Question

Three taps A, B and C can fill a tank in 10, 18 and 6 hours, respectively. If A is open all the time and B and C are open for one hour each alternatively, starting with B, the tank will be full in:

• A. $$6\frac{1}{2}$$
• B. 5
• C. 6
• D. $$5\frac{1}{2}$$

Answer 1

Answer:

B. 5

Solution: Given: Tap A can fill the tank in 10 hours. Tap B can fill the tank in 18 hours. Tap C can fill the tank in 6 hours. B and C run alternatively for one hour each, starting with B, and A runs continuously. Concept Used: Time = Total Work/Efficiency Calculation: Efficiency of A = 1/10 Efficiency of B = 1/18 Efficiency of C = 1/6 According to given condition, Efficiency for 1st hour = 1/10 + 1/18 = 28/180 Efficiency for 2nd hour = 1/10 + 1/6 = 16/60 = 48/180 Now, Work done in 4 hours: ⇒ 2 × (28/180 + 48/180) ⇒ 2 × (76/180) ⇒ 152/180 Remaining work = 1 - 152/180 = 28/180 28/180 will done in next 1 hour by A & B together. So, Total time required to fill the tank: ⇒ 4 + 1 = 5 hours ∴ The tank will be full in 5 hours.