Mathematics
One year ago, a man was 8 times as old as his son. Now his age is equal to square of his son's age. Find their present ages.
Quadratic Equations
24 Likes
Answer
Let one year ago,
son's age = x years
So, man's age = 8x years
Present age of,
son = x + 1
man = (x + 1)2
We can write,
⇒ (x + 1)2 - 1 = 8x
⇒ x2 + 1 + 2x - 1 = 8x
⇒ x2 + 2x - 8x = 0
⇒ x2 - 6x = 0
⇒ x(x - 6) = 0
⇒ x = 0 or x - 6 = 0
Since, age cannot be zero,
∴ x = 6
∴ x + 1 = 7 and (x + 1)2 = 72 = 49.
Hence, present age of son = 7 years and man = 49 years.
Answered By
14 Likes
Related Questions
The product of the digits of a two digit number is 24. If it's unit's digits exceeds twice it's ten's digit by 2; find the number.
The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304 ?
The age of a father is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.
Mr. Mehra sends his servant to the market to buy oranges worth ₹ 15. The servant having eaten three oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.