At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Given:

At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age.

To do:

We have to find the present ages of both Asha and Nisha.

Solution:

Let the present age of Nisha be $x$ years.

This implies, 

The present age of Asha$=x^2+2$ years

It is given that when Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha.

Difference in their ages$=x^2+2-x$ years.

Therefore, after $x^2+2-x$ years Nisha will grow to Asha's age.

Asha's age after $x^2+2-x$ years$=x^2+2+x^2+2-x=2x^2-x+4$ years

Nisha's age after $x^2+2-x$ years$=x+x^2+2-x=x^2+2$ years

According to the question,

$2x^2-x+4=10x-1$

$2x^2-10x-x+4+1=0$

$2x^2-11x+5=0$

Solving for $x$ by factorization method, we get,

$2x^2-10x-x+5=0$

$2x(x-5)-1(x-5)=0$

$(2x-1)(x-5)=0$

$2x-1=0$ or $x-5=0$

$2x=1$ or $x=5$

$x=\frac{1}{2}$ or $x=5$

Therefore, the value of $x$ is $5$.

$x^2+2=(5)^2+2=25+2=27$

The present age of Nisha is $5$ years and the present age of Asha is $27$ years.

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Updated on: 10-Oct-2022

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