Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The third pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time ( in minutes ) in which the tank will be filled up is -
A. 12 minutes
B. 24 minutes
C. 30 minutes
D. 36 minutes
Answer: Option B
Solution(By Examveda Team)
A . . . . . (+) 30 minutes
B . . . . . (+) 45 minutes
C . . . . . (-) 36 minutes
⇒ Water filled by (A + B) in 12 min
= 12 × (6 + 4)
= 12 × 10 = 120 liters
⇒ Remaining capacity
= 180 - 120 = 60 liters
⇒ After 12 minutes emptied pipe C is also opened
⇒ Total capacity (A + B - C)
= (6 + 4 - 5) = 5 liters/minutes
⇒ Time taken by (A + B - C) with capacity 5 liters/minutes to fill the remaining part
$$ = \frac{{60\,\,{\text{liters}}}}{{5\,\,{\text{liters/minutes}}}}{\text{ = 12 minutes}}$$
⇒ Therefore, total time in which the tank will be filled up is
= 12 + 12
= 24 minutes
This Question Belongs to Arithmetic Ability >> Pipes And Cistern
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Related Questions on Pipes and Cistern
A. $$\frac{5}{{11}}$$
B. $$\frac{6}{{11}}$$
C. $$\frac{7}{{11}}$$
D. $$\frac{8}{{11}}$$
A. $$1\frac{{13}}{{17}}$$ hours
B. $$2\frac{8}{{11}}$$ hours
C. $$3\frac{9}{{17}}$$ hours
D. $$4\frac{1}{2}$$ hours
A. $$4\frac{1}{3}$$ hours
B. 7 hours
C. 8 hours
D. 14 hours
For 12 min, work of A+B = 12/30 + 12/45 = 2/3 part
So the remaining work is 1/3 part
1 min work of A+B+C = 1/30 + 1/45 - 1/36 = (6+4-5)/180 = 1/36 part
Finally total required time = 12 + (1/3)*36 = 24 min