MBAOver30 GMAT Quant Practice #2

May 31, 2013

Admissions Help, Business School, GMAT, GMAT Quant

You asked for it, and now you’ve got it. Our friends over at Target Test Prep have sent this year’s GMATTers two more juicy quant problems to chew on. Here’s to your 680-750+ next GMAT attempt!

Here’s bath #2!

PRACTICE PROBLEM 2A

Jamshid can paint a fence in 50 percent less time than Taimour can when each works alone. When they work together, they can paint the fence in 8 hours. How long would it take Taimour to paint the fence alone?

(A) 6 hours (B) 8 hours (C) 14 hours (D) 24 hours (E) 32 hours

Solution:

We have a relative comparison between Jamshid’s work time and Taimour’s work time. Jamshid takes 50 percent less time to paint the fence. If we let Taimour’s time equal t hours, Jamshid’s time would be 0.5t hours. This means that the men’s respective rates would be

$\Rightarrow \text{Wor}{{\text{k}}_{Taimour}}=\frac{1}{t}\frac{fence}{hour}\times 8\text{ hours}=\frac{8}{t}\text{ fences}$

$\Rightarrow \text{Wor}{{\text{k}}_{Jamshid}}=\frac{2}{t}\frac{fence}{hour}\times 8\text{ hours}=\frac{16}{t}\text{ fences}$

The two men working together take 8 hours to finish a fence; we can use this figure to determine the contribution of each to a particular job:

	Rate	Time	Work
Taimour	$\frac{\mathbf{1}}{\mathbf{t}}\frac{\mathbf{fence}}{\mathbf{hour}}$	8 hours
Jamshid	$\frac{\mathbf{2}}{\mathbf{t}}\frac{\mathbf{fence}}{\mathbf{hour}}$	8 hours

Now we can calculate the work contributed by each:

$\Rightarrow \text{Wor}{{\text{k}}_{Taimour}}=\frac{1}{t}\frac{fence}{hour}\times 8\text{ hours}=\frac{8}{t}\text{ fences}$

$\Rightarrow \text{Wor}{{\text{k}}_{Jamshid}}=\frac{2}{t}\frac{fence}{hour}\times 8\text{ hours}=\frac{16}{t}\text{ fences}$

	Rate	Time	Work
Taimour	$\frac{1}{t}\frac{fence}{hour}$	8 hours	$\frac{\mathbf{8}}{\mathbf{t}}\mathbf{ fences}$
Jamshid	$\frac{2}{t}\frac{fence}{hour}$	8 hours	$\frac{\mathbf{16}}{\mathbf{t}}\mathbf{ fences}$

It must be true that the work completed by Jamshid plus the work completed by Taimour equals the total work done (1 fence).

$\Rightarrow Wor{{k}_{Taimour}}+Wor{{k}_{_{Jamshid}}}=Wor{{k}_{Total}}$

$\Rightarrow \frac{8}{t}\text{ }+\frac{16}{t}=1$

$\Rightarrow \frac{24}{t}=1$

$\Rightarrow t=24\text{ hours}$

Thus it would take Taimour 24 hours to paint the fence alone.

Answer: D

PRACTICE PROBLEM 2B

A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Solution:

The information provided by the problem is the number of infielder cards = 8, the number of outfielder cards = 12, and, by extension, the total number of cards = 8 + 12 = 20. We can let x represent the number of cards that will be removed from the deck. With this information, we can set up an equation to determine the lowest number of outfielder cards that need to be removed from the deck. Remember that any outfielder cards removed are also removed from the total, so x needs to be subtracted from both the number of outfielder cards in the pack and the total number of cards. It is very important to see that in this problem, the part-to-total ratio is being altered

$\Rightarrow \frac{\text{outfielder cards}}{\text{total cards}}=\frac{12-x}{20-x}\le \frac{2}{5}\to 5\left( 12-x \right)\le 2\left( 20-x \right)$