Mason's 1 h work = 1/12
Mason's 6 h work = 6/12 = 1/2
Remaining work = 1 - (1/2) = 1/2
Remaining work can be finish in 5 h total work can be finished in 2 x 5 = 10 h.
(1/12) + (1/B) = (1/10)
? 1/B = (1/10) - (1/12) = 1/60
Boy can complete the work in 60 h.
Let the number of the men at the beginning = N
Given, M1 = N, M2 = N - 12, D1 = 18 and D2 = 30
According to the formula
M1D1 = M2D2
? N x 18 = (N - 12) x 30
? 3N = 5N - 60
? 2N = 60
? N = 30
B's 4 days work = (1/6) x 4 = 2/3
? Remaining work = 1 - (2/3) = 1/3
A's 1 day's work = 1/10
? 1/3 work is finished by A in
(10 x 1/3) = 31/3 days
A's and B,s 1 h work= 1/10
B's and C,s 1 h work = 1/15
and A's and C's 1 h work = 1/12
? A's B's and C's 1 h work
= (1/2) x [(1/10) + (1/15) + (1/12)] = (1/2) x (1/4) = 1/8
Hence B's work in 1 h = (1/8) - (1/12) = 1/24
? B independently can complete the work in 24 h.
Let the required days be x.
A works for (x - 2) days, while B work for x days.
According to the question
(x - 2)/10 + x/20 = 1
? 2x - 4 + x = 20
? 3x = 24
? x = 8 days
A's 1 day's work = 1/12
B's 1 day's work = 1/24
(A + B) 1 day's work = 1/12 + 1/24 days
= (2 + 1)/24 = 3/24 = 1/8
? (A + B)'s will finish the work in 8 days.
Time take by A to complete a work = N days
? Time taken by B to complete the same work = 2N days
According to the question,
(1/N) + (1/2N) = 1/12
? 3 / 2N = 1/12
? N = (3 x 12)/2 = 18
Time taken by B to complete the same work = 2N= 2 x 18 = 36 days
Let A takes x and B takes 3N days to finish the job.
According to the question,
3N - N = 30
? N = 30/2 = 15 days
? B's time to complete the work = 3N
= 3 x 15 = 45 days
? (A + B)'s 1 day's work
= (1/15) + (1/45) = (3 + 1)/45 = 4/45
? (A + B) will finish the work in 4/45 days
? Required time = 4/45 = 111/4
Work done by (x + y + z) in 1 day
= 50% = 1/2
Work done by x in 1 days = 20% = 1/5
Work done by y in 1 days = 25% = 1/4
? Work done by z in 1 day
= (1/2) - (1/5) - (1/4)
= (10 - 4 - 5)/20
= 1/20 = 5%
4 men = 6 women = 10 children
? 1 man = 5/2 children
1 women = 5/3 children
Now, 1 couple + 5 children = 1 man + 1 women + 5 children
= (5/2) + (5/3) + 5 = 55/6 children
According to the formula
M1D1 = M2D2
? 10 x 5 = (55/6) x D2
? D2 = 60/11 days
= 55/11 days
Part of field grazed by 4 goats in 1 day = 1/50
Part of field grazed by 1 goat in 1 day = (1/50) x 4 = 1/200
? 4 g = 6 s [here, g = goats, and s = sheep]
? 1 s = (4/6) g = (2/3) g
Now, 2 g + 9 s = 2 g + 9 x (2/3) g
= 2 g + 6 g = 8 g
? 8 goats can graze the field in (1/8) / 200 = 25 days
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