Question

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.

Answer 1

Step-by-step explanation:

this is a question of linear equations in 2 variable

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The sum of the ages of a man and his son is 45 years.

Let assume that,

Man present age be x years.

So,

Son present age be (45 - x) years.

Further given that,

Five years ago, the product of their ages was four times the man’s age at the time.

Now, 5 years ago,

Man age be x - 5 years

Son age be 45 - x - 5 = 40 - x years.

So,

$\rm \: (x - 5)(40 - x) = 4(x - 5)$

$\rm \: (x - 5)(40 - x) - 4(x - 5) = 0$

$\rm \: (x - 5)[40 - x - 4]= 0$

$\rm \: (x - 5)[36 - x]= 0$

$\rm\implies \:x = 36 \: \: or \: \: x = 5 \: \{rejected \}$

Hence,

Man present age = 36 years.

and

Son present age = 45 - 36 = 9 years.

$\rule{190pt}{2pt}$

Verification :-

Man present age = 36 years

Son present age = 9 years.

So, The sum of the ages of a man and his son is 45 years.

Now, Five years ago,

Man age = 36 - 5 = 31 years.

Son age be = 9 - 5 = 4 years.

So, the product of their ages = 31 × 4 which was four times the man’s age at the time.

Hence, Verified