Bob and John form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years.

How old are Bob and John?

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If we let the age of Bob be B and the age of John be J, we can set up our math.

ReplyDeleteWhen Bob was half as old as the sum of their current ages, he had reached the age of (B + J) / 2 years. This is now B - (B + J) / 2 = B/2 - J/2 years ago. At that moment, John was J - (B/2 - J/2) = 3/2 J - B/2 years old. If Bob is twice as old as John at that moment, then Bob is (3/2 J - B/2) * 2 = 3 J - B years old. That moment is 3 J - 2 B years from now. Then John will be J + 3 J - 2 B = 4 J - 2 B years old. And that is exactly the age of Bob, so B = 4 J - 2 B which gives that 3 B = 4 J.

The second fact is that John is as old as Bob was when John was half as old as he will become over ten years.

When John had half the age as he will have over ten years, he was (J+10)/2 years old. This is now J - (J + 10) / 2 = J/2 - 5 years ago. At that moment Bob was B - (J/2 - 5) years old. According to the second fact, John is now as old as Bob was at that moment, so J = B - (J/2 - 5). It now follows that 3 J / 2 = B + 5.

Summarizing, we end up with two equations.

3 B = 4 J

B + 5 = 3 J / 2

Multiplying the bottom equation with -3 gives after adding the top equation -15 = (4 - 9/2) J

Solving this equation gives J = 30. And now it easily follows that B = 40.

So John is 30 years old and Bob 40.