Source: Official Guide for the GMAT 13th Ed. Problem Solving; #81 Official Guide for the GMAT 2015 14th Ed. Problem Solving; #81

5

Working simultaneously at the respective constant rates,

Working simultaneously at the respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms f x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

3 Explanations

1

Simon Gerstmann

Is it possible to use numbers instead of the algebra approach?

Thank you for your help :)

Feb 19, 2016 • Comment

Cydney Seigerman, Magoosh Tutor

Hi Simon :)

For this question, since the answer choices are in terms of x and y, the most efficient way to solve this question is to keep everything in terms of those variables. Once we put in real numbers and carry out the various operations, like addition, we can't determine the exact relationship between x and y. And this means, we can't evaluate the answer choices.

That being said, we could plug in numbers to check the answer. Let's say that x = 20 hours and y = 80 hours. This would mean that the rate of A and B working together is

800 nails/20 hr = 40 nails/hr

while the rate of A working alone is

800 nails/80 hours = 10 nails/hr

As Mike mentions in the explanation video, the combined rate is the sum of the individual rates:

rate A+B = rate A + rate B
40 nails/hr = 10 nails/hr + rate B
rate B = 30 nails/hr

We can now use that rate to determine the number of hours it takes B to make 800 nails:

800 nails * hr/30 nails = 26 2/3 hr

And we can check if this is the same answer as we get using the correct algebraic expression:

xy/(y-x) = 20*80/(80-20) = 26 2/3 hr

And it is :D

Again, while we're able to check the answer by plugging in numbers, the best way to solve the question is by keeping x and y as variables and solving for the expression algebraically.

I hope this helps :)

Feb 20, 2016 • Reply

3

Machine·····W·······T
A·············800······y
B·············800······b
A+B·········800······x

Since the work done is equal in all the cases, we can use:

T(A+B) = [T(A) * T(B)] / [T(A) + T(B)]

x = [y*b] / [y + b]

xy + xb = yb

xy = yb - xb

xy = (y - x)b

(xy) / (y - x) = b

Jan 25, 2015 • Comment

3

Gravatar Mike McGarry, Magoosh Tutor

Dec 27, 2013 • Comment

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