The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is

### Concept & Strategy This is a classic weighted average problem. To find the mean of the remaining group, subtract the total values of the known subgroups (the top and bottom percentiles) from the overall total value, then divide by the remaining percentage of students. ### Step-by-Step Solution * **Assume a base:** Let the total number of students be $100$. Overall mean score = $52$. Total score of all students = $100 \times 52 = 5200$. * **Subgroup 1 (Brightest 20%):** Number of students = $20$. Mean score = $80$. Total score of this group = $20 \times 80 = 1600$. * **Subgroup 2 (Dullest 25%):** Number of students = $25$. Mean score = $31$. Total score of this group = $25 \times 31 = 775$. * **Subgroup 3 (Remaining 55%):** Remaining students = $100 - (20 + 25) = 55$. Remaining total score = $5200 - (1600 + 775) = 5200 - 2375 = 2825$. * **Calculation:** Mean of the remaining students = $2825 / 55 \approx 51.36$. This is approximately $51.4$. ### Exam Strategy & Shortcut Use the concept of Net Deviation from the average. Overall average is $52$. Top $20\%$ deviation = $+28$ per student. Total deviation = $20 \times 28 = +560$. Bottom $25\%$ deviation = $-21$ per student. Total deviation = $25 \times (-21) = -525$. Net deviation of these two groups combined = $+560 - 525 = +35$. For the overall average to remain $52$, the remaining $55$ students must balance this $+35$ deviation with an exact $-35$ deviation. Deviation per student for the remaining $55$ = $-35 / 55 = -0.636$. Therefore, their average is $52 - 0.636 = 51.364$, which safely rounds to $51.4$. ### Common Pitfall A common mistake is simply averaging the averages (i.e., taking the straight mean of $80$ and $31$) without accounting for the respective weights (percentages) of the groups. Always use the weighted sum to preserve accuracy. ### Final Answer **Therefore, the correct answer is 51.4 approx.**