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There are two pipes T and V which can empty the tank and one pipe S can fill the tank. Pipe S can fill the tank in 5 hours. Pipe T can empty half of the tank in 5 hours. Pipe S is opened for 1 hour and 15 min then Pipe T and Pipe V are also opened. Then the (remaining) tank gets filled in 10 hours. In what time pipe V can empty the 1/4th filled tank?

There are two pipes T and V which can empty the tank and one pipe S can fill the tank. Pipe S can fill the tank in 5 hours. Pipe T can empty half of the tank in 5 hours. Pipe S is opened for 1 hour and 15 min then Pipe T and Pipe V are also opened. Then the (remaining) tank gets filled in 10 hours. In what time pipe V can empty the 1/4th filled tank? Correct Answer 10 hours

Given:

Pipe S's 1 hour's work = 1/5

Pipe T's 1 hour's work = 1/10             (half tank in 5 hours so full tank in 10 hours)

Calculation:

Time for pipe S is opened = 1 + 15/60 = 5/4 hour

Pipe S's 1.25 hours work = 5/4 × 1/5 = ¼

⇒ Remaining work = 1 – ¼ = 3/4

Let pipe V's 1 hour's work be 1/V.

Pipe (S + T + V)’s 1 hour's work = 1/5 – 1/10 – 1/V = 1/10 – 1/V = (V – 10)/10V

Then,

⇒ ¾ × = 10

⇒ 3V = 4V – 40

⇒ V = 40

Pipe V's 1 hour's work = 1/40

∴ Time taken by pipe V to empty 1/4th tank = ¼ × 40 = 10 hours

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