In 2002, Meenu's age was one-third of the age of Meera, whereas in 2010, Meenu's age was half the age of Meera. What is Meenu's year of birth?

Correct Option (B)

Let Meera's age in 2002 be $x$ years.

According to the problem statement, Meenu's age in 2002 was one-third of Meera's age.

• Meenu's age in 2002 = $\frac{1}{3}\times x=\frac{x}{3}$ years.

Eight years later, in 2010, their ages would be:

• Meera's age in 2010 = $(x+8)$ years.
• Meenu's age in 2010 = $(\frac{x}{3}+8)$ years.

In 2010, Meenu's age was half the age of Meera. This can be expressed as:

$$\frac{x}{3}+8=\frac{1}{2}(x+8)$$

To solve for $x$:

$$\frac{x}{3}+8=\frac{x}{2}+4$$

Rearranging the terms to isolate $x$:

$$8-4=\frac{x}{2}-\frac{x}{3}$$

$$4=\frac{3x-2x}{6}$$

$$4=\frac{x}{6}$$

$$x=24$$

Therefore, Meera's age in 2002 was 24 years.

Meenu's age in 2002 was $\frac{1}{3}\times 24=8$ years.

To find Meenu's year of birth, subtract her age in 2002 from the year 2002:

Meenu's year of birth = 2002 - 8 = 1994.

Incorrect Options:

The calculated year of birth for Meenu is 1994. Options A (1992), C (1996), and D (1998) are inconsistent with the derived ages based on the given conditions.