Let the men do the work in a days and the boys in b days.
? (3/a) + (4/b) = 1/8 ..... (i)
Now, 6/a + 8/b = 2[(3/a) + (4/b)] = 2 x 1/8 = 1/4
So, 6 men and 8 boys can do the same work in 4 days.
A's 1 day's work = 1/6
B's 1 day's work = 1/12
(A + B)'s 1 day's work = 1/6 + 1/12
= (2 + 1)/12 = 3/12 = 1/4
Hence, both will complete the work in 4 days.
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
(A + B)'s 1 days work = 1/12
B's 1 days work = 1/30
? A's 1 days work = 1/12 - 1/30 = (5 - 2)/60 = 1/20
? A alone can finish the work in 20 days.
A's 1 day's work = 1/18
B's 1 day's work = 1/9
(A + B)'s 1 day's work = 1/9 + 1/18 = 3/18 = 1/6
let the work done be 1 work.
Here M1=15, D1=16 W1=W2 =1
M2 =24 D2= ?
? According to the formula
M1 D1 W2 = M2 D2 W1
? 15 x 16 x 1 = 24 x D2 X 1
? D1 = 15 X 16/24 = 10
Therefore 10 days are required to complete the work
A's 1 day's work A = 1/x
B's 1 day's work B = 1/3x
(A + B)'s 1 day's work = (1/x) + (1/3x) = 4/3x
And given one day work of both A and B = 1/12
? 4/3x = 1/12
? 3x = 48
? x =16
X's 1 day's work = 1/12
(X + Y)' s 1 day's work = 3/20
? Y's 1 day's work = (3/20) - (1/12) = 4/60 = 1/15
? Number of day's taken by Y to complete the work = 15 days
Given M1 = 10
D1 = 8, M2= ? and D2 = 1/2
From M1 D1 = M2 D2
? 10 x 8 = M2 x 1/2
? M2= 10 x 8 x 2
? M2 = 160
(3 x 6) men = (5 x 18) women
18 men = 90 women
? 1 man = 5 women
? 4 men + 10 women
= 4 x 5 + 10 = 30 women
Given, M1 = 5, M2 = 20, D1 = 18 ,
W1 = W2 = 1 and D2 = ?
According to the formula, M1D1W2 = M2D2W1
? 5 x 18 x 1 = 30 x D2 x 1
? D2 = 5 x 18/30 = 3 days
3 men = 6 children
? 1 man = 2 children
? 4 men + 4 children = 4 men + (4/2) men = 6 men
Given, M1 = 3, M2 = 6, D1 = 18, W1 = W2 = 1 and D2= ?
According to the formula,
M1D1W2 = M2D2W1
? 3 x 18 x 1 = 6 x D2 x 1
? D2 = 3 x 18/6 = 9 days
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