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Machine A and Machine B - Problem Solving

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Machine A and Machine B - Problem Solving

by Fdelpesce » Sun Jan 13, 2013 10:52 am
Could someone explain how to do this problem below? When I originally attempted it, the formula that ran through my head was a x b/a + b. But do you not use that because it is not asking for the combined total rate?

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. 5x-10z/2xz
B. 2xz/5x-10z
C. 5xz/x+z
D. xz/x+z
E. xz/x+2z


Thank you!
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by Brent@GMATPrepNow » Sun Jan 13, 2013 11:52 am
Fdelpesce wrote:Could someone explain how to do this problem below? When I originally attempted it, the formula that ran through my head was a x b/a + b. But do you not use that because it is not asking for the combined total rate?

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. 5x-10z/2xz
B. 2xz/5x-10z
C. 5xz/x+z
D. xz/x+z
E. xz/x+2z
Here's the algebraic solution:

Determine the rates per hour.

Rate = output/time

Machine A, working alone at a constant rate, can complete a certain production lot in x hours.
Output = 1 lot
Time = x hours
So, hourly rate = 1/x lots per hour

Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours.
Output = 1/5 lots
Time = y hours
So, hourly rate = (1/5)/y lots per hour
= 1/(5y) lots per hour

Machines A and B, working together, can complete 1/2 of the same production lot in z hours.
Output = 1/2 lots
Time = z hours
So, hourly rate = (1/2)/z lots per hour
= 1/(2z) lots per hour

IMPORTANT: the hourly rate for machine A + the hourly rate for machine B = combined hourly rate for A and B
So, 1/x + 1/(5y) = 1/(2z) . . . solve for y
Multiply both sides by 10xyz to get: 10yz + 2xz = 5xy
Rearrange to get: 2xz = 5xy - 10yz
Factor right side to get: 2xz = y(5x - 10z)
Isolate y to get: 2xz/(5x - 10z) = y

Answer = B

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by GMATGuruNY » Sun Jan 13, 2013 1:27 pm
Fdelpesce wrote:Could someone explain how to do this problem below? When I originally attempted it, the formula that ran through my head was a x b/a + b. But do you not use that because it is not asking for the combined total rate?

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. 5x-10z/2xz
B. 2xz/5x-10z
C. 5xz/x+z
D. xz/x+z
E. xz/x+2z

Thank you!
Let the lot = 20 units.

Let x = 5 hours.
Rate for A alone = w/t = 20/5 = 4 units per hour.

Let y = 4 hours.
This is the time for B to produce 1/5 of the lot (4 units).
Thus, B's rate = w/t = 4/4 = 1 unit per hour.

Half of the lot = 10 units.
Combined rate for A+B = 4+1 = 5 units per hour.
z = the time for A+B to produce 1/2 of the lot = w/r = 10/5 = 2 hours.

The question asks for the value of y (4 hours). This is our target.
Now we plug x=5 and z=2 into the answers to see which yields our target of 4.
Only answer choice B works:
2xz/5x-10z = (2*5*2)/(5*5 - 10*2) = 20/5 = 4.

The correct answer is B.
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by Fdelpesce » Sun Jan 13, 2013 2:54 pm
Why is it that you multiple both sides by 10xyz?[/quote]
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by alexander.vien » Tue Jan 29, 2013 1:06 pm
That's the common denominator of all the terms. When you have a bunch of fractions on one side of an equation an easy way to simplify is to multiply both sides by the product of all the denominators, which in this case would be 10xyz.
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by chiccufrazer » Sat Feb 02, 2013 3:41 pm
Aha so this is not a hard problem at all..ok
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by Tommy Wallach » Sat Feb 02, 2013 6:22 pm
Hey Chiccufrazer,

Actually this would certainly be considered a hard question (7-800). Most GMAT explanations look very straightforward in retrospect, but both plugging in and algebra here are pretty time-consuming, with lots of steps where one could make a mistake.

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