Mathematics
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.
Quadratic Equations
Answer
Let present age of son be x years
Two years before age of,
Son = (x - 2) years
Man = 3(x - 2)2 years
Man's present age = 3(x - 2)2 + 2
In three years,
Son = x + 3 years
Man = 4(x + 3) years
We can write,
⇒ 3(x - 2)2 + 5 = 4(x + 3)
⇒ 3(x2 + 4 - 4x) + 5 = 4x + 12
⇒ 3x2 - 12x + 12 + 5 = 4x + 12
⇒ 3x2 - 12x - 4x + 17 - 12 = 0
⇒ 3x2 - 16x + 5 = 0
⇒ 3x2 - 15x - x + 5 = 0
⇒ 3x(x - 5) - 1(x - 5) = 0
⇒ (3x - 1)(x - 5) = 0
⇒ 3x - 1 = 0 or x - 5 = 0
⇒ x = or x = 5.
x ≠ as in this case (x - 2) will be negative and age cannot be negative.
Man's present age = 3(x - 2)2 + 2 = 3(5 - 2)2 + 2 = 3(3)2 + 2 = 29 years.
Hence, son's present age = 5 years and man's present age = 29 years.
Answered By
Related Questions
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
In an auditorium, seats were arranged in rows and columns. The number of rows was equal to number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator , the fraction is decreased by . Find the fraction.
A two digit number contains the bigger at ten's place. The product of the digits is 27 and the difference between two digits is 6 . Find the number.