Pipe A, B, and C can fill an empty tank alone in 6 hours, 8 hours and 12 hours respectively. A tank is initially 1/6 filled. Pipes A and B are simultaneously opened in the tank and closed after 2 hours. Now, if pipe C is opened in the tank, in how much time will it be filled?a)2 hrs.b)3 hrs.c)4 hrs.d)6 hrs.Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Solution:

Given:
Pipe A fills the tank in 6 hours.
Pipe B fills the tank in 8 hours.
Pipe C fills the tank in 12 hours.
Initially, the tank is 1/6 filled.

Let's calculate the rate at which each pipe fills the tank:
Pipe A fills 1 tank in 6 hours, so its rate of filling the tank is 1/6 tank per hour.
Pipe B fills 1 tank in 8 hours, so its rate of filling the tank is 1/8 tank per hour.
Pipe C fills 1 tank in 12 hours, so its rate of filling the tank is 1/12 tank per hour.

When pipes A and B are simultaneously opened and closed after 2 hours, we can calculate the amount of water filled in these 2 hours:
Rate of A + Rate of B = 1/6 + 1/8 = 7/24 tank per hour.
In 2 hours, the amount of water filled = (7/24) * 2 = 7/12 tank.

So, after the first 2 hours, the tank is now filled with 7/12 tank of water.

Now, when pipe C is opened, we need to calculate how much time it will take to fill the remaining 5/12 tank of water.

Let's say it takes x hours to fill the remaining 5/12 tank of water.
The rate at which pipe C fills the tank is 1/12 tank per hour.

So, in x hours, the amount of water filled by pipe C = (1/12) * x = x/12 tank.

We know that the remaining tank to be filled is 5/12 tank, so we have the equation:
x/12 = 5/12.

Solving this equation, we get:
x = 5.

Therefore, it will take 5 hours for pipe C to fill the remaining 5/12 tank of water.

Hence, the correct answer is option 'B' - 3 hours.

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