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Work Rate Problem

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Work Rate Problem

by saadishah » Tue Sep 27, 2016 10:37 am
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant
rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working
together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on
its own?

1. 1/2
2. 2
3. 3
4. 5
5. 6

I solved it this way. Since the change is because of the change in Machine A's rate, it's old rate would be equal to combined new rate - combined old rate i.e., 1/2 - 1/3 = 1/6

So Machine A would take 6 hours to finish the job at its old speed. Is this correct?
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by [email protected] » Tue Sep 27, 2016 10:51 am
Hi saadishah,

Yes, the approach that you used absolutely works. There are actually a number of different ways to answer this question (including TESTing THE ANSWERS) - so you might want to think about whether your way was the fastest and easiest method or not.

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by GMATGuruNY » Tue Sep 27, 2016 12:03 pm
saadishah wrote:Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant
rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working
together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on
its own?

1. 1/2
2. 2
3. 3
4. 5
5. 6
Let 1 widget = 6 units.

Since A and B take 3 hours to produce a 6-unit widget, the rate for A+B = w/t = 6/3 = 2 units per hour.
Since 2A and B take 2 hours to produce a 6-unit widget, the rate for 2A+B = w/t = 6/2 = 3 units per hour.
Rates can be ADDED and SUBTRACTED.
Thus:
A's rate = (2A+B) - (A+B) = 3-2 = 1 unit per hour.
At a rate of 1 unit per hour, the time for A to produce a 6-unit widget = w/r = 6/1 = 6 hours.

The correct answer is E.
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by Matt@VeritasPrep » Thu Sep 29, 2016 7:01 pm
saadishah wrote:I solved it this way. Since the change is because of the change in Machine A's rate, it's old rate would be equal to combined new rate - combined old rate i.e., 1/2 - 1/3 = 1/6

So Machine A would take 6 hours to finish the job at its old speed. Is this correct?
Yes, and that is a very clever approach! Really nice work.
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