Sonu can do a piece of work in 20 days. He started the work and left after some days, when 25% work was done. After it Abhijeet joined and completed it working for 10 days. In how many days Sonu and Abhijeet can do the complete work, working together?

Efficiency (X : Y) = 5 : 1
No. of days (X : Y) = 1 : 5

Efficiency of Ajit : Bablu = 3 : 1
No. of days 1 : 3

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.

Efficiency of A = 4.16%
Efficiency of B = 1.6 x 4.16 = 6.66%

? Number of days required by B = 100/6.66 = 15 days

One day's work of A and B = 1/14 + 1/21 = 5/42
? Required number of days = 42/5 = 8.4 days

Work of (A + B + C) in 1 H = 1 / 6
Work done in 2 h = 2/6 = 1/3
Remaining part of the tank = 1 - 1/3 = 2/3
Time taken by A + B to fill this 2/3rd part of the tank = 8 h
A and B together can fill tank in 12 h (i.e., 8 x 3/2).
Now, we have A, B and C together filling the tank in 6 h and A and B together fill tank in 12 h.
Work done by C alone in 1 h = 1/6 - 1/12 = 1/12
Hence, C alone can fill the tank in 12 h.

Alen's one day's work =1/21
Border's one day's work = 1/42
Alen and Border's two days work(working alternatively) = (1/21) + (1/42) = 1/14
So, Alen and Border do 1/14 work in 2 days
So, they complete the work in 14 x 2 = 28 days.

12 men =18 women
? 1 man = 18/12 women
? 8 men = 18/12 x 8 = 12 women
Given m₁ = 18 M₂ = 12 +16 = 28,
D₁ , D₂ = ? and W₁ = W₂ = 1
According to the formula
M₁D₁W₂ = M₂D₂W₁
? 18 x14 x1= 28 x D₂ x1
? D₂ = (18 x 14)/28 = 9 days

Given, M₁ = 10 D₁ = 8, M₂ = ? and D₂ = 1/2
From M₁D₁ = M₂D₂
? 10 x 8 = M₂ x 1/2
? M₂ = 10 x 8 x 2
? M₂ = 160

Given, M₁ = 6, M₂ = 8 , T₁ = 16 h,
and T₂ = ?, W₁ = W₂ = 1
According to the formula.
M₁T₁W₂ = M₂T₂W₁
? 6 x 16 x 1 = 8 x T₂ x 1
? T₂ = 16 x 6/8 = 2 x 6 = 12 h