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 ✓ UPSC's answer
Why the answer is (d)
• Set X: 20 pipes fill 70% in 14 min, so 1 pipe fills 70/(20*14) = 0.25% per minute.
• Set Y: 10 pipes fill 3/8 (37.5%) in 6 min, so 1 pipe fills 37.5/(10*6) = 0.625% per minute.
• Set Z: 16 pipes empty 50% in 20 min, so 1 pipe empties 50/(16*20) = 0.15625% per minute.
• Active pipes: 10 from X (filling), 5 from Y (filling), 16 from Z (emptying).
• Net rate = (10 * 0.25) + (5 * 0.625) - (16 * 0.15625) = 2.5 + 3.125 - 2.5 = 3.125% per minute.
• Time to fill 50% = 50 / 3.125 = 16 minutes.
Why the other options are wrong
- (a) 8 minutes
- 8 minutes is incorrect because it likely results from ignoring the emptying effect of set Z or miscalculating the net rate.
- (b) 10 minutes
- 10 minutes is incorrect because it does not account for the specific contribution of the reduced number of pipes from sets X and Y against the full set Z.
- (c) 12 minutes
- 12 minutes is incorrect because it fails to correctly compute the net filling rate after subtracting the emptying rate of set Z.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.