Pump A starts drawing water and after some time pump B joins in and draws water at the same rate as A. Both the pumps stop drawing water at the same time. B accounted for how much percent of all water drawn?
(1) 75% time of A’s drawing time is 50% more than B’s drawing time
(2) A drew the water for 80 minutes.
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DS work and time question
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tohellandback
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IMO A.
1) lets say time for A is t
and time for B is t1
3/4 *t= t1+t1/2
t=2t1
lets say the rate for both pumps be 'r'
for A- water drawn=rt
for B- rt1
we need to fing rt1/(rt+rt1) *100---- we can find this using equation t=2t1
SUFF
2) we need to know the time for B. Not suff
1) lets say time for A is t
and time for B is t1
3/4 *t= t1+t1/2
t=2t1
lets say the rate for both pumps be 'r'
for A- water drawn=rt
for B- rt1
we need to fing rt1/(rt+rt1) *100---- we can find this using equation t=2t1
SUFF
2) we need to know the time for B. Not suff
Last edited by tohellandback on Thu Aug 13, 2009 5:53 pm, edited 1 time in total.
The powers of two are bloody impolite!!
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mehravikas
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@tohellandback,
Agree with your answer but I think your equation is not correct.
3/4 *t= t+t/2
it should t1 on R.H.S
Agree with your answer but I think your equation is not correct.
3/4 *t= t+t/2
it should t1 on R.H.S
tohellandback wrote:IMO A.
1) lets say time for A is t
and time for B is t1
3/4 *t= t+t/2
t=2t1
lets say the rate for both pumps be 'r'
for A- water drawn=rt
for B- rt1
we need to fing rt1/(rt+rt1) *100---- we can find this using equation t=2t1
SUFF
2) we need to know the time for B. Not suff
