Question

(1) Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.

Answer 1

Answer:

Son = 12 years

Mother = 42 years 





Explanation - 

Six years ago let the son age = x

Mother = x² 



Present ages now:

Son = x + 6

Mother = x²  + 6



3 years later:

Son = x + 6 + 3 = x + 9

Mother =  x²  + 6 + 3 =  x²  + 9


Solving x - 

x²  + 9 = 3(x + 9)

x²  + 9 = 3x + 27

x²  - 3x - 18 = 0

(x - 6)(x + 3) = 0

x = 6 or x = -3

Take x = 6. As the other value is (- value)

Finding their Present age -

Son = x + 6 = 6 + 6 = 12 years.

Mother =  x²  + 6 = (6)² + 6 = 42 years.



Hence : 

Son = 12 years

Mother = 42 years

Answer 2

Answer:

12 yrs, 42 yrs

Step-by-step explanation:

Given Six years before, the age of mother was equal to the square of her son's age.

Let the age of mother be x years and age of the son be y years.

six years before means it will be x - 6 and square of the age of her son will be (y - 6)^2

So x - 6 = (y - 6)^2

we have the formula (a -b)^2 = a^2 - 2ab + b^2

x - 6 = y^2 - 12 y + 36

or x = y^2 - 12 y + 42

Given three year hence her age will be thrice the age of her son . So the equation will be  

x + 3 = 3(y + 3)

x = 3y + 6

Substituting in x we have

3y + 6 = y^2 - 12 y + 42

y^2 - 15 y + 36 = 0

y^2 - 12y - 3y + 36 = 0

y(y - 12) - 3(y - 12) = 0

(y - 12)(y - 3) = 0

y = 12, 3

We take y = 12 since 6 yrs ago it will be 3 - 6 = -3 yrs.

taking y =12

Given x = 3y + 6 = 3(12) + 6 = 42

Now we get mother's age x = 42 yrs and son's age y = 12 yrs