Question

9. The sum of present ages of a man and his son is 45 years. 5 years ago product of their ages was 4 times that of age of man. Find their present ages.

Answer 1

let the present ages of man and son be x and y respectively
x+y = 45

5 yrs ago , age of man = (x-5)
age of son = (y-5)
(x-5)(y-5) = 4(x-5)
(y-5) = [4(x-5)] / (x-5)
y-5 = 4. [(x-5) got cancelled out]
y = 9 yrs


x+y = 45 yrs
x+9 = 45
x = 45-9 yrs => 36 yrs


hope this helps

Answer 2

Solution

Let present age of son is X years and present age of man ( father) is Y years.

Then according to question,

X + Y = 45

$⟹y = (45 - x)..(1)$

5 years ago son's age = ( X - 5) years

And father's age = ( Y - 5) years

$⟹(45 - X - 5) \: years \:$

[from eqn. (1)]

$⟹(40 - x) \: year$

According to question,

$(x - 5)(40 - x) = 4.(40 - x)$

$⟹40x - {x}^{2} - 200 + 5x = 4(40 - x)$

$⟹45x - {x}^{2} - 200 = 160 - 4x$

$⟹ {x}^{2} - 45x - 4x + 200 + 160 = 0$

$⟹ {x}^{2} - 49x + 360 = 0$

$⟹{x}^{2} - 40x - 9x + 360 = 0$

$⟹x(x - 40) - 9(x - 40) = 0$

$(x - 40)(x - 9) = 0$

$⟹x - 40 = 0 \\ or \\ x - 9 = 0$

$∴ \: \: x = 40 \\ or \\ x = 9$

When X = 40, then Y = 45 - X

⟹ 45 - 40 = 5

It is not possible, because it can't be happened that the age of son is 40 and age of father is 5 years old.

Therefore, on taking X = 9

Y = 45 - X

⟹45 - 9 = 36

Thus, the age of man is 36 year and his sons age is 9 years