A piece of work was to be completed in 40 day | vedclass

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(B) Let $x$ be the number of men employed at first.$x$ men complete $\frac{1}{2}$ of the work in $24$ days.Therefore,$1$ man would complete the whole

Vedclass · Accepted Answer

$32$

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(B) Let $x$ be the number of men employed at first.$x$ men complete $\frac{1}{2}$ of the work in $24$ days.Therefore,$1$ man would complete the whole work in $24 	imes 2 	imes x = 48x$ days.Now,$(x + 16)$ men do the remaining work $\left(1 - \frac{1}{2} = \frac{1}{2}ight)$ in $(40 - 24 = 16)$ days.Therefore,$1$ man would complete the whole work in $16 	imes 2 	imes (x + 16) = 32(x + 16)$ days.Since the total work is the same,we equate the two expressions:$48x = 32(x + 16)$$48x = 32x + 512$$16x = 512$$x = 32$.Thus,$32$ men were employed at first.

$A$ piece of work was to be completed in $40$ days. $A$ number of men employed upon it did only half the work in $24$ days. $16$ more men were then set on,and the work was completed in the specified time. How many men were employed at first?

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