A can complete a piece of work in 18 days and B can complete the same work in half time taken by A. Then, working together, what part of the same work they can complete in a day?

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

Work done by (A + B) in 1 day = 1/24
Work done by B alone in 1 day = 1/(3 x 12) = 1/36
? Work of A for 1 day = 1/24 - 1/36 = 1/72
After 12 days the remaining work = 1 - 1/3 = 2/3
? 1/72 work is done by A in 1 day
? 2/3 work is done by A in (1 x 72)/(1 x 2/3) = 48 days

Let B takes N days to do the work
? A takes 2 x (3/4) x N = 3N/2 days to do it
? (A + B)'s 1 day's work = 1/18
? 1/N + 2/3N = 1/18
? N = 30

Radio of times taken = (1/3) : (1/4) = 4 : 3

? Mohan moves the whole lawn in x hours.
? mohan moves in 2 hours = 2/x part of the lawn.
? Unmoved part = 1 - 2/x = (x - 2)/ x part

(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.

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'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.

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 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