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adkikani
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 pump can fill the tank in 6 hours.
How many hours would it take for the small pump to fill the tank working alone?

A. 8
B. 9
C. 12
D. 15
E. 18

1/x + 1/2x = 1/6
3/2x = 1/6
x = 18/2 = 9

2x = 18

IMO E
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Basically Small machine is doing x part of the work and Large marchine is doing 2x part of the work, if both work the same time...
Now if large machine is removed, the smaller machine would have to do all the 3x part of the work along......
Now...that means ......first it worked for x part for 6 hours...now it has to complete additional 2x parts ... it will take 12 hours...total time alone it will take is 18 hours E

adkikani
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 for the small pump to fill the tank working alone?

A. 8
B. 9
C. 12
D. 15
E. 18­
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I assumed work done to be 60
and made eqn
60/x+60/2x = 6

90=6x
x=15

where am i going wrong
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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 for the small pump to fill the tank working alone?

A. 8
B. 9
C. 12
D. 15
E. 18­

Let the large pump take x hours, so the small pump takes 2x hours.

Their combined rate is:

1/x + 1/(2x) = 1/6

3/(2x) = 1/6

18 = 2x

x = 9

So the small pump takes:

2x = 18 hours

Answer: E.

laborumpossimus
I assumed work done to be 60
and made eqn
60/x+60/2x = 6

90=6x
x=15

where am i going wrong

60/x + 60/(2x) is the combined work rate, in units per hour. So it should equal 60/6 = 10, not 6.

Thus:

60/x + 60/(2x) = 10

which gives x = 9 hours for the large pump, so the small pump takes 18 hours.

Hope this helps.
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i understand now
thank you so much

Bunuel
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 for the small pump to fill the tank working alone?

A. 8
B. 9
C. 12
D. 15
E. 18­

Let the large pump take x hours, so the small pump takes 2x hours.

Their combined rate is:

1/x + 1/(2x) = 1/6

3/(2x) = 1/6

18 = 2x

x = 9

So the small pump takes:

2x = 18 hours

Answer: E.



60/x + 60/(2x) is the combined work rate, in units per hour. So it should equal 60/6 = 10, not 6.

Thus:

60/x + 60/(2x) = 10

which gives x = 9 hours for the large pump, so the small pump takes 18 hours.

Hope this helps.
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