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 4th 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
∴ Percentage of work done by tap $A = 70%$Thanks