A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the 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.

Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5

Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively

They can do work for 12 days.

=> Total work = 12 x 16x = 192x

Now, the required time taken by Q to complete the job alone = 162x8x = 24days.