# CAT Practice : Pipes cisterns, Work time

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Classic Work - Time Question!!

## Amar, Akbar and Anthony

Q.24: Amar, Akbar and Anthony set out to complete a work. Anthony being the eldest would take 1 day less than Amar if he were to complete the work alone. All three together could complete the work in a day. However, Anthony was kidnapped by Shakal. Amar and Akbar began the work in his absence. After a day, Amar was also kidnapped. Akbar took 3 more days to finish the work. How much portion of the work Anthony could do in a day?
1. $\frac{2}{3}$
2. $\frac{1}{6}$
3. $\frac{1}{2}$
4. $\frac{1}{3}$

Choice C. $\frac{1}{2}$

## Detailed Solution

Let Amar take x days and Akbar take y days to finish the work. So, Anthony would take x-1 days to complete the work alone.

Therefore, as per question,

$\frac{1}{(\frac{1}{x} + \frac{1}{(x-1)} + \frac{1}{y})} = 1$

=) $\frac{1}{x} + \frac{1}{(x-1)} + \frac{1}{y} = 1$ …………………(1)

Also given, $(\frac{1}{x} + \frac{1}{y}) + \frac{3}{y} = 1$

=) $\frac{1}{x} + \frac{4}{y} = 1$

=) $(\frac{1}{y} = \frac{1}{4}) + \frac{1}{4x} = 1$…………………(2)

Putting in equation (1),

=) $\frac{1}{x} + \frac{1}{(x-1)} + \frac{1}{4} - \frac{1}{4x} = 1$

=) $\frac{1}{x} + \frac{1}{(x-1)} - \frac{1}{4x} = \frac{3}{4}$

Now using option analysis,

a) Work done by Anthony in 1 day = $\frac{1}{(x-1)} = \frac{2}{3} = ) 3 = 2x - 2 = ) x = \frac{5}{2} = 2.5$

L.H.S of (3) = ) $\frac{1}{2.5} + \frac{1}{1.5} - \frac{1}{10} = 0.4 + 0.66 – 0.1 ≠ 0.75$

b) $\frac{1}{(x-1)} = \frac{1}{6} =) x = 7 L.H.S of (3) = \frac{1}{7} + \frac{1}{6} - \frac{1}{28} ≠ \frac{3}{4}$

c) $\frac{1}{(x-1)} = \frac{1}{2} =) x = 3 L.H.S of (3) = \frac{1}{3} + \frac{1}{2} - \frac{1}{12} = \frac{(4 + 6 – 1)}{12} = \frac{9}{12} = \frac{3}[4]$

Correct Answer: $\frac{1}{2}$

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## More questions from Pipes Cisterns, Work Time

Mathematicians are creatures of habit. This topic has been called Pipes and Cisterns because someone named it like it long ago. Pipes and Tanks would be much better - at least makes it sound like wartime.