Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?a)30 %b)35 %c)45 %d)50 %e)None of the AboveCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Pipe A’s work in % = 100/10 = 10%
Pipe B’s work in % = 100/20 = 5%
Pipe C’s work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours, tap C is closed + After the 4th hour, tap B is also closed = 100
⇒ (10+5+2.5)*2 + (10+5)*2 + X = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100-65 = 35%

Answer

Pipe A’s work in % = 100/10 = 10%
Pipe B’s work in % = 100/20 = 5%
Pipe C’s work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours, tap C is closed + After the 4th hour, tap B is also closed = 100
⇒ (10+5+2.5)*2 + (10+5)*2 + X = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100-65 = 35%

Answer

Problem:
Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning, all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?

Solution:
Let us assume that the capacity of the tank is ‘L’. We know that the time taken to fill the tank by pipes A, B and C are 10 hours, 20 hours and 40 hours respectively. Hence, the amount of work done by each pipe in one hour can be calculated as follows:

Pipe A can do 1/10th of the work in one hour.
Pipe B can do 1/20th of the work in one hour.
Pipe C can do 1/40th of the work in one hour.

Let us assume that all the pipes are opened simultaneously and after 2 hours, pipe C is closed. Hence, the amount of work done in two hours can be calculated as follows:

Amount of work done by pipe A in 2 hours = 2*(1/10) = 1/5th of the work.
Amount of work done by pipe B in 2 hours = 2*(1/20) = 1/10th of the work.
Amount of work done by pipe C in 2 hours = 2*(1/40) = 1/20th of the work.

Hence, the total amount of work done in 2 hours = 1/5 + 1/10 + 1/20 = 7/20th of the work.

After 2 hours, pipe A and B are kept running and pipe C is closed. After 4 hours, pipe B is also closed. Hence, the amount of work done in 4 hours can be calculated as follows:

Amount of work done by pipe A in 4 hours = 4*(1/10) = 2/5th of the work.
Amount of work done by pipe B in 4 hours = 4*(1/20) = 1/5th of the work.
Amount of work done by pipe C in 2 hours = 0.

Hence, the total amount of work done in 4 hours = 2/5 + 1/5 = 3/5th of the work.

After 4 hours, pipe A is the only pipe running. Hence, the amount of work done by pipe A in the remaining time can be calculated as follows:

Amount of work done by pipe A in the remaining time = 1 – 3/5 = 2/5th of the work.

Therefore, the percentage of the work done by pipe A alone = (2/5) / 1 * 100 = 40%.

Hence, the correct answer is option (b) 35%.

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