The average of $10$ numbers is $40.2$. Later it is found that two numbers have been wrongly added. The first is $18$ greater than the actual number and the second number added is $13$ instead of $33$. Find the correct average.

### Concept & Strategy To correct an average when items are incorrectly recorded, we determine the net deviation (the total difference between the correct values and the incorrect values) and divide it by the total number of items. $$\text{Change in Average} = \frac{\text{Net Deviation}}{\text{Total Numbers}}$$ ### Step-by-Step Solution * **Given:** * Total numbers ($n$) = $10$ * Initial Average = $40.2$ * Error 1: The wrong number is $18$ greater than the actual number. * Error 2: The wrong number added was $13$, but the actual number was $33$. * **Calculation:** 1. Analyze the first error: The recorded number is $18$ too high. Therefore, to correct it, we must subtract $18$. Deviation = $-18$. 2. Analyze the second error: The recorded number was $13$, but it should be $33$. The recorded sum is too low. To correct it, we must add $20$ (since $33 - 13 = 20$). Deviation = $+20$. 3. Calculate the net deviation: $-18 + 20 = +2$. 4. Calculate the change in the average: $\frac{+2}{10} = +0.2$. 5. Calculate the new average: $40.2 + 0.2 = 40.4$. ### Exam Strategy & Shortcut **Mental Math:** Interpret the wording carefully. "First is 18 greater" means a $-18$ correction. "$13$ instead of $33$" means a $+20$ correction. Net correction is $+2$. Divided by $10$ numbers is $+0.2$. Original average $40.2 + 0.2 = 40.4$. Solved in 5 seconds without writing anything down. ### Common Pitfall Misinterpreting the phrasing "The first is $18$ greater than the actual number". Students sometimes add $18$ instead of subtracting it. Always map out whether the *correction* requires an addition or subtraction to the total. ### Final Answer **Therefore, the correct answer is 40.4.**