Let 1 man's 1 day work = x

and 1 woman's 1 days work = y.

Then, 4x + 6y = 1/8

and 3x+7y = 1/10

solving, we get y = 1/400 [means work done by a woman in 1 day]

10 women 1 day work = 10/400 = 1/40

10 women will finish the work in 40 days

To calculate the answer we need to get 1 man per day work and 1 woman per day work.

Let 1 man 1 day work =x

and 1 woman 1 days work = y.

6x+5y = 1/6

and 3x+4y = 1/10

On solving, we get x = 1/54 and y = 1/90

(9 men + 15 women)'s 1 days work =

(9/54) + (15/90) = 1/3

9 men and 15 women will finish the work in 3 days

Work done by A in l0 days = (1/25) ×10 = 2/5

Remaining work = 1 - (2/5) = 3/5

(A+B)s 1 days work = (1/25) + (1/20) = 9/100

9/100 work is done by them in 1 day.

hence 3/5 work will be done by them in (3/5)×(100/9)

= 20/3days.

Total time taken = (10 + 20/3) = 16×(2/3) days

1 woman's 1 day's work = 1/70

1 Child's 1 day's work = 1/140

5 Women and 10 children 1 day work =

So 5 women and 10 children will finish the work in 7 days.

A's 1 day work = 1/16

B's 1 day work = 1/12

As they are working on alternate day's

So their 2 days work = (1/16)+(1/12)

= 7/48

[here is a small technique, Total work done will be 1, right, then multiply numerator till denominator, as 7×6 = 42, 7×7 = 49, as 7×7 is more than 48, so we will consider 7×6, means 6 pairs ]

Work done in 6 pairs = 6×(7/48) = 7/8

Remaining work = 1-7/8 = 1/8

On 13th day it will A turn,

then remaining work = (1/8)-(1/16) = 1/16

On 14th day it is B turn,

1/12 work done by B in 1 day

1/16 work will be done in (12×1/16) = 3/4 day

So total days =

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