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1. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?

A. 7

B. 8

C. 9

D. 10

**Answer : ****10**

From the problem, A = B +2, B = 2 x C and A+B+C = 27

--> A + B + C = 27

--> (B + 2) + B + (B/2) = 27

--> 2B + 2 + (B/2) = 27

--> 2B + (B/2) = 25

--> 4B + B = 25 * 2

--> B = 10

2. Father’s age is 4 times that of his son. 5 years back, it was 7 times. His age now is

A. 30

B. 35

C. 40

D. 45

**Answer : ****40**

Let Father’s age --> F and Son’s age --> S

From the problem,

Now, F = 4S

5 years back, (F – 5) = 7 (S – 5)

---------------------------------------------

Solving above equations,

F = 40 (Father’s age)

S = 10 (Son’s age)

3. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child ?

A. 2 years

B. 4 years

C. 6 years

D. 8 years

**Answer : ****4 years**

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

i.e. 5x = 20 and x = 4.

So, the age of the youngest child (x) is 4 years.

4. The ratio between the present ages of P and Q is 6 : 7. If Q is 4 years old than P. What will be the ratio of the ages of P and Q after 4 years?

A. 3 : 4

B. 3 : 5

C. 4 : 3

D. 7 : 8

**Answer : **

Let P’s age and Q’s age be 6x years and 7x years respectively

Then, 7x - 6x = 4

x = 4

Required ratio = (6x + 4) : (7x + 4) = 28 : 32 = 7 : 8

5. A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father (in years) is

A. 44

B. 32

C. 22

D. 45

**Answer : ****44**

Let son’s age = x. Then father’s age = 2x

12 (x - 20) = (2x - 20)

10x = 220

x = 22

Father’s present age = 44 years.

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