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Work question

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by Cybermusings » Sat Apr 28, 2007 7:07 am
Working alone, a small pump takes twice as long
as a large pump takes to fill an empty tank.
Working together at their respective constant rates,
the pumps can fill the tank in 6 hours. How many
hours would it take the small pump to fill the tank
working alone?

a) 8
b) 9
c) 12
d) 15
e) 18

OA coming in after a while
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Source: — Problem Solving |

by jayhawk2001 » Sat Apr 28, 2007 10:17 am
If t = time taken for large pump to fill the tank

1/t + 1/2t = 1/6
Hence t = 9

Small pump takes 2*t = 18 hours
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by ns88 » Sat May 05, 2007 4:48 pm
S=small

L=large


S= 2L

1/S+ 1/L=1/T

1/2L + 1/L= 1/6

times each by two to get rid off the two in denominator

1/L+2/L=2/6

L=9

S=18
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