Aptitude and Reasoning Questions

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

Let total population = p
number of females = 5p/9
number of males =(p-5p/9) = 4p/9
married females = 30% of 5p/9 = 30x5p/100x9 = p/6
married males = p/6
unmarried males =(4p/9-p/6) = 5p/18
Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) +  (Prob. that brother is not selected) × (Prob. that sister is selected)

= 15*23+45*13= 25

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 =5*35:8*35=  3:245  

A : B : C = 2 : 3 : =>  = 10 : 15 : 24

 B =245 = 30