NIELIT 2022 April Scientist B | Section A | Question: 35

Question

Four years ago a man was $6$ times as old as his son. After $16$ years he will be twice as old as his son. What is the present age of man and his son?

• A. $34, 9$
• B. $33, 7$
• C. $35, 5$
• D. $36, 6$

Answer 1

Assume the present age of the father is $x$ year, son's age is $y$ year.

$4$ year ago father age=$(x-4)$

$4$ year ago son age =$(y-4)$

now it is give that: father's age=6(son’s age)

$\therefore (x-4)=6(y-4)$

$\implies x-4=6y-24$

$\implies x-6y=-20 ------  (i)$

Now after $16$ year father’s age =$(x+16)$

son’s age after $16$ year= $(y+16)$

now it is given that father age=2(son’s age)

$\implies (x+16)=2(y+16)$

$\implies x+16=2y+32$

$\implies x-2y=16------  (ii)$

from equation $(i) \& (ii)$ we get son’s age $y=9$ year, father’s age $x=34$ year.

Option (A) is correct.