(Q) A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.

Ans:  Let’s assume B alone takes X days to finish the work.

(Note: Here, we take X for B because we need to find time taken by B)

Therefore, A alone can finish it in (X – 6) days.

(Note: it was given that A takes 6 days less than B)

Therefore, A’s one day’s work = \frac{1}{\times}

and B’s one day’s work = \frac{1}{\times - 6}

Next, it is given that A + B, both together finish the work in 4 days,

Hence, (A + B)’s one day’s work = \frac{1}{4}

Now, work done by A in a day and work done by B in a day will be equal to work done by both in a day.

therefore, (A’s one day’s work) + (B’s one day’s work) = (A + B)’s one day’s work

Lets put the values:

\Rightarrow \frac{1}{\times - 6}+\frac{1}{\times}=\frac{1}{4}

\Rightarrow \frac{\times + (\times - 6)}{\times(\times-6)}=\frac{1}{4}

  \Rightarrow \frac{2\times - 6}{\times^2 - 6\times}=\frac{1}{4}

\Rightarrow 8\times - 24=\times^2 - 6\times

\Rightarrow \times^2 - 14\times + 24=0

\Rightarrow \times^2 - 12\times - 2\times + 24=0

\Rightarrow \times (\times - 12) - 2 (\times - 12)=0

\Rightarrow (\times - 12)(\times - 2)=0

\Rightarrow \times = 12~or~\times = 2

Here we reject X = 2 because if X = 2, X – 6 will be – 4 but number of days can not be negative,

Therefore, X \neq 2 and only X =12. 

Hence, B alone can finish the work in 12 days.

Check:

If B can finish the work in 12 days, his one days’ work is \frac{1}{12}

A will finish the work in 12 – 6 i.e. 6 days, hence his one days’ work is \frac{1}{6}

Now \frac{1}{12} + \frac{1}{6} = \frac{3}{12} = \frac{1}{4} = one day’s work together

Since, here both finish the work in 4 days, hence our answer is correct. 

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