What is the age of A if A is elder to B by 2 years, and F is twice as old as A, while B is twice as old as his sister S, given that the age difference between Father F and Sister S is 40 years?

Let the age of A be a years.

According to the question:

  • A is elder to B by 2 years, so B’s age is b = a – 2.
  • Father F is twice as old as A, thus f = 2a.
  • B is twice as old as his sister S, leading to s = b / 2 or s = (a – 2) / 2.

We also know the age difference between Father F and Sister S is 40 years:

f – s = 40

Substituting the values of F and S into the equation:

2a – (a – 2) / 2 = 40

Now, multiply through by 2 to eliminate the fraction:

4a – (a – 2) = 80

Distributing gives:

4a – a + 2 = 80

This simplifies to:

3a + 2 = 80

Subtracting 2 from both sides yields:

3a = 78

Dividing both sides by 3 gives:

a = 26

Thus, the age of A is 26 years.

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