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in Commercial Mathematics by (49.2k points)
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Three pipes A, B and C can fill a cistern in 10h, 12h and 15h respectively. First A was opened. After 1 hour, B was opened and after 2 hours from the start of A, C was also opened. Find the time in which the cistern was just fill.

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Best answer

Let it take x hours to fill the tank by all the three pipes. Then,

Work done by {A in x hours + B in (x – 1) hours + C in (x – 2) hours} = 1

\(\Rightarrow\) \(\frac{X}{10}\) + \(\frac{(X-1)}{12}\)\(\frac{(X-2)}{15}\) = 1

\(\Rightarrow\) 6x + 5(x – 1) + 4(x – 2) = 60

\(\Rightarrow\) 15x – 13 = 60

\(\Rightarrow\) 15x = 73

\(\Rightarrow\) x = \(\frac{73}{15}\) = 4\(\frac{13}{15}\) = 4 hours + \(\frac{13}{15}\) x 60 min = 4 hours 52 min.

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