Q. A set (X) of 20 pipes can fill 70% of a tank in 14 minutes.
Another set (Y) of 10 pipes fills 3/8th of the tank in 6 minutes. A third set (Z) of 16 pipes can empty half of the tank in 20 minutes. If half of the pipes of set X are closed and only half of the pipes of set Y are open, and all pipes of the set (Z) are open, then how long will it take to fill 50% of the tank?
(a) 8 minutes
(b) 10 minutes
(c) 12 minutes
(d) 16 minutes
Correct Answer: (d) 16 minutes
UPSC Prelims 2025 CSAT
Explanation :
1. Compute each set’s filling or emptying rate (in “tank‐fractions per minute”):
• Set X: 20 pipes fill 0.70 of the tank in 14 min
– Combined rate of X = 0.70 ÷ 14 = 0.05 tank/min
– Rate per X-pipe = 0.05 ÷ 20 = 0.0025 tank/min
• Set Y: 10 pipes fill 3/8 = 0.375 of the tank in 6 min
– Combined rate of Y = 0.375 ÷ 6 = 0.0625 tank/min
– Rate per Y-pipe = 0.0625 ÷ 10 = 0.00625 tank/min
• Set Z: 16 pipes empty 0.5 of the tank in 20 min
– Combined emptying rate of Z = 0.5 ÷ 20 = 0.025 tank/min
– Rate per Z-pipe = 0.025 ÷ 16 = 0.0015625 tank/min
2. In the new scenario:
– Half of X’s pipes are in use ⇒ 10 X-pipes ⇒ filling at 10×0.0025 = 0.025 tank/min
– Half of Y’s pipes are in use ⇒ 5 Y-pipes ⇒ filling at 5×0.00625 = 0.03125 tank/min
– All Z-pipes are open ⇒ 16 Z-pipes emptying at 16×0.0015625 = 0.025 tank/min
3. Net filling rate = (X filling) + (Y filling) – (Z emptying)
= 0.025 + 0.03125 – 0.025 = 0.03125 tank/min
4. Time to fill 50% = required fraction ÷ net rate
= 0.50 ÷ 0.03125 = 16 minutes
Therefore the correct answer is 16 minutes.
