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successstory
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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate 01 Feb 2007, 15:33
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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate of the other (and both have constant rates). How many hours would it take for the faster pump if it
worked alone?

Answer: 20/3



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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate 01 Feb 2007, 15:46
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According to my calculation the answer should be 50/3

Let say, pump 1 has a rate of x c.c./hr
Therefore, pump 2 will have a rate of 1.5x c.c./hr

So, if both pump were to work together for an hour, they would fill:

x + 1.5x c.c.

So, in 1 hour both the pumps together can fill x + 1.5x c.c.,
Hence, in 4 hours both the pumps can fill 4(x + 1.5x)c.c. = 10x c.c.

Now if the faster pump was to work alone it has to fill 10x c.c.
Now the rate for the faster pump is: in 1 hour it fills 1.5x c.c.
Hence, to fill 10x c.c. it will take 10x/1.5x = 100/15 = 20/3
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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate 01 Feb 2007, 17:15
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Hi, the answer is from gmatprep -- 20/3.

50/3 is not an option, though your solution looks good :)

Anyone arrive at 20/3? this is a good rate problem. Thanks!
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Sumithra
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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate 01 Feb 2007, 20:34
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successstory
Hi, the answer is from gmatprep -- 20/3.

50/3 is not an option, though your solution looks good :)

Anyone arrive at 20/3? this is a good rate problem. Thanks!
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1/a + 1/b = 1/4 =>ab/a+b=4

1/a=1.5/b =>b=1.5a
substituting in the first eqn.
1.5a*a/1.5a+a = 4
solving
1.5a/2.5 = 4
a=10/1.5=20/3
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2 pumps together take 4 hrs. rate of 1 is 1.5 times the rate 02 Feb 2007, 16:41
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work rate = job / time ; R = J/T

If we make job constant, we can conclude that work rate(R) is proportional to 1/time (1/T)

Thus, R1/R2 = T2/T1

Rate Pump 1 = 1.5 x Rate Pump 2

Rate(Two Pump) = 1.5R + R = 2.5R and Two pumps spend 4 hours. => R1 = 2.5R , T1 = 4
Rate of the faster pump = 1.5R => R2 = 1.5R, T2 = ?

R1/R2 = T2/T1

2.5R / 1.5R = T2/4

T2 = 4 * 2.5/ 1.5 = 4 x 25 / 15 = 4 * 5 / 3 = 20/3



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