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A work/rate problem

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A work/rate problem

by Woozler » Sat Dec 04, 2010 11:17 pm
A ship developed a leak and started filling up with water (at a stable rate in inches/minute). Five minutes later a pump was turned on. However, it was not powerful enough and 10 minutes later the level of water was higher by an additional 20 inches. Then a second pump of the same power was turned on and in 5 minutes the water level fell by 10 inches. That's when the leak was stopped. How much time (in minutes) will it take for the two pumps to pump out all the remaining water?

(A) 5
(B) 10
(C) 12
(D) 15
(E) 20
Last edited by Woozler on Sun Dec 05, 2010 12:09 am, edited 1 time in total.
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by limestone » Sun Dec 05, 2010 12:06 am
Let's call:

The rate of leaking : x (inches/minute)
The rate a pump draining water out: y

Leaking + one pump in action in 10 minutes, water level rose by 20 inches. Or:

(x -y)*10 = 20, or
x -y = 2 (I)

Leaking + two pumps in action in 5 minutes, water level decreased by 10 inches. Or:

(x-2y)*5 = -10, or
x-2y = -2 (II)

From (I) & (II), we can figure out x and y:
x=6, y=4

The next step is to calculate water level at the time the leak stopped.

Water level of:

First five minute: 6*5 = 30
Ten minutes later: (6-4)*10 + 30 = 50
Five minutes later: (6 - 4*2)*5 + 50 = 40

Then at the time the leak stopped water level was 40 inches high. Time to drain out all the water:

40/(4*2) = 5 (minutes)

Hence pick A.
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by Rahul@gurome » Sun Dec 05, 2010 12:15 am
Solution:
Let ri inches of water be filling up the ship every minute.
Let ro inches of water be pumped out by one pump every minute.
So in initial 5 minutes ri*5 inches is filling up the ship.
Now the pump is being turned on.
So in 10 minutes, water is rising by (ri - ro)*10 = 20 inches.
Next a second pump is turned on.
So in 5 minutes, water fell by (2ro - ri)*5 = 10 inches.
Solving the equations (ri - ro)*10 = 20 and (2ro - ri)*5 = 10, we get ro = 4 and ri = 6.
The remaining water in the ship is (10 + 6*5) inches = 40 inches.
This water will be removed by both the pumps in 40/(2*4) = 5 minutes.

The correct answer is A.
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