A can complete a piece of work in $14$ days. He worked for $2$ days and remaining work was completed by B in $12$ days. If they start the work together, then in what time they can complete the work?

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Solution

A (1 day's) $= \frac{1}{14}$ work.
He worked for 2 days, i.e., he completed
$(2 \times \frac{1}{14})$ work $= \frac{1}{7}$ work
$∴$ Remaining work $= 1 - \frac{1}{7} = \frac{6}{7}$
$B (12 d a y^{'} s) = \frac{6}{7} w o r k$
$∴ B (1 d a y^{'} s) = \frac{6}{7 \times 12} = \frac{1}{14} w o r k$
Hence, A + B (1 day's) $= \frac{1}{14} + \frac{1}{14} = \frac{1}{7} w o r k$
$∴$ (A + B) can complete the whole work in 7 days.

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