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Word problems based on quadratic equations - class-X
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Two years ago Sam's age was $\displaystyle 4 \frac{1}{2}$ 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 ?
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A
$10$
💡 Explanation:
Let $m$ be the age of Sam and $s$ be the age of his son
Then, 2 years ago,
$(m -2) = \dfrac{9}{2} (s-2)$
$2m - 4 = 9s - 18$
$2m = 9s - 14 ...(i)$
$6$ years ago,
$(m - 6) = 2(s- 6)^2$
$\left(\dfrac{9s-14}{2} - 6\right) = 2(s-6)^2$
$9s - 14 - 12 = 4(s^2 - 12s + 36)$
$9s - 26 = 4s^2 - 48s + 144$
$4s^2 - 57s + 170 = 0$
$s = \dfrac{57 \pm \sqrt{529}}{8}$ = $\dfrac{57 \pm 23}{8} = 10, 4.25$
Neglect the 4.25 which is a fraction.
Hence, age of his son is $10$ years.