A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent ?

Let p be the required number of days.

From the given data,

12 x 14 = 12 x 5 + (12+3) x p 

12 x 14 = 12[5 + 3p]

14 = 5 + 3p

3p = 9

p = 3 days.

Hence, more 3 days all of them take to complete the remaining work.

Let workdone 1 boy in 1 day be b

and that of 1 girl be g

From the given data,

4(5b + 3g) = 23

20b + 12g = 23 .......(a)

2(3b + 2g) = 7

6b + 4g = 7 ........(b)

Solving (a) & (b), we get

b = 1, g = 1/4

Let number og girls required be 'p'

6(7 x 1 + p x 1/4) = 45

=> p = 2.

Hence, number of girls required = 2

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

 Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.