Given:
A is 9 years older than B.
In 10 years, A will be twice as old as B was 10 years ago.
Solution:
Let the present age of B be x. Then A = x + 9.
In 10 years, A will be (x + 9 + 10) = x + 19 years.
B was x - 10 years old, 10 years ago.
Given, A's age in 10 years = 2 * (B's age 10 years ago).
=> x + 19 = 2 * (x - 10).
=> x + 19 = 2x - 20.
=> x = 39.
Thus, A's present age = x + 9 = 48 years.
English
100 Questions
100 Marks
90 Mins English English100 Questions100 Marks90 Mins English English10 Questions10 Marks5 Mins English
5 years ago, the age of a father was 14 years more than twice his son's age. After how many years from now will he be twice his son's age?
The ratio of the present ages of A to B is 16:9, respectively. If the age of A eight years hence is twice the age of B two years hence, then find the present age of B (in years).
If 3 times the mother's present age is 36 years more than 5 times her daughter's age, and 3 times the daughter's age is 4 years less than the mother's age, then what is the difference (in years) between the present ages of the mother and the daughter?
