According to the formula
M1D1T1/W1= M2D2T2/W2
Given M1 = 105 D1 = 25 T1 = 8 W1 = 2/5
Now let the additional men be N
Then, M2 = 105 + N, T2 = 9, D2 = 25 and W2 =1 - (2/5) = 3/5
On putting these value in the above formula
(105 x 25 x 8) / (2/5) = [(105 + N) x 25 x 9] / (3/5)
? (105 x 8) / 2 = (105 + N) x (9/3)
? 105 x 4 = (105 + N) x 3
? 105 x 4 = 105 x 3 +3N
? 3N = 105
? N = 35 men
According to the question,
5 men's 1 day's work = 1/15 .....(i)
? 5 men's 5 day's work = (1/15) x 5 = 1/3
? Remaining work = 1 - (1/3) = 2/3
5 men + 10 women can do 2/3 of the work in 5 days.
? 5 men + 10 women can do the whole of the work in 15/2 days.
? (5 men + 10 women's) 1 day's work = 2/15 .......(ii)
10 women 1 days work = (2/15) - 5 men's 1 day's work
= (2/15) - (1/15)
= 1/15
? 10 women can finish the work in 15 days
According to the question,
1 man = 2 women
? 10 men + 8 women = (20 + 8) women = 28 women
4 men + 6 women = (8 + 6) women = 14 women
Given, M1 = 28, M2 = 14, D1 = 5, D2 = ?
and W1 = W2 = 1
According to the formula, M1D1W2= M2D2W1
? 28 x 5 x 1 = 14 x D2 x 1
? D2 = (28 x 5)/14 = 10 days
Let time taken by B = N
Then, time taken by A = 2N
and time taken by C = 3N
According to the question,
(1/N) + (1/2N) + (1/3N) = 1/12
? 1 + (1/2) + (1/3) = N/12
? (6 + 3 + 2)/6 = N/12
? 11 = N/2
? N = 22
? Required number of days = 2N = 2 x 22 = 44 days
According to the question,
2 children = 1 man
? (16 children + 24 men) = (8 + 24) men = 32 men
Given M1 = 32, M2 = 24, D1 = 18, D2 = ?
and W1 = W2 = 1
According to the formula, M1D1W2 = M2D2W1
? 32 x 18 x 1 = 24 x D2 x 1
? D2 = (32 x 18)/24 = 8 x 3 = 24 days
Work done in first in 3 days
= (1/18) + (1/24) + (1/36) = (4 + 3 + 2)/72 = 1/8
? Time taken to complete 1/8 part of work = 3 days
? Time taken to complete the whole work = 3 x 8 = 24 days
Here, a1 = 14, b1 = 12, p = 4, a2 = 8, b2 = 16 and q = 5
One days work of 1 man/One days of 1 boy
= (qb2 - pb1) / ( pa1 - qa2)
= [5 x 16 - 4 x 12] / [ 4 x 14 - 5 x 8]
= (80 - 48) / (56 - 40)
= 32/16
= 2
Let extra hours a day are N.
According to the formula,
(M1D1T1) / W1 = (M2D2T2) / W2
? [1 x 1 x (6 + 4)] / 1 = [1 x 1 x (6 + 6 + N)] / 11/2
? (3/2) x 10 = 12 + N
? 15 = 12 + N
? N = 15 - 12 = 3
Extra hours of works everyday are 3.
Since, 3 women + 18 children complete work in 2 days. Therefore, (3 x 2) women + (18 x 2) children complete work in 1 days i.e, 6 women + 36 children complete work in 1 day.
Work of 36 children for 1 day = 1 - (1/3) = 2/3
[? work of 6 women for 1 day = 1/3]
? 36 children do 2/3 part of the work in 1 day.
36 children can do the work in 3/2 days.
9 children can do the the work in (3/2) x 4 = 6 days
Given, M1 = 120, M2 = 120 - n, D1 = 64, D2 = 60
W1= 2/3 and W2 = 1/3
According to the formula,
(M1D1) / W1 = (M2D2) / W2
? [120 x (64/2)] / 3 = [(120 - n) x 60] / (1/3)
? (120 x 64) / (2 x 60) = (120 - n)
? (120 - n) = 64
? n = 120 - 64 = 56
Now, 56 men can be discharged to finish the work in time.
Remaining work = 1 - (4/7) = 3/7
Remaining period = (92 - 66) = 26 days
Let the number of addition men = N
Given, M1 = 234, D1 = 66, T1 = 16, W1= 4/7,
M2 = (234 + x), D2 = 26, T2 = 18, W2 = 3/7
According to the question,
M1W2T1D1= M2W1T2D2
? 234 x (3/7) x 16 x 66 = (234 x N) x (4/7) x 18 x 26
? 234 + N = (3 x 66 x 16 x 234) / (4 x 26 x 18)
? 234 + N = 36 x 11 = 396
? N = 396 - 234 = 162
Additional men to be employed = 162
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