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by amanjena » Wed Apr 23, 2014 8:16 am
A certain liquid leaks out of a container at the rate of k liters for every x hours. If the liquid costs $6 per liter, what is the cost, in dollars, of the amount of the liquid that will leak out in y hours?

A) ky/6x
B) 6x/ky
C) 6k/xy
D) 6ky/x
E) 6xy/k

I think it is D
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Source: — Problem Solving |

by theCodeToGMAT » Wed Apr 23, 2014 8:20 am
Yes..

x hours = k litres

1 hr = k/x litres

y hours = ky/x litres

Cost of 1 litre = 6$

Cost of ky/x ==> 6ky/x
R A H U L
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by GMATGuruNY » Wed Apr 23, 2014 9:18 am
A certain liquid leaks out of a container at the rate of k liters for every x hours. If the liquid costs $6 per liter, what is the cost, in dollars, of the amount of liquid that will leak out in y hours?

A. ky/6x
B. 6x/ky
C. 6k/xy
D. 6ky/x
E. 6xy/k
Let k=10 and x=2, implying that 10 liters are lost every 2 hours.
Thus, the rate of leakage = 10/2 = 5 liters per hour.
Since each liter costs $6, the amount of money lost per hour = 6*5 = 30.
Let y = 4 hours.
Over 4 hours, the amount of money lost = 4*30 = 120. This is our target.
Now we plug k=10, x=2, and y=4 into the answers to see which yields our target of 120.
Only answer choice D works:
6ky/x = (6*10*4)/2 = 120.

The correct answer is D.

Algebraically:
The number of liters lost per hour = k/x. (k liters per x hours).
Since each liter costs $6, the amount of money lost per hour = 6(k/x).
Since the leakage lasts for y hours, we multiply by y:
y*6(k/x) = 6ky/x.
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by Jeff@TargetTestPrep » Mon Jan 08, 2018 4:55 pm
amanjena wrote:A certain liquid leaks out of a container at the rate of k liters for every x hours. If the liquid costs $6 per liter, what is the cost, in dollars, of the amount of the liquid that will leak out in y hours?

A) ky/6x
B) 6x/ky
C) 6k/xy
D) 6ky/x
E) 6xy/k
The leak rate is k/x; thus, in y hours, ky/x liters will leak out, which will cost (6)(ky/x) = 6ky/x dollars.

Answer: D

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