M can do a piece of work in 40 days. He works at it for 5 days and then N alone finishes the remaining work in 42 days. The two together could complete the work in how many days?

Created: 1 month ago | Updated: 3 weeks ago

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M can do in 5 days = $\frac{5}{40}$ = $\frac{1}{8}$ Part of the work

Remaining work = (1 - $\frac{1}{8}$ ) = ( $\frac{8 - 1}{8}$ ) = $\frac{7}{8}$ Part

Now, N does $\frac{7}{8}$ Part of the work in = 42 days

$∴$ 1 or complete part of the work done by N in = ( $\frac{42 \times 8}{7}$ )=48 days

Now, (M+N)'s 1 day's work = ( $\frac{1}{40} + \frac{1}{48}$ )= $(\frac{88}{40 \times 48}) = \frac{11}{240} p a r t$

They together do $\frac{11}{240}$ part of the work in =1 day's

$∴$ They I or complete part of the work in = $\frac{240}{11}$ = 21 $\frac{9}{11}$ days

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