Work Rate Problems - All in One Topic!

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 n-5, you can use the knowledge that every GMAT question is do-able 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 2-3 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.

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 one-fourth, or 25%, of the bread is baked. For each baker’s per-hour 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.

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 counter-intuitive 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.

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.

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.

Cross-multiply 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 problem-types 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.

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 no-no 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 minimum-work-solution 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!

• Theory on Work/Rate Problems
• Data Sufficiency Work/Rate Questions
• Problem Solving Work/Rate Questions

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.

I'm sorry, but you needed to start this entry WTIH THIS post please. This was what i needed to read first.

Total Cost of Pear = Total Cost of Apple
...............Q1 * P1 = Q2 * P2 = Quantity * Price is the same

Q1/Q2 = P2/P1

X/5 = Y/1.5Y (Y is the price of apples)
X = 10/3 pounds

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

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) yz(yz+xz−xy)
yz cancels xy in the denominator giving us yz/xz=y/x=2/4=1/2

(E) (yz+xz−xy)
yz cancels xy in the numerator giving us xz/yz=x/y=4/2=2
-------
how can it can assumed that Y cancels out Z? from where can you make that assumption ?
also, if I plug in x=1, y=3, z=3, then it just don't work, can someone explain me why?
Thanks in advance

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

It's getting displayed properly in Chrome and Safari. Where are you seeing the issues ?

You can think of this problem as Two pumps filling an empty container that can hold 300 gallons.

Thus you'll end up adding the RATES of both the pumps.

- One pump will be filling the 300-gallon tank from some unknown water source
- The other pump will be filling the 300-gallon thank using the 500-gallon water source.

Ideally, if one pump is filling a tank, it is similar to "taking out" water from some other tank/river/xyzSource.

Hey, I had a small doubt on a question. The question is 'Pat bought 5 lbs. of apples. How many pounds of pears could Pat have bought for the same amount of money?'. The formula used is Quantity*Unit Price = Total Cost. Now, if I say the number of pounds of pear brought is X pounds, while the number of apples brought stands at 5 pounds. Now the quantities of the two are in the ratio 5/X. Using the reciprocal formula, the prices of the two should be in the ratio X/5. Now, in the second statement, it says that the price of a pear/pound is 1.5 times greater than the price of an apple/pound. So, I make the equation, X/5 = 1.5Y/Y. The Y cancels out and I get X = 7.5 Pounds. But this answer is different that the correct answer. Where am I going wrong? Please help. VeritasKarishma Bunuel

We have detailed discussion of this question here: https://gmatclub.com/forum/pat-bought-5 ... 54823.html Hope it helps.

[quote="KarishmaB"][quote="Bunuel"]


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?



Hi KarishmaB Bunuel - Please help me with the below queries which is bugging me since I have read the above brilliant way of solving -

Query 1 : In below question why did we have to calculate speed by using formula when time was just calculated using ratios

-----------------

The speed of bus A is 20% more than the speed of bus B. Bus B takes 2 hours longer than bus A to travel 600 miles. What is the speed of bus A?

Ordinarily, people would make equations and solve them to get to the answer. But we can do it quickly and orally.

Speed of bus A is 20% more than speed of bus B. This means that speed of bus A : speed of bus B is 120:100 i.e. 6:5. To travel the same distance, time taken by bus A: time taken by bus B will be 5:6. This difference of 1 in the ratio of time taken is actually given to be 2 hours. Hence, the multiplier is 2. Time taken by bus A to travel the 600 miles must be 5*2 = 10 hrs and time taken by bus B to travel the 600 miles must be 6*2 = 12 hrs.

Speed of bus A = 600/10 = 60 mph

------------------------

Query 2: In below question Why do we calculate distance using speed ratio and not time ratio (As they Distance is directly proportional to both , I am confused why we select distance over time here )

-------------------------

Two trains, A and B, traveling towards each other on parallel tracks, start simultaneously from opposite ends of a 250 mile route. A takes a total of 3 hours to reach the opposite end while B takes a total of 2 hours to reach the opposite end. When train A meets train B during the journey, how far is train A from its starting point?

Since A takes 3 hours to travel the same distance for which B takes 2 hours, time taken by A and B is in the ratio 3:2 so their speeds must be in the ratio 2:3. Hence, A will cover 2/5th of the total distance of 250 miles and B will cover the rest of the (3/5)th of the total distance. Therefore, A will be 100 miles from its starting point when A and B meet.

---------------------------

Please help!!!