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? 

Given that three athletes can complete one round around a circular field in 16, 24 and 36 min respectively.

Now, required time after which they met for the first time = LCM of (16, 24 & 36) min

Now, LCM of 16, 24, 36 = 144 minutes = 2 hrs 24 min.

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.