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The ages of two friends X and Y differ by 2 years. X’s father A is twice as old as X and Y is twice as old as her sister B. If the ages of A and B differ by 40 years, then what is the age of X (in years)?

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26 or 2713

25 or 2513

26 or 2513

25 or 2713

answer is A.

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Detailed Solution

It is given that the ages of two friends X and Y differ by 2 years such that X’s father A is twice as old as X and Y is twice as old as her sister B.
Let’s consider the ages of X and Y be

x

and

y

years.
According to the question,

x − y = ± 2 ⇒ x − y = 2 (1) & x − y = − 2 (2)

Now, the age of A will be

2 x

years, and the age of B will be

y 2

years.
Clearly, B is younger than A.
Therefore according to the question,

2 x − y 2 = 40 4 x − y = 80 (3)

Subtract equation (1) from equation (3), and we get,

3 x = 78 ⇒ x = 26

Again, subtract equation (2) from equation (3), and we get,

3 x = − 82 ⇒ x = 2713

Therefore the age of X (in years) is either

26 or 2713 .

Hence option (1) is correct.

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