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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-
Answer & Solution
Correct Answer:
Option
C
Let the required mean score be x
Then,
$$\eqalign{ & 20 \times 80 + 25 \times 31 + 55 \times x = 52 \times 100 \cr & \Leftrightarrow 1600 + 775 + 55x = 5200 \cr & \Leftrightarrow 55x = 2825 \cr & \Leftrightarrow x = \frac{{2825}}{{55}} \approx 51.4 \cr} $$
Then,
$$\eqalign{ & 20 \times 80 + 25 \times 31 + 55 \times x = 52 \times 100 \cr & \Leftrightarrow 1600 + 775 + 55x = 5200 \cr & \Leftrightarrow 55x = 2825 \cr & \Leftrightarrow x = \frac{{2825}}{{55}} \approx 51.4 \cr} $$
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