Mathematics
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.
Quadratic Equations
3 Likes
Answer
Let age of son be x years so father's age = (45 - x) years.
Five years ago age of,
son = x - 5
man = 45 - x - 5 = 40 - x
According to question,
⇒ (x - 5)(40 - x) = 124
⇒ 40x - x2 - 200 + 5x = 124
⇒ 45x - x2 - 200 - 124 = 0
⇒ x2 - 45x + 324 = 0
⇒ x2 - 36x - 9x + 324 = 0
⇒ x(x - 36) - 9(x - 36) = 0
⇒ (x - 9)(x - 36) = 0
⇒ x - 9 = 0 or x - 36 = 0
⇒ x = 9 or x = 36.
Father's age = 45 - x
Substituting x = 9 we get,
Father's age = 45 - x = 45 - 9 = 36
Substituting x = 36 we get,
Father's age = 45 - x = 45 - 36 = 9.
This is not possible as father's age cannot be less than son's age.
Hence, age of son = 9 years and that of father = 36 years.
Answered By
2 Likes
Related Questions
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
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.
Two trains leave a railway station at same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
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.