###
**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**

## Show Answer

Solution(By Apex Team)

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

## Related Questions On Average

### Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average19 people went to a hotel for combine dinner party 13 by 8 runs. His new average is:

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C. 28

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### Find the average of all the numbers between 6 and 34 which are divisible by 5.

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### If the arithmetic mean of 0, 5, 4, 3 is a, that of -1, 0, 1, 5, 4, 3 is b and that of 5, 4, 3 is c, then the relation between a, b, and c is.

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C. 4a = 5b = c

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