2 Quant Basics Arithmetic + Word Problems Part 2

Summary

The video, created by Top One Percent GMAT and GRE, offers a comprehensive guide to understanding arithmetic and word problems, focusing on Venn diagrams, two and three sets, and practical applications in everyday problems such as speed, time, and distance problems. The video breaks down complex concepts into relatable examples and interactive problem-solving sessions, providing a methodical approach to mastering these topics. It emphasizes the importance of understanding through examples rather than theoretical formulas, ensuring that viewers can apply these concepts in diverse scenarios, especially in quantitative aptitude tests like GMAT and GRE.

Highlights

• Break down problems into smaller components for easier understanding and solution finding. 🔍
• Use visual tools like Venn diagrams for clarity in set-related problems to reduce confusion. 🎨
• Take a methodical approach, using tables and structured formats to organize data logically. 💼
• Formulas are useful, but focus more on examples to see how concepts are applied in real-world scenarios. 🌎
• Speed, time, and distance problems can be tackled with simplification strategies to avoid getting bogged down in details. 🚗
• Understanding the relationship between different variables in word problems enhances problem-solving skills. 🔄

Key Takeaways

• Understanding Venn diagrams can simplify complex set problems, turning potential headaches into straightforward solutions. 🎨
• Use matrix formats for two-set problems to visualize and solve them more efficiently. 🧮
• Clarify upstream and downstream concepts with speed adjustments caused by the stream's flow, simplifying problem-solving. 🚤
• A blend of visual aids and problem-solving helps solidify learning and reinforce the application of concepts. 💡
• Working together: Understand how different rates of work contribute to overall task completion, especially in group efforts. 🤝
• Arithmetic problems can often be solved faster with pattern recognition and ratio applications, avoiding cumbersome calculations. 🧠

Overview

The video kicks off by addressing common challenges students face with arithmetic and word problems, especially regarding Venn diagrams and set theory. It takes a deep dive into using Venn diagrams for two and three sets, explaining how to effectively distinguish between different groups and their intersections. This approach is vital for solving complex problems in tests like GMAT and GRE, providing a clear visual representation of abstract concepts.

As the session progresses, the instructor emphasizes the importance of understanding upstream and downstream problems, a common obstacle in speed, time, and distance questions. Through engaging examples, viewers learn how to navigate these tricky waters. This part of the session is designed to simplify problems by showing solutions step-by-step, using relatable scenarios that paint a clear picture of the problem-solving process.

The session concludes with a variety of examples and practice problems that highlight the utility of methods taught. From understanding the dynamics of group work in task completion to unraveling complex word problems using ratio and proportion, viewers are equipped with many tools to enhance their quantitative reasoning skills. The instructor encourages focusing on the examples discussed, rather than merely memorizing formulas, to gain a deeper understanding and ability to tackle similar problems independently.

Chapters

• 00:00 - 00:30: Introduction and Greetings The chapter titled 'Introduction and Greetings' begins with a cordial atmosphere as the participants exchange evening greetings. It sets a polite and respectful tone for the proceeding conversation or presentation.
• 00:30 - 04:30: Venn Diagrams and Matrix Format The chapter 'Venn Diagrams and Matrix Format' introduces Venn diagrams as a tool to visually represent the relationships between different sets. In this specific context, it uses the example of people who play football and cricket. It prompts the reader or listener to analyze and understand the overlap and distinctions within these sets, focusing on interpreting the Venn diagram to identify who participates in each sport.
• 04:30 - 14:30: Solving Problems Using Venn Diagrams The chapter focuses on problem-solving using Venn diagrams, particularly emphasizing the distinction between groups based on shared and exclusive activities. The provided example involves categorizing individuals who only play football and those who only play cricket, illustrating the concept of 'complement' in a set-theoretical context.
• 14:30 - 23:00: Three-Sets Problems and Matrix Format The chapter delves into three-set problems and their representation in matrix format. It explores the intricacies of determining regions or portions that denote different categories or groups based on set belongings, using the example of sports participants.
• 23:00 - 25:00: Solving Three-Sets Problems The chapter discusses solving problems involving three sets by visualizing them as regions labeled A, B, C, and D. It introduces the concept of interpreting these regions in a matrix format, exemplified by categorizing people playing different sports, starting with football.
• 25:00 - 30:00: Basic Arithmetic and Time Problems This chapter explains basic arithmetic through the context of sports, particularly focusing on cricket and football. It explores relations between various groups of people, such as those who play either of the sports or both.
• 30:00 - 38:00: Time and Work Problems This chapter discusses the distribution of people participating in sports activities, specifically focusing on cricket and football. It breaks down the categories into those playing both sports, neither sport, and each sport exclusively. The summary elaborates on understanding and calculating these groupings to find the total number of participants.
• 38:00 - 59:00: Men and Work Problems This chapter discusses the distribution of people engaged in different activities, specifically focusing on how many people are playing cricket. It states the equation for calculating the number of people playing cricket as 'b + c'. It also touches upon calculating those not playing cricket as 'a + t'. The chapter further mentions that a total count can be achieved by adding the vertical segments, described as 'a + interest'. Additionally, it hints at taking a horizontal perspective, looking at the number of people playing football.
• 59:00 - 85:00: Speed, Distance, and Time Concepts The chapter explains the concepts of speed, distance, and time using Venn diagrams. The primary focus is on how to use these diagrams to solve problems. An example is given where the total is calculated as a sum of different components, represented as a + b + c + d, and further explained using a question to illustrate the application of the concept.
• 85:00 - 93:00: Average Speed and Related Problems This chapter discusses problems related to the concept of average speed, with examples that include scenarios involving apartments. It involves interpreting and solving questions to calculate average speeds and understanding the implications within given contexts.
• 93:00 - 115:00: Meeting Point Problems In this chapter, the concept of using Venn diagrams is introduced to solve meeting point problems involving apartments. The chapter discusses categorizing apartments based on features such as windows and hardwood floors. It explains how to visually represent these features and their absence using a Venn diagram, providing a clear method for analyzing and solving such problems.
• 115:00 - 135:00: Upstream and Downstream Problems This chapter delves into the intricacies of upstream and downstream problems using a mathematical approach. It begins with a scenario involving apartments in a building, analyzing the relationship between window presence, flooring types, and percentages. First, 50% of the apartments feature both windows and hardwood floors, setting the stage for further exploration. Then, a particular focus is given to the apartments without windows, 25% of which have hardwood floors. Moreover, 40% of apartments do not possess hardwood floors, prompting a deeper investigation into the distribution and characteristics of those with windows and the prevalence of hardwood flooring among them.
• 135:00 - 153:00: Speed and Ratio Methods The chapter titled 'Speed and Ratio Methods' discusses problems involving percentages and assumptions in relation to total amounts. The transcript presents a scenario about assuming 100 as a total number for the sake of calculation. It mentions that if the total number is considered 100, then 50 percent of the apartments in a building have windows and hardwood floors. The strategy involves dividing the number into portions to simplify calculations with percentages like 50%, 25%, and 40%.
• 153:00 - 168:00: Two-Digit Number and Ratio Problems This chapter explores the topic of two-digit number and ratio problems, focusing on figuring out percentages and total amounts from given parts. The example discussed involves apartments without windows, marking a portion of the total based on specific criteria. The transcript refers to this part of the topic by using an unknown variable, x, representing the total amount. From this total, 25% is calculated for specific features like hardwood floors.
• 168:00 - 171:00: Solving Income and Expenditure Problem using Ratios This chapter discusses the method of solving problems related to income and expenditure using ratios. It draws an analogy with apartments, specifically focusing on the 25% of them that have hardwood floors to explain the concept. The discussion includes understanding how 40% of apartments without hardwood floors fits into the overall problem-solving strategy.
• 171:00 - 177:30: Train Speed and Meeting Time Problems This chapter discusses solving problems related to train speed and meeting time, focusing on understanding proportion and percentage. An example problem breaks down the components of dividing a total by percentages: if one part is 40% of the total 100, then another part must be 60%. Additionally, it poses a calculation scenario requiring the determination of what 25% of a variable 'x' would be, highlighting practical applications in problems such as calculating total numbers of items, like hardwood floors.
• 177:30 - 185:00: Conclusion and Closing Remarks The conclusion synthesizes previous calculations regarding apartments with hardwood floors, establishing that when 25% of a variable x equals 10, then x calculates to 40. The total in context is 100, which helps to further determine values across variables, ensuring coherence in percentages. The final considerations rest on logical deductions from arithmetic progressions and finishing thoughts.

2 Quant Basics Arithmetic + Word Problems Part 2 Transcription

• 00:00 - 00:30 good evening guys good evening sir okay let's start
• 00:30 - 01:00 see venn diagram for two sets see these are the number of people who play football and these are the number of people who play cricket okay can you tell me about these guys
• 01:00 - 01:30 who are these guys only we play football only play football okay all right so i'll be writing it like this these are the people who play football and not cricket okay similarly these will be the guys who will play cricket and not football okay do you know this is a compliment
• 01:30 - 02:00 right ticket and not football what will be these guys outside of the ball electric neither football and neither cricket neither football nor cricket and these will be the guys playing both the games football and cricket so how many portions are there how many regions how many different
• 02:00 - 02:30 regions are there suppose i marked this region as a this region as b this region as c and this region as d the same thing can be interpreted in a matrix format like this if i make if i write football here people playing football i'll write to
• 02:30 - 03:00 the right hand side number of people not playing football similarly if i write people claim playing cricket here i'll write people who are not playing cricket here okay just help me out while filling this table yeah so people playing both cricket and football will be this portion or this block
• 03:00 - 03:30 people playing playing cricket and football which is bb people playing neither cricket nor football d d people playing cricket and not football c c all right ticket and not football i'll write here and people playing football but not cricket which is eight okay this will be your total okay total
• 03:30 - 04:00 will be written here so if i ask you people playing cricket how many people are playing cricket b plus c similarly people who are not playing cricket will be a plus t plus so your total vertically would be a plus interest okay so when you see this thing horizontally as in number of people playing football
• 04:00 - 04:30 it will be a plus b similarly not playing football would be c plus dn again your total would be a plus b plus c plus d okay so the venn diagram for two set can be interpreted like this okay this helps you while solving the question let's see this with the help of a question
• 04:30 - 05:00 first read the question okay so the question says 50 of the apartments the question is talking about apartments
• 05:00 - 05:30 so in a venn diagram format this would have been your total apartments okay have windows and hardwood floors that means if i have to write if this is window this will be apartments without the windows if i write here hardwood floors i'll write here right here not hardwood floor
• 05:30 - 06:00 all right 50 of the apartments in a certain building have windows and floors hardwood floors 25 percent of the apartments without windows okay this should be read together have hardwood floors if 40 percent of the apartments do not have hardwood floor what percent of the apartments with windows this should be read together apartments with windows have hardwood
• 06:00 - 06:30 floors okay i do not know the total number of apartments but see here it is 50 percent 25 percent 40 percent so i am assuming the total to be 100 if total is 100 50 of the apartments in a certain building have windows and hardwood floors this will be this box and this will be this will correspond to 50 percent of the total which is going to be
• 06:30 - 07:00 50 okay 25 of apartments without windows where is this block located apartments without windows left of total the left of total this one so i don't know the value of it right now so let's just assume this value is x so 25 percent of x are hardwood floors so
• 07:00 - 07:30 apartments without windows okay so these two were apartments without windows but i need the hardwood floor one which is going to be the first box are you getting my point this will be 25 of x so here am i clear to everyone yes sir yes yes if 40 of the apartments do not have hardwood floors what am i talking about
• 07:30 - 08:00 this box this will be 40 of the total which is 40. if this is 40 and total is 100 this box will become 60. are you getting my point if this is 60 and this is 50 what will be 25 percent of x equal to and see the total number of hardwood floors
• 08:00 - 08:30 and apartments with hardwood floors is 16 so this is going to be 10. so 25 of x is 10 that is 1 4th of x is 10 so x comes out to be 40 if x is 40 if this thing is 40 this thing will become 60 because your total is going to be 100 if this is 60 this is going to be 10 and this is going to be 30 okay so read the last statement what percent
• 08:30 - 09:00 of apartments without windows that means my base here is apartments without windows what is the value of apart windows sorry yeah yeah so my base would be apartments with windows okay so what will be my base here apartments with windows is 15 60. so what percent of apartments with windows have hardwood floors
• 09:00 - 09:30 okay so this is five by six and i know the value of five by six if we don't know i do one minus one by six which is hundred percent minus sixteen point six six percent which is going to be eighty three point three three percentages all right so this for two sets it's better to use this tabular format or matrix format let's see the three i have a question
• 09:30 - 10:00 okay uh you said is it 25 percent of x or is it just 25 because 25 plus 30 won't be 40 right this is 20 see 25 percent of apartments without windows these are apartments without windows so this is 25 of x this is equal to 10 okay 50 plus 10 is 60 10 plus 30 is 40. yeah so excuse me how did you get 60
• 10:00 - 10:30 here if you added 50 with something how did 25 of x got to 10 see uh 40 percent of the apartments do not have hardwood floors if you're so it should be 0.4 right just like in 0.25 40 percentage of the apartments means 40 percentage i assume the total to be 100 so 40 percent of 100 would be okay 14 like that one 40 so the
• 10:30 - 11:00 remaining thing will be 60 i know this is 50 so this box will be 10 and this box also has a value which is 25 percent of x that is why i equated 25 percent of x as 10. so in these percentage related problems where they are asking the final percentage we can consider it safe to assume a total value right like you did 100 here yeah yeah if you know if the total was
• 11:00 - 11:30 given you shouldn't assume a value but if it's not given and the question asks for a percentage then you can assume a value for sure thank you sir okay so for three sets let's say there are people who may be who may like or not like football cricket okay there are three sports how many cases are possible how many cases are possible then say for example
• 11:30 - 12:00 in a society only three types of games are available football cricket and hockey so how many persons would be there who are interested or not interested how many kinds of persons eight seven okay let me let me give you an example for example there would be someone who is not liking any of the sports there would be someone who is liking all the sports eight cases
• 12:00 - 12:30 or what can be the other cases please tell me one like like footballing cricket but not hockey okay so so these were two cases there would be other cases like people who like only one sport as in maybe football or cricket or hockey and didn't like the other right yes there may be someone who likes two spoons two spots like this like this
• 12:30 - 13:00 or like this and didn't like the others so how many cases are there one two three four five six seven eight so for three sets it's better to use a venn diagram i don't know how to place this in a matrix format because there are eight possibilities okay so if i name this thing name this
• 13:00 - 13:30 thing is a b c d e f g and h what does region h corresponds to and this is going to be your total region h implies no no first scenario this is h what would be all
• 13:30 - 14:00 liking all the spots [Music] for the people who are liking just one sport okay so if this were football cricket and hockey so this would have been a this would have been b and this would
• 14:00 - 14:30 have been liking hockey c and this would be exactly to d e and f football and cricket would be region e similarly cricket and hockey cricket and hockey cricket and hockey f and similarly it will be d football and cricket this and g is the intersection of all the three
• 14:30 - 15:00 sports that's that is people who like playing all the sports okay so there are a total of eight regions all right let's do this thing with the help of a question let's understand it i could explain uh the edge part in the previous slide so this is football cricket and hockey okay and these are people outside football
• 15:00 - 15:30 cricket and hockey sets so these people do not like football or cricket or hockey so neither call it ticket okay first of all read the question i'll draw the diagram here you know we'll be framing equations but the equations equations will look a bit tough but they are not actually tough once you
• 15:30 - 16:00 solve more questions you will get acquainted with the equations okay three types of clubs are there and the total of how many members so that's why i am writing a 59 here okay
• 16:00 - 16:30 what does this statement mean is required to sign up for a minimum of 1 and a maximum of 3 academic clubs there will be no none so what about the region outside 0 this will be 0 because everyone has to sign up for a minimum of 1 and a maximum of 3 so this region will become 0 that means everything i mean this total
• 16:30 - 17:00 is going to be 59 so first of all i'll write a b c d e f and g so could you tell me the value of a plus b plus c plus d plus e less f plus g this is going to be 59 59 the three clubs to choose from are these three clubs poetry history and
• 17:00 - 17:30 writing a total of 22 students sign up for the poetry club poetry club has 22 students a total of 22. so how can i write this 22 a plus d plus g a plus d plus e plus g this yes a plus d plus e plus g is going to be 22 27
• 17:30 - 18:00 students for the history club b this region plus this region plus this region plus this region the whole circle for or this whole set for history b plus e plus f plus g and similarly for writing which is 28 and this is c d f and g c d
• 18:00 - 18:30 f and g okay if six students sign up for exactly two clubs what does it mean exactly e plus d plus f what is what is what are the regions for exactly to d e so d plus e plus f is equal to 6 i know this value and i i can use this value in any equation all right d plus e plus f
• 18:30 - 19:00 so only include g in this exactly to exactly two oh okay g has three all three so if i put the value of d plus e plus f as six here what will i get a plus b plus c plus g is equal to if this 6 comes to the
• 19:00 - 19:30 right hand side 59 minus 6 which is 53 okay i'll say this is my equation 1 okay what more should i do see i have to eliminate a plus b plus c okay because i need the value of g i need to find out g so i'll have to eliminate a plus b plus c using these three equations
• 19:30 - 20:00 can you make something yes yes a plus b plus c if i look at these three equations at the same time there are two d's two e's and two f's that means if i sum it up i get two times of d plus e plus f and i know the value of b plus e plus f okay similarly there is a b and c if i add it up that will be
• 20:00 - 20:30 a plus b plus c plus 2 times of b d e n f f b e f plus 3 g g plus g plus g 3g is equal to what would be the sum see 27 plus 28 is 55 plus 22 is 77
• 20:30 - 21:00 what is the value of 2 times of d plus e plus f well but if we subtract d e f and d f eliminates right how okay no no you know the value of d e n f b e and f so you need to use that i mean what if you subtract you won't get anything i mean you get b minus e something or d minus
• 21:00 - 21:30 f something um i have a question uh how can we multiply two by adding all d e f together won't it be like two d separately two e separately and two f separately instead of adding all and then multiplying by 2 2 d plus 2 e plus 2 f plus 3 g if if you add all the three equations this will be the result right a plus b plus c plus 2 d plus 2 e plus 2
• 21:30 - 22:00 f plus 3 g is equal to 77 can't you take a 2 common from here ok right two times of d plus e plus f and the value you know this will be two times of six which is twelve so i'm writing it in on another page this will be
• 22:00 - 22:30 a plus b plus c plus 2 times of d plus e plus f which is 6 plus 3 g is equal to 77 and one more equation was there a plus b plus c this equation a plus b plus c plus g is equal to 53 so if you take this to the right hand
• 22:30 - 23:00 side this will be 77 minus 12 which is going to be 65 and a plus b plus c plus three times of g so these are my two equations and i need to eliminate a plus b plus c what would you do right now b plus 53 you can just subtract so plus 53 is equal to 65
• 23:00 - 23:30 either subtract you know you can subtract right if you subtract what will happen to the left hand side this a plus b plus c and a plus b plus c will get cancelled out and on the left hand side there will be 3g minus g on and on the right hand side there will be 65 minus 53 this is going to be 12. so 2g is 12 so your answer g comes out to be 6. all right
• 23:30 - 24:00 have a look at it again the total was 59 but i know there are no one who is playing neither of the sports because everyone has to sign up for a minimum of one and a maximum of three so the outer region or the value of h is zero okay and hence i know that these a plus b plus c plus d plus e plus f plus g will be equal to 59 that's my first equation and then the question
• 24:00 - 24:30 gave me idea about the number of students in poetry club which was 22 which is equal to a plus d plus e plus g and similarly for history and similarly for right writing yeah okay so i need to use the value of b plus e plus f hence i put the value of d plus c plus f as 6 in the in this equation which gave me the result a plus b plus c
• 24:30 - 25:00 plus g is equal to 53 i need to eliminate a plus b plus c and i have these three equations i saw that there are two d's two e's and two f's involved and that is where i get the clue to add all the three equations so if i add these three equations i got a plus b plus c plus two times of d plus two times of e plus two times of s plus three is 77 i put the value of d plus e plus f as six so 2 times of 6
• 25:00 - 25:30 became 12 i sent that 12 to the right hand side and so your equation became a plus b plus c plus 3 g is equal to 65 i had another equation a plus b plus c plus z is 53 and if you subtract your answer is going to be 2g is equal to 12 and g comes out to be 6. in this way you can solve the two set and three set questions and more questions will be discussed on the next class which is going on
• 25:30 - 26:00 saturday so we'll take the class can i move forward okay so this is a very important topic time and verb time speed and distance let me ask you a few things say for example when i say a can do or a can read 20 books in
• 26:00 - 26:30 five days okay so if i ask you how many books can a read in one day four four see 20 books in five days so in one day 20 by five which is going to be four books please try to understand you know uh in my lecture what should you focus upon focus on the examples that i'm giving you do not
• 26:30 - 27:00 focus on the formulas for focus on the questions that i'm giving you and through questions i'm coming to a formula that formula is not as important as the examples that i am discussing okay so please focus on the examples so four books in one day four books in one day okay if i write books per day if i write days here
• 27:00 - 27:30 if i write books here what is the relationship between the three things that i have written weighted average weighted average no relation i'm just asking a relationship there is 20 there is 5 and there is 4. direct personality ocean i'm not here you are making it too difficult see i'm saying i'm saying in five days
• 27:30 - 28:00 20 books and books how many books per day four books what's the relation between five four and twenty ratio one time and uh division five so this is kind of time this is kind of your rate in time speed distance this becomes your
• 28:00 - 28:30 speed and this is your work done which is distance in your time speed distance so time into rate is verb done okay when i say four books per day when i say two units per day what do i mean in one day i read four books in one day
• 28:30 - 29:00 i read or i do two units of work okay please be very clear about it are you getting my point so i can say if the number of days were given and books per day it can be said as it can be said as rate or efficiency and if this was not given and if this work done is also not given and if i had to assume the value of work
• 29:00 - 29:30 what values can i assume of the work done a multiple of five okay you can either assume a one or an x or a y okay but i would want to assume a value which makes my calculations easier so i see that there is a five here this will be five and two something multiple of 5 that is why i'll take the value here as a multiple of
• 29:30 - 30:00 5 okay you guys are getting this point very good so let's see with the help of an example see in the question it is given i'll always write here days units per day and units this will be my format just to make sure that you don't miss anything between okay so when i say a does a piece of work in five days did i tell you about
• 30:00 - 30:30 the work done anything about work done how many units or what amount of work no i am saying b does the same amount of work same amount of work that means the work done is same whatever is here is here a does a piece of work in five days and b does the same piece of work in 15 days okay if they work
• 30:30 - 31:00 together okay doing the same work how long will it take for a and b to do the same work okay is someone's answer 20 or 10 i mean a plus b what will be the instantaneous answer that you can give
• 31:00 - 31:30 or what can you interpret from this data how long are they going to take not 20. no no no i am not asking you the answer interpreting this data if i can do a piece of work in five days and one of my friend can do the same piece of work in 15 days how much time when we work together will be less than 15 days less than 5 for
• 31:30 - 32:00 sure than 5 less than 15 won't it be less than 20 20 so i am doing a work in five days and if i take help of someone i'll do the same work in 20 days i mean is he making my work or destroying my work i will take less time because i alone can do this work in five days and someone who is helping me out
• 32:00 - 32:30 while we are working together we can do the same work in less than five days that is why i am taking his help are you getting my point this is just the interpretation okay you know answer will come so all these three blocks are the same same work that means whatever that is going to come here will be a multiple of 5 and a multiple of 15 i'm assuming it so a multiple of 5 and a
• 32:30 - 33:00 multiple of 15 can you name some multiple of 5 and 15 some common multiples 15 3 18 see someone someone is saying three it's a factor 15 15 45 yeah yeah good good good okay so i'll assume the minimum value 15 15 okay so what will come here
• 33:00 - 33:30 for eight three three and one three and one any in your mind it should be 5 into something is a 50 so this will be 3 in your mind 15 into something is a 15 this will be one okay let me share something very valuable and important and i don't think you have you know observed it so if i were to
• 33:30 - 34:00 multiply something to three to get to four what should i multiply by three four by three four by three if i were to multiply see this is a star and this is a hash if i were to multiply something to a star that means if you understand this thing
• 34:00 - 34:30 your calculations in this topic will be so so easy okay let's see if they are working together see in one day a does how many units in one day see three units per day if i interpret this data in one day a does three units of work similarly in one day b does one unit if they work together
• 34:30 - 35:00 in a single day how much work will they do four units if here it is four if here it is 15 what will come here see yes yes very good see i want to get a 15 and here it is something either you can say x into 4 is 15 so x is 15 by 4 very good well and nice but i don't want to use this thing i want to save my time a bit see here it is 4 that means whatever
• 35:00 - 35:30 is here 4 is not coming to the right hand side that means 4 must have gotten cancelled with something that is why in the denominator there has to be a 4 and i want 15 to the right hand side that means in the numerator there has to be a 15 so your answer comes out to be 15 by 4 and is 15 by 4 less than 5 yes if you were to calculate the value of 15 by 4 quickly can someone tell me the method how to
• 35:30 - 36:00 calculate this value less than four or something around three plus one four is equal to one plus eleven by four one plus eleven or something twelve by four plus three five four four four minus four minus one by four see either this is sixteen minus one either this is twelve plus three okay this is four minus 0.25 this is 3 plus 0.75
• 36:00 - 36:30 whichever seems easier to and but this should happen in your mind correct so if you understand this thing this topic is going to be a cakewalk for you a does a piece of work but yes there will be certain levels to the questions okay it does a piece of work in 10 days number of days it will be units per day and here it will be units okay it is a piece of work in 10 days we does
• 36:30 - 37:00 the same piece of work same piece of work i am giving you that information b does the same piece of work in 20 and c does the same piece of work in 30 days how long will a and b take to finish the work finish the same work how long will everyone take when they are working together to finish the same work can someone tell me the method by which
• 37:00 - 37:30 we solve this question [Music] okay all right i'm telling you i am asking you the method method whatever we are discussing right now see this box per day 10 units 60 60 see whatever is here will
• 37:30 - 38:00 be a multiple of 10 multiple of 20 and multiple of 30 so why can't we just take my common multiple of 10 20 and 30 which is 60 you can also take 120 that will give you the same answer but just for simplicity i'm taking this as 60 60 60. here it will be 3 here it will be sorry here it is 6 here it is 3 and here it is
• 38:00 - 38:30 2 okay whenever a and b are working together how many units will they do and what one day nine units nine units here it is sixty so your answer is going to be sixty by nine sixty units of work to be done by a b and c how many units work this single day 6 plus 3 9 plus 2 11 so this value is going to be 60 by 11 are you getting my point
• 38:30 - 39:00 if someone has a problem please say okay i'll move forward next question a builds a wall in 50 days b destroys the same wall in hundred days
• 39:00 - 39:30 if both are working together how many days in how many days will the wall be completed is it possible or not possible
• 39:30 - 40:00 the other guy would have to wait 50 more days or like would it start after that like do they do it simultaneously they're doing it together yeah it's possible actually yes possible a minus b see this is days this is
• 40:00 - 40:30 units per day and this is units all right so a builds the wall in 50 and b destroys it in 100 days and i have to find out when they both work together the same wall can be built in how many days is it possible or not yes is it yes it is possible why because a is speed or efficiency is more with respect to b okay here it will be a multiple of 50
• 40:30 - 41:00 multiple of 100 common multiple of 50 100 would be 100 b builds the wall and sorry a builds the wall and b destroys the wall that is why i am taking a negative sign here is everyone getting my point here because b is destroying the wall is building a wall
• 41:00 - 41:30 okay so if they work together in one day how many units of work will they do two minus one the other one unit one unit so when they work together they'll take 100 days to build the same so um mathematically it's making sense but uh if you could just theoretically like explain on the 50th day won't be be destroying
• 41:30 - 42:00 the wall that is making so that you won't be able to complete it a hundred percent the case which you are taking is of alternatives i mean you are first assuming that a is building the wall and then b is destroying a is building b is just trying okay let's take let's take just as in like not not as an alternate case but like you know how let's say how we do integration where you believe that some percentage happens per day so if i do let's say 20
• 42:00 - 42:30 of a wall in a day and b is destroying let's say 10 percent of it won't possibly i don't know maybe i'm wrong but no no i mean this this is a story for another day i mean you know uh so for now a builds of all in 50 we destroy is the same in 100 a is doing positive b is doing negative work okay that's that's it i mean that's all the question wants to ask you okay thank you all right
• 42:30 - 43:00 okay i'll ask you um so i'm sorry i'm just repeating his question again but is it even logically possible i mean a is doing a positive okay he's building something two units and b is destroying as in okay all right all right let's let's take an example there is a monkey okay and this is a poll of
• 43:00 - 43:30 okay i don't know a monkey takes in one jump in one jump a monkey is here for example and in one jump he goes three meters up okay three meters up and he comes two meters down in a single jump this happens
• 43:30 - 44:00 what did i say in a single jump he moves three meters up and since there is less friction he kind of slips and comes down two meters okay if the pole's length is of 15 meters how many jumps okay okay you have asked me the question uh uh i mean those two guys i mean there's
• 44:00 - 44:30 this boy and this this girl have asked me this question will answer this question okay nobody nobody will answer this question how many jumps are required i mean this is the only case that i can take right now to you know make you explain otherwise i'll think about something afterwards and then tell you 15 inches 15 jumps okay come on the other one
• 44:30 - 45:00 come on come on come on 15 sir 15 jumps okay any any other answers 13 jumps 13 zips okay so in a single jump the monkey moves one meters up in five jumps the monkey will move five
• 45:00 - 45:30 meters up in 10 jumps the monkey will move 10 meters up in 12 jumps the monkey moves 12 meters up now he is now he is on 12 meters how many meters are remaining three meters what will happen in the next jump
• 45:30 - 46:00 the thirteenth one he will beat at the top why would he slide now okay you are saying you are assuming a case wherein the work gets completed and b builds some extra wall and then this aid builds this wall and b destroys it and again the work comes to normal that is not the case a will be the last one to work on that
• 46:00 - 46:30 wall are you getting my point otherwise there'll not be any work done okay so in this case the monkey doesn't strike down two meters in the last jump yeah why would he he has reached at the top okay thank you sir [Music] okay the similar case is with pipes and tanks or you can say pipes and cistern okay there is a pipe say for example
• 46:30 - 47:00 this is a tank there is a pipe inlet pipe p1 there is an also a pipe p2 and this is my outlet pipe p3 okay p1 fills a tank in 10 minutes let's say p2 fills the same tank in 20 minutes and p3 empties the same tank in
• 47:00 - 47:30 let's say 60 minutes if all the pipes are open together these two are filling and this is an outlet pipe okay if all the pipes are open together how long will it take to fill the tank
• 47:30 - 48:00 i have a great program for you what is it 75 75 i i don't know the answer i mean i have just randomly taken the values
• 48:00 - 48:30 what will be the lcm 60 60 60 it will be plus 6 here plus 3 here and minus 1 here so i'll have to fill 60 and 6 plus 3 nine minus one eight what will come here sixty by sixty by eight which is nothing but fifteen by two i guess and which is
• 48:30 - 49:00 seven point five minutes so these were only the basic questions okay just just ah let me add a few more twist i would like to ask you some things if four men do a piece of work in 11 days four men takes 11 days to finish a work one man will take how many days
• 49:00 - 49:30 44 24 11 by 4 or 44 11 eleven by four okay so four men are working and they finish a work in 11 days if only one man works will take 11 by four days the reason that you know you some guys
• 49:30 - 50:00 are saying this answer because they are assuming that 1 kg of rice is 10 rupees so 2 kg of price would be 20 this is direct release is this a direct relation between between men number of men and time i'm not saying between work and men and work and time relationship between men and the time taken if more men then less time so this will
• 50:00 - 50:30 be inverse proportion okay that is why the answer would be 11 into 4 are you getting my point so one man will take 44 days to finish the book are you guys getting my point [Music] okay that's the same question i think
• 50:30 - 51:00 but just before this if i were to draw our chart this is units this is units per day and this is the number of days you know when you are so efficient in calculating just forget all these values and just calculate but for now for starters you make this chart and you perform the calculations okay see four men working together
• 51:00 - 51:30 all day can finish a piece of work in 11 days that means here it is an 11 can you tell me what will be here 7 i don't know 4 by 11 why i i still don't know the number of units right now and there are no value no other value to take lcm right now
• 51:30 - 52:00 if this is 11 and see four men are working do i know the efficiency or the rate of a single men no either you take the value of x as the rate or x units per day by a single min or you take one unit this is the efficiency of a single man so what will be the efficiency of four men
• 52:00 - 52:30 see one men can do one unit per day so four men can do four into one units so here if this is four into one how many units of work will be done 24 44 units okay now read and answer the question but two of them having other engagements
• 52:30 - 53:00 that means out of these four men two have other engagements and they work only half time as in in one day they are working only half a day the one one guy is working half a day that means he is doing how many units per day one by two units per day and the other men one side and the other two men
• 53:00 - 53:30 so together in a single day how much work they are doing 2 3 by 4. this is see this is 0.75 and 0.75 is 3 by 4 so 8 plus 11 by 4 so if they are working together okay i mean now the conditions have changed and they are doing 11 by 4 units per day
• 53:30 - 54:00 and they have to complete this one these 44 units of work so what will come here see there is 11 here and there is 44 here okay so if i multiplied this thing by 4 1 4 and 4 got cancelled that means i need another 4 here so your answer is going to be 16 your answer will be 16 days
• 54:00 - 54:30 so i i didn't really get this one can we just summarize again yes 16 first of all make this chart this will be units this is units per day and this will be number of days how many men are there for men do i know
• 54:30 - 55:00 the efficiency i just know that 11 days and four men are there and they are they can finish a piece of work in 11 days that's the information i have while reading the first statement all right now another statement says okay forget about another statement let's see do i know the value right now see do i know the value of efficiency or the
• 55:00 - 55:30 rate right now how many units per day i don't know i'm just saying if the efficiency of a single min was x units then the efficiency of four men would have been 4x i am saying in place of x since nothing is given i am assuming this value to be one that each men one men is doing one unit per day
• 55:30 - 56:00 so four men will do four into 1 units and that is why i have written 4 into 1 here that means a total of 44 units of work is there does anyone have a problem with this this is an ideal situation what happens after that but this didn't happen but two of them having other engagements as in the rest of the two
• 56:00 - 56:30 will work one unit each as in two units per day what about the other two these two these two people but two of them having other engagements can work only half time so in a full day if a single person is working is doing one unit so in half days he'll do one by two and in quarter of the day he'll do one by four so in a total of one day how much of how many units will get
• 56:30 - 57:00 done by all these four men 2 plus 0.5 2.5 plus 0.25 which is 2.75 which is nothing but 11 by 4 and they have to do the same piece of work which is 44 okay either you take this thing as x or of y so y times of 11 by 4 is equal to 44 so 11 times 4 so y comes out to be 16 this is one way or
• 57:00 - 57:30 you try it was you know you tried to do this right here this is 11 and here it is 44 so 11 into 4 might have been 44 but there is also 4 here and since these could have gotten cancelled you multiply another 4 here so y is equal to 16 is going to be your answer can i move forward thank you all right uh do you get my point it's just the method of the presentation
• 57:30 - 58:00 right which is that so it's basically 44 divided by 11 by four right yeah yeah but that is from i mean that is from you know my saying and my calculating there in this box right now says that you know division i find division difficult i find multiplication easier that is why i'm doing like this i mean i want yes yes that is only excuse me sir how how did you get
• 58:00 - 58:30 eleven by four can you please explain once again okay so uh two men were working as it is one units per day out of the four but the other two since they are having other engagements one is working only half time in a single days he is working half time that is in a full day one unit in a half day it will be one by two units in quarter
• 58:30 - 59:00 of a day it will be one fourth so the total units in a single day while having these conditions would be two units plus one by two plus one by four one by two plus one by four is what 3 by 4 so this is 2 plus 3 by 4 4 into 2 8 plus 3 11 by 4 so this was 11 by 4.
• 59:00 - 59:30 so so i have a doubt in the first case we gave four into one that is four men uh puts up for one unitary yeah so uh that piece of information was obtained from the first line of the question yeah yes okay then in the second line how did we conclude that two of them are doing uh a one unit of work per day because the question talks about the
• 59:30 - 60:00 other two men those who are having the engagements rest of the things are going to be same as the first thing right only though those two persons are having other engagement that is the other two persons are working one units each okay so if such an information is not given we can uh conclude it from the previous statement right if the question maker wants you to have it he would have given you the information about all the four individuals okay and sir are these questions from
• 60:00 - 60:30 the solved examples of the human guide yeah yeah yeah okay thank you continuously okay solve this question first of all read the question i will just this way of presentation can always be used for this kind of case right yeah thank you
• 60:30 - 61:00 see 20 men can complete a piece of work in 20 10 days sorry so number of days given is 10 do i know the efficiency of every man yes i do not i don't know the efficiency of these men i am assuming the efficiency to be one units per day for a single man
• 61:00 - 61:30 for one man the efficiency is one unit per day so for 20 men the efficiency would be 20 into one so in a single day 20 min though 20 units is everybody getting my point so the total units that i would have to work will be of the total work done will be 200 units
• 61:30 - 62:00 is everybody getting my points this is in the ideal situation i didn't get the 20 how did you get the 20 if i assume the efficiency of one man here it would have been 20 x okay but i am assuming x to be one i am saying if the if i assume that in a single day a single person does one unit of work so 20 men will do how much work
• 62:00 - 62:30 20 units so so can you uh change or can you edit the terrible you know heading as work into unit i mean or number of men into work done i'm not understanding your question see what we did in the table was 20 men into num uh into the number of
• 62:30 - 63:00 units i mean the unit did for a day yeah so a man number of men in due unit right unit per day yes that should be the terrible heading right instead of just unit per day no no it will be units per day but this is for 20 min not for a single man okay this is 200 read the question after it but after
• 63:00 - 63:30 every four days what is happening five men are called off how can you interpret this information this is new information after every four days that that means for four days how many men work for now initially 10 15 20 they all 20 started the work but
• 63:30 - 64:00 after 4 days 15 15 after 4 days 10 after 4 days 5 so won't that be 14 14 every time five men are called off five men so four days first four days 20 men will work so everyone has an efficiency of one
• 64:00 - 64:30 unit so 20 units so how much of work will get done 80 units so how much of the work am i remaining with see i have to do a total work of 200 units out of which 80 units has been done how much of the work is remaining 120 all right after after these fours four days the next
• 64:30 - 65:00 four days how many men are going to work 18 15 how much work will get completed 60 out of the 120 so i am left with 60 units okay after after these four days for four days how many men will work so i'll have 40 and 20 remaining 20 units
• 65:00 - 65:30 okay after that what will happen how many men five so five units if i work for four days now the work will get completed so i have completed the work see 80 plus 60 140 plus 40 180 plus 20 is 200 so how much time 16 days sorry
• 65:30 - 66:00 are you guys getting my point okay let's solve this question do this by yourself if you practice 10 or 15 such questions this topic is going to be a cakewalk just practice 10 15 questions
• 66:00 - 66:30 is [Music] they were all started at the same time
• 66:30 - 67:00 okay so uh first of all let's assume the value of work what will be the value lcm off or the common multiple of 15 20 and 30 which is going to be 60 60. so 60 60 60 there will be two here there'll be three here and there'll be four here they were all started at the same time
• 67:00 - 67:30 okay after five minutes that means they were all they were all working for how many minutes five minutes see for five minutes they are working together and after five minutes something happened right that means they work together for five minutes in one minute together they can do how much how many units of work 99 units that means they have already
• 67:30 - 68:00 done 45 45 units of work and they are left with just 15 units which pipes are the first two pipes that is a and b were turned off so only c was remaining and i know in one minute c does how many units of work so what will come here seven point five seven point five seven point five see sometimes in the
• 68:00 - 68:30 question in what time will the time be filled okay so if the question talks about total time 12.5 total point total time is 12.5 and the time after that five minutes would be 7.5 the question will specify in what time in what total time will the time be okay the question will be clear but you know read the question carefully in what more time
• 68:30 - 69:00 in what total time will the time be filled the question will be like that okay and you find that okay abc five abc uh five these five five minutes yeah uh do you want me to explain okay okay uh it's just that i'm a bit lost with with this table it looks like super easy to find answer i did like one divided by 15 plus r1 divide by 21 out of 30 i went through like a whole
• 69:00 - 69:30 process like taking like 10 minutes but through your way is much more faster yes yes i mean these two are basically the same process uh process but the difference is in assuming the values there you are assuming the work to be one units and here i am assuming the work to be a multiple of the time which is being given to me the 9 you found by some by this summing the 4 plus 3 plus 2 is it
• 69:30 - 70:00 yeah in 1 minute a does four units of work in one minute b does three units and then all right okay and in the case of time we can't add those right because it will be inversely proportional yeah yeah you can't add the time i mean you can add the efficiency one minute in one minute they are working together and they are doing some units you can add that you can't add the time so we get five in the question after five minutes okay okay
• 70:00 - 70:30 the two pipes were turned off that means the three pipes were working together for how many minutes five minutes all right yes thank you okay see like time into rate is was equal to the work done time into speed is equal to the distance
• 70:30 - 71:00 okay so if the speed is in miles per hour the time will be in r and the distance will be in miles if the speed is in meters per second the time will be in seconds and distances in meters okay so three important form i won't say these are formulas same i have a doubt can you go to the last previous
• 71:00 - 71:30 in the a plus b plus c column how did the 45 come under units just a doubt they are working together for five minutes and in one minute they do nine units of work so five into nine is a forty five okay and for a c see what so out of sixty units they did 45 units they are left with 15 units and the question says after five minutes
• 71:30 - 72:00 the first two pipes a and b are turned off that means we are left with c so c has to do the remaining work yes [Music] okay so 50 miles suppose my speed is i'm sorry 50 miles per hour and if i travel a distance of two hours how much did that travel
• 72:00 - 72:30 in one hour i travel 50 so in 2 hours 2 into 50 okay so from point a to point b this was my speed this was my time please try to understand this is a very important concept okay so the distance travelled by me is 100 miles so from a to b this was scenario one and please try to explain my examples they are going to help you immensely
• 72:30 - 73:00 if you remember my examples you won't have to remember any form never i mean at least i think like that okay so distance so your speed was while going from a to b your speed was 50 miles per hour time is 2 hours so the distance travelled by you is 100 miles while coming back from b to a this is you this is your scenario 1 this
• 73:00 - 73:30 is your scenario 2 while moving from b to a your speed is 25 miles per hour can you please tell me what would be the time taken in order to reach a or the distance is going to be the same right a to b b to a so that if the distance is 100 miles the time taken would be 4
• 73:30 - 74:00 r is everybody getting my point till now now i am going to ask you something in both the scenarios what is constant in both distance distance is a constant let's look at the ratio of speeds what is the ratio of s 1 is to s 2 2 1 by 2 by 2 s 1 is 2
• 74:00 - 74:30 1 what would be the ratio of t1 is to t2 see if the distance was constant if the distance was constant look at the ratio of speed and look at the ratio of time okay only multiplication gives you the constant value see if the speeds were in the ratio to b and in the question if it is given the question won't tell you that distance is constant
• 74:30 - 75:00 that will be a subtle information that you would have to interpret if the distance is a constant and if speeds are in the ratio s one is sorry a is to be then the time taken will be in the ratio please do it yeah that is one important information okay i'm sorry can you please start over i had a network issue okay uh in one in one scenario you are moving from a to b you have a speed 50 time two hours that
• 75:00 - 75:30 means you are traveling a distance of 100 miles okay this is your first scenario in the second scenario you are moving from point b to point a okay and your speed is 25 miles which is given 25 miles per hour and the time taken is 4 hours or forget about the time if the distance if you know this distance is 100 miles the time taken would be 4 hours 100 by 25 which is 4 hours so comparing both these scenarios i am
• 75:30 - 76:00 saying that distance here is a constant and since distance is constant i check the ratio of speeds which was 2 is to 1 and then i checked the ratio of time taken which was 1s to 2 so i can interpret this information as when distance is constant the speed of the ratio so the ratio of the speeds will be if the ratio speeds of s1 and s2 is a is to be the time taken will be in the ratio b is to it
• 76:00 - 76:30 please remember coming to the second case you are moving from point a with a speed of 50 kilometers per hour or 50 miles per hour okay and you are moving for 20 hours how much distance you'll cover in one hour 50 kilometers so in 20 hours 20 into 50. so your distance comes out to be
• 76:30 - 77:00 1000 kilometers you're moving from point a to point b okay now this was your scenario when now you're moving from point b to point a and in between this an accident has happened and that is why you had to change your route and it took you 25 hours to reach a your speed was 50 kilometers per hour
• 77:00 - 77:30 could you please tell me the distance traveled kilometers from b to a one two five fifty into twenty five which is one twenty five at a zero so the distance traveled is going to be 1 2 5 0 kilometers in both of those scenarios what is the thing which is constant speed so if speed is a constant look at the ratio of
• 77:30 - 78:00 time taken t1 is to t2 is it look at the ratio of the distances this is also positive so when speed is constant the ratio of time taken and the ratio of distances are going to be the same so we can say any of these three uh if there is one constant the two others behave on the opposite on the ratio side is it no see these are not behaving opposite
• 78:00 - 78:30 these are same the ratios are the same okay i'm sorry in case of distance they are behaving inversely okay a is to be became b's 2a in the first case don't you think it's too particular for example if in uh s2 from going b to a if we if the speed is not 25 miles per hour c it is 60 miles per hour then it won't be in inverse even if the distance is still constant how can it be possible then you know if
• 78:30 - 79:00 distance is 100 in both the cases how can the speed be in 60 how can the speed be 60. you're getting my point i am saying i have reduced the speed to half that means the speed is definitely 25 and you know the distance is definitely hundred so obviously your time will be four hours the speed has halved
• 79:00 - 79:30 so your basic assumption here is the speed will be half no no i'm saying the speed is half this is the information which is given to you i am saying that the speed is half okay you take the speed as one this is 50 your speed is 50 you take the speed to be 10 here okay
• 79:30 - 80:00 your time initially was two hours okay so your distance was 100 while moving from b to a this becomes 10 so your distance is again 100 check the ratio this is 5 is to 1 check the ratio this is 1 is to fight okay okay cool now thanks yeah so in case number three
• 80:00 - 80:30 what was case number three yes so you are moving from point a to point b all right the distance is given okay distance is given 300 kilometers i put an alarm of i asked the bus driver how long is it going to take he told me five hours so i put an alarm for five hours after five hours i reached b okay what was my speed
• 80:30 - 81:00 60 kilometers per hour okay this is my first scenario okay while moving from b to a i assume that the time is again 5 hours i put the alarm again but when i woke up i realized that i have moved to a point c
• 81:00 - 81:30 which is hundred kilometers further from point a what went wrong in my assumption in the second case i was traveling at a more at a higher speed and what was the speed because the distance traveled was 400 kilometers i getting my point now check i assume that the time was
• 81:30 - 82:00 going to be constant so in both the scenarios what is constant here time check the ratio of distances it is threes to four check the ratio of speeds it is also three is to four so what can you say what can you interpret if the time is constant or if the speed is constant whatever the
• 82:00 - 82:30 ratio of time the same will be the ratio of distances when time is a constant whatever will be the ratio of distances will be the same ratio as the ratio of speeds but when distance is constant here you have to pay special focus when distance is constant if the ratio of speeds is 2 is to 1 the ratio of time taken will be 1 is to 2 they are having inverse relation okay
• 82:30 - 83:00 that's what you should remember okay so average speed we have discussed it earlier average speed is total distance traveled upon total time taken let's give you a simple scenario in which you are moving from point a to point b to point c can you see my screen guys okay
• 83:00 - 83:30 okay yes you are moving from point a to point b and then to point c if the distance here is d1 and if the distance here is d2 and the time taken from a to b is t1 and from b to c is t2 can you please tell me what would be your average speed d1 plus d1 plus d2 upon e1 plus t2 let's consider let's make this question a bit tougher now
• 83:30 - 84:00 you are traveling from a to b you have traveled d kilometers from b to c you have traveled d kilometers from c to point d you have traveled d kilometers okay with speed s1 from a to b with speed s2 from b to c and with speed s 3 from c to d what will be your average speed
• 84:00 - 84:30 my first t1 will be s1 by d then s2 plus s2 d plus s3 s1 by d are you sure sorry total distance what is the total distance traveled 3d upon total time what is the time taken in moving from a to b
• 84:30 - 85:00 we are not directly given the time but i know since distance is speed into time can i say time is nothing but distance by speed can i say from point a to point b the time will be d by s one from b to c the time will be b by s two from point c to d the time will be d upon s three can i cancel out d i mean if i take d common
• 85:00 - 85:30 and there is 3d yeah so we have only s1 plus s2 plus s3 1 upon s1 plus 1 upon s2 plus 1 upon s3 so if there were two such stretch stretches what will be your answer if there were three stretches your answer was 3 upon 1 upon s1 plus 1 upon s2 if there were two stretches 2 into s1 s2 by s1 or you can write or
• 85:30 - 86:00 you can remember like this 2 upon 1 upon s1 plus 1 upon s2 that will give you the same result but i am not you know uh saying to because you would then say if for two stretches the result is this then for 3 the answer would be 3 s 1 s 2 s 3 upon s 1 plus s 2 plus s 3 and there you will go wrong so it's better that you remember this or you if you wish to
• 86:00 - 86:30 remember this you can remember this but just for this case okay for three stretches your answer is going to be 3 upon 1 upon s 1 plus 1 upon h 2 plus 1 upon s 3 are you getting my point and and for one simple stretch is d 1 plus d 2 over t 1 plus e2 for for one single stretch it would be just s1 that will be your average in the in the theory is two times
• 86:30 - 87:00 uv divided by u plus v yeah that's the same as this thing see for see your answer is the same your answer is the same see it is d from b to c it is d this is d miles or d kilometers a to b and b to c and your speed is s1
• 87:00 - 87:30 here and your speed is s2 here so average speed is total distance upon total time what is the time taken from a to b it is d by s one what is the time taken from b to c d by s two this result is same as what we discussed in statistics while moving from a to b i travel at speed s one and while moving from b to a i travel with s2 this is the same result
• 87:30 - 88:00 this will give you 2 upon 1 upon s 1 plus 1 upon s 2 which is if you simplify it you'll get 2 s 1 s 2 upon s 1 plus s 2 i am saying for three stretches you will need to look at this thing 3 upon 1 upon s 1 plus 1 upon s 2 plus 1 upon s 3 if you looked at this thing
• 88:00 - 88:30 your answer would have been 3 s 1 s 2 s 3 upon s 1 plus s 2 plus s 3 which is not correct why you solve it what you get is 3 s 1 s 2 s 3 upon s s1 s2 s2 s3 plus s1 s3 this is your correct answer not this answer so better to use better to have this version in order to find this version
• 88:30 - 89:00 getting my point yes thank you okay let's consider other case wherein from a to b the time taken is t from b to c the time taken is c sorry t from point c to d the time taken is also t from a to b your speed is s1 from b to c your speed is s2
• 89:00 - 89:30 and from c to d your speed is s3 can you comment on the average speed what will be your answer s1t plus s2t plus s3t by 32. see total distance what is the distance from a to b see i'm not i'm not directly given the distance but i know which is s1t
• 89:30 - 90:00 from b to c the distance would be s2t from b from c to d the distance would be s three t divided by what total time which is three t t plus t plus t so your answer is going to be s one plus s2 plus s3 upon 3 so average speed would be 3 so for 4 stretches when the time was constant your answer would have been
• 90:00 - 90:30 s1 plus s2 plus s3 plus s4 by 4 for n such stretches s1 plus s2 till sn upon n in a way i am writing what arithmetic mean yes yeah
• 90:30 - 91:00 all right that is it for average speed okay can you flip the page for a minute yeah thank you all right all right okay so this is a very important point very very very very very important point okay
• 91:00 - 91:30 just listen to the example that i'm giving you okay there are two persons a and b separated by 100 meters a has a speed of 5 meters per second b has a speed of 3 meters per second both of them are wearing apple watches and both watches are showing time
• 91:30 - 92:00 8 a.m okay they are starting simultaneously okay and both are going in the same direction you need to tell me in okay you you shouldn't tell me that can a and b meet at some point or can they not yes yes yes why why why can they meet because this one
• 92:00 - 92:30 because a is changing and a speed is more than the speed of b that is why they will meet okay so there are two scenarios here okay a moving to point c this is your scenario one and b moving to point c this is your scenario two you need to tell me what is the thing which is constant here see so don't give me don't give me the
• 92:30 - 93:00 answer right now a has a speed of 5 meter per second b has a speed of 3 meters per second okay which one is travelling a larger distance so distance traveled by a and distance traveled by b is definitely not constant for example they reach point c and in these watts the time is 8 15 am so in a is what the time will
• 93:00 - 93:30 be 8 16 8 10 8 5 what date the same time right that is time is constant here time is constant okay so from b to c b took t time and from a to c a also took t time so can you please tell me the distance travelled by a when this when the speed is 5 and the
• 93:30 - 94:00 time is t speed into time is 5 that means this whole distance from a to c it is five times of t distance traveled by b speed into time if this is five t and this is 3t the distance can i say 5 t minus 3t is equal to 100 meters
• 94:00 - 94:30 this distance is nothing but the difference of 5 t minus 3 t can i say t common 5 minus 3 is equal to 100 so t comes out to be 100 upon 5 minus 3. okay this is your scenario sorry this is your case one when they are moving in the same direction see when they are moving in the same direction there is a
• 94:30 - 95:00 minus sign here could you tell me the answer when they both are moving towards each other say for example a speed is 5 meter per second and b speed is 3 meters per second and if the distance is 100 meters between them could you please tell me if this is the midway would you please tell me where will they meet on the left hand side or on the right hand side right
• 95:00 - 95:30 right hand side why because a is faster than b okay and remember in this question both are starting simultaneously a starts at 8 am and b also starts at 8 am so what will be constant in this case right see the speeds are speeds are different the distance travelled by a will be greater than the distance traveled by b so what is constant
• 95:30 - 96:00 both of them are meeting here at 8 15. i am just kind i am just saying that they are meeting at 8 15 they are not exactly meeting at 8 15 but just in order to let you understand that time is constant a is travelling for say for example 15 minutes and b is also traveling for 15 minutes so time is a constant so the distance travelled by a till point c will be if they are having if they are taking the same time the
• 96:00 - 96:30 distance will be speed into time which is 5 t and from b to c the distance will be 3 into t can i say 5 t plus 3 t is equal to 100 can i say t common 5 plus 3 is equal to 100 so can i say t and this t is the meeting time meeting time they are meeting this t is the meeting time this will be 100 upon
• 96:30 - 97:00 5 plus 3 in both the cases there is just a sign difference since they are traveling in opposite direction there will be a positive sign and since they are moving in the same direction there is a negative sign so if i were to find the meeting time i'll say the formula would be the initial separation between them or the initial distance between them what was the initial distance in both
• 97:00 - 97:30 the cases 100 meters and 100 meters upon speed of a plus minus speed of b plus is the case when they are traveling in from opposite directions moving towards each other and minus is the case when they are moving in the same direction either to the right hand side or to the left hand side this is for same direction am i getting my point
• 97:30 - 98:00 yes sir this is a very important thing again examples are important not the formula okay if you can please explain this again again okay in the first case what was happening there was a distance of 100 meters between them a
• 98:00 - 98:30 and b and if both are traveling in the same direction with the speed of 5 meters per second and 3 meters per second and then reaching to point c so a has traveled this much distance from b to c this is this much distance and since time is same see both started at the same time you know please don't forget that they are starting from the same time or
• 98:30 - 99:00 simultaneously at the same time eight o'clock eight o'clock eight o'clock and if they are reaching here at 8 15 that means both have traveled for 15 minutes that's just an example to give you that means both are traveling for the same amount of time so if i assume the time to be d the distance traveled would be speed into time t this will be speed into time t again in the second case it was 100
• 99:00 - 99:30 a was travelling from the left to the right with a speed of 5 meter per second and b with a speed of 3 meters per second will they meet halfway no they'll meet somewhere here so from here to point c and they are starting again simultaneously at the same time that means time will be a constant here also in this case
• 99:30 - 100:00 the speed so the distance travelled by a to c would be 5 times t and the distance traveled by b to c will be 3 times of t so the total distance would be what the total distance in terms of value is given to me as hundred meters can i say a to c is five t and b to c the distance is three t can i say five t plus 3 t is equal to 100 and here 5 t is the total distance and from b to c
• 100:00 - 100:30 it is 3t so from here to here it will be 5 t minus 3t so phi t minus three t is equal to hundred so t comes out to be hundred upon five minus three and here it would be hundred upon five plus three so when they are traveling in the same direction minus sign when they are traveling from opposite directions plus sign and mind you this is the meeting time meeting
• 100:30 - 101:00 all right so until unless it's mentioned in the question about the direction we will have to assume both the directions together no no no the question will tell you in which direction the objects are moving okay can i move forward can you show the formula once again
• 101:00 - 101:30 okay thank you so upstream downstream is also as in the same case suppose the speed of the river the river this direction is called as
• 101:30 - 102:00 downstream in which the river is flowing in the opposite direction this will be upstream if the speed of river is 10 kilometers per hour okay and if a boat is there in one hours could you please tell me how much the boat will move 10 kilometers 10 kilometers but suppose
• 102:00 - 102:30 the boat have a speed 20 kilometers per hour 10 30 kilometers so in one hour how much will the boat travel 30 kilometers see with respect to the speed of the boat as in in one hour the boat will travel 20 kilometers but the river is also helping the boat in one hour the river
• 102:30 - 103:00 gives the boat a push of 10 kilometers that means in total the boat travels 20 plus 10 that means the river is helping the board in case of downstream that is the overall downstream speed overall downstream speed i am saying d is equal to speed of the boat plus speed of the river are you getting my
• 103:00 - 103:30 point while in upstream say for example the boat is moving upstream with the speed of b and the river has a speed say for example 10 kilometer per hour so what should be the minimum value of boat in order to move upstream 11 kilometers more than 10 more than 10 it may be 10.01 or 10.1 or 10.2
• 103:30 - 104:00 at least at least the speed should be 10.0 something or 10.1 something otherwise if the speed were 10 what would happen the bird would be still if the speed were eight what will happen the boat will travel downstream right so the speed of the boat has to be greater than 10 so for example
• 104:00 - 104:30 the speed of boat is 20 kilometers per hour so in one kilometer sorry in one hour could you please tell me how long how much will the boat travel [Music] the boat itself travels 20 kilometers but the river pushes this boat down in 1 hour by 10 kilometers so overall the boat will travel just 10 kilometers so with respect to speed the upstream speed would be the speed of
• 104:30 - 105:00 the boat minus the speed of the river okay so these are the two things that you should have an idea of getting my point everyone yes okay okay now i'm going to discuss a few
• 105:00 - 105:30 methods which i haven't discussed till now not few methods just the ratio method that we have used in the last class uh please try to solve this question first of all using the variables using variables now writing the data down speed is four
• 105:30 - 106:00 kilometers per hour whatever my usual time was with respect to that time i am reaching 30 minutes late whatever the time was so if you are you know taking the time to be in ours you should divide this 30 by a 60 but i am assuming right now the time is in minutes all right this is your first scenario if i walk at
• 106:00 - 106:30 the rate of five kilometers per hour five kilometers per hour i reach 30 minutes soon whatever the actual time i reached i am taking 30 minutes less what can be framed from this two things what is what is constant here first of all tell me distance from office to home or from home to office distance is
• 106:30 - 107:00 constant so can i say if distance is constant what is the distance in the first case can i say it will be 4 into t plus 30 speed into time yes and in the second case the distance would be 5 into 5 3 minus 30 but please remember the time that will come from here will be in minutes okay not in hours because this value is in minutes
• 107:00 - 107:30 and this value is also in minutes all right so while solving what you will get is 4 t plus 120 is equal to 5 t minus 150 so the value of t comes out to be 217 17 this is 270 minutes okay so in ours this value would be 270 by 60. okay and you need to find out how far is your destination so your total distance is you take any case that you want to have
• 107:30 - 108:00 suppose you are taking the first case so in that case your speed is 4 so distance is speed into time the value of t came out to be 270 by 60 see again this is in this is 30 minutes and you are first of all you can say t is 270 minutes and this is t plus 30 so this is 300 minutes now you need to convert this 300 minutes into ours that is why you are dividing this by
• 108:00 - 108:30 60 okay so 60 times 5 is 300 200 degrees comes out to be 20 kilometers is everybody getting my point the important thing is distance is constant so i have used this formula distance is speed into time so speed into time this distance should be equal to the this distance because distance is equal or constant
• 108:30 - 109:00 here i took the value i found out the value of t but it is in minutes so i shouldn't forget that i should convert this into rs so p plus 30 minutes i am taking these values to find the distance so distance would be speed which is 4 into time which is t plus 30 which is equal to 300 minutes and now i am dividing this minutes in order to convert it into ours which is 60 minutes so your distance comes out to be 20 kilometers is
• 109:00 - 109:30 everyone clear with this hello yes is clear okay this scenario is not as same to uh where we could have suppose that speed and time are in opposite ratio right we can we can i'm about to tell you that so uh have you you know understood this method because right now i'll tell you the ratio approach
• 109:30 - 110:00 yes so there are two scenarios s1 the speed is 4 kilometer per hour and s2 the speed is 5 kilometers per hour but you need to understand it first see there is some exact time at which i reach the office but if you are walking at the rate of 4 kilometers per hour 4 kilometers an hour you reach your destination
• 110:00 - 110:30 30 minutes late with respect to the usual timing t is your usual timing are you getting my point with respect to t you are 30 minutes late so this is your p1 time p1 okay if i walk at the rate of 5 kilometers per nr 5 kilometers an hour that is s2 i reach about 30 minutes too soon with respect to your usual timing so your t2 is 30 less than
• 110:30 - 111:00 the usual time could you please tell me the difference between t1 and t2 the difference between 60 minutes okay i know that the distance is constant here does anyone have a problem with distance being constant i am traveling from point a to point b
• 111:00 - 111:30 again in the second scenario i am traveling from the same point to the same point so distance is constant so can i say whatever the ratio of speeds was 4 is to 5 the ratio of time would have been the inverse which is inverse to 4 what is the difference right now 5 minus 4 is 1 if i assume t1 to be 5 minutes and t2 to be 4 minutes the difference is 1 minutes i want
• 111:30 - 112:00 60 minutes has the difference what should i do i multiply this ratio by a 60 so t1 i get 300 minutes and t2 i get 240 minutes okay so t1 comes out to be 300 minutes and if i want to find out the distance the distance would be speed into time which is 4 into 300 i'll have to divide this by 60 since the
• 112:00 - 112:30 time is in minutes so the same answer will come 20 kilometers you want to move forward guys yes sir yes sir yes sir did you get this method i mean
• 112:30 - 113:00 this is the same as ratios yes sir but we discussed last tuesday so if the uh timings were different per se 30 minutes late when going to the office with four kilometers and let's say 45 minutes with five so whatever the difference 30 plus 45 so 75 would have been your difference okay so we could have multiplied by that only yeah
• 113:00 - 113:30 okay this is an easy one a man rose 18 kilometers down the river in four hours so downstream he travels a distance of 18 kilometers in 4 hours what would be his speed and the speed will be a combination of speed of boat plus speed of river that will be distance upon time which is 18 by 4 which is
• 113:30 - 114:00 9 by 2 which is 4.5 this is my equation 1 and returns in 12 hours so he's moving from this point to this point and then he is going to the same point but in how many hours 12 hours all right so how much distance did he travel while going upstream 18 kilometers so could you please tell
• 114:00 - 114:30 me his upstream speed 18 by 12. which is 3 by 2 which is 1.5 that will be b minus r so these are two equations you need to find the speed of both and the speed of the stream okay
• 114:30 - 115:00 so if you add it what you get is two times of b is equal to 6 and if you subtract it what you get is r minus of minus r which is equal to 2r so 4.5 minus 1.5 is 3. so the speed of boat comes out to be 3 and the speed of river comes out to be 1.5 i getting my point
• 115:00 - 115:30 this is an easier one this question was an easier one i just put this question on the slide just to make you understand the concept i mean b plus rb minus f it's given in your solved examples guys shall i move forward yes sir okay yes sir all right this is a good question
• 115:30 - 116:00 first of all read the question see a b and c
• 116:00 - 116:30 can walk change my pen color at the rates of the speed of a is given to me as three kilometers per hour the speed of b is four kilometers per hour and the speed of c is five kilometers per hour okay they start from the same point a starts at one o'clock
• 116:30 - 117:00 b starts at two o'clock and c starts at three o'clock so at three o'clock could you please tell me the positions of a b and c b
• 117:00 - 117:30 so c is here so a has traveled a distance of six kilometers and b has travelled a distance of four kilometers so b has traveled a distance of four kilometers and a has travelled a distance of six kilometers what happens after that we have interpreted the statement till here when b catches up with a please tell me when will b catch up with a now
• 117:30 - 118:00 now focus on b and a what's the distance between two dna and both are traveling in the same direction same direction okay so the meeting time is given by the initial distance initial distance between b and a or the initial separation between b and a with respect to this relative speed of b and k so the relative speed would be
• 118:00 - 118:30 four minus minus three oh one kilometer that will be two hours are you getting my point so when will b catch up with a i mean b will move here in two hours and a will move from here to here in two hours are you getting my point yes sir so b is here at three o'clock
• 118:30 - 119:00 now what's the time five o'clock okay so a and b are here right now at five o'clock and could you please tell me where will c be where will 10 kilometers see c is omega x c is here at three o'clock and now the time is five o'clock that means he must have traveled from here
• 119:00 - 119:30 for two hours the speed of c is five kilometer per hour and if he travels two hours that means he travels for 10 kilometers i getting my point so i'll make the diagram again and b has traveled how long how long has b traveled from b to c the speed of b is 4 from here the speed of b is 4 8
• 119:30 - 120:00 kilometers and the time taken is two so eight kilometers from here eight kilometers from here so 8 plus 4 the distance was 12 kilometers and c has traveled 10 kilometers so c is here and a and b are here and the distance between
• 120:00 - 120:30 them is two kilometers two kilometers okay b sends a back b sends a back with a message to c now a is coming here and c is also moving forward so they'll meet somewhere here when will c get the message c is also moving right c is also moving and
• 120:30 - 121:00 a is also moving what is the speed of c i know the speed of c is five kilometers per hour and the speed of a is three kilometers so the meeting time will be given by the initial separation two divided by five plus three kilometers per hour which is eight so two by eight is going to be 1 by 4 which is 15 minutes so at this point what is the correct
• 121:00 - 121:30 time this is at five o'clock after 15 minutes that is five o'clock plus 15 minutes as in 5 15 a.m would be your answer if anyone has a problem please let me know
• 121:30 - 122:00 a quick question on the first point um when a meets b 2 divided by 4 minus 3. understand the 4 minus 3 is a different of the speeds okay now the 2 is supposed to be the gap right um yeah two kilometers yes so um how did you find these two kilometers see uh one works at three the other one
• 122:00 - 122:30 at four kilometers per hour right so there is one kilometer difference here so see a traveled for two hours that means a travelled six kilometers similarly b traveled for one hour because they start from one o'clock two o'clock three o'clock okay so b has traveled for one hour and the speed of b is four kilometer per hour the time taken is one so the distance traveled by b is going
• 122:30 - 123:00 to be four into one which is four kilometers so b will be here and c is again right now at three o'clock he will be here at the origin you can see still here okay so i should use a tweak the three o'clock as a reference okay yeah three o'clock is the reference that means the distance between b and a right now is two kilometers okay in order to meet them what would be the meeting time since they are traveling in
• 123:00 - 123:30 the same direction that will be initial separation divided by by sum of the displease yeah the difference because they are traveling in the same direction okay after that both have come here now the time is five o'clock that means c should have moved for two hours that means c has moved for two hours c has traveled a distance of 10 kilometers now c is at 10 kilometers
• 123:30 - 124:00 and a and b are at 12 kilometers that means the difference or the distance between them is 2 kilometers so i again use the formula two kilometers but since b has sent a back so now both are traveling in the opposite direction or moving towards each other that means that meeting time will be two kilometers divided by 5 plus 3 which is 1 by 4 which is 15 minutes so after 5 o'clock
• 124:00 - 124:30 again there will be a 15 minute time wherein c will get the message so your answer is going to be 5 hours and 15 minutes okay okay so uh excuse me uh i actually do not understand at five o'clock you're saying that uh a basically covered six plus uh the leftover two hours right a basically covered okay at five o'clock
• 124:30 - 125:00 how much distance a covered yeah so a covered six plus sorry this is 4 and this is 8 kilometers so 12 kilometers hello how see with respect to three o'clock when i mean when the time was three o'clock where was he
• 125:00 - 125:30 a covered six kilometers when it was three o'clock so it was here at three o'clock right after that a travelled for two hours hello yes yes i'm listening a travelled for two hours with a speed of what three kilometers so again he traveled a distance of six kilometers from here to here it is also six
• 125:30 - 126:00 kilometers and from here to here it is also six kilometers so at five o'clock he's on twelve kilometers okay okay and b catches up to him uh a so b also covered twelve kilometer all right thanks if b catches up that means we have travelled a larger distance but right now they are at 12 kilometers or total of 12 kilometers from point from this point
• 126:00 - 126:30 okay uh just a question out of curiosity is this an official gmat question i mean i don't know whether it's an official question or not i mean this is given in your examples you can expect this kind of question it can come so but without you know sort of drawing a diagram yeah so would it be solved within a minute and a half or two yeah yeah it
• 126:30 - 127:00 can be it can be definitely solved you just need to practice i mean you know i since i'm explaining it to you guys that is why i'm taking this much of time and this much of diagrammatic representation you can you know if he has to get the message you need to focus on a so just focus on a forget about b just focus on a and c that will reduce your time right now you know diagrammatic representation is important
• 127:00 - 127:30 so as to let you know i at what points of time the events are happening you solve this question once or twice and you know there is again a question like this at the end you need to try that out okay so can i move can i move yes okay so framing equations nothing nothing in this theory wise so if a two digit number is there two digit number for example 12
• 127:30 - 128:00 while you know making equation i should write this x y as see this is my units place so y into 1 plus tens place which is 10 into x so x y can be written in equation form as n x plus y what would be the reverse of this y x how can i write this y x 10 times of y plus x
• 128:00 - 128:30 if i say sum of two digit number sum of a two digit number and the number obtained by reversing its digits is 99 how will i frame this equation sum of the two digit num sum of a two
• 128:30 - 129:00 digit number and the number obtained by reversing its digits is 99 so if your two digit number was x y this was your case n x plus y plus when you reverse the numbers or reverse the digits your equation was 10 times of y plus x and the sum is going to be a 99. so if you solve your answer would be 11
• 129:00 - 129:30 times of x plus 5 will be an entity so some of the digits should be nine are you getting my point yes sir okay solve this question then three consecutive digits
• 129:30 - 130:00 that means if this digit is x this is x plus 1 this is x plus 2 or a better way or a better way would be if this is x this is x plus 1 and this is x minus 1. why is this a better way because i mean in cases of arithmetic rotation when there are three consecutive terms
• 130:00 - 130:30 and if somewhere the sum is given and if you add this what you get is 3 times x whereas when you add this it gives 3 x 3x plus 3. yeah but here it might not be the same case all right let's frame the equation in the units place being the greatest of the three so
• 130:30 - 131:00 we have assumed this to be the greatest so you know let's take this the number formed by reversing the digits the number found by reversing the digits let's reverse x plus 1 x and x minus 1 okay and how to write this number 100 into x plus 1 plus 10 x plus x minus 1 okay the number form exceeds the
• 131:00 - 131:30 original number and what was the original number 100 x minus 1 100 x minus 1 10x plus x plus 1 plus 22 this would be plus 22. can you find the value of x yes see here it is 100x plus 100 plus 10x plus x minus 1 and here it is 100x minus 100 plus
• 131:30 - 132:00 10x plus x plus 1 plus 22 which is 23 okay 100x 100x will get cancelled okay 10x 10x will get cancelled okay what am i missing here everything is getting cancelled so it's 22 times uh it's not it's 32 times yes yes yeah yeah come on i mean i have
• 132:00 - 132:30 uh yeah it's very sorry sorry okay eraser it would be 22 times the sum of the digits it's multiplication sorry yeah so 22 times of sum what would be the sum of the digits 3x the sum of the digits is 3x right yes so this will be 22 times of 3x
• 132:30 - 133:00 so 100x and 100x will get cancelled here 100 will come and here minus 100 will come similarly 10x and 10x will get cancelled x and x will get cancelled there is minus 1 here there is plus 1 here and there is 66 x here okay so 100 minus 99 is what
• 133:00 - 133:30 i'm sorry 100 minus 1 is 99 and this is also 99 so what would be your answer 198 this is 198 and this is 66 x so the value of x comes out to be 3 if the value of find the original number okay so this is your original number so if you put the value of x here as 3 it will be 4 and it will be 234
• 133:30 - 134:00 234 will be your answer okay i just wanted to tell you how to frame the equation okay we are not bothered here about the answer right now okay what what part does uh 20 uh the units place digit being the greatest place here in the question overall it doesn't help us with anything right yeah it it helps you if the information was not given
• 134:00 - 134:30 you could have assumed like this right x x plus 1 x plus 2 your answer could have been 4 32. are you getting my point yeah okay this is a question from ratios you need to find the equation find the equation form solve this question the ratio of incomes and ratio of expenditure is given and saving is given the value of saving
• 134:30 - 135:00 is given and you need to find the monthly income you know every question given in your question bank has importance okay if it is there then
• 135:00 - 135:30 it is important see the ratio of incomes is given income of a income of a is to income of b which is 9 is to 7 if i have to assume a value can i say if the income of a is 99 the income of b will be 7i yes similarly if i talk about
• 135:30 - 136:00 expenditures okay this is 4 is to 3 okay if the expenditure of a is 4a for e expenditure of b would be 3e what is if my salary is 100 rupees and if my expenditure is 20 rupees what can you comment about my saving 18 how did you find 80
• 136:00 - 136:30 in the minus expenditure 100 minus 20 is saving so can i say 9 i minus 4 e is equal to 200 and 7 i minus 3 e is also equal to 200 and can i solve both these equations for values of i and e and get the answer okay this is your first method
• 136:30 - 137:00 and the second method that i am about to tell you for this particular kind of question might not work if you are good with your practice then you can think about this method okay if you are not just apply this method you know in in case of exam if you don't know if you are not you know understanding the approach just take regular values as in just take a variable and solve okay so wouldn't that be in our linear equation
• 137:00 - 137:30 when we solved both the equations together yeah these are linear equations these are normal equations a plus b is equal to 10 how do we get a unique value these are two two different equations why won't it give you a different why won't it give you a different value okay if you wish to solve just you know just multiply this equation by 3 and multiply this equation by 4 okay and then you subtract it why
• 137:30 - 138:00 subtract because here minus 12 will come and here also minus 12 will come okay okay so i am saying this is equation 1 and 2 so you do equation 2 minus equation 1. so on the right hand side 800 minus 600 which is 200 and on the left hand side these two values will get cancelled when multiplied by 3 and 4 respectively so in the left hand side 4 into 7
• 138:00 - 138:30 which is 28 i minus 27 i will come so the value of i comes out to be okay what wrong did i do here now see 9 i minus 4 e was 200 okay minus of minus and 7 i minus 3 e was also 200
• 138:30 - 139:00 yes 9 into 3 27 and 28 so the value of i comes out to be 200 and the saving is also 200 no what am i doing wrong here their income will be 9 i right 9 i and 7 and not high thank you thank you thank you thank you so much so nine times of i is 20 times
• 139:00 - 139:30 200 into nine is eighteen hundred and seven into two hundred is fourteen hundred all right i thought you know the saving is 200 and also the income is coming out to be wanted okay so actually i is a constant here okay i shouldn't have taken i as a constant nine times of i would be eighteen hundred and seven times of i would be 1400 okay so the monthly incomes will be 1800 and 1400
• 139:30 - 140:00 respectively in order to do this question with the help of ratios how would you do it can anyone tell me the ratio of incomes is given to me as 9 is to 7 and the ratio of expenditures is 4 is to 3 okay and their saving has to be the same like 200 in both the cases if i assume the income
• 140:00 - 140:30 to be 9 rupees and seven rupees and the expenditure to be four rupees and three rupees is the saving of both of them same right now say for example this is seven rupees three rupees same thing of four saving of file are they same right now no sir that means i cannot apply the ratio method right now because they will have to be the same in order to apply this approach so i'll think about something see this
• 140:30 - 141:00 is 9 is to 7 this is 4 raised to 3 now the difference is 5 and 4 you know just first of all i double this i get 18 and 14. say for example 18 and 14 and this is still 4 and 3 because multiplying by ratio by a constant won't change the ratio but this is again 14 and this is again 11 if i multiply the second ratio if you multiply the second ratio by 2
• 141:00 - 141:30 what will you get this will be 8 and this will be 6. i'm multiplying this ratio by 2. i can multiply right ratio by constant now you see the difference in the income is 1 and the difference in the income is 1. now i can apply the ratio approach here in our ratios i am saving one rupees each in both the cases
• 141:30 - 142:00 so if i have to save 200 rupees what should i do i multiply everything by 200 so this ratio will get multiplied by 200 and this ratio should also get multiplied by 200 so income of a will come out to be 1800 and income of b will come out to be 1400 i got getting my point guys
• 142:00 - 142:30 yes i am getting a point down okay shall i move forward yes okay solve this question i would like you guys to answer this question
• 142:30 - 143:00 see the question itself states that these are moving in the same direction okay 3.5 okay one answer is 3.5
• 143:00 - 143:30 3.5 okay how did you solve it
• 143:30 - 144:00 relative speed is uh 40 minus 30 10 okay just wait just just just a sec what's the initial separation initially 25 25 miles x is 25 miles ahead train way so train y is here right now and train y has a speed of 40 and train x has a speed of 30 okay so uh
• 144:00 - 144:30 when will they meet or you know at the be at the same place would you tell me the time when they are at the same place 2.5 3.5 or 2.5 at the same place 2.5 so meeting time is given by the initial separation which is 25 miles upon relative speed 40 minus 30 that will give you 2.5 hours
• 144:30 - 145:00 all right so from here to here y took and x also took 2.5 hours now they are here what happens after that i just took this question and this is a very good good question in order to understand now what's going on now you have to find find out after how many hours will it
• 145:00 - 145:30 be until train y but as in from here train y has traveled something from here train x has traveled something now the distance between them has become 10 miles so can you say this is just an you know a reverse case see they are traveling from here to here and now the separation is 10 miles can i say or can i think of it reverse that they both are here
• 145:30 - 146:00 and they are moving to the left hand side and they are again coming here isn't the time going to be the same from here to here or from here to here isn't the time going to be the same so the initial difference between them is 10 miles so this was meeting 0.1 and this will be i'll say meeting point 2
• 146:00 - 146:30 that will be distance between them upon the ratio of the relative speeds which is 10 your answer is going to be one hour if you don't understand this logic in the exam it's also all right you know in how many years you know how many hours will train y be 10 miles ahead so time taken from here to here will be equal to time taken from here to here
• 146:30 - 147:00 okay so if the speed of y was 40 so the distance traveled by y was 40 t and the distance travelled by x was 30 so can i say 40 t minus 30 t is equal to 10 miles can i say 10 t is equal to 10 so can i say t is equal to 1 r that means from here to here 2.5 hours and from here to here
• 147:00 - 147:30 1r that means a total of 3.5 hours you're getting my point even if you had to do this question with the help of options could you have done it with the help of options is it possible you know you just you just have to put the values and check
• 147:30 - 148:00 say for 1.5 hours how much has y traveled y is here and x is here okay x speed is 30 and the speed of y is 40 and you have to prove that train y is 10 miles ahead of train x if you multiply 40 with a 1.5 what would be your answer and and the distance between them is 25
• 148:00 - 148:30 kilometers or 25 miles this will be what 60 from here to here and if you multiply 13 to 1.5 it will be 45 45 so there is a gap of i am doing something 45 plus 25 will be around 80 so 60 20 kilometer cap is still there
• 148:30 - 149:00 yeah see this was 20 this was 25 and a total of 16 that means so he traveled 35 and this guy traveled 45 that means a distance of what am i doing in just a second first one traveled
• 149:00 - 149:30 to 60 he traveled 45 so his total become 80 so there is a gap of 20 right just a sec yeah now tell me yeah the first one travelled 60 km 60 miles the first one traveled 60 miles and the second one travelled 45 but he was already ahead 25 miles so his total gap was 80. so this one's 60 so there is
• 149:30 - 150:00 still gap of 20. just just just a sec to change the pin all right see here y was here and y traveled for y has a speed of 40 and 1.5 hours that means 60 kilometers similarly x was 25 miles ahead and he traveled for 1.5 hours
• 150:00 - 150:30 so x traveled how much 45 45 so from here he traveled 45 82 so x is ahead in this case i am trying to say i getting my point y traveled a total of 60 he traveled
• 150:30 - 151:00 from here 45 are you getting my point but the actual question is train y has to be ahead of train x but here what it says train x is ahead of train y that means option a can't be the answer similarly try for option b similarly try for options c and d you will easily eliminate these options
• 151:00 - 151:30 and whenever train y is ahead and train x is beside i mean before y and whenever it satisfies that the distance between them is 10 miles there will be your answer are you getting my point you're basically saying just to put time and speed and take out the distance if the difference between a distance is 10 then
• 151:30 - 152:00 that's our very answer yeah that will be your answer yes you can you could have done it otherwise just go by the method it won't take that long you have to apply the meeting well approach meeting time approach two times and you'll get to the answer uh sir can you explain the second meeting time how did you get that 10 by 10 okay
• 152:00 - 152:30 all right uh till this point of time you understood yes the initial distance is 25 and because then they are in the same direction yeah so this is the difference see after that how many hours will it be until train y is 10 miles ahead of train x
• 152:30 - 153:00 train y is ahead now train excess here and there is a gap of 10 miles i am saying from here to here isn't the time going to be the same are you getting my point that the time will be the same yeah so think of this scenario the reverse of the scenario say for example x and y both were here and they were
• 153:00 - 153:30 moving from right to the left to this point what time should we take that would be initial separation between them divided by the relative speed what was the speed of y 40 and what was the speed of x 30 so 40 minus 30 n by 10 if you don't understood it if you don't understand it just say aft from this point to this point the time is going to be the same
• 153:30 - 154:00 so if x travels 30 t why travel 40 so what would be this distance the difference would be 10 y 10 t so t is equal to 1 from here you can also get the answer all right yes okay so this was about your arithmetic portion and today we discussed uh the two set three set
• 154:00 - 154:30 time speed distance time and work and some equation framing okay solve all the questions and the questions are going to be discussed on saturday so we'll take the session all right and if anyone has doubts please send me the doubts on whatsapp i'll clear those doubts okay finish the pre-work before the class all right this section is very important and questions come from this section
• 154:30 - 155:00 all right guys uh does anyone have any questions uh yeah one question is there what is the class schedule because i am from the 10th april batch and i'm not understanding the basic class schedule that was informed to me was tuesday and thursday so is there a class on saturday also see tuesday tuesday and thursday were optional sessions for the for the upcoming class on saturday okay basic sessions yeah
• 155:00 - 155:30 and the main class will be on saturday these are the doubt i i would say the basic sessions basic concept sessions which are optional main class on saturday that means uh for those whose complete batch is starting from saturday actually there is a current batch going on so this is optional for the ones who will be joining from 10th but this is for the ones which is happening like at present like the batch which is
• 155:30 - 156:00 taking classes okay okay thank you thank you thank you thank you too all right all right guys okay thank you thank you okay so how was the session today i mean uh did you learn something new yes sir the work done type of questions so the table is really helping me to solve the questions that was easy to understand okay solve solve everything and whenever you
• 156:00 - 156:30 face any doubts you contact me and then we'll deal with with those doubts together all right okay so thank you so much sir hello guys good night guys