Interview Q&A / Aptitude

1. How do you solve ratio and age problems?

   Write the present ages as multiples of one unknown using the given ratio, such as 4x and 5x, then add or subtract the same number of years to both and set up an equation from the second condition. Both people age by the same amount, so the difference between their ages never changes.

   BeginnerIn-depth4 min read
2. A father is three times as old as his son. In 15 years he will be twice as old as his son. What are their present ages?

   Son 15, father 45. Let the son be x and the father 3x. In 15 years: 3x + 15 = 2(x + 15), so x = 15.

   BeginnerQuick answer1 min read
3. Rohan's age 6 years from now will be 5/3 of his age 6 years ago. What is his present age?

   24 years. If his age is x, then x + 6 = (5/3)(x - 6). So 3x + 18 = 5x - 30, giving x = 24. Check: 30 is 5/3 of 18.

   BeginnerQuick answer1 min read
4. The sum of the ages of 5 children born at intervals of 3 years each is 50 years. What is the age of the youngest child?

   4 years. If the youngest is x, the others are x + 3, x + 6, x + 9 and x + 12. So 5x + 30 = 50 and x = 4. The ages are 4, 7, 10, 13 and 16.

   BeginnerQuick answer1 min read
5. Ten years ago, A was half as old as B. If the ratio of their present ages is 3 : 4, what is the sum of their present ages?

   35 years. Present ages 3x and 4x. Ten years ago: 3x - 10 = (4x - 10)/2, so 6x - 20 = 4x - 10 and x = 5. Ages are 15 and 20.

   IntermediateQuick answer1 min read