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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: (approx.)?

Question

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: (approx.)?

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Solution

To solve this problem, we can use the concept of weighted averages.

  1. First, we know that the average score of all students is 52.

  2. We also know that the top 20% of students have an average score of 80 and the bottom 25% of students have an average score of 31.

  3. We need to find the average score of the remaining 55% of students.

  4. We can set up the following equation using the concept of weighted averages:

    (20% * 80) + (25% * 31) + (55% * x) = 100% * 52

  5. Solving for x gives us:

    x = (52 - (0.20 * 80) - (0.25 * 31)) / 0.55

  6. Calculating the above expression gives us:

    x = (52 - 16 - 7.75) / 0.55 = 28.25 / 0.55 = 51.36 (approximately)

So, the mean score of the remaining 55% of students is approximately 51.36.

This problem has been solved

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