Time & work concept question

Question

If 3 men or 5 women can finish a work in 43 days. Then in how many days 5 men and 6 women together do it ?

Answer 1

here the answer..

$3M=5W$

$1M=5/3W$

Then Formula of this $(Man1 * days * hours)/work=(man2 * days * hours )/work$

$5(5/3) + 6 +D = 5* 43$

$D=15$

Answer 2

Hint:

Let $a$ be the amount done by a man in one day and $b$ be the amount done by a woman in one day. Then if there are $d$ days and $m$ men and $w$ women, the total amount done $T$ is presumably $$T=d \left(am+bw\right)$$

You can calculate

• the value for $a$ when $T=1, d=43, m=3, w=0$
• the value for $b$ when $T=1, d=43, m=0, w=5$

and (assuming they do not change) use these to find

• the value for $d$ when $T=1, m=5, w=6$

Answer 3

A man does $\frac{1}{3\times 43} = \frac1{129}$th of the job in one day, whereas a woman does $\frac{1}{5\times 43} = \frac{1}{215}$th of the job.

Working together, $5$ men and $6$ women do $$\frac{5}{129}+\frac{6}{215}=\frac{1}{15}\text{th}$$ of the job. Thus they require 15 days.

Answer 4

If $3$ men can finish a work in $43$ days, then $1$ man can finish a work in $43\times3=129$ days. so a man can finish $\frac{1}{129}$ of a work in a day.

If $5$ women can finish a work in $43$ days, then $1$ woman can finish a work in $43\times5=215$ days, so a woman can finish $\frac{1}{215}$ of a work in a day.

So, 5 men and 6 women can finish $\frac{1}{129}\times5+\frac{1}{215}\times6=\frac{1}{15}$ work in a day, so it requires $15$ days to finish the work with 5 men and 6 women.