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 t…

Question

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?

1. A. 8 minutes
2. B. 10 minutes
3. C. 12 minutes
4. D. 16 minutes (Correct answer)

Correct Answer

Option D — 16 minutes

Detailed Solution & Explanation

The correct answer is 16 minutes.

Key Points

• Set X (20 pipes) fills 70% in 14 min → 20 pipes fill 1 tank in 14×10/7 = 20 min → 1 pipe fills in 400 min.
• Set Y (10 pipes) fills 3/8 in 6 min → 10 pipes fill in 16 min → 1 pipe fills in 160 min.
• Set Z (16 pipes) empties half in 20 min → 16 pipes empty in 40 min → 1 pipe empties in 640 min.
• Active: 10 X-pipes (rate: 10/400=1/40), 5 Y-pipes (5/160=1/32), 16 Z-pipes emptying (16/640=1/40).
• Net fill rate = 1/40 + 1/32 − 1/40 = 1/32 per min.
• Time to fill 50% = 16 minutes. Answer: 16 minutes.

Additional Information

• Convert every set to a per-pipe rate before combining, since the question changes how many pipes of each set are open.
• Set X: 20 pipes fill 70% in 14 min, so 20 pipes fill the whole tank in 20 min and one pipe takes 400 min. Set Y: 10 pipes fill 3/8 in 6 min, so 10 pipes fill it in 16 min and one pipe takes 160 min. Set Z: 16 pipes empty half in 20 min, so 16 pipes empty it in 40 min and one pipe takes 640 min.
• With 10 X-pipes, 5 Y-pipes and all 16 Z-pipes: 10/400 + 5/160 − 16/640 = 1/40 + 1/32 − 1/40 = 1/32 of the tank per minute. Note that the X inflow and Z outflow cancel exactly.
• Filling 50% therefore takes (1/2) ÷ (1/32) = 16 minutes. Emptying pipes always enter the sum with a negative sign, and if the net rate came out negative the tank could never fill.