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Question

Two years ago my age was 412 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 412x-2=92x-2  years and,
six years ago, 2(x – 6)years.
Thus, we get,
92x-2-2x-62=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 = 174 
Since, the age cannot be in fraction,
Thus, the present age of my son is 10 years.

Mathematics
Mathematics (Algebra)
Standard X

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