What are the present ages of a father and his son if two years ago the father was three times as old as his son, and two years hence twice his age will be equal to five times that of his son?

To find the present ages of the father and son, let’s set up the problem step by step.

Let the present age of the son be S years. Then, the present age of the father can be represented as F years.

From the information given:

1. Two years ago, the father’s age was three times the son’s age:
2. (F – 2) = 3 * (S – 2)

Expanding this gives us:

F – 2 = 3S – 6

Thus, we can rearrange it to:

F = 3S – 4         (1)

1. Two years hence, twice the father’s age will equal five times the son’s age:
2. 2 * (F + 2) = 5 * (S + 2)

Expanding this we get:

2F + 4 = 5S + 10

We can rearrange this to:

2F = 5S + 6         (2)

Now, we have a system of equations:

1. F = 3S – 4 (1)
2. 2F = 5S + 6 (2)

Substituting equation (1) into equation (2):

2(3S – 4) = 5S + 6

Which simplifies to:

6S – 8 = 5S + 6

By rearranging, we find:

6S – 5S = 6 + 8

S = 14

Now substituting S back into equation (1) to find F:

F = 3(14) – 4 = 42 – 4 = 38

Thus, the present ages are:

• Father’s age: 38 years
• Son’s age: 14 years