Book: RS Aggarwal - Mathematics

Subject: Maths - Class 10th

Q. No. 42 of Exercise 10E

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42
Two years ago, a man's age was three times the square of his son's age. In three years’ time, his age will be four times his son's age. Find their present ages.

Let son’s age 2 years ago be x years, Then

man’s age 2 years ago be 3x2 years

son’s present age = (x + 2) years

man’s present age = (3x2 + 2)years

In three years’ time :

son’s age = (x + 2 + 3) = (x + 5) years

man’s age = (3x2 + 2 + 3)years = (3x2 + 5) years

According to question

Man’s age = 4 son’s age

3x2 + 5 = 4(x + 5)

3x2 + 5 = 4x + 20

3x2 – 4x – 15 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 3 b = – 4 c = – 15

= 3. – 15 = – 45

And either of their sum or difference = b

= – 4

Thus the two terms are – 9 and 5

Difference = – 9 + 5 = – 4

Product = – 9.5 = – 45

3x2 – 9x + 5x – 15 = 0

3x(x – 3) + 5(x – 3) = 0

(x – 3) (3x + 5) = 0

(x – 3) = 0 or (3x + 5) = 0

x = 3 or x = – 5/3 (age cannot be negative)

x = 3

son’s present age = (3 + 2) = 5years

man’s present age = (3.32 + 2) = 29years

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