(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.
B's one day's work = (1/6 - 1/9) = 1/18
? B alone can finish it in 18 days.
A's 1 day's work = (1/15 - 1/20) = 1/60
? A alone can finish it in 60 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.
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
? [(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.
(A + B)'s 1 day's work = (1/30 + 1/40) = 7/120
? Time taken by both to finish the work = 120/7days = 171/7 days
? 1/3 of work is done by A in 5 days.
? Whole work will be done by A in 15 days.
? 2/3 of work is done by B in 10 days.
Whole work will be done by B in 10 x (5/2) i.e., 25days
? (A + B)'s 1 day's work = (1/15 + 1/25) = 8/75
So, both together can finish it in 75/8 days, i.e., 93/8days.
B's 9 day's work = 9 x (1/12) = 3/4
Remaining work = (1 - 3/4) = 1/4
1/4 work is done by A in = 20 x (1/4) = 5 days.
Required time taken by B to complete the work = xy / x - y
[Here x = 12 and y = 8]
? required time = (12 x 8) / (12 - 8) = 96/4 = 24 days
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