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(a) 18 days

(b) 15 days

(c) 12 days

(d) $ 7\dfrac{1}{2} $ days

Answer
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It is given that A and B can do a piece of work in 6 days. Let us suppose the amount of work as X.

If A and B both can finish X amount work in 6 days then, we get that the work done in one day is $ \dfrac{1}{6}th $ of the total amount of work, i.e.

Work done in one day by both A and B = $ \dfrac{X}{6} $

It is also given that A finishes the work in 9 days, hence we get that if A finishes work in 9 days then work done by A in 1 day is $ \dfrac{1}{9}th $ of the total work, so we get

Work done by A alone in 1 day = $ \dfrac{X}{9} $

Now, we know that work done by B alone in one day would be equal to

= (work done by both in 1 day) – (work done by A alone in 1 day)

= $ \dfrac{X}{6} $ - $ \dfrac{X}{9} $

By taking LCM we get,

= $ \dfrac{9X-6X}{9\times 6} $

= $ \dfrac{3X}{54} $

= $ \dfrac{X}{18} $

So, work done by B alone in 1 day is = $ \dfrac{1}{18}th $ of the total amount of work.

Multiplying by 18 on both sides of above expression we get,

Work done by B alone in 18 days = total amount of work

Hence B alone can do the work in 18 days

This question can also be solved by considering amount of work directly for 6 days instead of 1,

For example,

X amount of work is done by both A and B in 6 days,

And $ \dfrac{X}{9} $ amount of work is done by A alone in one day

So, $ \dfrac{6X}{9} $ is the amount of work done by A alone in 6 days

Amount of work done by B in 6 days = work done by both in 6 days – work done by A alone in 6 days

= $ X-\dfrac{6X}{9} $

$ \begin{align}

& =\dfrac{9X-6X}{9} \\

& =\dfrac{3X}{9} \\

& =\dfrac{X}{3} \\

\end{align} $

Hence amount of work done by B alone in 6 days is $ \dfrac{1}{3}rd $ of total work,

So, total work done will be done by B alone in (6 $ \times $ 3) days i.e. 18 days.

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