# GMAT Work Rate Problems

GMAT rate problems might seem intimidating, but they’re not so bad once you get the fundamental concepts down. We’ll show you how to master this common type of GMAT word problem in this post, then give you some practice problems with answers and explanations!

## GMAT Rate and Work Rate Problems: Main Concepts

### Big Idea #1: The “ART” Equation

You may be familiar with the distance equation, D = RT (“distance equals rate times time”), sometimes remembered as the “dirt” equation. It turns out, that equation is just a specific instance of a much more general equation. In that equation, R, the rate, is distance per time, but in non-distance problems, rate can be anything over time — wrenches produced per hour, houses painted per day, books written per decade, etc. In these cases, typical of work problems, we are no longer concerned with “distance” per time, but with the amount of something produced per time. We use A to represent this amount (the number of wrenches, the number of houses, etc.), and the equation becomes A = RT. Sometimes folks remember this as the “art” equation.

Here’s a simple mnemonic. When you travel, you are moving on the Earth, which is made of dirt, so for traveling & distance you use D = RT. Work problems involve machines, and machines make things —– making is creation, and creation is the essence of art, so use the A = RT equation. (I know, I know, what comes out of most machines is hardly worthy of aesthetic elevation, but it works for a mnemonic!)

### Big Idea #2: Rates are Ratios

The word “rate” and the word “ratio” have the same Latin root: in fact, they also share a Latin root with the “rationality” of our minds, but that’s a discussion that would bring up to our noses into Pythagorean and Platonic philosophies. The point is: a rate is a ratio, that is to say, a fraction. Technically, any fraction, any ratio, in which the numerator and the denominator have different units is a rate. Fuel efficiency (mpg) and price per unit and most baseball fractions (ERA, BA, OBP, SLG, etc.) are rates. Currency rates and exchanges rates are common financial market rates that, ironically, almost never appear on the GMAT —- go figure! Most GMAT rates have time in the denominator, and it’s a rate of how fast work is being done or how fast something is being produced or accomplished.

The fact that rates are ratios means: we can solve these problems by setting up proportions and using proportional thinking! As you will see in the solutions below, that’s an extremely powerful strategy for solution.

### Big Idea #3: Add Rates

The vast majority of work problems on the GMAT involve two people or two machines and comparisons of their individual production to their combined production. The questions will often give you information about times and about amounts, and what you need to know is: you can’t add or subtract times to complete a job and you can’t add or subtract amounts of work; instead, you add and subtract rates.

(rate of A alone) + (rate of B alone) = (combined rate of A & B)

Here A and B can be two people, two machines, etc. The extension of this idea is that if you have N identical machines, and each one works at a rate of R, then the combined rate is N*R.

### Big Idea #4: Understand Speed and Average Speed

Rate is another word for speed. One common source of errors with GMAT rate problems is that all three variables have to be in the same units. If you travel at 30 mph for 10 minutes, you do not go 30*10 = 300 miles!

Here’s an example:

$$\frac{30 \text{miles}}{\text{hour}} = \frac{30 \text{miles}}{60 \text{minutes}} = \frac{x \text{miles}}{10 \text{minutes}}$$
$$\frac{1 \text{mile}}{2 \text{minutes}} = \frac{x \text{miles}}{10 \text{minutes}}$$

Cross-multiply, and you get 10 = 2x. So, x must equal 5 miles.

Many trickier rate questions ask about “average speed” or “average velocity” (for GMAT purposes, those two are identical). The formula for average speed is:

$$\text{Average speed}=\frac {\text{Total distance}}{\text{Total time}}$$

For a single trip at one speed, there’s nothing particularly mysterious about this question. This concept becomes much trickier in two-leg trips, especially trips in which the car travels at one speed in one leg, and at another speed in another leg. You can never simply average the two velocities given, and that will always be a tempting incorrect choice on the GMAT. You always need to apply D = RT separately in each leg of the trip, and then you need to add results from the individual legs to find the total distance and the total time.

With just these four ideas, you can unlock any GMAT work rate problem. At this point, you may want to go back and give another attempt at those three practice questions. Follow carefully how they are applied in the solutions below.

## Practice GMAT Rate Problems

For practice, here are some GMAT rate problems for you! The last two are challenging.

1) A car drives 300 miles at 30 mph, and then 300 miles at 60 miles per hour. What is the car’s average speed, in mph?

In order to figure out the average velocity, we need to know both the total distance and the total time. From the question, we know the total distance is 600 miles. We need to figure out the time of each leg separately. In the first leg, T = D/R = 300/30 = 10 hr. In the second leg, T = D/R = 300/60 = 5 hours. The total time is 10 + 5 = 15 hours. The average velocity, total distance divided by total time, is 600/15 = 40 mph. Answer = B.

2) A car drives for 3 hours at 40 mph and then drives 300 miles at 60 mph. What is the car’s average speed, in mph?

In the first leg, we know time and rate, so find distance: D = RT = (3)*(40) = 120 miles. In the second leg, we know distance and rate, so find time: T = R/D = 300/60 = 5 hours. Total distance = 120 + 300 = 420 miles. Total time = 3 + 5 = 8. Average velocity = 420/8 = 210/4 = 105/2 = 52.5 mph. Answer = D.

3) For the first 150 miles of a trip, a car drives at v mph. For the next 200 miles, the car drives at (v + 25) mph. The average speed of the whole trip is 35 mph. Find the value of v.

The distance of the first leg is 150 miles, and the rate is v, so the time of the first leg is:

$$\text{t}_1 = \frac{150}{\text{v}}$$

The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is:

$$\text{t}_2 = \frac{200}{\text{v}+25}$$

The total distance was 350 miles, and the average speed was 35 mph, so the total time of the trip must have been T = D/R = 350/35 = 10 hours. At this point, the algebra becomes hairy, so I will just plug in numbers from the answer choices.

Choice A. If v = 20 mph, then v + 25 = 45 mph. The first leg takes 150/20 = 7.5 hours, and the last leg 200/45 takes way more than three hours, so this total time is well over 10 hours. This choice is not correct.

Choice B. If v = 25, then v + 25 = 50. The first leg takes 150/25 = 6 hours. The second leg takes 200/50 = 4 hours. The total is 10 hours, which is the correct value, so this is the correct answer choice. Answer = B.

4) A car travels at one speed for 4 hours, and then at twice that speed for 6 hours. The average velocity for the whole 10-hour trip is 40 mph. Find the initial speed in mph.

If the average velocity for the 10 hour trip is 40 mph, that means the total distance is D = RT = (40)*(10) = 400 miles. The distance in the first leg is d1 = RT = 4v. The distance in the second leg is d2 = RT = (2v)*(6) = 12v. The total distance is the sum, 4v + 12v = 16 v. Set this equal to the numerical value of the total distance.

400 = 16v → 100 = 4v → 25 = v

So the initial speed is v = 25 mph. Answer = A.

5) Running at the same rate, 8 identical machines can produce 560 paperclips a minute. At this rate, how many paperclips could 20 machines produce in 6 minutes?

“Running at the same rate, 8 identical machines can produce 560 paperclips a minute.” That 560 is a combined rate of 8 machines —- 560 = 8*R, so the rate of one machine is R = 560/8 = 70 paperclips per minute.

“At this rate, how many paperclips could 20 machines produce in 6 minutes?” Well, the combined rate of 20 machines would be Rtotal = 20*70 = 1400 pc/min. Now, plug that into the “art” equation: A = RT = (1400)*(6) = 8400 pc. Answer = C.

6) Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks. If they both work together, how many weeks will it take for them to produce 15 handcrafted drums?

Method I: the rates solution

“Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks.” Jane’s rate is (1 drum)/(4 weeks) = 1/4. Zane’s rate is (1 drum)/(6 weeks) = 1/6. The combined rate of Jane + Zane is

R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12

That’s the combined rate. We need to make 15 drums — we have a rate and we have an amount, so use the “art” equation to solve for time:

T = A/R = 15/(5/12) = 15*(12/5) = (15/5)*12 = 3*12 = 36

BTW, notice in the penultimate step, the universal fraction strategy: cancel before you multiply (Tip #3: https://magoosh.com/gmat/2012/can-i-use-a-calculator-on-the-gmat/. Jane and Zane need 36 weeks to make 15 drums. Answer = B.

Method II: the proportion solution

“Jane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks.” Let’s find the LCM of 4 and 6 — that’s 12 weeks. In a 12 week period, Jane, making a drum every 4 weeks, makes three drums. In a 12 week period, Zane, making a drum every 6 weeks, makes two drums. Therefore, in a 12 weeks period, they produce 5 drums between the two of them. If they make 5 drums in 12 weeks, they need triple that time, 36 weeks, to make 15 drums. Therefore, Jane and Zane need 36 weeks to make 15 drums. Answer = B.

7) Machines P and Q are two different machines that cover jars in a factory. When Machine P works alone, it covers 1500 jars in m hours. When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours. In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars?

This is a particularly challenging, one because we have variables in the answer choices. I will show an algebraic solution, although a numerical solution (https://magoosh.com/gmat/math/word-problems/variables-in-gmat-answer-choices-algebraic-approach-vs-numerical-approach/) is always possible.

“Machines P and Q are two different machines that cover jars in a factory. When Machine P works alone, it covers 1500 jars in m hours. When Machines P and Q work simultaneously at their respective rates, they cover 1500 jars in n hours. In terms of m and n, how many hours does it take Machine Q, working alone at its constant rate, to cover 1500 jars? ”

Since the number “1500 jars” appears over and over, let’s arbitrarily say 1500 jars = 1 lot, and we’ll use units of lots per hour to simplify our calculations.

P’s individual rate is (1 lot)/(m hours) = 1/m. The combined rate of P & Q is (1 lot)/(n hours) = 1/n. We know

(P’s rate alone) + (Q’s rate alone) = (P and Q’s combined rate)

(Q’s rate alone) = (P and Q’s combined rate) – (P’s rate alone)

(Q’s rate alone) = 1/n – 1/m = m/ (nm) – n/ (nm) = (m – n)/(nm)

We now know Q’s rate, and we want the amount of 1 lot, so we use the “art” equation.

1 = [(m – n)/ (nm)]*T

T = (mn)/(m – n)

8) Working together, 7 identical pumps can empty a pool in 6 hours. How many hours will it take 4 pumps to empty the same pool?