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Asked by aruni_maiyer 18th May 2010, 7:42 PM

Given 2 8/11 = 30/11

Let q1, q2 be the flow rates from two pipes, and cistern volume be V.

t1 = V/q1 and t2 = V/q2

t1 + 1 = t2              ..... given than one pipe takes 1 min more to fill the cistern.

t = V/(q1 + q2)             ... time to fill the cistern together.

30/11 = V/(q1 + q2)

30/11 = 1/(q1/V + q2/V)

30/11 = 1/(1/t1 + 1/(t1 + 1))

30/11 = t1(t1 + 1)/(2t1 + 1)

60t1/11 + 30/11 = t1(t1 + 1)

60t1/11 + 30/11 = t12 + t1

t12 - 49t1/11 - 30/11 = 0

Solving this quadratic equation we find,

t1 = (49/11 ± 61/11)/2

t1 = 110/22 = 5 min and hence t2 = 6 min.

Hence the times in which each pipe will fill the cistern are 5 and 6 min respectively.

Regards,

Team,

TopperLearning.

Answered by Expert 19th May 2010, 11:38 AM
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