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Two pipes A and B can fill a cistern in 3721тАЛ minutes and 45 minutes, respectively. Both pipes are open. The cistern will be filled in just half an hour, if the B is turned off after:
9 min
5 min
10 min
15 min
9 min
Calculate work rates: A = 2/75 per min, B = 1/45 per min. If A works for 30 mins, fill = 30 * (2/75) = 4/5. Remaining work = 1/5, which B does in (1/5) / (1/45) = 9 minutes.
Pipe A fills in 37.5 minutes (75/2 minutes), Pipe B fills in 45 minutes, total time to fill is 30 minutes (half hour).
T=EAтАЛ+EBтАЛWтАЛ
Calculate work rates: A = 2/75 per min, B = 1/45 per min. If A works for 30 mins, fill = 30 * (2/75) = 4/5. Remaining work = 1/5, which B does in (1/5) / (1/45) = 9 minutes.
Confusing 'half an hour' as the time pipe B works, rather than the total time the cistern takes to fill.
Define Work Rates
LCM of 37.5 (which is 75/2) and 45 is 225. Rate of A = 225 / 37.5 = 6 units/min. Rate of B = 225 / 45 = 5 units/min.
EAтАЛ=6,EBтАЛ=5,Total┬аCapacity=225
Set up the equation
Pipe A works for the full 30 minutes. Pipe B works for x minutes. The total work done equals the capacity.
(6├Ч30)+(5├Чx)=225
Solve for x
Calculate the work done by A: 180 units. Remaining work for B: 225тИТ180=45. Thus, B works for 45/5=9 minutes.
x=545тАЛ=9
A is correct because Pipe B must be turned off after 9 minutes to fill the cistern in exactly 30 minutes.
This logic is identical to mixture problems where one component is added or removed over a fixed duration.
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