12 men ? 18 women
? 1 man ? 18/12 women ? 3/2 women
? 8 men ? (3/2) x 8 = 12 women
? 8 men + 16 women = 12 women + 16 women = 28 women
? 18 women can do the work in 14 days .
? 1 woman can do the same work in (14 x 18) days.
? 28 women will do the same work in (14 x 18)/28 days.
? Required number of days = (14 x 18)/28 = 9 days
Let 15 men work for m days.
Work done in 1 days = m/20
Remaining work = (1 - m/20)
25 men's 1 days work = 1/20
1 man's 1 days work = (1/20) x (1/25) = 1/500
10 men's 1 days work = 10/500 = 1/50
10 men's 75/2 days work = (1/50) x (75/2) = 75/100 = 3/4
? (1 - m/20) = 3/4
? m/20 = 1/4
? m = (1/4) x 20 = 5
Clearly, 15 men leave after 5 days.
Let man's 1 day's work = 1/m
and boy's 1 day's work = 1/n
1 day's work man and boy = 1/24
Man's 6 day's work = 6/m
Now, for 20 days, both man and boy do the work and for last 6 days, only man does the work.
According to the question,
1/m + 1/n = 1/24
? 20(1/m + 1/n) + 6/m = 1
? 20 x (1/24) + 6/m = 1
? 6/m = (1 - 20/24) = 4/24 = 1/6
? 1/n = 1/36
Now from eq. (i)
1/m + 1/n = 1/24
1/36 + 1/n = 1/24
? 1/n = (1/24) - (1/36) = 1/72
Hence, the boy alone can do the work in 72 days.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.