Question

3 men or 7 women can do a piece of work in 32 days. the number of days required by 7 men and 5 women to do a piece of work twice as large is

Answer 1

$\huge\red{Answer}$

Let work done by 1 man in 1 day =m

work done by 1 woman in 1 day =w

Work done by 3 men in 1 day =1/32

3m=1/32

m=1/96

Work done by 7 men in 1 day =1/32

7w=132

w=1/224

Work done by 7 men and 5 women in 1 day =7m+5w

=7/96+5/224

=49+15/672

=64/672

=2/21

Therefore, 7 men and 5 women can do the same work in 2/21 days

To do work twice as large, time required 

=21/2×2

=21 days

HOPE IT HELPS YOU !!

Answer 2

The work would be completed in 21 days by 7 men and 5 women

Given that 3 men or 7 women can do the work in same 32 days.

Hence, efficiency of 3 men is equal to efficiency of 7 women.

So, one man can do the work done by 7/3 women.

Hence, 7 men and 5 women would be equal to 7*7/3 women and 5 women

i.e. 64/3 women.

Now, 7 women can do the work in 32 days.

New work is twice as large, hence time required would be

64 days for 7 women

=> 64*7 days for one woman

=> 64*7 /(64/3)  days for 64/3 women

=> 21 days for 64/3 women

=> 21 days for 7 men and 5 women