The GMAT Quant section loves word problems. Here are six problems of six different genres, easy to medium difficulty.

1) Sam’s car was fined when he gave Joe and Peter a ride, so they decided to help Sam pay the fine. Joe paid $3 more than 1/4 of the fine and Peter paid $3 less than 1/3 of the fine, leaving pay $4 less than 1/2 the fine to complete the payment. What fraction of the fine did Sam pay?

- (A) $13

(B) $15

(C) $20

(D) $28

(E) $48

2) Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator?

- (A) $275

(B) $325

(C) $425

(D) $575

(E) $625

3) A rectangular garden is surrounded by a 3 ft. wide concrete sidewalk. If the length of the garden is 4 ft more than its width, and if the area of the sidewalk is 60 sq ft more than the area of the garden, then what is the length of the garden?

- (A) 6

(B) 8

(C) 10

(D) 12

(E) 14

4) A vendor sells only Product A, for $6, and Product B, for $21. If Q% of the products sold are Product B, and if T% of the total revenue comes from sales of Product B, find Q in terms of T.

5) A machine, working at a constant rate, manufactures 36 staplers in 28 minutes. How many staplers does it make in 1 hr 45 min?

- (A) 125

(B) 128

(C) 135

(D) 144

(E) 150

6) Pump X takes 28 hours to fill a pool. Pump Y takes 21 hours to fill the same pool. How long does it take them to fill the same pool if they are working simultaneously?

- (A) 7 hr

(B) 10 hr

(C) 12 hr

(D) 14 hr

(E) 18 hr

Solutions will follow this article.

## Word Problems

Word problems come in several different varieties. Here are a few blogs that cover different kinds of word problems

1) Distance and Rate: the D = RT formula

3) Work rate

5) Simple and Compound Interest

You may find some hints in this blogs, and of course, if the problems are difficult for you, you should study the solutions below carefully.

## Practice problem solutions

1) Call the fine F. Joe paid (1/4)F + 3 and Peter paid (1/3)F – 3, leaving (1/2)F – 4 left. If we add those three up, they should add up to F.

F = [(1/4)F + 3] + [(1/3)F – 3] + [(1/2)F – 4]

F = (1/4)F + (1/3)F + (1/2)F – 4

Multiply all terms by 12 to clear the fractions.

12F = 3F + 4F + 6F – 48

12F = 13 F – 48

–F = – 48

F = 48

Well, if the fine cost $48, then Sam paid the part not covered by Joe or Peter. Half the fine is $24, and Sam paid $4 less than this: $20. Answer = **(C) **

2) This is a relatively easy word problem. Let A be Alice’s share, and B, Bruce’s share. Clearly A + B = 900. We also are told that 2A = B + 75. To solve, use substitution. Solve this for B:

B = 2A – 75.

Now, plug this into the first equation:

A + (2A – 75) = 900

3A = 900 + 75

A = 300 + 25 = $325

Alice’s share is $325. Answer = **(B)**.

3) This is a challenging geometry problem. Let the width of the garden be x, so the length is (x + 4). The garden itself would have an area of x(x + 4). Now, for the area of the sidewalk, consider this diagram:

Notice, the sidewalk can be subdivided into convenient pieces. There are two of the long horizontal rectangles at the top and bottom; each is 3(x + 4). There are two vertical side rectangles: each is 3x. Finally, there are four corner squares, each 3 x 3 = 9. The total area of the sidewalk is:

area of sidewalk = 6(x + 4) + 6x + 4*9 = 12x + 24 + 36 = 12x + 60

Now, we are told that

(area of sidewalk) = (area of garden) + 60

12x + 60 = x(x + 4) + 60

The solution x = 0 doesn’t make sense in the problem, so the only solution this gives is x = 8. This means, the garden is 8 x 12, and the length of the garden is 12. Answer = **(D)**.

4) Let’s say the vendor sells N products in total. We know (Q/100)*N is the number of Product B sold, and this generates 21(Q/100)*N dollars in revenue. We know that ((100 – Q)/100)*N is the number of Product A sold, and this generates 6*((100 – Q)/100)*N dollars in revenue. The sum of those two revenue terms is the total revenue, and the ratio of 21(Q/100)*N over the total revenue is T/100, the ratio of the total revenue that comes from sales of Product B. Thus,

Cancel the factors of N, and multiply the numerator and denominator of the big fraction by 100.

Cross-multiply:

15QT + 600T = 2100Q

Divide all terms by 15

QT + 40T = 140Q

Put all the terms with Q on one side, then factor out the Q, then divide by that factor to isolate Q.

Answer = **(D)**.

5) Change 1 hr 45 min to 105 min. For this, we need to set up a simple proportion of staplers per time

The absolutely worst thing you could do at this point in the problem is to cross-multiply. That would be a supremely unstrategic move. Instead, **cancel before you multiply**. For what we can and can’t cancel in a proportion, see this post. We can cancel the factor of 4 in the 36 and 38.

We absolutely cannot cancel any factors between the 9 and 105. That kind of “cross-canceling” is 100% illegal in a proportion. We can cancel the common factor of 7 in the two denominators.

Now that the fraction is entirely simplified, we can cross-multiply.

S = 9*15 = 135

The machine would be 135 staples in 1 hr 45 min.

6) Express the rates of pumps in terms of (pools/hours). Pump X operates at rate of 1/28, and pump Y operates at 1/21. When the pumps work together, we add the rates.

So, combined, the pumps operate at a rate of 1/12, which means they can fill a pool in 12 hr. Answer = **(C)**

Can you please explain how will we go about solving question 4 if we follow the numerical approach here.

I’ll be happy to help. But Nikhar, can you tell me a little more about what you mean by “the numerical approach”?

Mike,

I think you misconstrued my last statement. I tried to imply that its imperative to be flexible in strategy – algebra or numbers – and I felt that plugging numbers judiciously was a pragmatic and time saving strategy,as opposed to algebra, for the problem you’ve posed.

Thanks,

Sriram.

Sriram,

Perhaps I did misconstrue your words: if so, my apologies. Yes, it’s a very important skill to make that call — algebra vs. picking numbers — on these problems that have variables in the answer choices. I am of the opinion that, if one’s algebra skills are sharp, the algebraic solution is almost always more efficient, but that does require, essentially, complete fluency in algebra. It’s very good that you are exploring these issues for yourself, and learning about your own strengths and what works best for you! Such discernment will serve you well! 🙂

Mike 🙂

Hi Mike,

I tried to deal with problem 4 using numbers, albeit it was not a easy task to find the right set. I wanted to make sure the total revenue was at-least some multiple of 100, if not 100 itself and settled for 6a + 21b = 300 with a=15 and b=10. Hence Q=40% and T=70% and simply plugged in T in all the choices. I must admit, I get a little queasy dealing with algebra for these kind of problems.

Dear Sriram,

Picking numbers is one way to go, my friend, but the more “stomach” you can develop for algebra, the more that will serve you on the GMAT. I wish you the best of luck.

Mike 🙂