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Q)

A tank can be filled by a pipe A in 2 h and pipe B in 6 h. At 10 am pipe A was opened. At what time will the tank be filled if pipe B is opened at 11 am?

**[SSC CGL 2012]**

Explanation

According to the question,

Pipe A fills the tank in 2 hours (in 1 hour it fills $\frac{1}{2}$ part of the tank)

Pipe B fills the tank in 6 hours (in 1 hour it fills $\frac{1}{6}$ part of the tank)

As pipe A was opened at 10:00 am and it worked alone till 11:00 am. In 1 hour it will fill $\frac{1}{2}$ part of the tank.

the remaining $\frac{1}{2}$ part of the tank will be filled by both Pipe A and B.

According to the formula, if pipe A can comletely fill a tank in x hours and pipe B can comletely fill a tank in y hours, then the time taken by both pipes together to fill the tank is $\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$ hours.

So, $\frac{2\times 6}{2+6}{=}\frac{12}{8}{=}\frac{3}{2}{}{\mathrm{part}}$ (both the pipes together will fill $\frac{3}{2}$ part of the tank)

So, Time taken by both pipes to fill $\frac{3}{2}$ part of the tank = 60 minutes

Time taken to fill $\frac{1}{2}$ part = $\frac{60\times 3}{2}\times \frac{1}{2}$ = 45 minutes

Therefore, the tank will be filled at 11:45 AM

**Hence option C is correct.**

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