GMAT Rate Questions: Tackling Problems with Multiple Components

A few posts ago, I tackled rate/work questions, which are invariably a source of consternation for GMAT test-takers. On the latest official practice tests that GMAC has released, these questions showed up with surprising frequency, so I thought it might be worthwhile to tackle a challenging incarnation of this question type: one in which a single machine begins a project and then multiple machines complete the partially-finished work.

To review, the key for dealing with this type of question is to apply the following rules:

1. Rate * Time = Work
2. Rates are additive in work questions.
3. Rate and time have a reciprocal relationship.

For the questions involving partially completed jobs, we’ll throw in the addendum that a completed job can be designated as “1”’

And that’s it!

Here’s a question I saw on my recent practice test:

Working alone at its constant rate, pump X pumped out ¼ of the water in a tank in 2 hours. Then pumps Y and Z started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Y, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump Z, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?

A) 6
B) 12
C) 15
D) 18
E) 24

Okay, deep breath. Recall our three aforementioned rules. Next, let’s designate the rates for the pumps as x, y, and z, respectively.

If pump x can pump out ¼ of the water in 2 hours, then it would take 4*2 = 8 hours to pump out all the water alone. If pump x can complete 1 tank in 8 hours, then x = 1/8.

If x removes ¼ of the water on its own, then all three pumps working together have to remove the ¾ of the water left in the tank. We’re told that together they can do this in 3 hours. If x, y, and z together can do ¾ of the work in 3 hours, then x + y + z = (¾)/3 = 3/12 = ¼.

We’re told that y, alone, could have pumped out the rest of the water in 18 hours – again, there was ¾ of a tank left, so y = (¾)/18 = 1/24.

To summarize, we know that x = 1/8, y = 1/24, and x + y + z = ¼;  Not so hard to solve for z, right?

1/8 + 1/24 + z = ¼

Multiply everything by 24, and we get:

3 + 1 + 24z = 6

24z = 2

z = 1/12.

That’s z’s rate. If rate and time have a reciprocal relationship, we know that it would take z 12 hours to pump out all the water of one tank alone. The answer is, therefore, B.

Takeaway: The joy of seeing new material from GMAC (Is joy the right word?) is the realization that no matter how many additional layers of complexity the question-writers throw at us, the old verities hold true. So when you see tough questions, slow down. Remind yourself that the strategies you’ve cultivated will unlock even the toughest problems. Then, dive in and discover, yet again, that these questions are never quite as hard as they appear at first glance.

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

Quarter Wit, Quarter Wisdom: Solving GMAT Critical Reasoning Questions Involving Rates

In our “Quarter Wit, Quarter Wisdom” series, we have seen how to solve various rates questions – the basic ones as well as the complicated ones. But we haven’t considered critical reasoning questions involving rates, yet. In fact, the concept of rates makes these problems very difficult to both understand and explain. First, let’s look at what “rate” is.

Say my average driving speed is 60 miles/hr. Does it matter whether I drive for 2 hours or 4 hours? Will my average speed change if I drive more (theoretically speaking)? No, right? When I drive for more hours, the distance I cover is more. When I drive for fewer hours, the distance I cover is less. If I travel for a longer time, does it mean my average speed has decreased? No. For that, I need to know  what happened to the distance covered. If the distance covered is the same while time taken has increased, only then can I say that my speed was reduced.

Now we will look at an official question and hopefully convince you of the right answer:

The faster a car is traveling, the less time the driver has to avoid a potential accident, and if a car does crash, higher speeds increase the risk of a fatality. Between 1995 and 2000, average highway speeds increased significantly in the United States, yet, over that time, there was a drop in the number of car-crash fatalities per highway mile driven by cars.

Which of the following, if true about the United States between 1995 and 2000, most helps to explain why the fatality rate decreased in spite of the increase in average highway speeds?

(A) The average number of passengers per car on highways increased.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

(D) The average mileage driven on highways per car increased.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

Let’s break down the given argument:

• The faster a car, the higher the risk of fatality.
• In a span of 5 years, the average highway speed has increased.
• In the same time, the number of car crash fatalities per highway mile driven by cars has reduced.

This is a paradox question. In last 5 years, the average highway speed has increased. This would have increased the risk of fatality, so we would expect the number of car crash fatalities per highway mile to go up. Instead, it actually goes down. We need to find an answer choice that explains why this happened.

(A) The average number of passengers per car on highways increased.

If there are more people in each car, the risk of fatality increases, if anything. More people are exposed to the possibility of a crash, and if a vehicle is in fact involved in an accident, more people are at risk. It certainly doesn’t explain why the rate of fatality actually decreases.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

This option tells us that the safety features in the cars have been enhanced. That certainly explains why the fatality rate has gone down. If the cars are safer now, the risk of fatality would have reduced, hence this option does help us in explaining the paradox. This is the answer, but let’s double-check by looking at the other options too.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

This option is irrelevant – why the average speed increased is not our concern at all. Our only concern is that average speed has, in fact, increased. This should logically increase the risk of fatality, and hence, our paradox still stands.

(D) The average mileage driven on highways per car increased.

This is the answer choice that troubles us the most. The rate we are concerned about is number of fatalities/highway mile driven, and this option tells us that mileage driven by cars has increased.

Now, let’s consider the parallel with our previous distance-rate-time example:

Rate = Distance/Time

We know that if I drive for more time, it doesn’t mean that my rate changes. Here, however:

Rate = Number of fatalities/highway mile driven

In this case, if more highway miles are driven, it doesn’t mean that the rate will change. It actually has no impact on the rate; we would need to know what happened to the number of fatalities to find out what happened to the rate. Hence this option does not explain the paradox.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

This answer choice tells us that on average, the trips were made more quickly, i.e. the speed increased. The given argument already tells us that, so this option does not help resolve the paradox.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Make Rate Questions Easy on the GMAT

I recently wrote about 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:

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

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: 6miles/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.

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:

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.

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.

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.

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

You Can Do It! 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:

Ra + Rb = 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:

Ra + Rc = 2/3.

Last, if B and C can fill tank in 2 hours, then their combined rate will be ½, and we’ll have:

Rb+ Rc = 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:

Ra + Rb = 5/6.

Ra + Rc = 2/3.

Rb+ Rc = 1/2.

Now, if we sum those equations, we’ll get the following:

2Ra + 2Rb + 2Rc = 5/6 + 2/3 + 1/2. This simplifies to:

2Ra + 2Rb + 2Rc = 5/6 + 4/6 + 3/6 = 12/6 or 2Ra + 2Rb + 2Rc  = 2.

Dividing both sides by 2, we’ll get: Ra + Rb + Rc  = 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.

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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By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

GMAT Tip of the Week

At any rate…

(This is one of a series of GMAT tips that we offer on our blog.)

Rate problems on the GMAT can take a variety of forms, and can often be time consuming for test takers. They also offer a challenge for students who rely on memorization, as it can be trickier than expected to apply the correct relationship between the rate, distance, and time variables.

For a foolproof way to remember the proper relationship, keep in mind that you will almost certainly encounter a rate on your drive to the test center: Miles per hour. Because the word “per” indicates division, you can remember that the rate is the distance (miles) divided by the time (hour). Accordingly, R = D/T.

For difficult rate problems, keep in mind that the challenge is usually created by having multiple variables. The key is that there will generally be a relationship between the distances and times of the two items. Suppose a question asks when a car will catch up to a bicycle; they can only catch up at the same time, and so there will be a relationship between the “T” variable in each rate equation. Furthermore, when they meet they will be at the same point, and there will be a relationship between the “D” variables, as well.

For more practice with rate problems, the Veritas Prep syllabus offers both a Problem Solving and Advanced Word Problems lesson.