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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-

A. 45 B. 50 C. 51.4 approx D. 54.6 approx Answer: Option C
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Solution(By Apex Team)

Let the required mean score be x Then, $\begin{array}{l} 20 \times 80+25 \times 31+55 \times x=52 \times 100 \\ \Leftrightarrow 1600+775+55 x=5200 \\ \Leftrightarrow 55 x=2825 \\ \Leftrightarrow x=\large\frac{2825}{55} \approx 51.4 \end{array}$

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