General Questions
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Required time taken to complete the work by both together = x y / x + y
(Here x = 8 and y = 4)
∴ Required time = (8 x 4)/(8 + 4) = 32/12 = 22/3 hours
One day's work of A and B = 1/x + 1/2x = 1/14
∴ x = 21
Since A is twice efficient as B so A will take half of the day taken by B.
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
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)
= 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
h, then B can do it in: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
days, Y alone can complete the job in: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
(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
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
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
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