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
(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
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
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
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
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.
P,s 1 day's work = 1/12
Q's 1 day's work = 1/8
(P + Q)'s 1 day's work = (1/12) + (1/8)
= (2 + 3)/24
= 5/24
(P + Q + R)'s 1 day's work = 1/4
? R's 1 day's work = (1/4) - (5/24) = (6 - 5)/24 = 1/24
? R will finish the work alone in 24 days.
? Required time = 24 days
A's 5 days' work = (1/15) x 5 = 1/3
Remaining work = 1 - (1/3) = 2/3
? (A + B)'s 1 day's work
= (1/15) + (1/10)
= (2 + 3(/30
= 5/30
= 1/6
? 2/3 work will be finished by (A + B) in (2/3) x 6 days or in 4 days .
? Total time taken = (5 + 4) days = 9 days
Let time taken by B = N
Efficiency ? 1/Time taken
So, if B is 100% efficient, then A is 80% efficient.
So, 80/100 = N /(15/2)
? N = 6 h
Ratio of time taken by X
and Y = 160 : 100 = 16 : 1
Suppose Y takes y days to the work.
Then, 1.6 : 1 = 16 : y
? y = (16 x 1)/1.6 = 10 days
8M + 12W = 4 days (whole work)
? 32M + 48W = 1 day ................ (1)
Again 6M + 14W = 5 days
? 30M + 70W = 1 day ................(2)
From eq. (1) and (2)
32M + 48W = 30M +70W
? 2M = 22W
? 1M = 11W
Now, 30M + 70W = 1 day
(30 X 11 X 70)W = 1 day
Therefore 400W requires 1 day to complete the whole work.
Thus 20W needs 20 days to complete the whole work.
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