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Work Rate Problems  All in One Topic!
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09 Sep 2015, 01:32
Work Rate Problems  All in One Topic!Cracking the Work Rate Problems Being comfortable with common ratios can save you a lot of time on the GMAT. Great application of ratios is work rate problems. An important relation that helps us solve work rate problems is: Work Done = Rate * TimeThis relation will lead a perceptive observer to draw a parallel with another very popular relation most of us have come across: Distance = Speed * TimeSpeed is the same as Rate of work i.e. how fast you cover some distance or how fast you complete some given work. So obviously, if we can use ratios to solve many Distance Speed Time problems, we should be able to solve many Work Rate Time problems using ratios too. Let’s look at some examples. Try and solve each one of them on your own before you go through the solutions. Example 1: A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?(A) 1/2 (B) 2/3 (C) 3/4 (D) 3/7 (E) 4/7 We are given that rate of work of 1 narrow pipe : rate of work of 1 wide pipe = 1:2 If we can find the ratio of rate of work of 2 wide pipes : rate of work of all pipes together, then we can easily get the ratio of time taken by 2 wide pipes : time taken by all pipes together. This is because ratio of time taken will be inverse of the ratio of rate of work since work done in both the cases is the same. (For a further explanation of this concept, check out the previous post) In ratio terms, rate of work of 3 narrow pipes is 1*3 and rate of work of 2 wide pipes is 2*2 Therefore, rate of work of 3 narrow pipes : rate of work of 2 wide pipes = 3:4 Or we can say rate of work of 2 wide pipes : rate of work of all pipes together = 4 : (3+4) = 4:7 Then, time taken by 2 wide pipes : time taken by all pipes together = 7:4 (i.e. inverse of 4:7) So all the pipes together will take 4/7 th of the time taken by the two wide pipes. Answer (E) This question is discussed HERE. Example 2: Working at their respective constant rates, Paul, Abdul and Adam alone can finish a certain work in 3, 4, and 5 hours respectively. If all three work together to finish the work, what fraction of the work will be done by Adam?(A) 1/4 (B) 12/47 (C) 1/3 (D) 5/12 (E) 20/47 It is given that: Time taken by Paul : Time taken by Abdul : Time taken by Adam = 3:4:5 Rate of work must be inverse of time taken. But how do you take the inverse when you have a ratio of 3 quantities? Does it become 5:4:3? No. Actually it becomes 1/3 : 1/4 : 1/5 (I will explain the ‘why’ for this when I take variation) Rate of Paul : Rate of Abdul : Rate of Adam = \(\frac{1}{3} : \frac{1}{4} : \frac{1}{5}\) Let’s multiply this ratio by the LCM to convert it into integral form. The LCM of 3, 4 and 5 is 60. Rate of Paul : Rate of Abdul : Rate of Adam \(= (\frac{1}{3})*60 : (\frac{1}{4})*60 : (\frac{1}{5})*60 = 20:15:12\) (I would like to remind you here that multiplying or dividing each term of a ratio by the same number does not alter the ratio) So if the total work is 20+15+12 = 47 units, Adam will complete 12 units out of it. Hence the fraction of work done by Adam will be 12/47. Answer (B) This question is discussed HERE. Example 3: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 In this question, you can think logically to arrive at the answer quickly. When machine A is working alone, it takes 64 extra minutes. Why? Because there is work leftover after t minutes. The work that would have been done by machine B in t minutes is leftover and is done by machine A in 64 minutes. Time taken by A : Time taken by \(B = 64:t\) ….. (I) Similarly, when machine B works alone, it takes 25 extra minutes to complete the work that machine A would have done in t minutes. Time taken by A : Time taken by \(B = t:25\) ……(II) From (I) and (II) above, \(\frac{64}{t} = \frac{t}{25}\) \(t = 40\) Time taken by machine A : Time taken by machine \(B = t:25 = 40:25 = 8:5\). Answer (B) This question is discussed HERE.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 01:41
Rates Revisited People often complain about getting stuck in workrate problems. Hence, I would like to take some 700+ level questions on rate today. I have discussed the basic concepts of workrate (using ratios) in a previous post. You might want to go through that post before you set out to work on these problems. Ensure that you are very comfortable with the relation: Work = Rate*Time and its implications: If rate doubles, work done doubles too if the time remains constant; if one work is done, rate = 1/time etc. Thorough understanding of these implications is fundamental to ‘reasoning out’ the answer. Question 1: Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?(A) 1/2 (B) 2 (C) 3 (D) 5 (E) 6 Solution: Tricky, eh? It is a little cumbersome if you get into variables. If you just try to reason it out, it could be done rather quickly and easily. Let’s see! Machine A and B together complete 1 work in 3 hrs i.e. together, they do 1/3rd work every hour. If machine A’s speed were double, they would do 1/2 work in 1 hour together. How come they do (1/2 – 1/3 =) 1/6th work extra in 1 hour now? Because machine A’s speed is double the previous speed. The extra speed that machine A has allows it to do 1/6th work extra. This means, at normal speed, machine A used to do 1/6 work in an hour (its speed had doubled so work had doubled too). Hence, at usual speed, it will take 6 hrs to produce 1 widget. Answer (E) This question is discussed HERE. Consider the amount of time and effort you would have spent on this question had you tried to use two variables to figure out the answer. You would have made equations like this: 1/a + 1/b = 1/3 and 2/a + 1/b = 1/2 and then you would have solved them simultaneously to get the value of a. Whereas in the solution above, we have done all the work orally! Question 2: One woman and one man can build a wall together in two hours, but the woman would need the help of two girls in order to complete the same job in the same amount of time. If one man and one girl worked together, it would take them four hours to build the wall. Assuming that rates for men, women and girls remain constant, how many hours would it take one woman, one man, and one girl, working together, to build the wall?(A) 5/7 (B) 1 (C) 10/7 (D) 12/7 (E) 22/7 Solution: This question is certainly quite tricky but if you understand the relation between work and rate, you can still solve this question easily. Mind you, we are using variables here only because I don’t want to write man, woman and girl again and again. Notice that there are no ‘=’ signs i.e. we are not making equations so we are not doing any algebraic manipulations. The question is long so take one line at a time and analyze it. We will keep condensing the information we get from each sentence and figuring out the implications of new and previous information as we go along. “One woman and one man can build a wall together in 2 hrs,” 1w + 1m > 2 hrs ……(I) “but the woman would need the help of 2 girls in order to complete the same job in the same amount of time.” 1w + 2g > 2 hrs …..(II) From (I) and (II), we can say that 1m is equivalent to 2g (i.e. 1 man does the same work as 2 girls do in the same amount of time; 1m ? 2g) “If 1 man and 1 girl worked together, it would take them four hours to build the wall.” 1m + 1g > 4hrs (Since 1m ? 2g, we can say that 3g will take 4 hrs to build the wall.) or 2m + 2g > 2 hrs …..(III) (If number of workers double, time taken to do the work becomes half) From (II) and (III), 1w ? 2m (i.e. 1 woman does the same work as 2 men do in the same amount of time) Hence, 1w ? 2m ? 4g “Assuming that rates for women, men and girls remain constant, how many hours would it take 1 woman, 1 man and 1 girl working together to build the wall?” 1w + 1m + 1g ? 4g + 2g + 1g ? 7g. Since 3g take 4 hrs to build the wall, 7g will take 3*4/7 = 12/7 hrs to complete the wall. Answer (D) This question is discussed HERE. We have done most of the work while reading the question only. Had we tried to solve it algebraically, we would have made 3 equations using 3 variables and then tried to solve them.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:19
WorkRate Using Joint Variation Let’s look at some workrate questions which use joint variation. Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?(A) 60 (B) 70 (C) 75 (D) 80 (E) 100 Solution: Can we say that 10 people can finish the work in 100 days? No. If that were the case, after 20 days, only 1/5th of the work would have been over. But actually 1/4th of the work is over. This means that ‘10 people can complete the work in 100 days’ was just the contractor’s estimate (which turned out to be incorrect). Actually 10 people can do 1/4th of the work in 20 days. The contractor fires 2 people. So the question is how many days are needed to complete 3/4th of the work if 8 people are working? We need to find the number of days. How is ‘no. of days’ related to ‘no. of people’ and ‘work done’? If we have more ‘no. of days’ available, we need fewer people. So ‘no. of days’ varies inversely with ‘no. of people’. If we have more ‘no. of days’ available, ‘work done’ will be more too. So ‘no. of days’ varies directly with ‘work done’. Therefore, ‘no. of days’ * ‘no. of people’/’work done’ = constant \(20*\frac{10}{(\frac{1}{4})} =\) ‘no. of days’*\(\frac{8}{(\frac{3}{4})}\) No. of days = 75 So, the work will get done in 75 days if 8 people are working. We can also do this question using simpler logic. The concept used is joint variation only. Just the thought process is simpler. 10 people can do 1/4th of the work in 20 days. 8 people can do 3/4th of the work in x days. Start with the no. of days since you want to find the no of days: \(x = 20*(\frac{10}{8})*(\frac{3}{1}) = 75\) From where do we get 10/8? No. of people decreases from 10 to 8. If no. of people is lower, the no of days taken to do the work will be more. So 20 (the initial no. of days) is multiplied by 10/8, a number greater than 1, to increase the number of days. From where do we get (3/1)? Amount of work increases from 1/4 to 3/4. If more work has to be done, no. of days required will be more. So we further multiply by (3/4)/(1/4) i.e. 3/1, a number greater than 1 to further increase the number of days. This gives us the expression \(20*(\frac{10}{8})*(\frac{3}{1})\) We get that the work will be complete in another 75 days. Answer (C) This question is discussed HERE. Let’s take another question to ensure we understand the logic. Question 2: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?(A) 1200 lt (B) 1555 lt (C) 1664 lt (D) 1728 lt (E) 4800 lt Solution: First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only 4/5 lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 companyequivalent cars. 4 cars running 10 hrs for 10 days consume 1200 lt of fuel 8 cars running 12 hrs for 6 days consume x lt of fuel \(x = 1200*(\frac{8}{4)}*(\frac{12}{10})*(\frac{6}{10}) = 1728\) lt We multiply by 8/4 because more cars implies more fuel so we multiply by a number greater than 1. We multiply by 12/10 because more hours implies more fuel so we multiply by a number greater than 1. We multiply by 6/10 because fewer days implies less fuel so we multiply by a number smaller than 1. Answer (D) This question is discussed HERE.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:30
Evading Calculations! The GMAT is not a calculation intensive exam. Whenever you land on an equation which looks something like this: \(\frac{60}{(n  5)}  \frac{60}{n} = 2\), you probably think that we don’t know what we are talking about! You obviously need to cross multiply, make a quadratic and finally, solve the quadratic to get the value of n. Actually, you usually don’t need to do any of that for GMAT questions. You have an important leverage – the options. Even if the options don’t directly give you the values of n or n5, you can use the knowledge that every GMAT question is doable in 2 mins and that the numbers fit in beautifully well. Let’ see whether we can get a value of n which satisfies this equation without going the whole nine yards. We will not use any options and will try to rely on our knowledge that GMAT questions don’t take much time. \(\frac{60}{(n  5)}  \frac{60}{n} = 2\) So, the difference between the two terms of the left hand side is 2. Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator. Say, if n = 10, you get \(\frac{60}{5}  \frac{60}{10} = 12  6 = 6\). The difference between them is much more than 2. \(\frac{60}{n}\) and \(\frac{60}{(n  5)}\) need to be much closer to each other so that the difference between them is 2. The two terms should be smaller to bring them closer together. So increase the value of n. Put n = 15 since it is the next number such that (15 – 5 =) 10 as well as 15 divide 60 completely. You get \(\frac{60}{10}  \frac{60}{15} = 6  4 = 2\). It satisfies and you know that a value that n can take is 15. Usually, you will get a solution within 23 iterations. This is enough for a PS question. Notice that this equation gives us a quadratic so be careful while working on DS questions. You might need to manipulate the equation a little to figure out whether the other root is a possible solution as well. Anyway, today we will focus on the application of such equations in PS questions only. Let’s take a question now to understand the concept properly: Question: Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?(A) 5 (B) 6 (C) 8 (D) 10 (E) 12 Solution: We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy. Machine A takes x hrs to make 40 widgets. Rate of work done by machine A = Work done/Time taken = \(\frac{40}{x}\) Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together. Time taken by machine B to make 40 widgets = x – 4 Rate of work done by machine B = Work done/Time taken = \(\frac{40}{(x  4)}\). We know the combined rate of the machines is 25/3 So here is the equation: \(\frac{40}{x} + \frac{40}{(x  4)} = \frac{25}{3}\) The steps till here are not complicated. Getting the value of x poses a bit of a problem. Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options). To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too. x cannot be 3 but it can be 6. If x = 6, \(\frac{40}{(6  4)} = 20\) i.e. the sum will certainly not be 20 or more since we have \(\frac{25}{3} = 8.33\) on the right hand side. The only other option that makes sense is x = 12 since it has 3 in it. \(\frac{40}{12} + \frac{40}{(12  4)} = \frac{10}{3} + 5 = \frac{25}{3}\) Answer (E) This question is discussed HERE. If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming. Another method could be to bring 3 to the left hand side to get the following equation: \(\frac{120}{x} + \frac{120}{(x  4)} = 25\) This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot of time.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:34
A Common Sense Approach to Work Rate Problems on the GMAT Combined work rate problems give many a headache at their mere mention. After all, you have to think in terms of that fourth dimension, “time” (cue the Twilight Zone theme). This alone puts it up there with Einsteinian Relativity in terms of difficulty. There are always three moving parts – time, work, and speed – and sometimes three or more machines or people working together. What really makes them difficult is the seemingly counterintuitive nature of what happens when two people work together. You probably get that if Kate can write thank you notes in 3 hours, and Will can write them it in 5 hours, when they work together it should go faster than Kate writing alone. Sure, there’s a formula out there that tells you to multiply (Kate * Will) then divide by the sum of (Kate + Will), but what does that really do? And it might work in a simple scenario, but what if Harry or Pippa join in to help? What if Kate takes a lunch break? Truth is, as with many high level concepts in life, there are a lot of factors playing against our common sense on how it all works. The main thing that plays against our common sense is that we are never given two equal work partners. Kate writes quickly, Will writes slowly, and 3 hours and 5 hours don’t divide evenly. So Will will speed up the work a bit, but it’s not obvious by exactly how much. The other thing playing against our common sense is that standard work rate solutions ask you to multiply Kate and Will in some form, say, by adding rates with different prime factors. This easily leads to the mistaken notion that when two people work together, their efforts are somehow multiplied. They are not! And we fall into this trap by our own experience. Ever had to move, clean, or do anything that would take a while? Remember preparing to spend most of your day on your own getting down and dirty? Then you managed to recruit a friend to help. Boom! How much faster it went with the two of you! With the motivation of working together, you managed to knock it out in only a couple hours. Indeed, effort really does multiply! Not on the GMAT it doesn’t. And the GMAT makers really want to punish the test taker that relies on a packaged formula or makes a faulty assumption, and reward the test taker that pieces an answer together quickly using common sense. Because common sense is not so common. There’s a simple switch we can flip in our minds to get us from hesitating at the cloudy intuition of combined work rate formulations, to diving in fearlessly with common sense. The adage that “two heads are better than one” applied to the classic work rate problem tells you that if you can do a job in two hours, with an equal partner you’ll be done in half the time. This is a good foundation, but to really crack the work rate problem on the GMAT, it’s going to take a little bit more simplification. So let’s get really, really simple. Kate can write their thank you notes in 3 hours; thus every hour Kate writes 33% of the notes. Will can write their thank you notes in 5 hours; thus every hour Will writes 20% of the notes. Combined, they write 53% of the notes every hour. They should take just under 2 hours to finish together. It’s really as simple as that. No setup necessary. No possibility for a calculation error, or a misconception. We’re off to a great start.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:39
A Common Sense Approach to Work Rate Problems on the GMAT: Part II We introduced the most common sense way of approaching a simple work rate problem above in Part I. No setup was necessary. There was zero possibility for a calculation error, or a misconception. The approach was to put simple work rates into percentages (rather than fractions). Kate writes thank you notes in 3 hours, or 33% per hour; William writes them in 5 hours, or 20% per hour. Together they write 53% per hour, or all their notes in under 2 hours.Yes, 1/3 and 1/5 is an OK way to put it, but it’s easy to get tangled up when we have to compare fractions like 1/3 and 1/5, even though it’s simple enough (and precise) to say 5/15 plus 3/15 equals 8/15 every hour. But if the numbers aren’t too difficult – and on the GMAT they usually won’t be – stick with percentages. They’re a little more natural for our minds to work with on equal terms. The common sense approach is basically this: break everything down into a unit of time that is easy to work with, and just figure out what happens during that time. Could be an hour, ten minutes, etc., depending on the question. Then add them up. With combined work rate, we’re really adding the efforts, never multiplying them. Here’s one where the formulation and concept is a bit trickier: Machines B and J working together can process one ton of ice cream in 30 hours. Machine B, working alone, takes 75 hours to process one ton of ice cream. In how many hours can Machine J, working alone, process one ton of ice cream?If B & J together take 30 hours, they process 3 and 1/3% per hour (100%/30hr). If alone Machine B takes 75 hours, it processes 1 and 1/3% per hour (100%/75hr). That’s a difference of 2% per hour. That difference is what Machine J is processing. At 2% per hour, alone, it should take Machine J 50 hours to process one ton of ice cream. In just three easy steps, we’ve got it. Let’s tackle one more, with tricky conditions: At a boulangerie, it takes 12 bakers working 4 hours each morning to bake the day’s bread. If on Saturday 12 bakers begin at 6am, and one assistant baker arrives to work every half hour, at what time will the day’s bread be baked?This problem is a bit tedious to work through, but common sense will get us there. With 12 bakers working, every hour onefourth, or 25%, of the bread is baked. For each baker’s perhour contribution, divide 25% by 12 – roughly 2%… or more precisely, each baker bakes 1/48 (12 bakers * 4 hours = 48) of the bread per hour. Since the question involves half hour increments, let’s frame it as each baker accounts for 1/96 of the bread per half hour. We have 96 ‘loaves’ to be baked, and each baker will make 1 of the ‘loaves’ per half hour. Or if you prefer to stick with the rough percentage, 1% per baker, per half hour. So at 6:00, 0 are baked. By 6:30, 12 loaves are baked (roughly 12%), then the first assistant joins. This will add 13 loaves during the next half hour – so 25 are baked by 7:00. A 14th baker means 39 by 7:30. A 15th means 54 by 8:00. By 8:30, 70, having added 16 loaves. And by 9:00, 87, having added 17. Not quite 96 yet, but close. At 9:00, an 18th baker joins and 9 ‘loaves’ remain – a further 15 minutes is all it takes. The bread is baked by 9:15. And common sense—not a work rate formula—got us there. It will every time.
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:48
Bringing Back the Lazy Genius to Solve GMAT Questions! Those of you who have seen the previous version of our curriculum would know that we had tips and tricks under the heading of ‘Lazy Genius’. These used to discuss innovative shortcuts for various questions – the way very smart people would solve the question – without putting in too much effort! Today, let’s bring back the beloved lazy genius through a question. Try to solve it lazily i.e. try to do minimum work on paper. This means making equations and solving them is a big nono and doing too many calculations is cumbersome. Question: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?(A) yz/(x+y+z) (B) yz/(yz + xz – xy) (C) yz/(yz + xz + xy) (D) xyz/(yz + xz – xy) (E) (yz + xz – xy)/yz Note that you have variables in the question and the options. Since we are looking for a lazy solution, making equations out of the variables is not acceptable. So then, should we plug in numbers? With three variables to take care of, that might involve a lot of calculations too. Then what else? Here is our minimumworksolution to this problem; try to think one of your own and don’t forget to share it with us. Plugging in numbers for the variables can be troublesome but you can give some very convenient values to the variables so that the effect of a pump and a drain will cancel off. There are no constraints on the values of x, y and z except z > x (drain Gamma empties slower than pipe Alpha fills) Let’s say, x = 2 hrs, y = 4 hrs, z = 4 hrs What did we do here? We made the rate of Beta same as the rate of Gamma i.e. 1/4 of the tank each. This means, whenever both of them are working together, drain Gamma cancels out the work of pump Beta. Every hour, pump Beta fills 1/4th of the tank and every hour drain Gamma empties 1/4th of the tank. So the entire tank will be filled by pump Alpha alone. Hence, if y = z, pump Alpha fills the entire tank i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1. In the options, put y = z and see which option gives you 1. Note that you don’t have to put in the values of 2, 4 and 4. We gave those values only for illustration purpose. If y = z, xy = xz. So in option (B), xz cancels xy in the denominator giving yz/yz = 1 Again, in option (E), xz cancels xy in the numerator giving yz/yz = 1 The other options will not simplify to 1 even though when we put y = z, the answer should be 1 irrespective of the value of x, y and z. The other options will depend on the values of x and/or y. Hence the only possible options are (B) and (E). But we still need to pick one out of these two. Now let’s say, x = 4, y = 2, z = 4.00001 ( z should be greater than x but let’s assume it is infinitesimally greater than x such that we can approximate it to 4 only) Rate of work of Gamma (1/4th of the tank per hour) is half the rate of work of Beta (1/2 the tank per hour). Rate of work of Gamma is same as rate of work of Alpha. Half the work done by pump Beta is removed by drain Gamma. So if pump Beta fills the tank, drain Gamma empties half of it in that time – the other half would be filled by pump Alpha i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1/2. Put x = z in the options (B) and (E). The one that gives you 1/2 with these values should be the answer. Again, you don’t need to plug in the actual values till the end. If x = z, yx = yz (B) \(\frac{yz}{(yz + xz  xy)}\) yz cancels xy in the denominator giving us \(\frac{yz}{xz} = \frac{y}{x}= \frac{2}{4} = \frac{1}{2}\) (E) \(\frac{(yz + xz  xy)}{yz}\) yz cancels xy in the numerator giving us \(\frac{xz}{yz} = \frac{x}{y} = \frac{4}{2} = 2\) Only option (B) gives 1/2. Answer (B) This question is discussed HERE. Even if you end up feeling that this method is complicated, try and wrap your head around it. If you do, you are on your way to becoming a lazy genius yourself!
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Re: Work Rate Problems  All in One Topic!
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09 Sep 2015, 02:52
How to Solve Relative Rate of Work Questions on the GMAT Today, we look at the relative rate concept of work, rate and time – the parallel of relative speed of distance, speed and time. But before we do that, we will first look at one fundamental principle of work, rate and time (which has a parallel in distance, speed and time). Say, there is a straight long track with a red flag at one end. Mr A is standing on the track 100 feet away from the flag and Mr B is standing on the track at a distance 700 feet away from the flag. So they have a distance of 600 feet between them. They start walking towards each other. Where will they meet? Is it necessary that they will meet at 400 feet from the red flag – the mid point of the distance between them? Think about it – say Mr A walks very slowly and Mr B is super fast. Of the 600 feet between them, Mr A will cover very little distance and Mr B will cover most of the distance. So where they meet depends on their rate of walking. They will not necessarily meet at the mid point. When do they meet at the mid point? When their rate of walking is the same. When they both cover equal distance. Now imagine that you have two pools of water. Pool A has 100 gallons of water in it and the Pool B has 700 gallons. Say, water is being pumped into pool A and water is being pumped out of pool B. When will the two pools have equal water level? Is it necessary that they both have to hit the 400 gallons mark to have equal amount of water? Again, it depends on the rate of work on the two pools. If water is being pumped into pool A very slowly but water is being pumped out of pool B very fast, at some point, they both might have 200 gallons of water in them. They will both have 400 gallons at the same time only when their rate of pumping is the same. This case is exactly like the case above. Now let’s go on to the question from the GMAT Club tests which tests this understanding and the concept of relative rate of work: Question: Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?(A) 5/(M+K) hours (B) 6/(M+K) hours (C) 300/(M+K) hours (D) 300/(M−K) hours (E) 60/(M−K) hours Solution: There are two tanks with different water levels. Note that the rate of pumping is given as K gallons per min and M gallons per min i.e. they are different. So we cannot say that they both will have equal amount of water when they have 350 gallons. They could very well have equal amount of water at 300 gallons or 400 gallons etc. So when one expects that water in both tanks will be at 350 gallon level, one is making a mistake. The two tanks are working for the same time to get their level equal but their rates are different. So the work done is different. Note here that equal level does not imply equal work done. The equal level could be achieved at 300 gallons when work done would be different – 200 gallons removed from tank X and 100 gallons added to tank Y. The equal level could be achieved at 400 gallons when work done would be different again – 100 gallons removed from tank X and 200 gallons added to tank Y. To achieve the ‘equal level,’ tank Y needs to gain water and tank X needs to lose water. Total 300 gallons (500 gallons – 200 gallons) of work needs to be done. Which tank will do how much depends on their respective rates. Work to be done together = 300 gallons Relative rate of work = (K + M) gallons/minute The rates get added because they are working in opposite directions – one is removing water and the other is adding water. So we get relative rate (which is same as relative speed) by adding the individual rates. Note here that rate is given in gallons per minute. But the options have hours so we must convert the rate to gallons per hour. Relative rate of work = (K + M) gallons/minute = (K + M) gallons/(1/60) hour = 60*(K + M) gallons/hour Time taken to complete the work \(= \frac{300}{60(K+M)} \ hours = \frac{5}{(K+M)}\) hours Answer (A) This question is discussed HERE.
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Re: Work Rate Problems  All in One Topic!
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Re: Work Rate Problems  All in One Topic!
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Updated on: 16 Sep 2015, 12:28
Bunuel Numerous math processing errors!! Can you please correct them Not sure what was wrong. These appear fine now. Perhaps, bad internet connection didn't load the scripts. My bad
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Re: Work Rate Problems  All in One Topic!
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15 Sep 2015, 12:04
swanidhi wrote: Bunuel Numerous math processing errors!! Can you please correct them It's getting displayed properly in Chrome and Safari. Where are you seeing the issues ?



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Re: Work Rate Problems  All in One Topic!
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17 Feb 2016, 06:01
How to Work on GMAT Work Problems Rate questions, so far as I can remember, have been a staple of almost every standardized test I’ve ever taken. I recall seeing them on proficiency tests in grade school. They showed up on the SAT. They were on the GRE. And, rest assured, dear reader, you will see them on the GMAT. What’s peculiar is that despite the apparent ubiquity of these problems, I never really learned how to do them in school. This is true for many of my students as well, as they come into my class thinking that they’re just not very good at these kinds of questions, when, in actuality, they’ve just never developed a proper approach. This is doubly true of work problems, which are just a kind of rate problem. When dealing with a complex work question there are typically only two things we need to keep in mind, aside from our standard “rate * time = work” equation. First, we know that rates are additive. If I can do 1 job in 4 hours, my rate is 1/4. If you can do 1 job in 3 hours, your rate is 1/3. Therefore, our combined rate is 1/4 + 1/3, or 7/12. So we can do 7 jobs in 12 hours. The second thing we need to bear in mind is that rate and time have a reciprocal relationship. If our rate is 7/12, then the time it would take us to complete a job is 12/7 hours. Not so complex. What’s interesting is that these simple ideas can unlock seemingly complex questions. Take this official question, for example: Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank.A) 1/3 B) 1/2 C) 2/3 D) 5/6 E) 1 So let’s start by assigning some variables. We’ll call the rate for p ump A, Ra. Similarly, we’ll designate the rate for pump B as Rb,and the rate for pump C as Rc. If the time for A and B together to fill the tank is 6/5 hours, then we know that their combined rate is 5/6, because again, time and rate have a reciprocal relationship. So this first piece of information yields the following equation: \(R_a + R_b = \frac{5}{6}\). If A and C can fill the tank in 3/2 hours, then, employing identical logic, their combined rate will be 2/3, and we’ll get: \(R_a + R_c = \frac{2}{3}\). Last, if B and C can fill tank in 2 hours, then their combined rate will be ½, and we’ll have: \(R_b+ R_c = \frac{1}{2}\). Ultimately, what we want here is the time it would take all three pumps working together to fill the tank. If we can find the combined rate, or Ra + Rb + Rc, then all we need to do is take the reciprocal of that number, and we’ll have our time to full the pump. So now, looking at the above equations, how can we get Ra + Rb + Rc on one side of an equation? First, let’s line our equations up vertically: \(R_a + R_b = \frac{5}{6}\). \(R_a + R_c = \frac{2}{3}\). \(R_b+ R_c = \frac{1}{2}\). Now, if we sum those equations, we’ll get the following: \(2R_a + 2R_b + 2R_c = \frac{5}{6} + \frac{2}{3} + \frac{1}{2}\). This simplifies to: \(2R_a + 2R_b + 2R_c = \frac{5}{6} + \frac{4}{6} + \frac{3}{6} = \frac{12}{6}\) or \(2R_a + 2R_b + 2R_c = 2\). Dividing both sides by 2, we’ll get: \(R_a + R_b + R_c = 1\). This tells us that the pumps, all working together can do one tank in one hour. Well, if the rate is 1, and the time is the reciprocal of the rate, it’s pretty obvious that the time to complete the task is also 1. The answer, therefore, is E. This question is discussed HERE. Takeaway: the most persistent myth we have about our academic limitations is that we’re simply not good at a certain subset of problems when, in truth, we just never properly learned how to do this type of question. Like every other topic on the GMAT, rate/work questions can be mastered rapidly with a sound framework and a little practice. So file away the notion that rates can be added in work questions and that time and rate have a reciprocal relationship. Then do a few practice questions, move on to the next topic, and know that you’re one step closer to mastering the skills that will lead you to your desired GMAT score.
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Re: Work Rate Problems  All in One Topic!
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15 Jul 2017, 03:16
How to Make Rate Questions Easy on the GMAT Above we discussed the reciprocal relationship between rate and time in “rate” questions. Occasionally, students will ask why it’s important to understand this particular rule, given that it’s possible to solve most questions without employing it. There are two reasons: the first is that knowledge of this relationship can convert incredibly laborious arithmetic into a very straightforward calculation. And the second is that this same logic can be applied to other types of questions. The goal, when preparing for the GMAT, isn’t to internalize hundreds of strategies; it’s to absorb a handful that will prove helpful on a variety of questions. The other night, I had a tutoring student present me with the following question: Question: It takes Carlos 9 minutes to drive from home to work at an average rate of 22 miles per hour. How many minutes will it take Carlos to cycle from home to work along the same route at an average rate of 6 miles per hour?(A) 26 (B) 33 (C) 36 (D) 44 (E) 48 Solution: This question doesn’t seem that hard, conceptually speaking, but here is how my student attempted to do it: first, he saw that the time to complete the trip was given in minutes and the rate of the trip was given in hours so he did a simple unit conversion, and determined that it took Carlos (9/60) hours to complete his trip. He then computed the distance of the trip using the following equation: (9/60) hours * 22 miles/hour = (198/60) miles. He then set up a second equation: 6 miles/hour * T = (198/60) miles. At this point, he gave up, not wanting to wrestle with the hairy arithmetic. I don’t blame him. Watch how much easier it is if we remember our reciprocal relationship between rate and time. We have two scenarios here. In Scenario 1, the time is 9 minutes and the rate is 22 mph. In Scenario 2, the rate is 6 mph, and we want the time, which we’ll call ‘T.” The ratio of the rates of the two scenarios is 22/6. Well, if the times have a reciprocal relationship, we know the ratio of the times must be 6/22. So we know that 9/T = 6/22. Crossmultiply to get 6T = 9*22. Divide both sides by 6 to get T = 9*22/6. We can rewrite this as T = (9*22)/(3*2) = 3*11 = 33, so the answer is B. This question is discussed HERE. The other point I want to stress here is that there isn’t anything magical about rate questions. In any equation that takes the form a*b = c, a and b will have a reciprocal relationship, provided that we hold c constant. Take “quantity * unit price = total cost”, for example. We can see intuitively that if we double the price, we’ll cut the quantity of items we can afford in half. Again, this relationship can be exploited to save time. Take the following data sufficiency question: Question: Pat bought 5 lbs. of apples. How many pounds of pears could Pat have bought for the same amount of money?(1) One pound of pears costs $0.50 more than one pound of apples. (2) One pound of pears costs 1 1/2 times as much as one pound of apples. Solution:Statement 1 can be tested by picking numbers. Say apples cost $1/pound. The total cost of 5 pounds of apples would be $5. If one pound of pears cost $.50 more than one pound of apples, then one pound of pears would cost $1.50. The number of pounds of pears that could be purchased for $5 would be 5/1.5 = 10/3. So that’s one possibility. Now say apples cost $2/pound. The total cost of 5 pounds of apples would be $10. If one pound of pears cost $.50 more than one pound of apples, then one pound of pears would cost $2.50. The number of pounds of pears that could be purchased for $10 would be 10/2.5 = 4. Because we get different results, this Statement alone is not sufficient to answer the question. Statement 2 tells us that one pound of pears costs 1 ½ times (or 3/2 times) as much as one pound of apples. Remember that reciprocal relationship! If the ratio of the price per pound for pears and the price per pound for apples is 3/2, then the ratio of their respective quantities must be 2/3. If we could buy five pounds of apples for a given cost, then we must be able to buy (2/3) * 5 = (10/3) pounds of pears for that same cost. Because we can find a single unique value, Statement 2 alone is sufficient to answer the question, and we know our answer must be B. This question is discussed HERE. Takeaway: Remember that in “rate” questions, time and rate will have a reciprocal relationship, and that in “total cost” questions, quantity and unit price will have a reciprocal relationship. Now the time you save on these problemtypes can be allocated to other questions, creating a virtuous cycle in which your time management, your accuracy, and your confidence all improve in turn.*GMATPrep questions courtesy of the Graduate Management Admissions Council.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Work Rate Problems  All in One Topic!
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