The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.

Step 1: Define variables

Let $x$ represent the marks the student needs to obtain in the annual examination. We are given the marks from the first and second terminal examinations.

Step 2: Set up the average marks inequality

To find the average marks, we sum the marks from the three examinations (first terminal, second terminal, and annual) and divide by the total number of examinations, which is 3. The problem states that the average should be at least 60 marks, so we set up an inequality where the average is greater than or equal to 60.

Step 3: Simplify the inequality

First, we sum the marks from the first two examinations: $62 + 48 = 110$. This simplifies the numerator of our inequality.

Step 4: Isolate x

To isolate $x$, we multiply both sides of the inequality by 3. This removes the denominator and prepares the inequality for solving for $x$.

Step 5: Solve for x

Finally, we subtract 110 from both sides of the inequality to find the minimum value of $x$. This gives us the minimum marks the student needs in the annual examination.

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