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Q)

Three pipes A, B and C can fill a tank in 6 hours. After three pipes worked for 2 hours C is closed. Then A and B filled the remaining part in 7 hours. calculate the number of hours taken by C alone to fill the tank is?

Explanation

The capacity of the tank can be obtained by finding LCM of 6, 2 and 7 which is 42.

Total parts of the tank is 42

A, B, C together can fill the tank in 6 hours

$\frac{42\mathrm{parts}}{6\mathrm{hours}}=\frac{7\mathrm{parts}}{1\mathrm{hour}}$

A,B and C can fill 7 parts in 1 hour

From the question A, B and C together are filling only for 2 hours

$=\frac{14\mathrm{parts}}{2\mathrm{hour}}$

Together they complete 14 parts in 2 hours

And the remaining 28 part will be filled by A and B in 7 hours

$\frac{28\mathrm{parts}}{7\mathrm{hours}}=\frac{4\mathrm{parts}}{1\mathrm{hour}}$

A and B together can fill 4 parts in 1 hour

From this we could find C’s filling parts per hour

(A+B+C)-( A+B)=C

7-4= 3 parts

Total 42 parts of the tank will be filled by C alone

$\frac{42}{3}=14\mathrm{hours}$

**Hence option D is correct.**

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