# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.9

Last Updated : 13 Jul, 2021

### Question 1: Ashu is x years old while his mother Mrs. Veena is  x2 years old. Five years  hence Mrs.  Veena  will be three times old as Ashu. Find their present ages.

Solution:

Ashu’s present age is =  x years

and Mrs. Veena’s present age is = x years

Now five years hence,

Mrs. Veena’s age will be = (x2 + 5) years

and Ashu’s age will be = (x + 5) years

So according to the question-

â‡’ Mrs. Veena’s age = Three times of the Ashu’s age

â‡’ x2 + 5 = 3(x + 5)

â‡’ x2 + 5 = 3x + 15

â‡’ x2 – 3x – 10 = 0

Now for factorizing above quadratic equation-

Break coefficient of x in difference form as constant term is negative-

â‡’ x2 – (5-2)x – 10 = 0

â‡’ x2 – 5x + 2x – 10 = 0

â‡’ x(x – 5) + 2(x – 5) = 0

â‡’ (x – 5)(x + 2) = 0

â‡’ Either x – 5 = 0       or     x + 2 = 0

x = 5            or      x = -2

So on discarding x = -2 (because age can not be negative)-

Mrs. Veena’s present age = x2 = 25 years

and Ashu’s present age = x = 5 years

### Question 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 manâ€™s age at that time. Find their present ages.

Solution:

Let present age of man = x years

And given,

Man’s present age + son’s present  age = 45 years

â‡’ Son’s present age = (45 – x) years.

Now Five years ago-

â‡’ Age of man = (x – 5) years

and age of his son = (45-x) – 5     years

â‡’ son’s age = (40-x) years

Now according to question (five years ago) –

Product of their ages = four times the manâ€™s age at that time

â‡’ (x – 5)(40 – x) = 4(x – 5)

â‡’ (x – 5)(40 – x) – 4(x – 5) = 0

â‡’ (x – 5)(40 – x – 4) = 0               [by taking common (x-5)]

â‡’ (x – 5)(x – 36) = 0

â‡’ either x – 5 = 0        or       x – 36 = 0

Now in first case if we take x = 5 then

Man’s age = 5 years

His son’s age = 40 years (It is not possible)

so x = 36

â‡’ Present age of Man = x = 36 years

â‡’ and present age of his son = (45 – x) = 9 years

### Question 3: 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.

Solution:

Let the present age of Shikha is = x years

So five years ago Shikha’s age was = (x – 5) years

And 8 years later Shikha’s age will be = (x + 8) years

Now according to question-

â‡’ (five years ago Shikha’s age) Ã— (8 years later Shikha’s age) = 30

â‡’ (x – 5)(x + 8) = 30

â‡’ x2 + 3x – 40 = 30

â‡’ x2 + 3x – 70 = 0

â‡’ x2 + (10 – 7)x – 70 = 0 [when constant term is negative always

break the coefficient of x in the difference form]

â‡’ x2 + 10x – 7x – 70 = 0

â‡’ x(x + 10) – 7(x + 10) = 0

â‡’ (x + 10)(x – 7) = 0                             [after factorization]

â‡’ Either x + 10 = 0             or          x – 7 = 0

x = -10            or          x = 7

â‡’ x = -10 is not valid as age can never be negative

So taking x = 7

Means present age of Shikha is 7 years.

### Question 4: 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.

Solution:

Let the present age of Ramu = x years

So,

Five years ago Ramu’s age was = (x-5) years

And Nine years later Ramu’s age will be = (x+9) years

Now according to question –

(Five years ago Ramu’s age)Ã—(Nine years later Ramu’s age) = 15

â‡’ (x-5)(x+9) = 15

â‡’ x2 + 4x – 45 = 15

â‡’ x2 + 4x – 60 = 0

Now performing factorizationâ€”

â‡’ x2 + (10 – 6)x – 60 = 0

â‡’ x2 + 10x – 6x – 60 = 0

â‡’ x(x + 10) – 6(x + 10) = 0

â‡’ (x + 10)(x – 6) = 0

â‡’ Either x + 10 = 0             or          x – 6 = 0

x = -10                  or          x = 6

So x = -10 is not valid as age can never be negative,

â‡’ Taking x = 6

Means present age of Ramu is 6 years.

### Question 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.

Solution:

Let A and B are two friends.

And age of A is = x years

Now According to the First condition –

(Age of A) + (Age of B) = 20

â‡’ x + (Age of B) = 20

â‡’ Age of B = (20 – x) years

So four years ago,

A’s age was = (x – 4) years

and B’s age was = (20 – x – 4)= (16 – x) years

Now coming to the Second condition –

(A’s age) Ã— (B’s age) = 48

â‡’ (x-4)(16-x) = 48

â‡’ 16x – x2 – 64 + 4x = 48

â‡’ -x2 + 20x – 112 = 0

â‡’ x2 – 20x + 112 = 0      ———————————–(1)

Now checking for Discriminant (D) = b2 – 4ac

Comparing  equation(1) with ax2 + bx + c = 0â€”

â‡’ a = 1, b = -20, c = 112

â‡’ D = (-20)2 – 4 * 1 * 112

â‡’ D = 400 – 448

â‡’ D = -48

â‡’ D < 0

So roots are imaginary means.

Therefore, above given situations are not possible.

### Question 6: 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.

Solution:

Let present age of girl is = 2x years

So present age of her sister = x years         [because girl is twice as old as her sister

so sister’s age will be half that of girl’s age]

Now four years hence-

Girl’s age will be = (2x + 4) years

and sister’s age will be = (x + 4) years

Now according to given condition â€”

â‡’ (Girl’s age) Ã— (sister’s) = 160

â‡’ (2x+4)(x+4) = 160

â‡’ 2x2 + 8x + 4x + 16 = 160

â‡’ 2x2 + 12x + 16 -160 = 0

â‡’ 2x2 + 12x – 144 = 0

â‡’ x2 + 6x – 72 = 0

â‡’ x2 + 12x – 6x – 72 =0

â‡’ x(x + 12) – 6(x + 12) = 0

â‡’ (x – 6)(x + 12) = 0

â‡’ Either x – 6 = 0      or      x + 12 = 0

x = 6       or          x = -12

But x =-12 is not possible as age can never be negative

So taking x = 6

â‡’ Present age of Girl = 2x = 12 years

and present age of her sister = x = 6 years.

### Question 7: 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.

Solution:

Let Rehman’s present age = x years

So three years ago Rehman’s age was = (x-3) years

and five years after Rehman’s age will be = (x+5) years

Now according to given condition-

â‡’ [1/(three years ago Rehman’s age)] + [1/(five years after Raehman’s age)] = 1/3

â‡’ [1/(x – 3)] + [1/(x + 5)] = 1/3

â‡’ (x + 5) + (x – 3) = (x + 5)(x – 3)/3 [after taking LCM and transferring denominator of LHS to the RHS]

â‡’ 3(2x + 2) = (x2 + 2x – 15)               [multiplying by 3]

â‡’ 6x + 6 = x2 + 2x – 15

â‡’ x2 – 4x – 21 = 0

â‡’ x2 – (7 – 3)x  – 21 = 0        [by factorization law]

â‡’ x2 – 7x + 3x – 21 = 0

â‡’ x(x – 7) + 3(x – 7) = 0

â‡’ (x – 7)(x + 3) = 0

â‡’ Either x – 7 = 0         or          x + 3 = 0

x = 7       or          x = -3

But x = -3 is not possible as age can never be negative.

So taking x = 7 â€”

Means present age of Rehman = 7 years.

### Question 8: If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?

Solution:

Let Zeba’s present age (actual age) is = x years

So her age if she were 5 years younger = (x – 5) years

Now according to condition-

â‡’ (x – 5)2 = 5x + 11

â‡’ x2 – 10x + 25 = 5x + 11

â‡’ x2 – 15x + 14 = 0

â‡’ x2 – (14 + 1)x + 14 = 0

â‡’ x2 -14x – x + 14 = 0

â‡’ x(x – 14) – (x – 14) = 0

â‡’ (x – 14)(x – 1) = 0

â‡’ Either x – 1 = 0           or          x – 14 = 0

But if x – 1 = 0 â‡’ x = 1, so in this case situation when she was 5 years younger is not possible.

Therefore, taking x – 14 = 0

â‡’ x = 14

means Zeba’s present age is = 14 years

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