Two pipes A and B can fill a cistern in $$37\frac{1}{2}$$ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
A. 5 min.
B. 9 min.
C. 10 min.
D. 15 min.
Answer: Option B
Solution(By Examveda Team)
Let B be turned off after x minutes.Then, Part filled by (A + B) in x min. + Part filled by A in (30 - x) min. = 1
$$\eqalign{ & \therefore x\left( {\frac{2}{{75}} + \frac{1}{{45}}} \right) + \left( {30 - x} \right).\frac{2}{{75}} = 1 \cr & \Rightarrow \frac{{11x}}{{225}} + \frac{{ {60 - 2x} }}{{75}} = 1 \cr & \Rightarrow 11x + 180 - 6x = 225 \cr & \Rightarrow x = 9 \cr} $$
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Sir ye Question konsi class k books SE hai plz tora sa guide krdy
Much simple way is the LCM method.
LCM of (75/2 and 45) = 225, assume this is total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins