Question

Two years ago Sam's age was $4\frac{1}{2}$ times the age of his son. Six years ago, his age was twice the square of the age of his son. What is the present age of Sam's son ?

• A. $20$
• B. $10$
• C. $15$
• D. $13$

Answer 1

Answer: (B)

$10$

Let $m$ be the age of Sam and $s$ be the age of his son
Then, 2 years ago,
$(m-2)=\frac{9}{2}(s-2)$
$2m-4=9s-18$
$2m=9s-14...\left(i\right)$
$6$ years ago,
$(m-6)=2(s-6{)}^2$
$(\frac{9s-14}{2}-6)=2(s-6{)}^2$
$9s-14-12=4(s^2-12s+36)$
$9s-26=4s^2-48s+144$
$4s^2-57s+170=0$
$s=\frac{57\pm \sqrt{529}}{8}$ = $\frac{57\pm 23}{8}=10,4.25$
Neglect the 4.25 which is a fraction.
Hence, age of his son is $10$ years.