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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

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प्रश्न

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.

योग
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उत्तर

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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अध्याय 4: Quadatric Euation - Exercise 4.4 [पृष्ठ ४२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 10
अध्याय 4 Quadatric Euation
Exercise 4.4 | Q 6 | पृष्ठ ४२

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