Aptitude and Reasoning Questions

 22 = 2 x 11

 54 = 2×33

108 = 22×33

135 = 33×5 

198 = 2×32×11

 ∴ L.C.M = 22×33×5×11=5940

Let’s break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 – sum(2, 2, 18) = 22
2, 4, 9 – sum(2, 4, 9) = 15
2, 6, 6 – sum(2, 6, 6) = 14
2, 3, 12 – sum(2, 3, 12) = 17
3, 4, 6 – sum(3, 4, 6) = 13
3, 3, 8 – sum(3, 3, 8 ) = 14
1, 8, 9 – sum(1,8,9) = 18
1, 3, 24 – sum(1, 3, 24) = 28
1, 4, 18 – sum(1, 4, 18) = 23
1, 2, 36 – sum(1, 2, 36) = 39
1, 6, 12 – sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 – sum(2, 6, 6) = 14
3, 3, 8 – sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

56% of 870 = 56x870/100 = 487.20

82% of 180 = 82x180/100 = 147.60

32% of 90 = 32x90/100

487.20 + 147.6 - 28.8  =  ?

? = 634.8  - 28.8

? = 606.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3*4C2) 

= 210. 

Number of groups, each having 3 consonants and 2 vowels = 210. 

Each group contains 5 letters. 

Number of ways of arranging 5 letters among themselves = 5! = 120 

Required number of ways = (210 x 120) = 25200.

Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.

So, 9 is wrong.

  4             3             4             7               15        ?
×1-1      ×1+1       ×2-1       ×2+1           ×3-1

15x3 - 1 = 45 - 1 = 44