A pupil's marks were wrongly entered as $83$ instead of $63$. Due to that the average marks for the class got increased by half. The number of pupils in the class is

### Concept & Formula The core principle of averages is that any change in the total sum is equally distributed among all the entities. $$\text{Total Change in Sum} = (\text{Number of Entities}) \times (\text{Change in Average})$$ ### Step-by-Step Solution * **Given:** * Incorrect marks = $83$ * Correct marks = $63$ * Increase in average = $\frac{1}{2}$ (or $0.5$) * **Calculation:** 1. Calculate the total artificial increase in the sum due to the data entry error. 2. Increase in total marks = $83 - 63 = 20$. 3. This extra sum of $20$ was distributed equally among all pupils, causing the average to rise by $0.5$ per pupil. 4. Let $n$ be the total number of pupils. 5. $n \times 0.5 = 20$ 6. $n = \frac{20}{0.5} = 40$. ### Exam Strategy & Shortcut **Reverse Distribution:** If an extra $20$ marks caused everyone's share (the average) to increase by $0.5$, how many people are there? Simply ask: How many times does $0.5$ fit into $20$? $20 \div 0.5 = 40$. ### Common Pitfall Getting confused because the initial average is not given. Students often think there is missing information. Remember that you don't need the original average to solve problems purely based on *changes* and *deviations*. ### Final Answer **Therefore, the correct answer is 40.**