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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 - Mathematics

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Question

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.

Sum
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Solution

Let Nisha’s present age be x years.

Then, Asha’s present age = x2 + 2  .....[By given condition]

Now, when Nisha grows to her mother’s present age

i.e., After [(x2 + 2) – x] year.

Then, Asha’s age also increased by [(x2 + 2) – x] year.

Again by given condition,

Age of Asha = One year less than 10 times the present age of Nisha

(x2 + 2) + {(x2 + 2) – x} = 10x – 1

⇒ 2x2 – x + 4 = 10x – 1

⇒ 2x2 – 11x + 5 = 0

⇒ 2x2 – 10x – x + 5 = 0

⇒ 2x(x – 5) – 1(x – 5) = 0

⇒ (x – 5)(2x – 1) = 0

∴ x = 5

[Here, x = `1/2` cannot be possible, because at x = `1/2`, Asha's age is `2 1/4` year which is not possible]

Hence, required age of Nisha = 5 years

And required age of Asha = x2 + 2 = (5)2 = 25 + 2 = 27 years.

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Chapter 4: Quadatric Euation - Exercise 4.4 [Page 42]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 10
Chapter 4 Quadatric Euation
Exercise 4.4 | Q 6 | Page 42

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