Question

Q29. Five years ago, a woman's age (in year ) was the square of her son's age. Ten years later from now her age
will be twice that of her son's age. find.
(a) The age of the son five years ago.
(b) The present age of the woman.

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Answer 1

Suppose the son age is p

Then the woman age be m

Equation will be like this :-

m - 5 = (p - 5)².................... (Eqn.1)

m + 10 = 2(p + 10).................... (Eqn. 2)

Solve Equation number (1)

m - 5 = p² - 10p + 25

m = p² - 10p + 30

Now solve Equ. (2) we get the value of m :-

m + 10 = 2p + 20

m = 2p + 20 - 10

m = 2p + 10

Substitute the solved value of (1),(2) in equation (1) and put it in an equation we get :-

2p + 10 = p² - 10p + 30

p² - 10p - 2p + 30 - 10 = 0

p² - 12p + 20 = 0

p² - 2p - 10p + 20 = 0

p(p - 2) - 10(p - 2) = 0

(p - 10)(p - 2) = 0

Here we get :-

p = 10 and p = 2 (Avoid the second value because age before 5 years is not applicable)

Therefore we get :-

Present age of Son = 10

a) Son age five years ago

= 10 - 5  

= 5 years

Put value of p in eq. (2) we get :-

m = 2p + 10

m = 2(10) + 10

m = 20 + 10

m = 30 years

Hence we get the age of woman

= 30 years