  Question

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

Solution

Let the present age of my son be x years. Then, his age two years ago was (x – 2) years and six years ago, (x – 6) years. Thus, my age two years ago was $4\frac{1}{2}\left(x-2\right)=\frac{9}{2}\left(x-2\right)$  years and, six years ago, 2(x – 6)2  years. Thus, we get, $\frac{9}{2}\left(x-2\right)-2{\left(x-6\right)}^{2}=4$  $⇒$ 9(x – 2) – 4(x – 6)2 = 8 $⇒$9x – 18 – 4(x2 – 12x + 36) = 8 $⇒$ 9x – 18 – 4x2 + 48x – 144 = 8 $⇒$ 4x2 – 57x + 170 = 0 On splitting the middle term –57x as –40x – 17x, we get:  4x2 – 40x – 17x + 170 = 0 $⇒$ 4x(x – 10) – 17(x – 10) = 0 $⇒$ (x – 10)(4x – 17) = 0 $⇒$ x – 10 = 0    or     4x – 17 = 0 $⇒$ x = 10   or   x = $\frac{17}{4}$  Since, the age cannot be in fraction, Thus, the present age of my son is 10 years.MathematicsMathematics (Algebra)Standard X

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