There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is a

Tap A can fill the tank in $25$ hrs.

Tap B cab fill the tank in $20$ hrs.

Tap C can fill the tank in $10$ hrs.

Total volume of the tank filled by $3$ pipes = LCM of $(25, 20, 10) = 100$ units

⇒ Pipe A can fill $4$ units of water in $1$ hr.

⇒ Pipe B can fill $5$ units of water in $1$ hr.

⇒ Pipe C can fill $10$ units of water in $1$ hr.

⇒ Pipe $(A + B + C)$ can fill $19$ units of water in $1$ hr.

According to question,

Pipe $(A + B + C)$ are opened for $1$ hr, then tap $C$ is closed.

⇒ $19$ units of water are filled in the tank.

After 4^th hr, tap B is also closed

⇒ for $3$ hrs, pipe A and B are open

⇒ $(4 + 5) × 3 = 27$ units of water are filled.

⇒ $100 - (19 + 27) = 54$ units of water are filled by tap A alone.

Total units of water filled by tap A is $4 × 1 + 4 × 3 + 54 = 4 + 12 + 54 = 70$ units

Thanks

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