Q75. 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?

First, calculate the time each full set of pipes takes to fill/empty the entire tank: - Set X: 20 pipes fill 70% in 14 min. So, 20 pipes fill 100% in (14 / 0.7) = 20 min. (Rate = 1/20 tank/min) - Set Y: 10 pipes fill 3/8 in 6 min.

So, 10 pipes fill 100% in (6 / (3/8)) = 16 min. (Rate = 1/16 tank/min) - Set Z: 16 pipes empty 50% in 20 min. So, 16 pipes empty 100% in (20 * 2) = 40 min. (Rate = -1/40 tank/min) Now, consider the modified scenario: - Half of pipes X (10 pipes): Their rate is (1/2) * (1/20) = 1/40 tank/min.

- Half of pipes Y (5 pipes): Their rate is (1/2) * (1/16) = 1/32 tank/min. - All pipes Z (16 pipes): Their rate is -1/40 tank/min. The combined rate of filling/emptying is (1/40) + (1/32) - (1/40) = 1/32 tank/min.

This means the tank will be filled at a rate of 1/32 of the tank per minute. To fill 50% (or 1/2) of the tank, the time required will be (1/2) / (1/32) = 16 minutes. This problem tests the application of work and time concepts, specifically pipes and cisterns, requiring careful calculation of individual and combined rates under varying conditions, a common type in CSAT.