There are 3 taps A, B, and C in a tank. These can fill the tank in 10 hours, 20 hours and 25 hours, respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours from the beginning, tap B is also closed. The remaining tank is filled by tap A alone. Find the percentage of work done by tap A itself.

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 4)

Answer 1

Calculations:

The capacity of the tank

= LCM(10, 20 and 25) = 100

Efficiency of A = 100/10 = 10

Efficiency of B = 100/20 = 5

Efficiency of C = 100/25 = 4

Tank filled in 2 hours by A, B, and C

= (10 + 5 + 4) × 2 = 38 unit

According to the question,

After 4 hours from the beginning, tap B is also closed

Tank filled in 2 hours by A & B

= (10 + 5) × 2 = 30 unit

Now, tap B is also closed

Remiang capacity of tank = 100 - 38 - 30 = 32 unit

These 32 unit is filled by A alone.

So, the total work done by A

= Work done in (1st 2 hr + next 2 hr + 32 unit)

= 10 × 2 + 10 × 2 + 32 = 72 unit

Percentage of a tank filled by A = (72/100) × 100 = 72%.

Hence, The Required value is 72%.

Mistake PointsNote that, B closed after 4 hr form the beginning, not after closing the C.