**Exercise 8.9**

**Q.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.**

**Sol:**

Given that Ashuâ€™s present age is x years and his mother Mrs. Veena is x^{2} years

Then, acc. to question,

Five years later, Ashu is (x + 5) years

And his mother Mrs. Veena is (x^{2} + 5) years

So,

x^{2} + 5 = 3(x + 5)

x^{2} + 5 = 3x + 15

x^{2} + 5 â€“ 3x â€“ 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

Since, the age can never be negative

Therefore, ashuâ€™s present age is 5 years and his motherâ€™s age is 25 years.

**Q.2:Â 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 manâ€™s age at the time. Find their present ages.**

**Sol:**

Let the present age of the man be x years

Then, present age of his son is = (45 – x) years

Five years ago, manâ€™s age = (x – 5) years

And his sonâ€™s age = (45 â€“ x – 5) = (40 – x) years

Then, acc. To question,

(x – 5)(40 â€“ x) = 4(x – 5)

40x â€“ x^{2} + 5x â€“ 200 = 4x â€“ 20

-X^{2} + 45x â€“ 200 = 4x – 20

-x^{2} + 45x – 200 â€“ 4x + 20 = 0

-x^{2} + 41x â€“ 180 = 0

x^{2} â€“ 36x – 5x + 180 = 0

x(x – 36) â€“ 5(x – 36) = 0

(x â€“ 36)(x – 5) = 0

x = 36Â orÂ x = 5

But, the fatherâ€™s age can never be 5 years

Therefore, when x = 36,

45 â€“ x = 45 â€“ 36 = 9

Hence, manâ€™s present age is 36 years and his sonâ€™s age is 9 years.

**Q.3:Â The product of Shikhaâ€™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.Â **

**Sol:**

let the present age of shikha be x years

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

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

Then, acc. To question,

(x – 5)(x +8) = 30

x^{2 }+ 8x â€“ 5x â€“ 40 = 30

x^{2} + 3x â€“ 40 â€“ 30 = 0

x^{2 }+ 3x â€“ 70 = 0

x(x – 7) + 10(x – 7) = 0

(x – 7)(x + 10) = 0

x = 7 or x = -10

Since, the age can never be negative

Heence, the present age of shikha is = 7 years.

**Â Q.4:Â The product of Ramuâ€™s age (in years) five years ago and his age (in years) nine years later is 15. Determine Ramuâ€™s present age.**

**Sol:**

let the present age of ramu be x years

Then, 9 years later, age of her = (x + 9) years

Five years ago, her age = (x – 9) years

Then, acc. to question,

(x – 5)(x + 5) = 15

x^{2 }+ 9x â€“ 5x â€“ 45 = 15

x^{2} + 4x â€“ 45 â€“ 15 = 0

x^{2} + 4x â€“ 60 = 0

x^{2} – 6x + 10x â€“ 60 = 0

x(x – 6) + 10(x – 6) = 0

(x – 6)(x + 10) = 0

x = 6Â orÂ x = – 10

Since, the age can be never be negative

Therefore, the present age of ramu is = 6 years

**Q.5: Is the following situation possible? if so, determine their present ages.**

**The sum of the ages of two friends is 20 years. four years ago, the product of their ages in years was 48.**

**Sol:**

let the present age of two friends be x years and (20 – x) years respectively

Then, 4 years later, the age of two friends will be (x – 4) years and (20 â€“ x – 4) years

Then, acc. To the question

(x – 4)(20 â€“ x – 4) = 48

(x – 4)(16 – x) = 48

16x â€“ x^{2} â€“ 64 + 4x = 48

– x^{2} + 20x â€“ 64 â€“ 48 = 0

x^{2} â€“ 20x + 112 = 0

Let the discriminant of the above quadratic eqn.

D = b^{2} â€“ 4ac

Here, a = 1, b = -20, c = 112

D = (-20)^{2} â€“ (4x1x112) = 400 – 448 = – 48

Since, D < 0

The above question does not have real roots.

Hence, the given situation is not possible.

**Q.6:Â A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.**

**Sol:**

let the present age of girl be x years then, age of her sister (x/2) years

Then, 4 years later, age of girl = (x + 4) years and her sisterâ€™s age be \((\frac{x}{2} + 4)\) years

Then, acc. to the question,

\((x + 4)(\frac{x}{2} + 4) = 160\)

(x + 4) (x + 8) = 160×2

x^{2} + 8x + 4x + 32 = 320

x^{2} + 12x â€“ 288 = 0

x^{2} – 12x + 24x â€“ 288 = 0

x(x – 12) + 24(x – 12) = 0

(x – 12) (x + 24) = 0

x = 12Â or x = – 24

Since, the can never be negative,

Therefore, the present age of the girl is = 12 years.

And her sisterâ€™s age will be,

\(\frac{x}{2} = \frac{12}{2}\) = 6 years.

**Q.7:Â The sum of the reciprocals of Rehmanâ€™s ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.**

**Sol:**Â let the present age of Rehman be x years

Then, 8 years late, age of her = (x + 5) years

Five years ago, her age = (x – 3) years

Acc. To question,

\(\frac{1}{(x – 3)} + \frac{1}{(x + 5)} = \frac{1}{3}\)

\(\frac{x + 5 + x -3}{(x – 3)(x + 5)} = \frac{1}{3}\)

\(\frac{2x + 2}{x^{2} + 5x – 3x – 15} = \frac{1}{3}\)

x^{2} + 2x â€“ 15 = 6x + 6

x^{2} + 2x â€“ 15 â€“ 6x â€“ 6 = 0

x^{2} â€“ 4x â€“ 21 = 0

x^{2} â€“ 7x + 3x â€“ 21 = 0

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7 or x = -3

Since, the age can never be negative

Therefore, the present age of Rehman be = 7 years.