At present,Asha's age (in years) is 2 more th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let Nisha's present age be $x$ years.Then,Asha's present age $= x^2 + 2$ years.The time taken for Nisha to reach her mother's present age is $(x^2

Vedclass · Accepted Answer

$5, 27$

Vedclass · Answer

(D) Let Nisha's present age be $x$ years.Then,Asha's present age $= x^2 + 2$ years.The time taken for Nisha to reach her mother's present age is $(x^2 + 2) - x$ years.After this time,Asha's age will be $(x^2 + 2) + (x^2 + 2 - x) = 2x^2 - x + 4$ years.According to the problem,this age is $10x - 1$.So,$2x^2 - x + 4 = 10x - 1$.$2x^2 - 11x + 5 = 0$.$2x^2 - 10x - x + 5 = 0$.$2x(x - 5) - 1(x - 5) = 0$.$(x - 5)(2x - 1) = 0$.Since $x = 1/2$ is not practical for the given conditions,$x = 5$.Nisha's present age $= 5$ years.Asha's present age $= 5^2 + 2 = 27$ years.

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 (in $years$).

Similar Questions

The value of discriminant $D$ of the quadratic equation $25x^{2} - 10x + 1 = 0$ is ..... .

The value of the discriminant of the quadratic equation $x^{2}-8x+15=0$ is ...... .

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$

Solve the following equation using the method of factorization: $(x+1)^{2} + x^{2} = 221$.

If the radius of a sphere is doubled,its volume becomes $\ldots \ldots \ldots \ldots$ times the original volume.

Vedclass Test Series

Exam Paper Generator

Online Exam Module