Question

Five years ago, a woman’s age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find : (i) The age of the son five years ago (ii) The present age of the woman.

Answer 1

Let Sons age = x
Let Womens age = y

y - 5 = (x - 5)² -------------1
y + 10 = 2(x+10)----------2

Equation 1
y - 5 =x² - 10 x + 25
y = x² - 10x + 30

Equation 2
y = 2x +10

Substituting The Value Of y in equation 1
2x + 10 = x² - 10x + 30
x² - 10x - 2x + 30 - 10 = 0
x² - 12x + 20 = 0
x² - 2x - 10x + 20 = 0
x(x - 2) -10(x - 2) = 0
(x - 10)(x - 2) = 0
x = 10 or x = 2[ x cant be 2 as five years ago it would be negative age]

therefore, sons present age = 10
1. So sons age five years ago = 5yrs
Substituting The Value Of x in equation 2
y = 2x + 10
y = 2 * 10 + 10
y = 20 + 10
y = 30 yrs
2. Women Present Age = 30 yrs

Answer 2

Answer:

Let Sons age = x

Let Womens age = y

y - 5 = (x - 5)² -------------1

y + 10 = 2(x+10)----------2

Equation 1

y - 5 =x² - 10 x + 25

y = x² - 10x + 30

Equation 2

y = 2x +10

Substituting The Value Of y in equation 1

2x + 10 = x² - 10x + 30

x² - 10x - 2x + 30 - 10 = 0

x² - 12x + 20 = 0

x² - 2x - 10x + 20 = 0

x(x - 2) -10(x - 2) = 0

(x - 10)(x - 2) = 0

x = 10 or x = 2[ x cant be 2 as five years ago it would be negative age]

therefore, sons present age = 10

1. So sons age five years ago = 5yrs

Substituting The Value Of x in equation 2

y = 2x + 10

y = 2 * 10 + 10

y = 20 + 10

y = 30 yrs

2. Women Present Age = 30 yrs

Step-by-step explanation: