Work done = 1/5
remaining work = 4/5
? 4 (20 x 75) = 40 x N
N = 150
Therefore 75 men should be increased.
Work done by 2 men = 3 woman = 4 boys
? 1 man = 2 boys
1 woman = 4/3 boys
? boys X days = 4 X 52 (boys - days)
Again 1 man + 1 woman + 1 boy = 2 + 4/3 + 1 = 13/3 boys
B1 x D1 = B2 x D2 B = boys, D = days
4 X 52 = 13/3 x D2
D2 = 48 days
Ratio of efficiencies of A,B and C = 6 : 5 : 4
? Share of C = [4/(6+5+4)] x 27000 = 7200
In 1 hour 314 weavers weave = 6594 x 6 shawls
In 1 hour 1 weaver weaves = ( 6594 x 6 ) / 314 shawls = 126 shawls
2 men = 7 boys
? 1 man = 7/2 days
5 women = 7 boys
? 1 women = 7/5 boys
7 men + 5 women + 2 boys = 7 x 7/2 + 5 x 7/5 + 2 = 67/2 boys
Now, B1 x D1 = B2 x D2
7 X 469 = 67/2 x D2
D2 = 98 days
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
? 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
? 10 children can do the work in 6 days.
Work done = 1/3
Remaining work = 2/3
So for double work in same days we need double number of people i.e. 40.
So, 20 men will be increased.
A's 1 day's work = (1/15 - 1/20) = 1/60
? A alone can finish it in 60 days.
B's one day's work = (1/6 - 1/9) = 1/18
? B alone can finish it in 18 days.
(A + B)'s 1 day's work = 1/18
(B + C)'s 1 day's work = 1/24
(A + C)'s 1 day's work = 1/36
Adding 2 (A + B + C)'s 1 day's work = (1/18 + 1/24 + 1/36) = 1/8
? (A + B + C)'s day's work = 1/16
Hence, all working together can finish it in 16 days.
? [(A + B) + (B + C) + (C + A)]'s 1 day's work = (1/12 + 1/15 + 1/20) = 1/5
? 2(A + B + C)'s 1 days work = 1/5
? (A + B + C)'s 1 day's work = 1/10
&rArr A's 1 day's work = (1/10 - 1/15) = 1/30
? A alone can finish it in 30 days.
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