Another application of quadratic equations is solving problems on ages. Many students get confused in understanding the actual application. Thatâ€™s why we at BYJUâ€™S have created the RD Sharma Solutions Class 10 containing step by step procedure for solving. Also, the students can access the RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.9 PDF given below.

## RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.9 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.9

**1. Ashu is x years old while his mother Mrs. Veena is x ^{2} years old. Five years hence Mrs. Veena will be three times old as Ashu. Find their present ages.Â **

**Solution:**

Given, Ashu’s present age is x years and his mother Mrs. Veena is x^{2}Â years.

After 5 years, Ashu age will be (x + 5) years

And his mother Mrs. Veena age will be (x^{2}Â + 5) years

Given relationship between their ages can be expressed as:

x^{2}Â + 5 = 3(x + 5)

x^{2}Â + 5 = 3x + 15 x^{2}Â + 5 – 3x – 15 = 0

x^{2}Â – 5x + 2x + 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x + 2) = 0

x = 5Â or x = – 2(neglected) since, the age can never be negative

Hence, Ashu’s present age is 5 years and his mother’s age is 25 years.

**2. The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at the time. Find their present ages.**

**Solution:**

Let the present age of the man be x years

Then, the present age of his son will be = (45 – x) years

Five years ago, man’s age = (x – 5) years

And, his son’s age = (45 – x – 5) = (40 – x) years

Given relationship between their ages can be expressed as:

(x – 5)(40 – x) = 4(x – 5)

40x – x^{2}Â + 5x â€“ 200 = 4x – 20

-x^{2}Â + 45x â€“ 200 = 4x – 20

-x^{2}Â + 45x – 200 – 4x + 20 = 0

– x^{2}Â + 41x – 180 = 0

x^{2}Â – 36x – 5x + 180 = 0 [By factorisation method]

x(x – 36) – 5(x – 36) = 0

(x – 36)(x – 5) = 0

x = 36Â orÂ x = 5,

But, the father’s age can never be 5 years

Thus, when x = 36, 45 – x = 45 – 36 = 9

Therefore, the man’s present age is 36 years and his son’s age is 9 years.

**3. The product of Shikha’s age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.**

**Solution:**

Letâ€™s assume the present age of Shikha be x years

So, 8 years later, age of her = (x + 8) years

Five years ago, her age = (x – 5) years

Given relationship between the ages can be expressed as:

(x – 5)(x +8) = 30

x^{2Â }+ 8x – 5x – 40 = 30

x^{2}Â + 3x – 40 – 30 = 0

x^{2Â }+ 3x – 70 = 0 [By factorisation method]

x(x – 7) + 10(x – 7) = 0

(x – 7)(x + 10) = 0

x = 7 or x = -10 (neglected)

Since, the age can never be negative.

Therefore, the present age of Shikha is 7 years.

**4. The product of Ramu’s age (in years) five years ago and his age (in years) nine years later is 15. Determine Ramu’s present age.Â **

**Solution:**

Let the present age of Ramu be x years

So, 9 years later, the age of him = (x + 9) years

And, five years ago, his age = (x – 5) years

Given relationship between the ages can be expressed as:

(x – 5)(x + 5) = 15

x^{2Â }+ 9x – 5x – 45 = 15

x^{2}Â + 4x – 45 – 15 = 0

x^{2}Â + 4x – 60 = 0

x^{2}Â – 6x + 10x – 60 = 0 [By factorisation method]

x(x – 6) + 10(x – 6) = 0

(x – 6)(x + 10) = 0

x = 6Â orÂ x = – 10(neglected) as the age can be never be negative.

Hence, the present age of Ramu is 6 years.