but not in set B so
• Segment 39: 47:30 - 50:00 that's just inside the a circle and then 10 which goes in the outside for the second part of the question we've been told we're going to select a number at random from the universal set and we need to find the probability it's in a then this symbol B this symbol is the union symbol sometimes we read this as a or b if we're to show the Union in a diagram it's both of the circles A and B combined so we need to count up how many numbers are in the union so there's 1 2 3 4 5 6 so there are six possible numbers that we could pick that are in the union and 10 possible numbers Al together so the probability of picking one that's in the union is 6 over 10 in this question we've been asked to multiply two numbers that have been given in standard form to do this we're going to multiply the number parts of them first so 8 * 3 that's 24 and then we multiply the P of 10 10 ^ 10 * 10 ^ 3 is 10^ 13 the reason for this is you add the powers 10 add 3 is 13 now the question does say to give your answer in standard form and this is not at the moment the number at the front there that 25 24 needs to be between 1 and 10 so we're going to divide that 24 by 10 which will give us 2.4 that is between 1 and 10 but we can't just divide this number by 10 that would change the number but if we also multiply it by 10 10 that would put it back so if we multiply the 10^ 13 by 10 we get an extra power of 10 so it's now 10 the^ 14 so both these two numbers are the same but the bottom one is written in standard form for the second part of the question we have a fraction to deal with we're going to just look at the numerator first 1 * 10 3 is 1,000 and 2 * 10 2 is 200 so the top is really saying 1,00 + 200 which is 1,200 we can write that in standard form that would be 1.2 * 10 ^ 3 so the numerator really looks like this and on the bottom we have 4.8 * 10 ^ -2 now
• Segment 40: 47:30 - 50:00 when you divide numbers in standard form you look at the number part again so we've got 1.2 / 4.8 now you should realize that 12 * 4 is 48 so this would actually simplify to give 1 qu or as a decimal 0.25 then we look at the powers of 10 10 3 / 10 2 when you divide powers of 10 you subtract the powers so we need to do three take away -2 if you subtract a
• Segment 41: 50:00 - 52:30 negative we actually need to add so it's 3 add 2 which is 5 so it's time 10 ^ 5 once again unfortunately this is not in standard form the first number is not above one so we're going to multiply this one by 10 this time which will give you 2.5 but again you can't just multiply numbers without consequence we also need to divide it by 10 to keep everything balanced if you divide this number by 10 you're reducing the amount of 10 you're multiplying by so it's now 10 ^ 4 in this question we're trying to find out which bottle is better value for money and we've been given this exchange rate 1 is equal to 1.17 with this exchange rate we can convert from pound to euros by multiplying by 1.17 or alternatively we can go from Euros back to pounds by dividing by 1.17 for this this question there are actually many different routes that can lead you to the right answer still I prefer to work in pounds so I'm going to convert the Euros from the bottle in Spain into pounds first so to do this I go from Euros to pound which is divide by 1.17 so if we take the 2.4 divide it by 1.17 we find the cost in pound which is2 and a bit more than 5 now that we've got both of them in pounds we need to work out which one's better value for money to work out which one better value for money I like to do the price in Pence and then divide this by the amount that we have so for the first bottle the one in the United Kingdom the price in Pence would be 299 p and I'm going to divide it by how many milliliters we have and 2 L is 2,000 milliliters I use my calculator at this point and I end up with this number here now I'm going to do exactly the same for the one in Spain so the price in pens which is a little bit more than 205 P divide by the amount that we have have which is 1,250
• Segment 42: 50:00 - 52:30 ml if you do this one in your calculator making sure to use the number 205 and all of the digits after it you end up with this this tells us that if we buy the one in the United Kingdom it costs us 01495 P for each milliliter of drink and the one in Spain is 0.641 P for each milliliter of drink so the one that's better value for money is the one that has a lower number here so that's the one in the United Kingdom to start this question we need to draw the front elevation of this shape that's
• Segment 43: 52:30 - 55:00 what you would see if you viewed the shape from the front you can see they've marked the front onto the question so we would see this face here so we just need to draw this face onto our grid we'll start on the left hand side you can see its height is 4 cm so let's draw a 4 cm line then going along the bottom we've got a 6 cm line horizontally so let's draw 6 cm line then going up the right hand side it's much smaller it's only 1 cm then we go to the left 2 cm and then we connect this point back up to the start now for the side elevation this is what you would see if you viewed the shape from the side which is also marked onto the diagram you would see this face here which is a rectangle which is 3 cm by 1 cm but you would also see this sloped face here now this one's a little more difficult the number 5 cm there may confuse some people the height of that is not 5 CM the total height of the shape if you look at the far left hand side is 4 cm so if we viewed the shape from this side the maximum height we would see would still be 4 cm so we just need to draw three more CM up and then complete the shape like this finally we're on to the plan the plan is what you would view if you saw the shape from above which would be these two faces here once again that five may cause some problems if you look at the horizontal width of the shape right at the bottom you can see it's 6 cm that would be the width of the rectangle you would see certainly not 5 + 2 which gives you 7 so we need to draw a rectangle which is six wide and then if you look at the 3 cm by the side that's going to be the height of this rectangle so it goes up three and then we complete the rectangle like this now Additionally you would see this line here which joins those two rectangular faces together that's 2 cm in from the right hand side so if we go in 2 cm we would draw this line here for Part B of this question we need to
• Segment 44: 52:30 - 55:00 work out the volume of the prism we're going to use the formula which is the volume of a prism is the area of its cross-section multiplied by its length the cross-section is the original shape we drew the front elevation here so we need to find the area of that shape it's actually quite easy to do this using the elevation we've drawn if we split the elevation into two shapes a rectangle and a triangle for the rectangle we can just count the squares there are six of those so it's 6 cm squar and for the triangle we can work work out the formula by doing the base time height and divide by two the base is four the
• Segment 45: 55:00 - 57:30 height of the triangle is three 4 * 3 is 12 and half of that is also 6 so the total area of that cross-section is 6 and 6 which is 12 now we multiply this area of the cross-section by the length of the prism which is 3 cm so we do 12 * 3 which is 36 and that's the answer to the question in this question we have to solve some simultaneous equations if we look at the equations in their current form then none of the corresponding coefficients are matching so we're going to need to scale the equations I'm going to try and make the X's match by multiplying the first equation by four and the second equation by five so for the first equation we do 5x * 4 which is 20x 3 y * 4 which is 12 Y and 5 * 4 which is 20 for the second equation it's 4X * 5 which is 20x x -2 y * 5 -10 Y and -7 * 5 -35 you can now see the coefficients of the X terms are matching since they're both 20 since the signs of those are both the same they're both positive we subtract the equations 20 x take away 20x is 0er 12 y subtract -10 Y is 22 y since you need to do 12 + 10 when you're subtracting a negative remember it's like adding and then 20 take away -35 once again we're subtracting a negative here so we need to do 20 + 35 which is 55 to get y you can just divide both sides by 22 so y = 55 / 22 this fraction will simplify nicely since they're both multiples of 11 it would give you 5 over 2 which is the same as 2.5 so now we found y we want to try and find X we need to substitute this y value back into one of the original equations I'm going to go for this first one here so I'm going to copy out that equation but whenever I get to a y I'm going to write 2.5 so we have 5x plus 3 lot of Y which is 2.5 = 5 3 LS of 2.5 is 7.5 so this is just 5x + 7.5 = 5 we can solve this by subtracting 7.5 from both
• Segment 46: 55:00 - 57:30 sides on the left that will give you 5x and on the right -2.5 and then divide both sides by 5 this gives you x = 0.5 so we have our two solutions X is
• Segment 47: 57:30 - 60:00 0.5 and Y is 2.5 in this question we need to calculate the surface area of the square base pyramid to find the surface area we need to work out the area of all of the faces This shape has five faces there's one square and four triangles let's start with one of the triangles the triangle Le would look something like this the base is 10 and both of the other two sides are actually 13 we can see that from the information we've been given about the sides we know that side EC must be the same as side e d now to find the area of this triangle we need its height if we just look at half of the triangle this part here we can use Pythagoras to work out that height since we've split the triangle in half the line along the bottom here must be 5 cm since that's half of 10 if we now do the hypotenuse Square take away that short side squared so 13 squ Take 5 squ we end up with 144 if you square root 144 you get 12 so the missing side here the height of the triangle must be 12 CM we can now use this to work out the area of the triangle the area of the triangle would be2 * the base which is 10 * the perpendicular height which is 12 if you do this you'll end up with 60 so each of the triangles is 60 cm squared now let's look at the square on the base the side length of that is 10 so the area of the square is simply 10 * 10 which is 100 so now we just need to add up all of these areas we've got four of the triangles so we'll take four lots of 60 and then add to this the square which is 100 when you do 4 * 60 you get 240 add 100 is 340 which is the answer to the question in this question we need to complete a distance time graph we've been told that Sova stays at the dentist for 45 minutes we can see from the part that's already drawn that she arrives there at 10:45 a.m. so if we do 10:45 plus 45 minutes we end up at halfast 11: now if she's staying at the dentist
• Segment 48: 57:30 - 60:00 for 45 minutes her distance from home won't change so we just need to draw a horizontal line for 45 minutes like this for the next section of the graph we're told that she drives home at a constant speed of 32 mph so we know the average speed for that journey home we also know the distance for that journey home because the distance that she lives from the dentist is the highest value
• Segment 49: 60:00 - 62:30 here 40 Mi so we can use the formula time equals distance / by speed to work out how long it's going to take since we know the distance and the speed the distance was 40 m and the speed was 32 mph so 40 / by 32 gives you 1.25 hours which is the same as 1 and a/4 hours so we know is going to leave the dentist at half 11: so we just add 1 and 1/4 hours to this or 1 hour and 15 minutes and we end up with 12:45 p.m. so we can now draw the final section of this graph if we start where we left off at the dentist at 11:30 we can then return home for 12:45 we're going to draw a straight line because it said she travels at a constant speed in this question we need to work out the value of x we're going to start by finding the total sum of all of the angles inside a p Pentagon we can do this using this formula the interior angle sum is the number of sides subtract 2 ultip 180 so the Pentagon has five sides so we do five takeway 2 * 180 that's the same as three lots of 180 which is 540 so this tells us the sum of all of the angles inside a pentagon is 540° but this pentagon is special it's a regular pentagon which means that all of the angles on the inside will be the same so we can do 54 divide by the five angles and we get 108 this means that all of the interior angles of a regular pentagon are 108° including this one here now we've also been told that AB FG is a square the interior angle of a square is 90° so this one here is 90° we can now find the angle X by doing 108° subtract 90° which is 18 so the answer to the question the value of x is 18 for this question we need to work out how much more money TI is going to make going with bank a compared to bank B so let's start with bank a in Bank a we've got an interest rate which is 2.5% and its compound interest we know that Tia
• Segment 50: 60:00 - 62:30 is going to invest 5,000 for 3 years first of all we need to work out what the multiplier is for a 2.5% increase so let's say 100% plus 2.5% which is 10.5% then if we divide that by 100 we get 1.025 so a multiplier for a 2.5% increase is
• Segment 51: 62:30 - 65:00 1.025 so if we take tier's money £5,000 and we multiply it by this multiplier but we need to do that three times one for each of the years so we put a power three if you type this into your calculator you'll get this number here now of course the answer here is monetary so we need to give it to two decimal places so it'll be £ 5384 45 now let's look at bank B for Bank B we've got two different interest rates the first year is 4% but all other years are 1% so we're going to take the same amount of money £5,000 for the first year it's going to increase by 4% that's a multiplier of 1.04 but then all the other years it's 1% so that's a multiplier of 1.01 but since there are 3 years in total we need two years at 1% because there's only one year at 4% and that makes 3 years total so we're going to square the 1.01 if you type all of this into your calculator you'll get this number here so you can see that bank a does give us more but the question says how much more so we're just going to subtract them when you subtract them you'll end up with this number here £79 93 in this question we need to work out the length of AC so we can label this onto our diagram as X this is a trigonometry question so we need to label all of the sides of the triangle let's start with the hypotenuse that's the longest one opposite the right angle here then the one that we call the opposite is opposite the angle that's been given so opposite the 81 is this one here and finally the last one is the one that's the adjacent next to the angle that's this one here now if you look at each of the sides you'll see that two of the sides have paired up information one of them doesn't though you can see the a is with the 11 and the O is with the X but the H is alone all by itself this tells us we're not going to need the hypotenuse for this
• Segment 52: 62:30 - 65:00 question now we write out the phrase so we need to pick the one of these that contains the A and the O that we've got left in the question that would be TOA which means we're going to need to use tan so we can write out that tan of the angle which is 81 is equal to the opposite which is X over The adjacent which is 11 this is just an equation to solve we can multiply both sides by 11 on the left this will give us 11 * T 81 but on the right the 11s will cancel so we just have X you can now just type this straight into your calculator and you'll find the value of x which is this number here the
• Segment 53: 65:00 - 67:30 question asked to give your answer to one decimal place so if you rounded this you get 69.5 this question is also a trigonometry question but this time we're trying to find the size of angle cab let's mark that on as X here once again we're going to label the triangle the hypotenuse first the longest one opposite the right angle then the opposite opposite the angle in the question and finally the adjacent next to the angle in the question once again we're going to cross out the piece of information that's by itself this time it's O so we don't need the opposite for this question if we now look at sakoa you can see we've got a and H left so we need to use C so we can write down that COS of the angle and the angle is X so COS of x equals the adjacent which is 8/ by the hypotenuse which is 13 for a question like this when you're finding an angle you need to use the inverse Tri functions so we can now write that x equals inverse COS of 83 if you type this into your calculator you'll get this number here once again we need to round to one decimal place so the final answer is 52.0 in this question we need to complete the table so we need to find the probabilities for blue and for green two probabilities we have already are red at 0.8 and yellow at 0.1 if we add these together we get 0.9 you can now subtract this from one which gives you 0.1 this means blue and green must add up to 0.1 but we're told in the question the number of blue cubes is the same as the number of green cubes so their probabilities must be the same so we can take 0.1 and split it in half if you do 0.1 divide by 2 you get 0.05 so the probability for blue is 0.05 and it's the same for
• Segment 54: 65:00 - 67:30 green in the second part of the question Kim claims that there are 75 cubes in the box and we need to explain why Kim must be incorrect well if we know there are 75 cubes in the Box we can multiply 75 by each of the probabilities to work out how many cubes there are of each color so if we do 75 * 0.8 for red this gives us 60 if we do 75 * 0.1 for yellow it's 7.5 and for blue and green it would be 75 * 0.05 which is 3.75 so if there were 75 in toal total there would be 60 red 7.5 yellow but 3.75 blue and green now clearly that
• Segment 55: 67:30 - 70:00 doesn't make sense because the number of cubes of each color needs to be a whole number an integer so we can now say the number of each color must be an integer and that concludes the question in this question we've been told the interior angle of a regular polygon is 168 and we need to work out the exterior angle if we draw a picture of what this shape might look like one of the angles would look like this where this is the interior angle of 168 to get the exterior angle we draw this line on and this angle here is the exterior angle you can see that both of these lines add together to make 180° so we just do 180 take away 168 which gives you 12° in the second part of the question we need to find out how many sides this polygon has to work out the number of sides of a regular polygon you just do 360 / by its exterior angle we worked out the exterior angle in the first part that's 12 so we do 360 ID by 12 which gives you 30 sides as the answer in this question we've been given a table of values but we're missing some of the Y values you can see we have all of the X values here though what we're going to do is substitute each of the X values into this equation here to give us the corresponding y value so let's start with x = -2 if x = -2 then we have y = x^2 which is -22 subtract three lots of X which is 3 lots of of -2 and then subtract 1 you can type this into a calculator and it will give you the answer 9 so when x = -2 y = 9 now we can skip over negative - 1 and move straight to 0o so we need to do 0^ 2 take away three lots of 0 and then take away one this gives you negative -1 so we can put that into the table as well then we move along to x = 1 we've got 1 2 take away three lots of one take away 1 which gives you -3 and the final one we need to do is when x = 4 so we do 4^2 take away three lots of four take away 1 and this gives you three and
• Segment 56: 67:30 - 70:00 that's the table complete for Part B of the question we need to draw the graph to do this we're going to plot each of the points of coordinates we start with -29 which goes here then we've got -13 which goes here 01 goes here 1 -3 goes here 2 -3 goes here 3-1 goes here and 43 goes here notice how these points don't form a straight line instead they form a smooth curve so we want to draw a nice
• Segment 57: 70:00 - 72:30 curve between them something that looks like this for the final part of the question we need to use the graph to estimate some solutions to an equation if you look at the left hand side of this equation you'll see it perfectly matches the equation of the graph but the right hand side however says two this means we need to know when the graph gives us a value of two to do this we can draw on the line y equal to find where these two graphs cross they cross it these two places here and we read off their X values so for the first one on the left I get about x = 0.8 and for the one on the right it's about X = 3.8 there'll be some slightly way in this answer depending on exactly how you drew that curve in this question we need to identify a region where the tower could be placed we're told first of all it's closer to point C than point B so what we're going to do is construct the perpendicular B B sector of the line from C to D to do this you take a pair of compasses and place them on C first of all and draw yourself a nice big Arc something that looks like this then you keep the compasses exactly the same size and draw the same Arc but place your compass point at D so it look something like this once you've drawn both of those arcs you can connect up their intersections to form this line here this line is the set of points that's exactly the same distance from C as it is to D now since the tower is closer to C than it is to point D it must be on the left of this red line the second thing we're told about the tower is it's within 350 M of point B and we're given a scale it says that 1 cm represents 50 m so if we divide 350 by 50 we find that it must be 7 cm on our diagram so the tower must be within 7 cm of point B so what we do is take a pair of compasses place them at B set them to 7 cm in width and then draw ourselves an arc
• Segment 58: 70:00 - 72:30 there something that looks like this so we must be inside this purple Arc so now we can find the region where the tower must be it must be to the left of the red line but it must be inside the purple Ark and of course inside the rectangle so the only region that's viable is this one here for the first part of this question we need to find the highest common factor of 75 and 210 to do this I'm going to write both of those numbers as a product of their prime factors and we can use a calculator to help with this so if you
• Segment 59: 72:30 - 75:00 take your calculator let's do 75 you press 75 so 7 5 then equals and your calculator will just say 75 but if we now go and press shift and then we press this button here you can see in yellow it says fact so if we press this button it will show us the prime factorization so it's 3 * 5^ 2 so we can now write that 75 is equal to 3 * 5^ 2 and 5^ S is just 5 * 5 I'm actually going to separate those out it will make this question a bit easier to do then you do exactly the same thing for 210 using your calculator type 210 then press equals then shift then fact and it will tell you that 210 is 2 * 3 * 5 * 7 now we're going to draw a v diagram the left Circle will be for 75 five and the right one for 210 we're going to look first of all for numbers that are in both of those prime factorizations I can see we've got a three in both of them so if I cross off this three and then put a three in the center I can see I've also got a five in both of them so if I cross off one five from each of them I can put that into the center now there's nothing else left that matches in the 75 I've got a five and the 210 I have a 27 so if I cross off the five I can place that in the 75 Circle and if I cross off the two and the Seven both of those can go in the 210 10 circle now all we need to do to find the highest common factor is multiply the numbers in the intersection this part here so you just do 3 * 5 which is 15 and the answer to the question for the second part of the question we need to find the lowest common multiple we can make use of that vend diagram once again to find the lowest common multiple you multiply all of the numbers you can find in the ven diagram here so we need to do five * 3 * 5 * 2 * 7 you can see we use all five of those numbers if you do this with your calculator you'll end up with the answer 1,50 the answer to the
• Segment 60: 72:30 - 75:00 question in this question we're told the volume of the cone is equal to 30% of the volume of the sphere so let's start by working out the volume of the sphere we've been given the formula for that it's the volume of the sphere is 4/3 piun R cubed so the volume of this sphere is 4/3 * by pi then we need to Times by the radius cubed you can see the radius of this sphere is five so it's time 5 cubed now for the volume of the cone the volume of the cone is 30% of this so if we take the volume of the sphere that we
• Segment 61: 75:00 - 77:30 had before and multiply this by a multiplier for 30% which is 0.3 this is the volume of the cone if you type this into your calculator it will come out as 50 Pi we've been asked to work out the height of the cone H imagine we were going to find the volume of the cone we would use this formula here which is 1/3 * pi then times the radius squ if you have a look at the cone you can see the radius will be half of 8 which is 4 so * 4^ 2 then we would multiply by the height but we don't know the height so we're just going to have to call it h and then we would get the volume which we know is 50 Pi so we just need to solve this equation to get the value of H if you type all of this part here into your calculator you'll get 16 over 3 Pi so we have 16 over 3 piun * H = 50 piun at this point we can divide both sides by pi with which will remove the pies so we end up with 16 over 3H = 50 now multiply both sides by three on the left that will give us 16 H and on the right 150 then divide both sides by 16 and you get H equals this number here 9.375 the question asks us to give the answer to one decimal place so it's 9.4 in this question we've been asked to work out the value of x we've been told that lines BD and AE are parallel this means we can be sure that this triangle here which is BCD is similar to this triangle here which is AC let's label on any information we know about these triangles so we know the length from B to D is 6 and we know from C to D is four for the larger triangle we know a to e is 18 and we can also work out the length from C to e by adding CD and D so it's x + 4 since the two triangles are similar we know that one of them is a scale factor enlargement of the other one so if we divided their corresponding sides we would find we always get the same number this means if we take C and CD and divide them that would give us the same
• Segment 62: 75:00 - 77:30 value as if we divided AE by BD so we end up with this equation here on the right hand side if you did 18 divide by 6 you get three you can now multiply both sides by four on the left this will give x + 4 and on the right it will give 12 then if we just subtract four from both sides we find that X is
• Segment 63: 77:30 - 80:00 equal to 8 and that's the answer to the question for the second part of the question we've been asked to write down the value of y remember from before lines AE and BD are parallel so if we draw those on but also draw on the line AC here we can see that angle y will be the same as the angle down here at 41° so so angle Y is also 41° for the final part of this question we need to give a reason for the answer to Part B so we need to explain why y was 41° since these lines are parallel lines and we end up with this upside down F shape we need to recognize that these are corresponding angles so all we would need to say is that corresponding angles are equal in this question we've been asked to find the equation of the line to find an equation of a straight line you need two bits of information you need the gradient and the Y intercept I'm going to start with the gradient to find the gradient you need to pick two points on the line and draw a gradient triangle I'm going to go for this point here and this point here and then I'll draw a gradient triangle between them that looks like this you then need to work out what those vertical and horizontal distances are vertically we can see we go from four to 1 so that's going down three and then horizontally we go from -2 to -1 so that's across one you need to be careful here and not just r on Counting squares you can see that's two squares but it's only actually one on the axis then we can work out the gradient by doing gradient equals the change in y over the change in X the change in y is the three that's in the vertical Direction but since we went down we're going to write -3 and then in the horizontal direction we went across one so the change in X is 1 if you do -3 IDE 1 you get -3 so the gradient is -3 the Y intercept is much
• Segment 64: 77:30 - 80:00 easier to find it's just where the line crosses the y-axis which is here so the Y intercept is -2 now we're ready to form the equation of the line the general equation of a straight line is y = mx plus C where M here represents the gradient and C represents the Y intercept so all we need to do is write this out again but instead of M WR -3 and instead of C WR -2 so it's y = -3x subtract 2 in this question we've been asked to show that this equation here is true in the question there's a trapezium and we've been told its area it's 24 cm squ
• Segment 65: 80:00 - 82:30 so let's work out the area of this trapezium to do this we need to use the formula for the area of a trapezium which is 1 12 of a plus b multiplied by the height A and B here are the parallel lines you can see that's line b c and a d it doesn't matter which way around we label them so let's label the top one a and the bottom one B and then of course this x - 3 here is the height so we just need to substitute all of those into the formula so it's 1/2 and then we need to add a and b together so I'm going to do quite a big bracket here and then we're going to put a which is x + 1 and then add to this B which is x + 3 so that's the a plus b part done and then we multiply this by the height which we can see is x - 3 so this is multiplied by xus 3 now this is an expression for the area of the trapezium which we know must be equal to 24 because it tells us so in the question the next thing we want to do is look at this bracket here we can do some simplification if you add X and 1 and x and 3 you get 2x + 4 so instead we can write this as 1 12 of 2x + 4 * x - 3 = 24 next I'm going to multiply this half by the first bracket here if you do half of 2x + 4 you get 1 x + 2 so it's x + 2 * x - 3 = 24 we can now expand the double bracket so let's do x * X that's x^2 x * -3 that's -3x 2 * X that's + 2 x and finally 2 * -3 is -6 and all of this equals 24 we can simplify the left hand side -3x + 2x is -1x so we've got x^2 subtract X subtract 6 = 24 now we just need to subtract 24 from both sides if you do this on the left you get x^2 - x and then we need to take 24 from -6 which is -30 and on the right hand side 24 take 24 is zero and you can see We've Ended up with what they asked us to in the question so we've shown it in the second part of this question we need to work out the value of x so let's return to that equation we had
• Segment 66: 80:00 - 82:30 before we need to solve this we can do this one by factorizing if you factorize on the left hand side you get X Take 6 x + 5 and then on the right hand side equal 0 this leads us to two solutions either
• Segment 67: 82:30 - 85:00 the first bracket equals 0 so x - 6 equal 0 or the second one does so x + 5al 0 if you solve both of these you end up with x = 6 or X = -5 now whilst this equation does have two solutions only one of them is valid in the context of this question remember the lengths of this shape are x + 1 we've got a height of x - 3 and then on the bottom we have x + 3 so let's imagine that X was equal to 6 what would happen well on the top here we do 6 + 1 which is 7 and then for the height it would be 6 - 3 which is 3 and then on the bottom 6 + 3 which is 9 that all looks okay but if it was xal -5 we do5 + 1 which is -45 subtract 3 which is8 and-5 + 3 which is -2 now clearly we can't have a length that's negative so we're not going to include the xal 5 solution so the answer to x is x = 6 in this question we're told the probability that she wins the 200 M Sprint is 0.9 this means the probability she does not win the Sprint is one takeway 0.9 which is 0.1 so we can complete the first part of the tree diagram by putting a 0.1 for the probability that she does not win the 200 M Sprint next we're told the probability she'll win the long jump is 0.8 this means the probability she does not win the long jump is one take away 0.8 which is 0.2 so we can now complete the long jump section of the probability tree diagram it's 0.8 for win and 0.2 for does not win and the same for the bottom Branch 0.8 and 0.2 in the second part of the question we need to work out the probability she wins exactly one of the events so she's going to win one of them but not win the other one there are two ways that this could happen first of all she could win the 200 M Sprint but then not win the long jump when we move across a tree
• Segment 68: 82:30 - 85:00 diagram like this we multiply the probabilities so we need to do 0.9 multipli by 0.2 which is 0.18 now the other option is for not winning the Sprint but we do win the long jump so 0.1 * 0.8 which is 0.08 we then just add these two probabilities together 0.18 plus 0.08 is 0.26 and that's the answer to this part of the question in the first part of this question we've been asked to work out the coordinates of the point D we've been told that point m is the midpoint of a b and CD
• Segment 69: 85:00 - 87:30 since we know the coordinates for a and for B we can work out the coordinates of M by finding its midpoint to find the midpoint of two points you just find the average of their corresponding coordinates so if you look at Point a the x coordinate is four and for point B the x coordinate is 10 so if we do 4 + 10 and then divide this by 2 we end up with 14 / 2 which is 7 this tells us the x coordinate of the midpoint is seven if we now do the same for the y-coordinates the y-coordinate of a is 10 and the y-coordinate of B is 2 so we just need to add those 10 add 2 and divide this by two this time it's 12 over 2 which is 6 so we know the coordinates of M are 76 next we've been told in the question that the line segment cm is parallel to the x-axis and that point C is on the y- AIS if cm is parallel to the x-axis we know it's a horizontal line and since C is on the y- axis point C and M must both have the same y-coordinate so we know the y-coordinate of C must also be six and since it's on the axis its coordinates are 06 now remember that m is also the midpoint of the line CD so this distance here which we can work out is seven since the x coordinates go from 0 to 7 must be the same as this distance here so that must also be seven so if you add seven and seven you get 14 which means the x coordinate of D must be 14 we know the y-coordinate is going to be six since that's a horizontal line so the coordinates of D are 146 and that's the answer to the first part in the second part of the question we need to prove that two triangles AMC and dmbb are congruent there are a few different ways of showing that triangles are congruent but for this one I'm going to use the rule s a s which stands for side angle side the first thing I'm going to write down is that cm is equal to MD that was the blue lines we had earlier we know this is true since m is the midpoint we also know that m is the
• Segment 70: 85:00 - 87:30 midpoint of the line AB in which case am is the same length as MB once again because m is the midpoint next we're going to look at this angle here angle AMC this is the same size as this angle over here angle bmd we can be sure that angle AMC equals angle bmd because they're vertically opposite angles we've now written down enough
• Segment 71: 87:30 - 90:00 information to show that these two triangles must be congruent we found two pairs and corresponding sides which must be the same length and also the angle in between them so we can write down that triangle AMC is congruent to Triangle dmbb due to SAS which means side angle side for the first part of this question we need to find the coordinates of the turning point of the graph the turning point is right at the bottom of the graph here so the coordinates for that one are -1 -3 for the second part of the question we need to write down the equation of the line of symmetry you can see the line of symmetry goes here the equation of this line that cuts the x-axis at -1 is x = -1 for the final part of the question we need to write down estimates for the roots of this equation since this equation matches the graph that's been drawn and it equals zero we're looking for where this graph crosses the x- axis there are two points where this is the case so we've got one here and one here if we read off those X values for the first one is X = -2.7 and the second one X = 0.7 in the first part of this question we're told that Y is directly proportional to X this means to get from y to X we're always multiplying by the same constant number so we just need to work out what that is if we look at the first set of values here we have a pair to get from 1 to four you're just multiplying by four so for the second set of values when Y is equal to 80 we can multiply 80 by 4 to get 320 now for the next two missing values we're actually going backwards we're going from X to Y in which case we need to reverse this and divide by four instead 10 / 4 is 2.5 and 0.5 ID 4 is 0.125 for the second part of this question we've been asked to sketch a graph of P against q and we've been told that P is inversely proportional to Q if
• Segment 72: 87:30 - 90:00 two variables are inversely proportional to each other their graph would look something like this the key point to understand here is that this is a nice smooth curve but it's never going to touch either of those axes it will get closer and closer and closer but will never actually get there so make sure when you sketch yours it doesn't touch the axis now this isn't the answer to the question but you may find it useful information if the question said directly proportional rather than inversely proportional you draw a straight line that goes up like this and it must go through the origin in this question we need to match
• Segment 73: 90:00 - 92:30 some equations to their corresponding graphs let's start with the first one y = x^2 an x s graph is always going to give a u-shape or an upside down U shape but since this one is a negative x^2 is actually the upside down U shape which is graph F then we move on to y x cubed there are two graphs that give cubic shapes graph e and graph I we just need to decide which one it is you can do this by substituting in some values let's think about what happens if we cube a positive number like if we Cube two if we Cube two we get eight what if we Cube -2 though that would be -2 * by itself and then by itself again which would give ne8 so we can tell it must be graph e because when you substitute X as a positive number you get y as a positive number and when you substitute X as a negative number you get y as a negative number so it's graph e for the next one you need to learn your trigonometric graphs there are two that it looks like it could be it could be graph a or it could be graph D when you do sign of zero you get zero which means this one must go through the origin so it must be graph D graph a was actually the graph of COS of X and finally we need to do 1 /x which could be graph C or graph G these are both reciprocal graphs but again this one if you think about positive values if you do one over a positive value you get a positive value back and the same idea for negatives one over a negative value will give you a negative value so it must be graph G graph C would be Y = -1x in this question we've been asked to work out the mass of the cubid in kilograms we've been given a ratio it's the volume of gold in the cuboid to the volume of silver in the cuboid so let's work out the total volume of the cuboid to find the volume of a cuboid you just need to multiply all of those lengths together 30 * 8 * 6 which is 1,440 now that we know the total volume
• Segment 74: 90:00 - 92:30 we can split it into the volume of gold and volume of silver using that ratio in the ratio there are three parts gold five parts silver so 3 + 5 gives eight parts in total so if we split this total volume into eight parts we do 1,440 divide by 8 which gives you 180 to every one part of the ratio is 180 since there are three parts for gold we multiply 180 by 3 which gives you 540 and since there are five parts for silver we're going to do 180 * 5 which is 900 so we now have
• Segment 75: 92:30 - 95:00 the volume of gold and volume of silver in the cuboid we've also been given their densities in the question so we can use this to work out the mass of gold and the mass of silver we can use the formula mass equals density time volume let's do this for gold first so the mass of gold is the density of gold which is 19.3 time the volume of gold which is 540 this gives you this number here then we'll do the same for silver so mass of silver the density of silver that's 10.5 Times by the volume of silver which is 900 gives you this number here we can add those two together to get the total mass of the cuboid which gives you this and the question does say to give it in kilogram so we need to divide that by 1,000 which is 19.87% and that's the answer to the question in this question we need to work out the length of one of the sides of the cube we've been given this formula pressure equals force over area so let's have a read of the question and see what information we have been given we've been told the force that's 320 Newtons and we've also been told the pressure the pressure is 500 Newtons per me squ so we can rearrange this formula to work out the area the area would be the force divid by the pressure so if we take the force from the question 320 Newtons and divide that by the pressure we end up with 0.64 m s so we found the area a 0.64 M squ that's specifically the area of the face that's in contact with the ground so this one here now in the question we're told this is a cube a cube has all of the sides the same length so if we just want to work out the side length of this Cube we can square < TK 0.64 to get 0.8 this works because 0.8 8 * 0.8 would give you the area of that square on the bottom 0.64 so the answer to this question is 0.8 m in this question we need to show that
• Segment 76: 92:30 - 95:00 lines L1 and L2 are parallel parallel lines have the same gradient so we're going to find the gradient of both of these lines and hope they're the same let's start with line L1 this one here so if we take this equation it's much easier to see its gradient if we write it in the form y mx plus c so we're going to rearrange it to get y as the subject let's first of all add 8X to both sides this will give us 2 y on the left and on the right 10 + 8 x then
• Segment 77: 95:00 - 97:30 divide both sides by two this gives you y = divide 10 by 2 and 8x by 2 that's 5 + 4x which we could also reorder as y = 4x + 5 now it's in the form y = mx + C you can see the gradient really clearly is the number in front of the X which is four so the gradient of line L1 is four now let's move on to the second line we've not been given its equation this time but we have been given two coordinates from the line we can find the gradient of this using the formula gradient equals change in y over change in X which some people write as Y2 over y1 - X2 over X1 in this case y1 is the y coordinate of the first point and Y2 is the y coordinate of the second Point X1 is the x coordinate of the first point and X2 is the X coordin coordinate of the second point so we just need to substitute these points in so Y2 - y1 would be 30 - 10 and X2 - X1 would be 8 subtract 3 30 take 10 is 20 and 8 take 3 is 5 so this is just 20 over 5 20 ID 5 is 4 so we've also found the gradient of line L2 now that's also four now we can just write a concluding statement lines L1 and L2 have the same gradient so they are parallel lines in this question we have a Fibonacci Sequence and we've been told some information about the fourth and fifth terms so let's work out the fourth and fifth terms of the sequence to work out the next term of a Fibonacci Sequence you add together the previous two terms so if we're going to work out the fourth term we need to add the second and third terms together that's these two here so if we added those together we do 2 b + 3 a + 2 B and if you simplify this you get 3 a + 4 B so that's the fourth term then to get the fifth term we would add the third and fourth terms together that's these two so we do 3 a + 2 b + 3 a + 4 B which would give you 6 a + 6 B and that's the fifth
• Segment 78: 95:00 - 97:30 term now we can return to these statements we've been told the fourth term is equal to 51 and the fifth term is equal to 84 if we write that down algebraically it's the fourth term 3 a + 4 B = 51 and then the fif term 6 a + 6 B = 84 these are some simultaneous equations we can solve I'm going to double the first equation to make the A's match if
• Segment 79: 97:30 - 100:00 I double the first equation I get double 3 a which is 6 a double 4 B which is 8 B and double 51 is 102 I can now subtract the equations 6 a take away 6 a is zero 8 B take away 6B is 2 B and 102 take away 84 is 18 divide both sides of this by two and you find that b is 9 we can now find the value of a by substituting this back into one of the equations I'm going to go for the second one here so we have 6 a plus 6 lots of B but we know B is now 9 = 84 6 9 are 54 so we've got 6 A+ 54 = 84 subtract 54 from both sides and you get 6 a = 30 divide both sides by 6 and you find that a is equal to 5 so we found the values of A and B A was 5 and B was 9 in this question we've been told that ABCDE is a pentagon and since we're dealing with its interior angles we're going to work out the interior angle sum using this formula here where n is the number of sides since a pentagon has five sides we do five takeway 2 * by 180 which is 540 we'll come back to that in a moment now we've been told that X to Y is in the ratio 1: 6 this means that Y is 6 times bigger than x X which we could write as Y = 6x so we can replace all of the Y's in the question with 6X like this now let's return to that 540 that's the sum of all of the angles in the shape so if we added up all of the angles we've got the 2 X's X Plus X and then we got the three y's but we just replace those with 6X so it's plus 6 X Plus 6X + 6x = 540° we can simplify the left side of this if you add up all of those X's you get 20x = 540 and then divide both sides by 20 you find that X is equal to 27 in this question we need to find X and Y though but remember y was equal to 6 lots of X so if you just multiply X by 6 you need to do 6 * 27 which gives you 162 so X is 27 and Y is 162 in this question we're trying to find the length of the arc DF I'm going to mark this Arc Conant the diagram so
• Segment 80: 97:30 - 100:00 you can see it it's this one here to find the length of an arc we use the formula Arc Length equals Theta over 360 multip by pi and multip by the diameter
• Segment 81: 100:00 - 102:30 so we just need to find the angle for this sector Theta and also its diameter let's start with the angle the angle we want to find is this red one here but we can't find that straight away instead we're going to find this blue angle here first we're going to look more closely at the triangle afg which looks like this you can see its hypot is 10 and one of the other sides is five and the angle we're after is in the bottom right so let's call that X we can find this using trigonometry the five is the opposite and the 10 is the hypotenuse so we can write down that sin of x equals the opposite which is 5 over the hypotenuse which is 10 to find X we would write x equals and then do inverse s of 5/ 10 which will give you 30° so we found this blue angle at 30° we can add that onto our diagram now if we look more closely at the the top sector we can see that lines Ag and CG are both a radius for that sector so the line CG must also be 10 this means that the triangle on the right hand side is actually congruent to the one on the left hand side so we can be sure that this angle in here is also 30° now we found all of the angles in that Circle apart from the red one so if we subtract those from 360 we'll find the red one so 360 take away the 230s and take away the 80 will give you 220 so we found the angle for that sector which is 220° we can replace the Theta in the formula with 220 now let's try and find the diameter we're going to find the diameter by first of all finding the radius the line FG here is a radius of the circle so we can try and find that line we already Drew out the triangle afg earlier so the line we're talking about is this one here we can find that length using Pythagoras so if we imagine it's called R we could write that r^ 2 is equal to 10 s take away 5^ s Square rooting both sides of this will give you R is equal
• Segment 82: 100:00 - 102:30 to < TK 75 so if the radius is < TK 75 the diameter must be double this so the diameter must be 2un 75 we can now replace the diameter in the formula with 2 < tk75 and all of this can be typed straight into a calculator and it'll give you this number here so we found the Arc Length it's this number the question wants us to give it to one decimal place so we'd round it to 33.3 which is the answer to the question in this question we've been asked to draw a box plot using the information that's given in the stem and leaf diagram to draw a box plot you need five bits of information we're going to start
• Segment 83: 102:30 - 105:00 with the minimum and maximum values the minimum value in the stem and leaf diagram is this one here which is nine and the maximum one is this one here which is 31 so we need to draw a vertical line at 9 and another one at 31 next we need to find the median and the lower and upper quartile notice in the question we have information about 23 mobile phones so we do 23 + 1 which is 24 and then divide this by 2 which gives you 12 and this tells you the 12 value is the median one if we divided 24 by 4 instead of two we would get six which means the six value is the lower quartile let's go and find those and then we'll come back and do the upper qual last so if we count through the data in order when we get to the SI value that will be the lower quartal so let's cross off the first second third fourth fifth and then the next one must be the sixth so that one there is the lower quartile so that's 13 so let's draw a line at 13 then if we keep going until we get to the 12th one that's the median so the 7th 8th 9th 10th 11th and there we go there's the 12th that's 19 so we need a line at 19 now we just need the upper quti to do this you can take the six which we got for the lower quile multiply it by three and this will tell us it's in the 18th position now there are two ways of doing this you could either count through until you get to the 18th one or you could just count backwards six from the end like this 1 2 3 4 5 and then the next one is six so the upper quarle is 27 we draw a line at 27 we connect together the lower quarle the median and the upper quartel to form this box and then we connect up the lowest and the greatest value to those box boxes with horizontal lines like this and there's your completed box plot in the second part of the question we've been asked to compare the distribution of battery lives for company a and Company B to do this we'll
• Segment 84: 102:30 - 105:00 make two comparisons one about the median and one about the interquartile range let's start with the one about the median so the median for company a is this line here which is 19 hours and the medium for Company B is this line here which is 17 hours so the medium battery life in hours was higher for company a than it was for Company B so we could write the medium battery life for company a was higher and then we need to explain what this means in context in
• Segment 85: 105:00 - 107:30 context this means the batteries lasted longer on average the phrase on average is particularly important here to make sure you get the marks it doesn't mean every battery will last longer for company a compared to Company B but the average battery will next we're going to look at the interquartile range to do this we need to do the upper quartile subtract the lower quartile for each one of the companies let's do it for company a you can see the Upp quartile here that's 27 and the lower quartile here that's 13 so for company a we need to do 27 takeway 13 which gives you 14 for Company B the upper quartiles here that's 21 and the lower quartiles here that's 15 21 takeway 15 gives you an inter quarar range of six so you can see here company a also has a larger inter quarar range so we could say the intercal range for company a was higher and then again we need to explain that in context so in the context of this question that would mean that their battery lives were more varied another phrase you could use here is you could say their battery lives were less consistent in this question we've been asked to work out the surface area of this solid shape the shape comes in two parts we've got a hemisphere on the bottom and a cone on the top let's start with the hemisphere this part here we've been given the formula for the surface area of a full sphere which is 4 pi r 2 where R is the radius we don't have the radius of this sphere so we're going to need to work that out but we do have this length here which will be the diameter that's 16 if we half that we get the radius so the radius is8 so if we worked out the surface area of this as though it was a full sphere it would be 4 * pi * the radius squar which we just said was 8 and if you type this into your calculator or work it out you'll get 256 Pi now this is for the full sphere but we only have a
• Segment 86: 105:00 - 107:30 hemisphere so we only have half of that surface area so if we divide this by two you end up with 128 Pi so that's the hemisphere done now let's have a look at the cone for the cone we need to find the curved surface area and we've been given its formula up here it's pi r l where L is the slant height of the cone that's this length here we don't have that at the moment so we're going to need to work it out let's draw on a radius from the center of the cone to the edge and we create a triangle which we can draw over here and we can label on the bits we know we know the base is eight since that's the radius and we also know the height that's on the diagram that's 15 and L is this one here the one we're trying to find we can find this using Pythagoras so we could say that the hypotenuse which is l^2 is
• Segment 87: 107:30 - 110:00 equal to 15 2 + 882 and then if we square root both sides of this we find that L is equal to < tk2 289 and if you do this you'll end up with 17 so we found that L is 17 we're now ready to use the formula the formula was pi r l so we need to do PI time the radius which is 8 * L which is 17 this gives you 136 Pi we're now ready to finish the question to get the total surface area of the whole shape we just add together the two surface areas we found 128 pi and 136 Pi if you add these together you end up with 264 Pi if you type 264 Pi into your calculator and convert it to a decimal you get this number here the question asks us to round it a three significant figures so the answer will be 829 you may be wondering why we didn't include the circular parts of the surfaces of those shapes since the cone and the hemisphere are attached together the circle Parts aren't actually on the outside surface of the shape so we don't need to include those if you were doing the surface area of just the cone though or just the hemisphere you would need to work out the circle as well and add that on for this question we need to expand and simplify some triple brackets to do this we're just going to expand the first two brackets first so if we expand x + 9 * x + 2 we do x * X which is x^ 2 x * 2 which is + 2 x 9 * X which is 9 x and 9 * 2 which is + 18 so if you expand the first two brackets we end up with this we can simplify these two terms here get x^2 + 11 x + 18 so we can now replace both of those first two brackets with this quadratic x^2 + 11 x + 18 so the question is actually x^2 + 11 x + 18 * YX - 3 now we need to multiply everything in the first bracket by everything in the second bracket so we do X2 * X that's X cubed x 2 * -3 -3x 2 11 x * X that's 11 x^2 11 x * -3 - 33x and then 18 * X that's + 18 x and finally 18 * -3 -54 there's a bit more simplifying to do
• Segment 88: 107:30 - 110:00 so we've got X cubed at the front then we've got two terms here in x^2 -- 3x2 + 11 x^2 is + 8 x^2 we've also got two terms in X we've got - 33x + 18x that's going to give you -5x and then finally -
• Segment 89: 110:00 - 112:30 54 at the end and that's your answer to this question in the first part of this question we need to work out the value of 16 to the^ 1/2 whenever you raise any number to the power of 1/2 you just need to find the square root of it so we need to do the square root of 16 which is equal to 4 in the second part of this question we need to work out 25 over 4 all to the^ 0 any number to the power 0 is always equal to one so it doesn't matter that it says 25 over 4 there the answer to this question is one in the final part of this question we've been told that 4 ^ n equal 8 we need to work out 100 ^ n to do this we're going to try and work out the value of n first so if we start with this 4 to the^ n equal 8 we're going to try and write both the left and right hand side as a power of two the number four is the same as 2^ 2 so we can rewrite the left hand side as 2^ 2 all to the power n and on the right hand side we have 8 which is equal to 2 ^ 3 on the left hand side here we can use the bracket to our power law to do this we just multiply the indices so this is equal to 2 the^ 2 N since we do two multipli by n and on the right hand side we'll still keep two to the^ 3 you can now see we've written both of the sides as a power of two since both sides are in this form we can see the indices must be equal so 2N here must be equal to 3 if 2 N is equal to 3 you can divide both sides by two and you find that n is equal to 3 / 2 now that we know the value of n we can work out 100 the^ n so since n is 3 /2 this is 100 ^ -3 /2 when we raise something to a fractional power like this we look at the denominator which is this two and that tells us what roote we need to do a power two there means to do the square root once we found this the power three on the top tells us to cube it so this is like doing the square root
• Segment 90: 110:00 - 112:30 of 100 and then cubing it but we've also got this negative power here which means we need to take the reciprocal so instead of this being the answer it's actually one over this now we can work this out the square root of 100 is quite straightforward that's 10 so on the bottom we have 1 over 10 cubed and then 10 cubed is 1,000 so it's 1 over 1,000 for this question we need to do an enlargement the first thing to do is Mark on the center of enlargement which is at -21 so let's put across there then
• Segment 91: 112:30 - 115:00 we're going to pick the first point on the shape that we're going to map it doesn't matter which one you go for but I'm going to go for this one here now in the question it says the enlargement must be scale factor -2 when you do a negative scale factor enlargement you start at your point and work out how you get from your point to the center of enlargement so you can see to get from the Red Cross to the black cross I just just need to go down four squares since the scale factor is -2 I then do this journey two more times but go beyond the center of enlargement so if I went down four to do this journey two more times I'd need to go down eight so if I go down eight from the center of enlargement I end up down here so this is where the red point would map to now let's pick another one let's go for this one here so to get from this point to the center enlargement I'd go this way which is one to the right and two down but I need to do this journey two more times from the center of enlargement so I need to go two right and four down that would take me to here so this is where the Green Point goes and finally let's do the last one so to get from this point to the center of enlargement I need to go this way which is three right and four down and I need to do this two more times so I need to go another six right and eight down if I go six right and eight down from the center of enlargement I end up here now that we' drawn all three of the points we can connect them up to make the triangle it says to label this one B so so we'll do that as well in this question we need to draw some inequality regions the first thing you should know about this is if we have either of these symbols greater than or equal to or less than or equal to the lines we draw must be solid lines if it's greater than or less than they must be dotted lines you can see in this question it's always greater than or
• Segment 92: 112:30 - 115:00 equal to or less than or equal to so we'll draw solid lines in all of the cases we're going to start with Y is greater than or equal to zero to help us draw this we going to draw y equal 0 instead the line y equal 0 is a horizontal line that crosses the y-axis at 0 it's actually just the xaxis this line here now we're interested in all of the points where Y is greater than zero so that would be everything above this line but at first this might seem counterintuitive we're going to shade out everything we don't want so we're going to shade out everything below the line leaving the points we do want above the line so it would look something like this now we'll move on to to the next inequality it's X is greater than or equal to -1 to help us with this inequality we're going to draw X is equal to1 this is a vertical line that goes through1 on the x- AIS it looks
• Segment 93: 115:00 - 117:30 like this now if we want when X is greater than or equal to1 we need all of the points to the right of this line but once again we're going to shade out the area we don't want so we don't want everything to the left of this line so we're going to shade this part as well then we move on to the line Y is less than or equal to x + 3 to help us do this one we're going to plot y = x + 3 there are a number of ways of doing this it doesn't matter as long as you end up with the right line I like to try and substitute in points to work out a few coordinates specifically when X or Y are equal to zero so let's first of all imagine that X is equal to0 if x is equal to0 then we can substitute Z into this equation here so it would say yal 0 + 3 which of course is 3 so the value of y is 3 so if x x is 0 and Y is 3 that's the coordinate 03 now let's also consider when y equals 0 if Y is equal to 0 then it would say 0 = x + 3 in which case X must be -3 so we end up with the point -3 0 if we plot both of these points onto the diagram like this we can draw a straight line between them so this is the line yal x + 3 now the inequality says Y is less than or equal to x + 3 so we're interested in the points below the line so once again we're going to shade out the points we don't want which are those above the line so it will look something like this then we move on to the final inequality to help us do this one we're going to draw the line 2x + y = 6 once again we're going to consider what happens when X and Y are equal to zero so let's do x equal 0 first if we substitute xal 0 into this it would be two lots of 0 which is 0 plus yal 6 in which case Y is equal to 6 if x is equal to 0 and Y is equal to 6 that's the coordinate 06 now let's have a look at when y = 0 if y = 0 we get 2x + 0 = 6 in which case 2x = 6 divide both sides of this by two and you find that X is equal to 3 so we
• Segment 94: 115:00 - 117:30 end up with a point 3 0 Let's plot both of those they go here and here and draw a straight line between them now in this one again it says 2x + y is less than or equal to 6 so we would like the points below this line so we're going to shade out the points above the line something like this now because we consistently shaded out the areas we don't want the area that's left is the area we do want
• Segment 95: 117:30 - 120:00 so we can label this region in the middle here Ah that's the region of points that satisfies all of these inequalities in this question we have Kenny who catches 350 fish from a lake Kenny's going to Mark each of those fish and then return them to the lake the next day he catches 17 5 fish from the lake and he finds that 70 of these fish have been marked we need to work out an estimate for the total number of fish in the lake this topic is called capture recapture and it relies on the idea that the proportion of fish that are marked on the second sample on the next day is the same as the proportion of fish that are marked in the entire Lake let's write down the proportion of fish that are marked in the entire Lake we know 350 of the fish have the mark and this is out of a total of all of the fish in the Lake that's the number we're trying to find we don't know this so we'll call it X this is going to be the same as the proportion of fish that are marked in the second sample in the second sample we catch 175 fish and 70 of them are marked so the proportion that are marked here is 70 out of 175 we can solve this equation by cross multiplying first we're going to multiply both sides by 175 and X this will give us 350 * 175 = 70x you can then divide both sides by 70 and we get 350 over 175 divide 70 is equal to X you can just type this straight into your calculator and it gives us the answer of 875 for the second part of the question Kenny returns all of those marked fish back to the lake and the next day he catches yet another 175 fish but this time he ends up with a different estimate we ended up with the estimate of 875 but this time Kenny is going to estimate that there are fewer than 700 fish in the lake we need to work out the lowest possible number of marks fish that Kenny could have caught so when we did part A we ended up with this
• Segment 96: 117:30 - 120:00 calculation here this time the 350 is still the same there still 350 marked fish and the 175 is still the same we' still caught 175 fish the question is how many were marked so we need to change the number 70 and see what happens to the answer if we increase 70 and just assume for a moment that 71 of those fish were marked this would change the answer from 875 to 862 676 if we increase it once again this changes the answer to this and we need to keep doing this until we get the answer to be below 700 now it does take a while to get this number below 700 so
• Segment 97: 120:00 - 122:30 I'm going to miss a few out and jump to 87 which gives you the answer 704 but we need it to be below 700 and if you go to 88 it does go below 700 at 6962 so the answer to the lowest possible number of mar fish that Kenny could have caught is 88 in this question we need to write the numbers in order starting with the smallest if we look at the first number here 2.03 to the^ 6 that's going to be relatively close to 2 to the^ 6 we can work out 2 to ^ 6 easily it's 2 * 2 * 2 * 2 * 2 * 2 which would give you 64 for the second one 7.95 s that's going to be a little bit lower than 8^ squ and we can work out 8 squ easily 8 s is also 64 but for the third one square < TK of 6,500 that will be a little bit bigger than the square < TK of 6,400 the sare < T of 6,400 is easy enough to work out that's going to be 80 since 80 * 80 is 6,400 and for the final one the cube root of 124,000 is going to be a little bit less than the cube root of 125,000 the cube root of 125,000 is going to be 50 since 50 * 50 * 50 is 125,000 so if we now look at each of these numbers in turn this one's going to be a little bit bigger than 64 this one's going to be a bit smaller than 64 this one's going to be a bit bigger than 80 and this one's going to be a bit less than 50 we can be sure then that the smallest one's going to be the cube root of 124,000 then it's going to be 7.95 squ then it'll be 2.03 to ^ 6 and finally the square root of 6,500 in this question we've got a manager who's going to select one player from each position to win an award and we need to to work out the number of different possible ways there are of picking a goalkeeper a Defender a midfielder and a striker to answer this question we just need to multiply the
• Segment 98: 120:00 - 122:30 number of goalkeepers which is three by the number of Defenders which is eight by the number of midfielders which is 10 and the number of Strikers which is five if you multiply all of these together you get the answer 1,200 in this question we need to sketch the graph of y = COS of x x in between 0 and 360 the graph of y = COS of X goes between1 and one on the y axis so let's mark those on first then we need to mark on some key points to help us draw this graph when you do c of zero you get one so we can plot the 01 when you do cos of
• Segment 99: 122:30 - 125:00 90 you get zero so we can plot 90 0 COS of 180 gives you negative 1 COS of 270 also gives you zero and COS of 360 gives you 1 so we need to connect these up with a smooth curve that look something like this for the first part of this question we need to complete a cumulative frequency table for the first group the money spent is between 0 and2 and you can see we actually have that value in the table above as well so the frequency will be the same as the cumulative frequency is this number here six so we can write six into our table the next group is from 0 to4 this means we need all of the people that spent between 0 and 20 and also the people from 20 to 40 so that's these 16 people here so if we add both of those together 6 + 16 is 22 then we can continue down the cumulative frequency table by adding in the next frequency each time so if we add in 29 as well now we've got 6 and 16 and 29 that gives us 51 and for the final group from 0 to 80 lb we need all of the frequencies if you add all of those up you get 60 for the second part of the question we now need to draw the cumulative frequency graph when you draw a cumulative frequency graph you're going to plot the points at the end of each of the intervals so we're going to plot them at 20 40 60 60 and 80 and we'll use the cumulative frequencies we just found so for the first one at the end point of 20 we need to plot a cumulative frequency of six so that goes here then we've got an end point of 40 with a cumulative frequency of 22 that goes here then an end point of 60 cumulative frequency 51 and finally the end point of 80 with a cumulative frequency of 60 we can now connect up these points either using straight lines like I've done here or you could draw a smooth curve between them both will be fine in your exam I would recommend doing straight lines just because I think it's
• Segment 100: 122:30 - 125:00 easier to do for part C of the question we need to use the graph to find the estimate for the median amount of money spent in the supermarket by the 60 visitors since the cumulative frequency goes up to 60 we'll find the median at the 30th value so if you go to 30 on your cumulative frequency that's halfway up the scale and then go across until you hit the graph you can then go down from this point here and then read off the value for the money I get the answer of £45.50 since I'm halfway through that square for the final part of the question we've been told that one of these 60 visitors is going to be selected to win a prize and we need to use the graph to estimate the probability that the visitor spent more
• Segment 101: 125:00 - 127:30 than £35 so what we can do is look for £35 on the money spent axis at the bottom go up to the graph and then read across we read off the value here of 18 this means that we would estimate that 18 people had spent up to £35 but in this question they want to know the probability of someone spending more than £35 so we'll subtract these 18 people away from the total of 60 which gives us 42 so we would estimate there are 42 people that have spent more than £35 so the probability of selecting one of them will be 42 over 60 in this question we have a geometric sequence and we've been asked to work out the difference between the third and the first term so if we take the sequence here if it's a geometric sequence then to get from the first term to the second term we must mult multiply or divide by some number we just need to work out what that number is so how do you get from 2un 5 to 10un 10 well to get from the 2 to the 10 we must multiply by 5 and to get from otk 5 to otk 10 we must multiply by < tk2 so to get from 2un 5 to 10un 10 you multiply by 5 < tk2 this means we can work out any term in the sequence if we just keep multiplying by 5 < tk2 we need the third term for this one so let's multiply the second term by 5 < tk2 if you multiply 10un 10 by 5 < tk2 you first of all multiply 10 by 5 which is 50 and then you multiply < TK 10 byun2 which isun 20 so this is the third term now we need to work out the difference between the third and the first term so let's take the third term and then subtract from this the first term so we've got 50 < tk20 subtract 2un 5 but the question wants us to give our answer in the form kotk 5 now at the moment it doesn't look like that form so what we're going to need to do is convert the root 20 into something that has a root 5 in it so the 50 < tk2 could be Rewritten as 50 * < TK 4 * < TK 5 since < TK * < 5 is < TK 20
• Segment 102: 125:00 - 127:30 and then of course we've got take away 2un 5 at the end now this squ < TK 4 here is actually just two so we could replace the square < TK 4 with a two and then we could multiply the 50 and the two to get 100 so the first term here is actually just 100 < TK 5 and then we've got this- 2un 5 at the end now we've got 100 < TK 5 subtract 2un 5 that's just 98 < TK 5 and now it's in the correct form kotk 5 where K is 98 so that's the answer to the question in this question we have y = a
• Segment 103: 127:30 - 130:00 ^ x where a is an integer and we need to work out the value of a to do this we're going to substitu in some values for X let's start with x = 1 so when x = 1 we've got y = a to the X so it's y = a ^ 1 a ^ 1 would just be a so we end up with Y = A now if we look at the graph specifically the point when xal 1 that's this one here you can see that when the x coordinate is 1 the y coordinate is 3 now we said that when X was equal to 1 y was equal to a and since Y is equal to 3 then a must be three so we can write down that a equals 3 this also works with the other points if you want to check them you can see that when xal 2 that's this point here we end up with Y is 9 and it makes makes sense because 3^ 2 or 3^2 = 9 for the second part of the question we need to complete the graph for X values from 0 to 4 now they've already done this for when x = 0 1 and 2 we just need to do it for when x = 3 and 4 so we now know the graph is y = 3 ^ x since we worked out that a was equal to 3 so when x = 3 y will be 3 ^ 3 which is 3 cubed which is 27 then when x equal 4 y will equal 3 to the^ 4 which is 81 so we can now plot these two points we've got x = 3 y = 27 that goes here and also x = 4 y = 81 that goes here this is an exponential graph so we're going to join it up with a nice smooth curve something that looks a bit like this in the first part of this question we've been asked to draw the graph of x^2 + y^2 = 36 36 is the same as 6^ 2 so we could rewrite this as x^2 + y^2 = s you should recognize this as the equation of a circle where this number here tells you the radius so in this one the radius is six so we just need to draw a circle which has Center at the origin with a radius of six you want to use a pair of compasses to do that and it should look something like this you'll see it goes through six and -6 on each of the axes for the second part of the question
• Segment 104: 127:30 - 130:00 we need to use the graph to find estimates to these equations here you'll notice the first equation is the one we drew and the second equation is the one that they've drawn for us this means all we're looking for is where these two graphs intersect and you can see there are two points there's one here and one here so we need to read off their coordinates if we start with the one at
• Segment 105: 130:00 - 132:30 the top we've got an x coordinate here and a y coordinate here the x1's about 1.4 and the Y one's about 5.8 and for the second one we go up here and across here and we read off the x value about -3.8 and the Y value about -4.6 so there are two pairs of solutions to these simultaneous equations firstly we've got the solution where X = 1.4 and Y = 5.8 and we've also got the solution where X = -3.8 and Y = -4.6 in this question we need to solve a quadratic equation we've been asked to give our answers to three significant figures this is a giveaway that we need to use the quadratic formula at the moment the equation isn't in the correct form for the quadratic formula we need the right hand side to be equal to to Z so if we take the equation you can see we have two terms on the right hand side there 3x and + 4 but we would rather the right hand side was zero so we're going to subtract 3x and subtract four from that right hand side and that will mean the right hand side is equal to zero but if you do that on the right hand side you must also do that on the left hand side so we subtract 3x and subtract 4 this will give us 6 x^2 subtract 3x subtract 4 equal 0 now it's in the right form now the quadratic formula looks like this this is something you need to memorize for your exam the letters a b and c can be found by looking at the quadratic equation a is the coefficient of x^2 that's this number six here B is the coefficient of x that's a -3 and C is the constant term at the end that's -4 we now write out what's inside the Box the quadratic formula but replace each of those letters a b and c with their numbers so we're going to start with x equals netive B well since B is already -3 that would switch to a positive 3 so so 3 plus or minus the square Ro TK of then we have b^ 2 which is -3 2 and you can see I put that in Brackets because it's
• Segment 106: 130:00 - 132:30 a negative number minus 4 * a which we know is 6 and then time C which is -4 once again I put that in bracket since it's a negative number and all of this is divided by 2 a so two lots of six before typing all of this into the calculator I like to do a little bit of tidying up so we're going to write it out again but simplify a bit so we have x = 3 plus or minus and then we get to the square root I would type all of this part here into my calculator to see what number it gives me if you do that it
• Segment 107: 132:30 - 135:00 will give you the number 105 so it's the square root of 105 and on the bottom two lots of 6 is 12 so we end up with 3 plus orus < TK 105 over 12 we can now separate this into the two solutions we have the version that has a positive and the version that has a negative so we've got x = 3 + < 1052 and x = 3 subtract < 1052 you can now type each of these into your calculator separately and you'll get the two solutions which are this and this now the question did say remember give it to three significant figures so we need to round both of these off so the first one is X = 1.10 and the second one X = 0.604 in this question we need to find the M term of a sequence so let's write out the sequence and look at the differences to get from 8 to 19 you add 11 to get from 19 to 34 you add 15 and if we keep going it's add 19 and add 23 since all of these differences aren't the same we know it's not a linear sequence so let's check the second differences what's the difference between 11 and 15 well that's plus 4 and it's the same for 15 to 19 and from 19 to 23 when the second difference is constant like this we know it's a quadratic sequence we take that second second difference and divide it by two which in this case gives us two and this tells us the coefficient of n^2 in the solution so we know it's going to be 2 n^ 2 what we do now is construct a table like this pop the sequence at the top and in the second row we're going to write out the numbers for 2 n^ 2 2 n^ s just means two lots of the square numbers so let's start by doing 1 squ which is one and then double this that gives you two then we do 2^ squ which is four and double this which is 8 3 squ is 9 and double this is 18 4 S is 16 and double that is 32 and 5 squ is 25 and double this is 50 we're now going to subtract the 2 n^ 2 from the sequence to leave us with the linear part so if we
• Segment 108: 132:30 - 135:00 do eight take away 2 that's 6 19 take away 8 that's 11 34 take 18 16 53 take 32 is 21 and 76 take 50 is 26 so what we're left with at the bottom is the linear part of the solution this means we can find ITS m term relatively easily we check the differences and you can see it's always + 5 which means it's 5 n but it doesn't start at five it starts at six so we need to add one so the linear
• Segment 109: 135:00 - 137:30 part of the solution is this bit here 5n + 1 so if we add the linear part to the quadratic part we get the answer to the question the nth term is 2 n^2 + 5 n + 1 for the first part of this question we need to work out the value of f of -5 to do this we're just going to substitute 5 into the F function which is this one here so the function is x^2 - 3 so we need to do -5^ 2 - 3 -5^ 2 is 25 so it's 25 subtract 3 which is 22 for the second part of this question we need to find the inverse function for H we can see the function h of X at the top that's x + 3 over 4 - x so we're going to write that down first of all then we swap the H of X for letter y and then we swap all of the Y's to x's and all of the X's to y's so instead of y equals it's going to be x equals and the two x's will turn into y's we now need to rearrange this to make y the subject I'm going to start by multiplying both sides by the denominator of the fraction on the right 4 - y If I multiply the left by 4 - y it's X lot so 4 - Y and on the right the 4 - y will cancel so I've just got the top y + 3 we can now expand the bracket on the left hand side x * 4 is 4X and x * y ISX Y and this equals y + 3 now what we want to do is get all of the terms that involve a y on one side and all of the terms that don't involve a y on the other side I'm going to do this by adding XY to both sides first if I add XY to the left that will cancel the XY that's there so I end up with 4X and if I add XY to the right I've got y + 3 and then I add XY next we're going to subtract three from both sides if we subtract it from the left it's 4x - 3 three and if we subtract three from the right that will cancel the three so it's y + XY let's take a copy of this and move up here we can now factorize the right hand side so we leave the left hand side alone and on the right hand side we're going to factorize out the Y so it's Y and then a bracket inside this bracket
• Segment 110: 135:00 - 137:30 would be 1 + x we can now just divide both sides by 1 plus X if we do this on the left it's 4x - 3 over 1 + x and on the right the 1 plus X will cancel so it's just y you can now replace y back with h inverse of X and that's the answer to the question so it's 4x - 3 over 1 + x for the final part of this question we have to solve an equation
• Segment 111: 137:30 - 140:00 it's GF ofx = F ofx + G of X so let's work out GF ofx first to work out GF ofx we need to place the whole of the F function inside the G function to do this I like to copy out the G function but replace the X with a bracket so G of X is 2x + 1 so I'm going to write 2 bracket + 1 then inside this bracket we're going to place the whole F function you can see the F function is x^2 - 3 so we're going to write x^2 - 3 inside that bracket you can expand this bracket if you do two lots of x^ s that's 2x^2 and 2 lot of -3 is -6 and then we have this plus one at the end we can simplify the -6 + 1 to give us 2x^2 - 5 so that's g f of x we can now write out the equation the equation says g f ofx = FX plus G of X so we're just going to write out all parts of that so g f ofx first that's 2x^2 - 5 is equal to F ofx Well that's right at the top of the question x^2 - 3 plus G of X which is also at the top of the question plus 2X + 1 now we have a quadratic equation to solve we're going to simplify first on the right hand side we have -3 + 1 that would be -2 now I'm going to try and get this so that the right hand side is equal to0 let's first of all subtract x^2 from both sides if we subtract it from the left we get x^2 - 5 if we subtract it from the right the x^2 will be cancelled and it's -2 + 2x now we can add two to both sides if we add two to the left hand side we get x^2 - 3 if we add two to the right the -2 will cancel so it's just 2x finally we can subtract 2x from both sides on the left this will give you x^2 - 2x - 3 and on the right 2X takeway tox is zero now that we have it in this form we can attempt to solve it this one will actually factorize so if we factorize the leftand side we get x - 3 x + 1 and on the right hand side equals z this will give us two solutions the first bracket gives us x = 3 and the second
• Segment 112: 137:30 - 140:00 bracket x = -1 in this question we've been asked to draw a histogram to do this we're going to add a column to our table this columns for frequency density then we need to work out the frequency density using the formula frequency density equals frequency / by class width if we look at the first group in the table the time goes from 0 to 10 that's a difference of 10 so the class width is
• Segment 113: 140:00 - 142:30 10 so for that one we do the frequency which is 23/ by the class width which is 10 and we get a frequency density of 2.3 we then do the same process as we go down the table the next group goes from 10 to 30 that's a difference of 20 so the class whip is 20 so we do 34 divide by 20 which gives you 1.7 the next group is from 30 to 35 that's a difference of 5 so 38 / 5 gives you 7.6 and on to the final group this time the class width is 15 and also the frequency is 15 so 15 ID 15 gives you 1 the next thing to do is label up and scale your axes the vertical axis is going to be for frequency density and we can see the maximum there is 7.6 so I need to do a scale that goes up to at least 7.6 so I'm just going to do one big Square for each one so it's 0 1 2 3 4 5 6 7 8 that's going to fit nicely and we should label this axis frequency density then along the horizontal axis we need to put the time you can see the times in the table range from 0 to 50 so we just need to make sure we go up to 50 this time I'm going to do two big squares to represent 10 minutes so it'll be 10 20 30 40 50 and again we label this axis this one is time now all we need to do is draw a bar for each of the groups in the table up to its frequency density so the first group goes from 0 to 10 minutes and it has a frequency density of 2.3 so we draw a bar that looks like this the next group is from 10 to 30 with a frequency density of 1.7 so it looks like this then we have from 30 to 35 but the frequency density this time is 7.6 and the final bars from 35 to 50 with a frequency density of one so it goes like this and there's your completed histogram there's no need to shade These Bars in I've just colored them in so it's a bit clearer for you to see in this question we're looking to convert 0.136 recurring to a fraction we're going to start by saying that X is equal to that number in the question
• Segment 114: 140:00 - 142:30 0.13 and then six recurring next what we do is multiply this by 10 if you multiplied X by 10 you'd get 10 x if you multiplied 0.136 recurring by 10 then all of the digits would move one place to the left so we'd end up with one point 3 six recurring then at this point we do a check we're looking to find a pair of numbers where each of the digits that comes after the decimal point matches so in this one it doesn't match you can see
• Segment 115: 142:30 - 145:00 the six is all match at the end of the number but the three doesn't match with the one and the six doesn't match with the three so we need to keep multiplying by 10 so we'll multiply it by 10 again this time we'd get 100x and all of the digits would move one place to the left once again and we'd end up with 13.6 occurring we do this check again and we can check all of those numbers and we still don't have a perfect match so we go once more now if we multiply by 10 again we get 1,000 x and it will be 136.6 recurring you can now see that in the top two numbers every single digit that comes after this decimal point matches this means these are the two numbers we need so what we do is subtract them we do 1,000 x take away 100 x which would give you 900 x and then we do 136.6 recurring take away 13.6 recurring now all of the sixes match up perfectly after that decimal point so they will all cancel nicely so we just need to do 136 take away 13 which is 123 now if we just divide both sides of this by 900 we find that X must be equal to 123 / 900 now you may remember that earlier in the question the first thing we wrote down was that X was equal to 0.136 recurring the number in the question but if x is equal to this fraction then they must both be equal to each other so this is the answer to the question the question does say to simplify the fraction though so we could divide both of these by three you would end up with 41 over 300 in this question we have a quadratic and we need to write it in the completed Square form so we're going to start with X inside a bracket and then all squared and we also need to put a number inside this bracket this number will be half of the coefficient of x if you look at the coefficient of x in the question it's 8 half of this is -4 next what we do is we imagine we expand this bracket we would get x^2 -
• Segment 116: 142:30 - 145:00 8X but we would also get a positive 16 from -4 * by itself so we're going to subtract 16 from this and then also bring down this positive 19 here at the end all we do now is simplify the final two terms here we've got x - 4 all 2 and then we have -16 + 19 which is a postive 3 you can see it now matches the form they ask for in the question so the answer is x - 4 all 2 + 3 for the second part of the question we've been asked to write down the coordinates of the turning point and we've been given a graph y = x^2 - 8x +
• Segment 117: 145:00 - 147:30 19 notice how the equation of this graph matches the quadratic expression from the first part of the question this means that if Y is equal to x^2 -8x + 19 it must also be equal to x - 4 all 2 + 3 when it's written in this form we can find the Turning Point really easily the x coordinate will be the opposite sign of the number that's inside the bracket so we've got a ne4 here so the x coordinate will be positive4 the y-coordinate will always be this number outside the bracket and that's a positive 3 so it's three in this question we're going to select two counters from a bag and we need to work out the probability that both of them are green the first line we need to spot in this question is the probability that the first counter is red is equal to 3 fifths we're told there are 25 counters in the bag so at the start of this 3 fifths of those must be red so let's just work out 3 fifths or 25 to do 1 F you do 25 divid by 5 that's 5 and then to get three fifths you multiply this by three which gives you 15 so right at the start there are 15 red counters in the bag so let's imagine we have a bag and we know it's got 15 red counters in it then we need to read this line here the probability that the first counter is blue and the second counter is red is equ Al to 1/10th now we don't know how many blue counters there are so I'm just going to call that b so the probability that I select a blue counter on the first pick is B out of the total number of counters which is 25 then we've been told we're going to select a red on the second counter so we would multiply this by the probability of selecting a red now the probability of selecting a red will be the number of red counters in the bag divided by the total number of counters in the bag now since we selected a blue on the first pick we can still be sure there are 15
• Segment 118: 145:00 - 147:30 Reds in the bag and then we divide this by the total number of counters however it says the counter is not replaced which means there are now one fewer counters in the bag than there were at the start so since we've already picked out a blue there must now be 24 counters in the bag in total and we know that this is going to equal 1/10th this is now an equation we can solve to find B which is the number of blue counters on the left hand side here if we multiply those fractions on the top we have B * 15 which is 15b and on the bottom 25 * 24 which is 600 and this equals 1/10th you can now multiply both sides by 600 on the left this will give you 15 B and on the right 600 * 1110 is 60 then divide both sides by 15 and you'll find that b is equal to 4 this
• Segment 119: 147:30 - 150:00 means there must be four blue counters in the bag at the start so let's stick four blue counters into the bag as well and now let's focus on green since in the question would told the counters must be red blue or green we can be sure the remaining counters must all be green so we have 15 red and four blue so we do 25 take 15 take four which gives us six so there must be six green counters in the bag we're now ready to work out the probability that both counters are green so we need to select a green on the first go there are six greens so it's six out of 25 and we also need to select a green on the second go but we've already selected one of the greens and remember we don't replace that counter so there are now five Greens in the bag and a total of 24 counters if you multiply these together you get 30 over 600 which you may simplify to 1 20 in this question we have ABCO o which is a sector and we need to find the area of the Shaded region to do this we're going to find the area of the whole sector and then find the area of the triangle and take it off so let's start with the area of the sector you can find the area of a sector using the formula Theta which is the angle for the sector over 360 * by pi * by the radius squar you can see the angle in this sector is 125 so let's replace the Theta with 125 and you can see the radius for this sector is 11 so let's replace r with 11 you can type this straight into your calculator and we find the area of the sector is equal to this next we need to work on the area of the triangle now normally you do area of a triangle equals 1/2 * the base time the perpendicular height but in a question like this it's much easier to use area equals 12 AB sin C where A and B are the sides either side of that angle and C is the angle itself so we can label on here A B and C now A and B are actually both equal to 11 because they're both the radius of
• Segment 120: 147:30 - 150:00 the sector so we would do 12 * a which is 11 * B which is also 11 * s of C which is s of 125 if you type this into a calculator you get this so all we need to do now is subtract the area of the the triangle from the area of the sector so we do 13 1.99 take away 49.5 58 and so on and we end up with this number here the question asks us to give it to one decimal place so if we rounded this off it's
• Segment 121: 150:00 - 152:30 82.4 in this question you're going to need to know your exact trig values so we're going to write this out again but instead of ss and coses we're going to work out those numbers it's important you do this without a calculator so the value of sin 60 is < tk3 /2 the value of sin 90 is 1 COS of 0 is also equal to 1 and COS of 30 is equal to < tk3 over2 we now just need to expand this double bracket we'll start with > 3 over2 ultip by 1 which is just > over2 then we come on to < TK 3/2 * < tk32 here we're just multiplying two fractions together so if we multiply the tops < tk3 * < tk3 that's just three and on the bottom 2 * 2 is 4 so this comes out at -34 then we have 1 * 1 which is obviously 1 and then 1 * < tk3 /2 which is > 3/2 now the first and last terms here will actually cancel out so we're just left with -3/4 + 1 which should give you 1 qu and that's the answer to the question in this question we're told that X is directly proportional to y this means we can write X = KY where K is some constant number we can find the value of K by substituting in this pair of numbers here so let's replace x with 1 and y with four so instead of x equals it's 1 equals and instead of K * Y is K * 4 if you just divide both sides of this by four you find that 1/4 is equal to K or k is equal to 1/4 now that we found the value of K we can write that in the formula at the top so instead of xal k y Y is X = 1/4 y we're also told in this question that Y is inversely proportional to z^ 2 this means that we can write Y is equal to some constant C over z^ 2 I've used the letter c here to differentiate from the K in the first one now we also have a pair of values here we've got when Y is four Zed is three so we can substitute those in so instead of y equals it's 4 equals then it's C over and then we have z s but we know Z is 3 so it's 3^2 which is 9 you
• Segment 122: 150:00 - 152:30 can multiply both sides of this by 9 and you find that c is equal to 36 so we can replace the C in that top formula with 36 in the question we've been asked to
• Segment 123: 152:30 - 155:00 find the value of x when Z is equal to < TK 6 to do this we're going to use both of these formulas here since we've been given the value of Zed we'll start with the one on the right and substitute Zed for < TK 6 so we have y = 3 6 over and then z^ 2 which is < TK 6^ 2 when you square < TK 6 you just get 6 so it's y = 36 over 6 and 36 / 6 is 6 now that we have the value of y we can substitute this into the first Formula it says x = 1/4 of Y but we know that Y is 6 so it's 1/4 of 6 which is 1.5 and that's the answer to the question in this question we've been given a line it has equation 6 y = 25 - 9x and we've also been told about the point a which is on the line and has coordinates k k we've been asked to find the value of K to do this we're going to take the equation of the line and substitute in the coordinates of that point a since both of the coordinates are K we can replace X and Y with K so instead of 6 y it's 6 K = 25 and then instead of - 9x it's - 99 K we can solve this equation to find the value value of K let's add 9k to both sides if you add 9k to the left 6K + 9 K that's 15K and if you add 9k to the right that'll cancel thetive - 9k so you've just got 25 divide both sides by 15 and you'll find that k equal 25 over 15 you can simplify this divide them both by five and you'll end up with 5 over3 so the answer to this question is K is 5 over 3 in the second part of the question we've been told about the coordinates of point B they're 7.58 and there's a second line which is L2 and it's perpendicular to line L1 and goes through that point B and we need to work out the equation of the line L2 to do this we're going to go back to the line L1 that's this one here and we're going to work out its gradient to do this we're going to write it in the form y = mx plus C so we need to get y as the subject to do this we'll divide both
• Segment 124: 152:30 - 155:00 sides by six if you divide by six on the left hand side that will give you y and if you divide the right hand side by six you can divide 25 by 6 which I'm just going to write is 25 6 and divide 9x by 6 so it's - 9x / 6 9 / 6 will simplify because 3 is a common factor there so that's 3 over2 and we can see from this the gradient is now -3 over2 so the
• Segment 125: 155:00 - 157:30 gradient of line L1 is 3/2 now if L2 is perpendicular to this its gradient must be the negative reciprocal of that so the gradient will be positive rather than negative and we do the reciprocal of 3 over2 which is 2 over 3 so the gradient of the line L2 must be 2 over 3 now we can write the equation of the line L2 as y = 2/3 x since we know that's the gradient but we don't know the intercept yet so we'll just write plus C but we can find The Intercept because we know the line L2 goes through this point here so if we substitute in those values into the equation of the line we'll be able to find the value of C so the x coordinate is 7.5 and the y coordinate is 8 let's replace y with 8 so 8 equals and then replace x with 7.5 so 2 /3 of 7.5 plus C this will give you 8 equals if you do 2/3 of 7.5 you get five so 5 + C and then subtract five from both sides and you'll find that c is equal to 3 we can now replace C in the equation to give us the equation of the line L2 y = 2/3 x + 3 in this question we have an equation and we've been asked to show this a solution between the values of 2 and three to do this we're going to take the left hand side of the equation and we're going to substitute in two and then three so let's substitute in two first so we need to do 2 cubed subtract two lots of two subtract six this will give you the value -2 then we'll do the same thing but with three instead so 3 cubed subtract two lots of three subtract 6 and this gives you 15 if you look at the two answers we got here you can see one's negative and one's positive so we've observed a change of sign this is sufficient to show show there's a solution here since the equation has a solution when it equals zero if we've gone from negative to positive there must be a solution somewhere between the values of 2 and three for the second part of the
• Segment 126: 155:00 - 157:30 question we need to show that this equation here can be rearranged to give this one so let's start with the equation and let's carefully one step at a time manipulate it so it ends up with the other equation let's start by adding 2x to both sides if we add 2x to the left hand side that will cancel the 2x that's already there so it's just X cub 6 if we add 2x to the right hand side 0 + 2x is 2x now we're going to add six to both sides that'll cancel the six on the left so it's just X cubed and then 2x + 6 on the right and finally we cube root both
• Segment 127: 157:30 - 160:00 sides if you cube root X cubed you get X and if you cube root the right hand side you're just going to have to write this as the cube root of 2x + 6 which is what we wanted to show in the third part of this question we've been given an initial value x0 = 2 and we've also been asked to use this iterative formula here three times to find an estimate to the solution of the equation so we're going to start with our first value x0 = 2 we're then going to substitute this into the formula to work out the next value which is X1 so X1 is going to equal then we write down the formula which is the cube root of 2xn + 6 but I'm going to replace that xn with a bracket and then inside that bracket I put the previous value which was x0 which was two if you type this into your calculator you'll get this as the answer now to get x2 we do exactly the same process but instead we put inside the bracket the previous number which was this 21544 and so on now if that number is already on your screen in reality you're probably not going to type that whole number out again but replace that with the answer button so you'll type the cube root of two brackets answer close bracket plus 6 and if you press equals you'll end up with this number here we then do this one more time remember it said three times in the question so to get X3 we do the cube root of two lots of this number so 2.17 and so on and once again this number will already be on your screen so you can use the answer button here and if you type all of this into your calculator and press equals you'll end up with this so we've done it three times now so this is our estimate to the solution of the equation for the final part of this question we've been asked to substitute the answer we got in part C into this expression here and then comment on the accuracy so what we're going to do is take this answer here from part C and
• Segment 128: 157:30 - 160:00 substitute it into this and see how close we get to zero so if we do 2.17 9477 396 cubed subtract two lots of that number and then subtract six using your calculator you'll get this number here which is very very close to zero therefore it would be a pretty good solution so if we're going to write a comment we could write our solution is accurate as this is close to zero in this question we've been given a histogram and we need to work out how many students revise for more than 1 hour whenever you're working out a frequency from a histogram we'll need to use this formula here the frequency is equal to the frequency density multiplied by the class width so if we're looking for students that revise
• Segment 129: 160:00 - 162:30 for more than 1 hour that's 60 minutes we're looking for these students here if we look at this bar you can see its frequency density is 0.2 since it goes up just one little square and its class width goes from 60 to 100 so that's 40 so we multiply the frequency density which is 0.2 by the width of that bar which is 40 and 0.2 * 40 is 8 so that bar is representing eight people so eight people revise for more than 1 hour for the second part of the question we need to work out an estimate for the interquartile range of the time spent revising to do this we're going to work out how many people there are in total so let's start with the first bar on the left you can see the frequency density here is four and the class width goes from 0 to 20 so that's 20 so we do four time 20 which is 80 so there are 80 people represented in that first bar now let's repeat this for the next bar the frequency density is 1.6 and its class width goes from 20 to 50 so that's 30 so if you do 1.6 multiply by 30 there are 48 people in that bar and now for the third bar the frequency density is 10.4 and its class width from 50 to 60 is 10 10.4 * 10 is 104 so there's 104 in that bar and for the final bar we've already done this it's 0.2 * 40 which was 8 so there's eight in the final bar if we add up all of these frequencies we find there are 240 people in total so if we want to find the inter quartile range we need the lower and upper quartiles to find the lower quartile we' do 1 quar of 240 which gives you 60 this means the 60th person is representing the lower quartile to do the upper quartile we do three quarters of 240 which gives you 180 so the 180th person will represent the upper quartile the question now is where are these people located so what we want to do is start at the left hand side and try and work out where the 60th person would be if we move from left to right well we know they're going to be in the first
• Segment 130: 160:00 - 162:30 bar since there are 80 people in the first bar if you chop the first bar in half there would be 40 people on either side of it if you then looked at the right hand side of this bar and Cho that in half as well you'd split that 40 into 2 20s and if you add 40 and 20 you get 60 so we can see there'll be 60 people to the left of this line here so if this is where the 60th person is then this value here on the bottom axis represents our estimate for the lower quartile that's halfway between 10 and 20 so we would say the lower quartile is equal to 15 now we need to do the upper quartile so this time we're going for 180 people
• Segment 131: 162:30 - 165:00 well if you add up these first two bars here you get 128 so it's going to be past the first two bars but how much more well if we do 180 take away 128 we get 52 so we need 52 more people which if you notice that's exactly half of the next bar since the bar is 104 and half of that is 52 so the person for the upper quarter will be located here if we read off that value we get an upper quartile of in between 50 and 60 that's 55 we can now work out the inter quartile range by doing the upper quartile subtract the lower quartile so 55 take 15 which gives you 40 minutes in the first part of this question we need to work out the length of the line ad we're going to look closely at the triangle ABD which is this one here we're going to label each of the sides of the triangle including labeling the one we're trying to find a as a so if we label ad as a and the other two as B and C we then label the angle opposite each of those sides with the same letter but a capital so like this now we can use the cosine rule to find the value of a we would write A2 = B2 + c^2 - 2 bc+ a now we substitute in any any of the values we know we don't know a so that's a 2 but we do know B and C so b^2 + c^2 would be 12 2 + 20 2 and then - 2 BC which is - 2 * 12 * 20 and then multiplied by COS of a where a is 30 so it multiplied by COS of 30 if you type this whole right hand side into your calculator you'll end up with a^2 equals this number here to find the value of a we now just square root both sides this will give you a equal the square root of that number and if you square root that using your calculator you'll get this number here the question asks us to round this one to one decimal place so we should give the answer 11.3 in the second part of the question we've been asked to work out the length of the line BC since the shape is a trapezium the top and bottom lines there
• Segment 132: 162:30 - 165:00 are parallel if we draw on this Zed shape here you should be able to spot that angle BDC is also 30° since it's an alternate angle to the other angle that's 30° angle ABD if we now look at the other triangle triangle BCD that's this one here we can label up these sides in the same sort of way we're going to label BC the side we're trying to find is a the angle that's opposite this is capital A the other side we're given 20 that's going
• Segment 133: 165:00 - 167:30 to be B and the angle opposite that is capital B we can now use the sign rule to find the value of a so the sign rule is a over S A is B over sin B let's substitute in the values we know we know capital A that's 30 we know little B that's 20 cm and we know capital B that's 36° to find a you just multiply both sides by sin 30 that will give you a = 20 sin3 over sin 36 type this straight into your calculator and you'll get this answer here once again rounded to one decimal place gives you 17.0 in this question we're trying to find the angle X we're going to look at the triangle on the base of this cuboid here triangle EG H we know the line from G to H is 6 cm you can see that on the diagram and we can also work out that eh must be 8 cm since it's the same as BC which is marked as 8 cm we can now use Pythagoras to find the length of EG so eg^ 2 must be 82 + 62 8 s/ 6 s is 100 so eg2 = 100 then if you square root both sides you find that EG equals 10 so from E to G must be 10 let's pop this on the diagram now we'll go to look at this triangle here triangle ECG we can tell the length from G to C is 20 and we just worked out the length from E to G is 10 and the angle we're trying to find X is down in the bottom left here we can now find the angle X using trigonometry if we label the sides we've got from E to C that's the hypotenuse from C to G is opposite and E tog is the adjacent in this question we're not going to need to use the hypotenuse since we don't have that information so we're going to need to use o and a if we write out soaka TOA o and a both appear in tan so we can write tan of the angle which is tan of X is equal to the opposite which is 20/ by the adjacent which is 10 if we now take inverse tan of both sides we find that X is equal to the inverse T of 20 / 10
• Segment 134: 165:00 - 167:30 type this into your calculator and you'll get this number here and this one wants us to round to three significant figures so it would be 63.4 in this question we've been told that is equal to 3.5 and that's correct to one decimal place and then R is equal to 0.08 correct to one significant figure the first thing we're going to need to do in a question like this is write down the error intervals for V and R let's do the one for V first so if V
• Segment 135: 167:30 - 170:00 is 3.5 to one decimal place it lower bound will be 3.45 and upper bound will be 3.55 then we have R which is 0.08 but correct a one significant figure the lower bound will be 0.075 and the upper bound 0.085 now we've been asked to work out the upper and lower Bound for the value of a and we have this formula for a which is V ^2 / R let's start by talking about the upper bound if we want to make a as big as possible then we want to make v as big as possible since that's the number on the top of the fraction here so we would use the upper Bound for V but once we've done v^2 we're going to divide it by R and if you're going to divide by something if it's quite a big number that will make your number small but if you divide by a small number you'll make your answer big so we want to divide by the smallest possible value for R so for R we're going to use the lower bound so if we do v^2 over R but make sure we use the upper Bound for V and the lower Bound for R we would do 3.55 s divide by 0.075 and that would give you this number here now let's do the lower Bound for the lower bound it'll be the other way around we'd want the lower Bound for V and the upper Bound for R so if we do the lower Bound for V but squared that's three 3.45 s divide by the upper Bound for R that's 0.085 we get this number here so these are the upper and lower bounds for a a must be somewhere in between these two values so we just write those on the answer line but we've been asked to give them to six significant figures so the first one upper bound will be 16833 and for the lower bound 14 0.029 in this question we have a and b which are consecutive integers if they're consecutive means they come after each other so if we had the two numbers A and B B would be the number that comes after a When you're counting
• Segment 136: 167:30 - 170:00 for example now if B comes immediately after a it must be one more than a so we could replace this B with an A + 1 now we've been asked to prove this statement here algebraically what we're going to do is write that out but every time we get to a b we're going to write a + 1 instead so let's start with a cubed plus a b but B is a + 1 plus B Cubed but B is a + 1 so we need to show that this statement is an odd number to do this we're going to expand it all out the a + 1 cubed at the end is a + 1 times by itself three times so let's replace that with a + 1 a + 1 a + one now let's start doing some
• Segment 137: 170:00 - 172:30 expanding so at the front we've got a cubed then if we expand the first bracket it's a * a which is a^ s and then a * 1 which is just a then if we move on to the triple bracket we're going to expand the first two of these brackets first and then write that inside this bracket here and then multiply by a + 1 so if we expand that double bracket we do a * a which is a 2qu we do a * 1 which is a and then 1 + a which is also a so there's 2 a and then 1 * 1 which is 1 now we can move on and expand the triple bracket so let's leave all of the first terms the same a cubed plus a s + a and then we would do a S * a which is a cubed a 2 * 1 which is a 2 2 a * a which is 2 a 2 2 A * 1 which is 2 a 1 * a which is a and 1 * 1 which is 1 now we can do some simplifying by collecting up the like terms we've got a cubed and a cubed that's 2 a cubed we've got a squ another a s and two more so that's 4 a 2 then we've got a + 2 a + a that's 4 a and finally + 1 now when you're trying to show that something's in odd number it's usually easiest to show it's one more than an even number so if we ignore this plus one on the end for the moment but factorize a two out of the other terms we' get two then inside this bracket a cubed + 2 a 2 + 2 a and then we do have this plus one at the end this section here must be an even number since it's two times some number it doesn't matter really what happens inside that bracket I know it's going to be an integer and when I double an integer it's always going to be even even so we can conclude the proof Now by saying that this is an even number so this which is the same thing plus one must be an odd number in this question we need to solve an inequality to do this we're going to pretend for the moment that it's not an inequality symbol but an equal symbol so
• Segment 138: 170:00 - 172:30 imagine we were trying to solve this instead you could solve this one by factorizing the only way to get 3x^2 is having 3x and x and the only way to get a ne5 is if we ' got a five and a one and one of them is going to be positive and one's going to be negative so it doesn't take too long to try numbers and see that you can factorize it like this this would give us two solutions if it were an equation we'd say 3x - 5 = 0 and also x + 1al 0 the first of these if
• Segment 139: 172:30 - 175:00 you add five then divide by 3 gives you X = 5/3 and the second one take away one from both sides X1 these are known as the critical values for this inequality if we imagine we drew a sketch of the graph 3x^2 - 2x - 5 it would look something like this we know it's going to cross the axis at these critical values then we need to consider are we interested in the part of the graph that's above the xaxis or below the x-axis since we want 3x^2 - 2x - 5 to be less than zero we're interested in the part that's below the x-axis which is this part here since it gives values that are less than zero so if we're interested in this part of the graph what x values give you this section of the graph well it's the X values that are in between these two critical values here so we can write the inequality solution as X is in between the values of -1 and 5/3 and that's the solution to your inequality if it were the other way around and we wanted a greater than we need it in two separate parts we need X to be less than -1 or X greater than 5/3 in this question we have shapes a b and c that are similar if shapes are similar it means they're each a scale factor enlargement of the other ones we're first of all going to look at the information we've been given about the heights you can see the height of shape B is 18 and the height of shape c is 63 we can use this information to work out the scale factor enlargement between shape B and C if we do 63 divide by 18 we end up with 3.5 this means all of the lengths of shape c are 3.5 times as long as they are for shape B so the scale factor enlargement for the heights is 3.5 but this won't be the same as the scale factor enlargement for the surface area or the volumes for the surface area we would need to square the scale factor that we use for the heights so this one
• Segment 140: 172:30 - 175:00 will be 3.5 squar and for the volume we would Cube the one we use for the height so it'll be 3.5 cubed now this does mean we can find out the surface area of shape c by doing 720 ultip by 3.5 s and that will give you 8,820 so we can put that in into the table we can't do anything with the volumes though because we don't know the volume of shape b or shape c yet now let's look at shapes A and B you can see we've got the surface area for both of them so let's work out the scale factor enlargement for the surface areas if you do 720 divid 320 you can see it's 2.25
• Segment 141: 175:00 - 177:30 times as big so the scale factor for the area is 2.25 we need to work out the corresponding scale factor for the height and the volume now remember earlier we said the scale factor for the area was the square of the scale factor for the height so if we square root 2.25 we get 1.5 so we could rewrite that surface area scale factor as 1.5 squ this would mean the height scale factor is 1.5 and the volume scale factor is 1.5 cubed we now have all the information we need to fill in the missing blanks in the table let's work out the height of shape a first so whatever this height here is multiplied by 1.5 will give us 18 so if we do 18 divide by 1.5 we get 12 which is the height of shape a then if we look at the volumes we've got the volume of shape a that's 1152 and if we multiply it by 1.5 cubed we'll find the volume of shape B which is 3,888 and finally if we take this volume and multiply it by 3.5 cubed we get the volume of C which is 1 1666 98 in this question we have a triangle that has side lengths 9 10 and 11 CM so it will look something like this we've been told some information about its angles a b and c a is the smallest then B and then C is the largest when you have a triangle like this you can be sure that the smallest angle will be opposite the smallest side so a must be opposite the 9 we must then have B opposite the 10 and the largest angle must be opposite the largest side so C must be opposite the 11 now we can label all of the sides as well as the angles the side opposite each of the angles will be the same letter but in lower case so the 9 will be a the 10 will be B and the 11 would be C we've been asked to show that COS of a is equal to 7 11 and we're not allowed a calculator for this one if you write out the cosine rule for angles it's COS of a is B2 + C S - A 2 over 2
• Segment 142: 175:00 - 177:30 BC so let's substitute in these numbers so we have COS of AAL b^ 2 which is 10 2+ c^ 2 which is 11 2 - A 2 which is 9 2 over two lots of BC which is 10 * 11 so 2 * 10 * 11 now let's work this out well 10 s that's 100 11 s is 1 21 9 s is 81 and 2 * 10 * 11 is 220 on the top here we can add 100 and
• Segment 143: 177:30 - 180:00 121 then subtract 81 and this will give us COS of a is equal to 140 over 220 there's a common factor of 140 and 220 of 20 so if you divide both of those by by 20 you get 7 over 11 which is what we needed to show in this question we need to work out the sides of angle ABC and we've been asked to give reasons for each stage of our working now there are a few different ways of doing this question I'm just going to do one of them but as long as you give very clear reasons you'll be able to get all of the marks still I'm going to mark on ABC the angle we're looking for first that's this red one here then if I mark on angle CDF which is already there at 53° that's the same as this angle over here angle DAC we call this the alternate segment theorem so angle DAC is also 53° so we're going to write this down and give reasons for it so we would say that angle CDF is equal to angle DAC due to the alternate segment theorem next I'm going to look at this angle here you can see this angle is inside a triangle and I have the other two angles so I can just add those ones up which gives me 137 and subtract this from 180 which gives me 43 so the blue angle here must be be 43 and we need to explain that step so we would say that angle CDA equals 43 because angles in a triangle add up to 180° we're now able to find the angle we're looking for angle ABC ab c d is a cyclic quadrilateral since all four of the corners of that shape are on the circumference of the circle there's a theorem that says the opposite angles in a cyclic quadrilateral must add to 180° so all we need to do is look for the angle that's opposite the one we're after so that's the blue one 43° and subtract this from 180 so 180 take away 43 is 137 and once again we'll explain this step so we would say that angle ABC equals
• Segment 144: 177:30 - 180:00 137° and the reason is opposite angles in a cyclic quadrilateral add to make 180 so the answer to this question is 137 in the first part of this question we need to describe a single transformation where points C and D are invariant but points A and B A not this means once we've done the transformation points C and D will remain in exactly the same place place the points A and B are going to move the only possible transformation that fits this description is if we reflect the shape in this line here you can see points C and D are both on this line of reflection therefore they won't move the shape will look like this when it's reflected and you can see that A and B do move so we could describe this one as
• Segment 145: 180:00 - 182:30 a reflection in the line y = -2 for the second part of the question we need to describe a transformation again but this time it's only point B that's invariant there's more than one option for this firstly you could do any rotation of the shape as long as the center of rotation is B and the angle you're going to rotate is not 360 so for example we could do this rotation here you can see point B remains in the same place but all of the other ones move so this is a rotation 90° clockwise about the point11 so once again any rotation is fine as long as the center is11 and it doesn't go 360° or it go back to where it started you could also do an enlargement for example if we enlarge the shapee like this this would be an enlargement scale factor -2 center11 so once again this works because we've chosen the center of enlargement to be the point B this will work for any enlargement as long as the scale factor isn't one in this question we need to solve an equation involving algebraic fractions we're going to multiply the denominators of the fractions together to get a common multiple so we would write both of these fractions again but over x + 5 x -1 so on the left fraction we had X on the numerator but we've multiplied the denominator by x - one so we must also multiply this X on the top by x - one for the second fraction we have a two on the top but we've multiplied the denominator by x + 5 so we multiply the two on the top by x + 5 as well and all of this is equal to -1 now that we have the denominators the same we can combine the fractions to have a common denominator so we've got x braet x -1 subtract two lots of x + 5 and this equals ne-1 now we're going to expand all of the brackets on this numerator here so let's write everything else the same
• Segment 146: 180:00 - 182:30 first then we would do x * X which is x^2 x * -1 which ISX -2 * x -2X and -2 * POS 5 -10 thex and -2X would simplify to give -3x and now what we're going to do is multiply both sides by both of these brackets here if you multiply the left side by these brackets they would cancel so we've just got x^2 - 3x - 10 and if we multiply the right hand side by those brackets we've got negative one lots of those brackets now we're going to leave the left hand side alone and we're going to expand that double bracket but write the
• Segment 147: 182:30 - 185:00 results inside a larger bracket so we've got negative one lots of whatever those brackets give us when we multiply them out so we do x * X which is x^2 x * 1 x 5 * X that's 5 x and 5 * 1-5 again we can simplify the X terms here x + 5x is + 4x now we'll continue to leave the left hand side as it is and on the right hand side we've got negative 1 lots of all of these terms if we multiply those terms by1 they'll remain exactly the same but a different sign so the positive x^2 becomes netive x^2 the positive 4X becomes NE 4X and the Nega 5 becomes positive 5 now let's take this and move it up to the top we want to get all of the terms on the left hand side here and the right hand side to be equal to zero so let's add X2 to both sides first if you added x^2 to the left we now have 2 x^2 and - 3x - 10 and on the right hand side that will cancel the X2 so we only have -4x + 5 now we can add 4X to both sides if we add 4X to the left we've got 2x^2 we add 4X to that - 3x which is just + 1 x and then - 10 and on the right hand side when you add 4X that will cancel then -4x so we have 5 then we subtract 5 from both sides this gives us 2 x^2 + x - 15 = 0 now that we've got the quadratic in this form we can try and solve it this one will actually factorize it would be 2x - 5 x + 3 = 0 and then we get two solutions either the first bracket is zero so 2x - 5 = 0 or the second one is x + 3al 0 to solve the left one we add five to both sides first so 2x = 5 and then divide both sides by 2 x will equal 2.5 for the right one subtract three from both sides and you end up with x = -3 there are two solutions to this equation x is 2.5 and x = -3 in this question we've been given some information about two ships A and B that are traveling directly to a port we're going to try and draw a diagram of what's going on so let's draw the port
• Segment 148: 182:30 - 185:00 first and call that P then we need to try and draw the ships if ship a is traveling on a bearing of 70° it must must be somewhere over here on the left because when we then draw a North Line at a the angle this makes when we draw a line from a to P needs to be 70° then we'll do the same thing for ship B so ship B would be somewhere down here since the angle this time when we draw a North Line at B and then connect it up to P needs to be 20° so this angle
• Segment 149: 185:00 - 187:30 in here is much smaller at 20° we then told the distance from ship B to the port is 35 so we can add this onto our diagram as well and the distance between ship A and B is 30 so if we connect up A and B with a straight line This is 30 and we've been asked to find the bearing of ship B from ship a so the bearing of ship B from ship a is the angle you would turn through if you were at ship a facing north and turn clockwise to face B if you are at a and turn clockwise you turn through that blue angle 70° but you'd have to keep going and turn through this red angle as well and then you'd be facing B so we just need to work at what this red angle is then we can add it to the blue angle to get the answer so how do we find this red angle what we're going to do is draw a North Line at p as well all of the north lines are facing in the same direction so they're parallel so if we look at the North line that's at a and the north line that's at P we have a set of Co interior angles which must add to make 180 so if we draw this angle here this must go with the blue angle to make 180 since the blue angle 70 this must be 100 10 we can now do the same process by looking at the North lines that are at B and P so if we draw in this angle here which is in purple the blue angle is co interior to the sum of the orange and the purple angles so they must all add to make 180 de as well so if we add 110 and 20 we get 130 which means there's 50 left to get to 180 so the purple angle here must be 50° we now have enough information to find the red angle if we look at this triangle here and we label the angle we're trying to find as capital A the side opposite this could be lowercase a the other angle in the question capital B and the side opposite that lowercase b we can now use the sign rule to find the value of capital A so we could write s of capital A over a is s of capital B
• Segment 150: 185:00 - 187:30 over B now let's substitute in the numbers we know lowercase a is 35 capital B is 50 and lowercase b is 30 if we multiply both sides by 35 we get sin a on the left and on the right 35 lots of sin 50 over 30 you can type this right hand bit into your calculator and it will give you this decimal here then to find the value of a we do the inverse sign of this number which comes out at this number here so we found the value
• Segment 151: 187:30 - 190:00 of the red Angle now it's about 63.3 de if we add this to the blue angle we get the answer to the question so 70 + 63.3 is for the first part of this question we need to find an estimate for the acceleration of the go-kart when T is equal to 10 we can find the acceleration of a speed time graph by finding the gradient at that point this graph is curved however so we're going to need to draw a tangent at the point when t equal 10 now everyone will draw a slightly different tangent but you want to draw a line that's straight that just touches the curve at the point when T is 10 so something that looks a little bit like this then we're going to draw a gradient triangle to try and work out this Line's gradient I'm going to draw quite a large one that looks like this and then I'm going to work out the vertical and horizontal distances let's start with the horizontal distance first you can see my triangle goes from zero all the way to 16 so that's a distance of 16 and vertically it goes from 8 to 16 so that's a distance of 8 We Now find the gradient by doing the change in y over the change in X the change in y here is 8 the change in X is 16 so it's 8 over 16 which is exactly equal to 1 12 so the answer here is 0.5 5 now you may get a slightly different answer here depending on which tangent you drew as long as your answer is somewhere near to 0.5 you're probably going to be okay for the second part of the question we need to work out an estimate for the distance that the go-kart travels in the first 15 seconds and we've been told to use three strips of equal width well if it's the first 15 seconds and we need to split that into three equal strips 15 ID 3 is 5 so we need to draw a strip every 5 seconds so if we draw lines that go up at 5 10 and 15 like this and then stop when they hit the curve we can then connect up the tops of those points like
• Segment 152: 187:30 - 190:00 this and here's our free strips the distance traveled is the area under this curve so we're going to find the area of those three shapes the first one here is a triangle so we do 1 12 times the base you can see the base is from 0 to 5 so that's five times the height and you can see the height of this one is about 9 so 0.5 * 5 * 9 is 22.5 now let's move on to the second strip this one want's a trapezium this time so we're going to use the formula for the area of a trapezium which is 12 times the sum of the parallel sides now the parallel sides are vertical on this trapezium so we need to read those vertical distances the one where the line goes up at five is N9 we just did that one before and when it goes up at
• Segment 153: 190:00 - 192:30 10 that's at about 13 so it's the sum of 9 and 13 and then multiplied by the height now perhaps confusingly the height of this trapezium is actually its width at five if you type this into your calculator you get the answer 55 then we do the same idea for the final Strip This is a trapezium again so it's 1 12 * the sum of the parallel sides again the parallel sides are vertical so it's the one that goes up from 10 and that's at 13 and the one that goes up at 15 goes to 14 so it's 13 + 14 and then again times the height which is the width of the strip which is also five type this into your calculator and you get 67.5 so all we need to do now is add up all of these areas if you add all of that up you'll end up with the answer 145 M the first thing we're going to do in this question is simplify < TK 500 < TK 500 is the same as < TK 100 * the < TK of 5 we know the squ < TK of 100 that's 10 so this is the same as 10 * < TK 5 or 10un 5 so let's rewrite the question but write 10un 5 instead of the < TK 500 like this now we're going to look at the denominator of this fraction it's < TK 5 + 1 this question requires us to rationalize the denominator so we're going to multiply the top and bottom of this fraction by that denominator apart from we're going to change the sign from a plus one to a negative 1 so we multiply both the top and the Bottom by root 5 subtract one to multiply these you put each of the numerators and denominators in Brackets like this so 6 - 10un 5 * < 5 - 1 and on the bottom < TK 5 + 1 * < 5 -1 and now we need to do some expanding so let's start with 6 * < TK 5 which is 6un 5 then 6 * -1 that's -6 -10 < TK 5 * < TK 5 to do this one you times the rooot 5S together < TK 5 * < TK 5 is < tk25 which is 5 and then times this by 10 so 50 but of course
• Segment 154: 190:00 - 192:30 it's negative since one of the signs was negative so 50 and finally -1 < 5 * -1 we're multiplying two negative so it becomes positive so it's positive 10un 5 on the bottom we have < TK 5 * < TK 5 which is 5un 5 * -1 that's < TK 5 1 * < TK 5 that's positive < TK 5 and 1 * -1 is -1 now we can do a bit of simplifying we have 6un 5 + 10un 5 which is 16 < TK
• Segment 155: 192:30 - 195:00 5 and we also have -6 subtract 50 which is 56 on the bottom we have five take away 1 which is four but we also have NE aunk 5 +un 5 which gives zero so those cancel out all that's left to do now is divide this numerator by 4 so if we do 16 < TK 5 divide that by 4 that's 4un 5 and- 56 IDE by 4 that's -4 you can now see it matches the form we were given at the top of the question aunk 5 minus B where A and B are integers so a would be four and B would be 14 in this question we need to find the coordinates of the turning point of the curve with this equation here to do this we're going to take the equation and we're going to complete the square on it we're going to factorize two out of these first two terms here so we'd have y equals two and then a bracket then inside this bracket we need X2 to get 2x^2 and we'd need plus 6 x to get 12 x and then we've also still got this- 7 at the end what we're now going to do is focus on what's inside the square bracket here so let's write everything else down the same so y equal 2 braet - 7 and now we'll take what's inside that bracket and we're going to complete the square on that so if we were to complete the square on this it would be X in a bracket all squared we would half the coefficient of x so half of 6 which is three we would then Square this number so 3 squared and subtract it so subtract N9 so these two expressions here are equivalent so we can replace the thing inside the square bracket with x + 3 all 2 - 9 you can now expand that bracket back so we're going to do two lots of that big bracket so it's y equal first of all 2 * x + 3 or^ 2 which is just 2 * x + 3 or 2 then 2 * 99 which is 8 and then we have this -7 at the end the final thing to do to complete the square here is just simplify these last two terms so we have Y = 2 lots of x + 3^2 and then -8 subtract 7 is
• Segment 156: 192:30 - 195:00 -25 now that it's in this form we can find the coordinates of the Turning Point the x coordinate will be the opposite sign of this number here so not positive3 but -3 and the y-coordinate will be this one here so -25 so the coordinates of the turning point of this graph are -3 -25 for this question we need to do some simplifying of algebraic fractions you'll notice the first two fractions
• Segment 157: 195:00 - 197:30 are divided when we divide fractions we multiply by the reciprocal of the second fraction so let's keep the first fraction the same and change this divide to a multiply but then take the reciprocal of the second fraction so a bit like flipping it upside down so the 6x2 - 20x goes on the bottom and the 5x2 - 5 goes on the top at the end it says + 2 x ^1 well X ^1 is 1 /x so this is the same as 2/x now we have loads of factorizing to do if you look at the first fraction on the numerator you should spot we have a difference of two squares that could be factorized as 3x + 10 3x - 10 the square numbers are a dead giveaway there on the bottom we can can also factorize and actually the brackets that are on the top give us a bit of a clue so it's actually 3x + 10 and x + 1 then we move to the second fraction on the numerator we could factorize out of five so it's five lots of x^2 - 1 but x^2 - 1 is also a difference of two squares it's x + 1 x - one so we could replace that with x + 1 x -1 on the bottom we can factorize into a single bracket we'd factorize out 2x then inside would be 3x - 10 and on the end we still have this plus 2X we can now multiply these two fractions together to make one huge fraction we just multiply both numerators and both denominators so we're just going to have lots and lots of brackets so the numerator will look like this and the denominator will look like this and we still have that plus 2 overx at the end now there's lots of cancelling that can be done we can cancel these two brackets and these two brackets and also these two brackets so all we're left with on the left fraction is just 5 bracket x -1 over 2x so let's write that down and plus 2X on the numerator of that left fraction we can expand the bracket so it's 5x - 5 over 2X and since we need to add these fractions together let's make a common denominator by multiplying the
• Segment 158: 195:00 - 197:30 second fraction by two on the top and the bottom so we end up with 4 over 2x we can now add them together we end up with 5x - 5 + 4 over 2x -5 + 4 is -1 so the final answer is 5x - 1 / 2X and you can see this matches the form they asked for in the question ax plus b over C where a B and C are integers so a would be 5 b is -1 and C is two in this question we have a graph that has this equation and also a straight line that has this equation and we've been asked to find the coordinates
• Segment 159: 197:30 - 200:00 of the points of intersection to do this we're going to solve them as simultaneous equations let's take the second equation first and rearrange it to get X as the subject to do this we can just Add 3 y to both sides if you add 3 y to the left you get X and if you add 3 y to the right hand side you get 4 + 3 y now we're going to take the first equation but rewrite this and instead of X write 4 + 3 y since X is equal to 4 + 3 y so instead of x^2 it's 4 + 3 y^2 then it's - y^2 = 48 this is now an equation in y that we can solve if we look at the bracket that's been squared there that just means * 4 + 3 3 y by itself so 4 + 3 y * 4 + 3 y then we have - y^2 = 48 let's expand this double bracket we do 4 * 4 which is 16 4 * 3 Y which is 12 y 3 y * 4 which is 12 Y and 3 y * by 3 Y which is 9 y^ 2 then we have - y^2 = 48 we can do some simplifying on the left hand side we've got 9 y^2 - 1 y^2 which is 8 y^ 2 12 y + 12 Y is 24 4 Y and then we also have + 16 = 48 since this is a quadratic equation we want the right hand side to be equal to zero so we subtract 48 from both the left and right if you subtract 48 from the left we get 8 y^2 + 24 Y and then we do 16 take away 48 which is -32 and this equals 0 we can now take this equation and you can actually divide both sides by 8 you'll notice that 8 24 and 32 are all multiples of 8 so if we divide the left hand side by 8 we get y^2 + 3 y - 4 and if you divide 0 by 8 that's still 0o we can solve this one by factorizing it would be y + 4 y - 1 = 0 which gives you two solutions the first one is y = -4 and the second one y = pos1 now remember in this question we're looking for coordinates of points of intersection so we have the Y values of those coordinates we need the corresponding X ones to do this you can substitute those y values back into one of our original equations but probably the easiest one to use will be this one
• Segment 160: 197:30 - 200:00 here since X is the subject so if we start with when y = -4 then X will be 4 + 3 lots of -4 which gives us8 then if we say that Y is equal to 1 we need to do 4+ 3 lots of 1 which is
• Segment 161: 200:00 - 202:30 s so we're almost there we just need to form these into coordinate pairs when y was equal to -4 x was equal to8 so the coordinate here will be when X is8 and Y is4 4 we also had when y was equal to 1 x was equal to 7 so the coordinate here will be when X is 7 and Y is 1 so we end up with two solutions the first one is8 with -4 and the second one is 7 with 1 for the first part of this question we've been asked to draw the graph of yal f ofx if you already have the graph of f ofx negative f ofx will be that graph reflected in the x-axis so we're going to take all of the key points of this graph and reflect them in the xaxis let's start with the one on the left that's only one square above the xaxis so if we reflect it it will go one square below the next one along is two squares up so that will go two squares below the next one is actually on the x-axis so that won't move then the next one is one below the x-axis so that will go one above and the final one is two above so it will go two below then you can join them up with straight lines just like the original function and it will look something like this in the second part of the question we have a curve and it has this equation here and we're going to transform it by this Vector here this Vector is telling us to go three to the left and then eight up let's consider this graph as a function so F ofx = 7 - 2x^2 if we're going to move this graph three units to the left that's doing the function f ofx + 3 whenever we move a function left or right the number that we adjust it by goes inside this bracket and it's also the opposite of what you might expect if we move left we actually put a plus three if we were to move right we would put a negative3 and then also we need to move this graph 8 up so we would put Plus 8 but this time it's in the y direction so we put that outside of the bracket of the function so we just need to work out
• Segment 162: 200:00 - 202:30 what this function would be to do F ofx + 3 in a bracket you'd write out the same function as F ofx but replace all of the X's with x + 3 so this is going to be equal to 7 - 2 lots of a bracket squared inside that bracket will be x + 3 and then we just have this Plus 8 on the end so all we need to do is expand this bracket out and simplify so we have 7 minus 2 lots of this bracket squared let's Square this bracket now remember think of this as a double bracket so it's X+ 3 * by itself this will give you x^2 + 3x + 3 x again and then + 9 and we also have this Plus 8 right at the end now we can multiply
• Segment 163: 202:30 - 205:00 through by -2 so we have 7 - 2x^2 - 6x - 6X again and - 18 and then this Plus 8 at the end all we need to do now is simplify and we probably reorder these to write the squared term first so we'd have -2X ^2 then we simplify the X terms that's -2X and then simplify the constant terms so we have 7 subtract 18 + 8 which gives you -3 so the answer to this question is - 2x^2 subtract 12 x subtract 3 in this question we've been asked to work out the equation of the line a to do this we're first of all going to draw on the line from o to B and then find its gradient to find the gradient of the line OB we can do the change in y over the change in X this is quite easy for this one because one of the points is the origin so to get from o to B we would need to go two up and then five across so the gradient is 2 over 5 now since the line AC is a tangent to the circle we know a tangent meets a radius at 90° so we have a right angle here this means the line o and the line AC are perpendicular when we have perpendicular lines we can work out their gradient by doing the negative reciprocal so the gradient of the line a must be the negative reciprocal of 2 5 which is 5/2 so we now know the gradient of the line we're after we just need its intercept so we could write down its equation in the form y equal -5 2x plus C to find the value of C we're going to need to substitute in a point that's on the line but we do know that point B is on this line and its coordinates are 52 so we're going to replace y with 2 and X x with 5 so let's replace the Y with 2 so 2 equals and then replace the X with 5 so5 over 2 lots of 5 plus C so we have two equals and then -5/2 * 5 is -12.5 and then plus C if you add 12.5 to both sides you end up with C being 14.5 so the equation of the line AC is y = 5 2x + 14.5 but the question requires us to
• Segment 164: 202:30 - 205:00 give the answer in a very particular form form ax + b y = c it doesn't look like this form yet let's multiply both sides of this by two to get rid of that fraction if we multiply the left by two we get 2 y then if we multiply 5 over 2x by 2 we get -5x and if you multiply 14.5 by 2 you get 29 if we now add 5x to both
• Segment 165: 205:00 - 207:30 sides we get 2 y + 5x = 29 which is now in the form of the answer if we just switch the X and Y terms round we have 5x + 2 y = 2 9 and you can see that's now in the form ax + b yal c where a B and C are integers in the second part of this question we're going to find the area of triangle a that's this triangle here to do this we need to find the coordinates of points A and C now we already have the equation of the line going from a to c that was in the previous part of the question 5x + 2 y = 29 so we know points A and C are both on this line now since a is on the y axis its x coordinate must be zero so if we say when x equal 0 and then substitute 0 into that equation of the line we'd get 5 * 0 which is 0 and then plus 2 y = 29 so we get 2 y = 29 which you can solve to get y = 14.5 this means the coordinates of this point a are 04.5 we can do a very similar process to find the coordinates of Point C but this time since it's on the x-axis the y coordinate must be zero so we say when y equal 0 now if you substitute 0 in for Y into that equation you get 5x + 0 = 29 so 5x = 29 divide both sides by 5 and you'll find that X is 5.8 so the coordinates of Point C are 5.80 we can now find the area of the triangle since we know the base must be 5.8 and the height must be 14.5 so the area is 1/2 time the base which is 5.8 * the height which is 14.5 use your calculator to do this and you end up with 42.5 in this question we've been asked to prove that these two angles here sum to make 90° there are many different ways of doing most proof questions I'm just going to show you one way here as long as your proof is fully explained and makes sense you'll still be able to get all of the marks even if it differs from this one so I'm going to start by marking on this angle here angle ACO and
• Segment 166: 205:00 - 207:30 I'm going to call it X then I have to look at the lines OC and O A they're both radi of this circle therefore they must be the same length so if I look at triangle AO that's an isoceles triangle so this angle over here must be the same size as X so we can label that one as X as well now when we're doing a proof we're going to need to explain all of the steps we do so for this step we could say OC equals OA as they're both radi which means angle AC
• Segment 167: 207:30 - 210:00 equals angle AOC as they're both the base angles in an isoc triangle next we're going to use the fact that the lines o a and CB are parallel if we draw on this Zed shape here going across OA and CB you can see we have a set of alternate angles this means the angle BCA down in the bottom left must also be equal to X since it's alternate to the angle O A at the top so we could say angle OAC equals angle ACB as alternate angles are equal next we're going to add those two angles in the bottom left together to get a bigger angle which has size 2x then if we look at the line OB this is also a radius of the circle so it must have the same length as the line OC so we can mark on this line to indicate they have the same length and then we can also spot another isoceles triangle the triangle OBC this means this angle here must also be 2x we can explain this by saying that OC equals OB as they're both radi and once again angle OCB equals angle OBC as they are base angles in an isocute triangle now if we look at the third angle that's inside that triangle triangle the one up here we could work that one out by adding the other two together and subtracting from 180 if you add the other two together you get 2x + 2x which is 4X so this one must be 180 take away 4X we can explain that by saying the angle Co B = 180 - 4x as angles in a triangle add to make 180° now the next part is possibly the hardest part to spot if we draw on the chord from C to B here this chord makes an angle at the center of the circle it's the one we just found Co which is 180 - 4x this chord also makes an angle at the circumference in the diagram though this one here the angle cab there's a circle theorem that says the angle at the circumference is half the angle at the center so if we half 180- 4X we can find
• Segment 168: 207:30 - 210:00 this angle here which gives you 90 - 2x we should explain this step as well so angle cab equal 90 - 2x as the angle at the circumference is half of the angle at the center now we've almost finished the proof we're proving that the sum of angles bco and cab equals 90° and we actually have expressions for those two angles in the question so the angle bco is in the bottom left that's 2X and the angle cab is the angle at the top we just found 90 - 2x so if we add them together angle bco plus angle cab
• Segment 169: 210:00 - 212:30 that must be equal to 2x + 90 - 2x and if you add those together the two x's cancel so it gives you 90 and that concludes the proof in this question we're told the speed of a car n seconds after its brakes are applied is given by SN we're also told the car is traveling at constant speed of 200 mph this means that the exact moment the brakes are applied the car must be still traveling at 200 mph and we could say this is 0 seconds after the brakes have been applied in which case we have s0 so s0 is equal to 200 we're also told in the question that the speed of the car Falls by 54% in the first two 2 seconds after the brakes have been applied if it's after 2 seconds then we would call this S2 and it's going to 4 by 54% but we know originally at s0 it's 200 so we just need to reduce 200 by 54% you could do this by multiplying 200 by 0.46 and this will give you 92 so we know immediately as the brakes are applied the car is traveling at 200 that's s0 and then 2 seconds later which is S2 the car is traveling at 92 mph we're then told the speed of the car n plus 1 seconds after those brakes have been applied is sn+ 1 and then we're given this formula here this is an iterative formula that we can use to work out the speed of the car at any point if we know the previous one so for instance if we were trying to work out S1 we would substitute the previous value s0 into this formula so that would be 0.8 lots of s0 minus K now we actually know what s0 is we wrote that down at the top it's 200 so we could take say that S1 is equal to 0.8 LS of 200us K we could expand this bracket you get S1 = 0.8 * 2 is 160 and then 0.8 * K is 0.8k we could then do the same process to find S2 using the value for S1 so we would say that S2 is equal to 0.8 lots
• Segment 170: 210:00 - 212:30 of the previous value which is S1 minus K now whilst we don't have the numerical value for S1 we do have it in terms of K is 160 - 0.8 K so let's replace S1 with this so we have S2 = 0.8 lots of S1 which is 160 - 0.8 K minus K we can simplify the K terms inside this bracket so we get S2 = 0.8 lot of 160 - 1.8k and
• Segment 171: 212:30 - 215:00 then we can expand this bracket as well so S2 = 0.8 * 160 that's 128 and then 0.8 * 1.8k that's 1.4 4 4K but earlier on in the question we worked out a numerical value for S2 it was 92 so we could replace the S2 with 92 and say that 92 is equal to 128 - 1.44k now we can solve this equation to find the value of K if you subtract 128 from both sides you get -36 = 1.44k and then divide both sides by NE 1.44 and you'll find that K is equal to 25 the question has asked us to work out the speed of the car 3 seconds after the brakes were applied so this will be S3 so we can write that S3 is equal to 0.8 lots of the previous value which is S2 minus K but we now know the value of S2 and the value of K so we'd say S3 equal 0.8 lots of S2 which is 92 subtract K which is 25 you can type this into your calculator and you'll get 53.6 which is the answer to the question for the first part of this question we need to find the vector a D in terms of A and B to get from a to d you could go from a to c first and then from C to D if we go from a to c we're going in the opposite direction of that Vector 3B so we can change its sign to minus 3B and then to get from C to D we go in the same direction as that Vector a so it's plus a in the second part of the question we've been given lots of information about the lines and points in the question and we've been asked to work out the value of K we're going to start by looking at this ratio here of ab to CD to EF let's mark on those lines now you can see that we know the vector for one of those lines the vector CD so if we write out the ratio and then write the vector for CD underneath the middle part of the ratio so underneath the three and that's a we can use this ratio to work out the other two vectors in terms of a to get from 3 to four you would divide by three and then multiply
• Segment 172: 212:30 - 215:00 by four so we could say that the vector a is 4/3 of a to get from 3 to 17 you divide by 3 and multiply by 17 so we could say the vector EF f is 173 of a let's add these on to our diagram so A to B is 4/3 A and E to F is 17/3 a now we're going to look at the part where it says BD is parallel to c e let's work
• Segment 173: 215:00 - 217:30 out the vector BD to get from B to D we could go from B to a then we could go from a to c and then from C to D and we know all three of those vectors So ba a is the opposite direction of ab which was 4/3 a so that's minus 4/3 a a to c is the opposite direction of that Vector 3B so - 3B and then C to D is that Vector a again so plus a we can simplify these two since they're in a - 4/3 a + a is - 1/3 a so - 1/3 a - 3 B so that's the vector from B to D let's put this onto our diagram as well now if the vector Vector C is parallel to this one then the vector CE must be some multiple of that Vector from B to D we don't know what multiple it is so I'm just going to call that n so it's n lots of the vector -3 a minus 3B which was the vector BD now we're going to look at these next two parts ADF is a straight line and AD to DF is in the ratio 1 to K if a d and DF are on a straight line then we can be sure that ad and DF are also parallel in which case DF is some multiple of AD we found the vector a in the first part of the question a was equal to -3b + a so we just need to work out the vector DF which is going to be some multiple of the vector a d now we actually know what this multiple is in the question we're told the ratio of a d to DF is 1 to K this means the vector DF is K times as big as the vector a d so we can write DF equal kots of the vector a d -3b + a now we could also work out the vector from D to F in a different way we could go the long way round which is from D to C then from C to e and then from E to F and we have all of those vectors as well so from D to C is the opposite direction of that Vector a so minus a from C to e is that complicated One n lots of negative A3 a subtract B and then from E to F is 17/3 of a so we have two ways of writing down the vector DF this way and this way but since they represent the same Vector they must be equal to each other so we could write this Vector is
• Segment 174: 215:00 - 217:30 equal to this Vector let's expand out the brackets so we have K lots of -3b that's -3 KB and then K lots of a so plus k a equal a and now let's expand this bracket too so this would give you
• Segment 175: 217:30 - 220:00 n / 3 a and then -3 NB and then we have this plus 17 over 3 a at the end now since these two vectors are equal we must have the same amount of a a and the same amount of B's on each side let's first of all look at the A's on the left hand side I have K A's and on the right hand side I have negative a subtract n over 3 a and also plus 17 over 3 a so if there are the same amount of A's on each side then I could write that K must be equal to -1 - n /3 + 173 we can tidy this one up a little bit by multiplying both sides by 3 we'd get 3K = -3 subtract n + 17 and then the -3 + 17 will give me 14 now let's do a similar process with the B's on the left hand side I have -3 KB and on the right hand side -3 NB this would mean that -3 K must be the same as -3n if you divide both sides by -3 you get k equal n so K and N are actually going to have the same value so if I look at this formula here that I got from the A's I could replace the N there with a K so it's 3K = 14 minus K we can now solve this equation to find the value of K if we add K to both sides we get 4k = 14 and then divide both sides by four we get K as 3.5 and this is the answer to the question thank you for watching this video if you found it useful please give it a like check out the video I think you should watch next And subscribe so you don't miss out on future videos