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?

The correct option is 1994.

Let us solve this problem step-by-step by setting up linear equations based on the relationships given for the ages of Meenu and Meera in the years 2002 and 2010.

Step 1: Define the variables for their ages
Let Meenu's age in the year 2002 be $x$ years.
Let Meera's age in the year 2002 be $y$ years.

Step 2: Translate the information for the year 2002
In 2002, Meenu's age was one-third of the age of Meera. We can express this mathematically as:
$x=\frac{1}{3}y$
Multiplying both sides by 3 gives:
$y=3x$ (Equation 1)

Step 3: Translate the information for the year 2010
The year 2010 is exactly 8 years after the year 2002 (since 2010 - 2002 = 8). Therefore, in 2010:
Meenu's age will be $x+8$ years.
Meera's age will be $y+8$ years.

According to the problem, in 2010, Meenu's age was half the age of Meera. This gives us:
$x+8=\frac{1}{2}(y+8)$
Multiplying both sides by 2 to clear the fraction:
$2(x+8)=y+8$
Expanding the left side:
$2x+16=y+8$ (Equation 2)

Step 4: Solve the system of equations
Substitute the value of $y$ from Equation 1 ($y=3x$) into Equation 2:
$2x+16=3x+8$
Subtract $2x$ from both sides:
$16=x+8$
Subtract 8 from both sides:
$x=8$
This means Meenu's age in the year 2002 was 8 years.

Step 5: Determine Meenu's year of birth
Since Meenu was 8 years old in the year 2002, her year of birth is:
$2002-8=1994$