What are the present ages of A and B if five years ago A was four times the age of B, and five years hence A will be twice the age of B?

Let the present age of A be A years and the present age of B be B years.

According to the problem:

  • Five years ago, A’s age was A – 5 and B’s age was B – 5.
  • At that time, the relationship was: A – 5 = 4(B – 5)

Expanding the equation:

  • A – 5 = 4B – 20
  • This simplifies to: A = 4B – 15 (Equation 1)

Now, let’s consider the second condition:

  • Five years from now, A’s age will be A + 5 and B’s age will be B + 5.
  • The relationship at that time will be: A + 5 = 2(B + 5)

Expanding this equation gives us:

  • A + 5 = 2B + 10
  • Simplifying this results in: A = 2B + 5 (Equation 2)

Now we have a system of two equations:

  • Equation 1: A = 4B – 15
  • Equation 2: A = 2B + 5

Since both equations equal A, we can set them equal to each other:

  • 4B – 15 = 2B + 5

Simplifying this equation:

  • 4B – 2B = 5 + 15
  • This leads to: 2B = 20
  • Thus, B = 10 years

Now, we can substitute the value of B back into one of our equations to find A’s age:

  • Using Equation 2: A = 2(10) + 5 = 20 + 5
  • A = 25 years

Therefore, the present ages are:

  • A is 25 years old.
  • B is 10 years old.

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