Kaplan 800 Rate Problem - Sprockets

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Kaplan 800 Rate Problem - Sprockets

by bml1105 » Thu May 29, 2014 4:53 pm
Machine A and Machine B are each used to manufacture 660 sprockets. It takes maching A 10 hours longer to produce 660 sprockets than Machine B. Machine B produces 10% more sprockets per hour than Machine A. How many sprockets per hour does machine A produce?

(A) 6
(B) 6.6
(C) 60
(D) 100
(E) 110


OA: A
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by GMATGuruNY » Thu May 29, 2014 6:45 pm
bml1105 wrote:Machine A and Machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than Machine B. Machine B produces 10% more sprockets per hour than Machine A. How many sprockets per hour does machine A produce?

(A) 6
(B) 6.6
(C) 60
(D) 100
(E) 110
We can PLUG IN THE ANSWERS, which represent A's rate.
It is stated that B produces 10% more sprockets per hour than A.
Thus, when the correct answer choice is plugged in, B's rate will be 110% of A's rate.

The answer choices imply the following times for A:
660/6 = 110 hours.
660/6.6 = 100 hours.
660/60 = 11 hours.
660/100 = 6.6 hours.
660/110 = 6 hours.

It is stated that A's time is 10 hours longer than B's time.
Thus:
Answer choices D and E are not possible.
Answer choice C seems unlikely, since it implies that B's time is 1 hour.
Answer choice B implies that B's time = 90, a value that does not divide easily into 660.
The correct answer almost certainly is A.

Answer choice A: A's rate = 6 sprockets per hour
Since A's time of 110 hours is 10 hours longer than B's time, B's time = 100 hours.
Thus, B's rate = 660/100 = 6.6 sprockets per hour.
(B's rate)/(A's rate) = 6.6/6 = 11/10 = 110%.
Success!

The correct answer is A.
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by GMATGuruNY » Thu May 29, 2014 6:59 pm
bml1105 wrote:Machine A and Machine B are each used to manufacture 660 sprockets. It takes maching A 10 hours longer to produce 660 sprockets than Machine B. Machine B produces 10% more sprockets per hour than Machine A. How many sprockets per hour does machine A produce?

(A) 6
(B) 6.6
(C) 60
(D) 100
(E) 110

OA: A
Alternate approach:

Since B produces 10% more sprockets per hour than A, for every 10 sprockets produced by A, B produces 11 sprockets.
Thus, the RATE RATIO for A and B = 10:11.
Since rate and time are reciprocals, the TIME RATIO for A and B = 11:10.
Since A's time is 10 hours longer than B's time, and 11:10 = 110:100, A's time = 110 hours, while B's time = 100 hours.
Thus, A's rate = w/t = 660/110 = 6 sprockets per hour.

The correct answer is A.
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by bml1105 » Thu May 29, 2014 8:10 pm
So the best way to answer this question really is just plugging in numbers or answer choices?

I kept trying to make it work with just equations, and I couldn't make it work.

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by GMATGuruNY » Thu May 29, 2014 8:22 pm
bml1105 wrote:So the best way to answer this question really is just plugging in numbers or answer choices?

I kept trying to make it work with just equations, and I couldn't make it work.
Here's an algebraic solution:

Let A = A's rate.
Since B produces 10% more sprockets per hour, B's rate = 1.1A.

A's time = w/r = 660/A.
B's time = w/r = 660/1.1A = 600/A.
Since A's time is 10 hours longer than B's time, we get:
660/A = 600/A + 10
60/A = 10
60 = 10A
A = 6.

The correct answer is A.
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by [email protected] » Thu May 29, 2014 8:38 pm
Hi bml1105,

Mitch's approach that focused on TESTing THE ANSWERS is perfect for these types of questions (and you'll find it to be applicable a few times on Test Day).

These questions usually include at least 1 clue to help you to avoid having to test all of the answers. Here, it's the respective rate and output of the two machines. Machine B produces more sprockets per hour AND takes 10 FEWER HOURS to produce the sprockets. That's a significant difference in "times." Machine A has to be producing sprockets at a low rate (compared to Machine B) so I'd be focused on the smaller answers.

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