Advertisements
Advertisements
प्रश्न
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Advertisements
उत्तर
Let the present age of the son be x years.
∴ Present age of man = x2 years
One year ago,
Son’s age = (x – 1) years
Man’s age = (x2 – 1) years
It is given that one year ago; a man was 8 times as old as his son.
∴ (x2 – 1) = 8(x – 1)
x2 – 8x – 1 + 8 = 0
x2 – 8x + 7 = 0
(x – 7)(x – 1) = 0
x = 7, 1
If x = 1, then x2 = 1, which is not possible as father’s age cannot be equal to son’s age.
So, x = 7.
Present age of son = x years = 7 years
Present age of man = x2 years = 49 years.
संबंधित प्रश्न
The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.
The sum S of n successive odd numbers starting from 3 is given by the relation: S = n(n + 2). Determine n, if the sum is 168.
The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Rs. 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Rs. 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs. 30 less. Find the original number of persons.
In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the 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:
- the number of rows in the original arrangement.
- the number of seats in the auditorium after re-arrangement.
Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, in how many days will Mohan alone complete the work?
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.
The ages of two sisters are 11 years and 14 years. In how many years’ time will the product of their ages be 304?
Two years ago the ages of Radha and Meena were in the ratio 5 : 2. If the sum of their present ages is 39 years, their ages, 2 years ago, were ______.
