Question 68
The average marks of 40 students was foundto be 68. If the marks of two students were incorrectly entered as 48 and 64 instead of 84 and 46 respectively, then what is the correct average?
Solution
According to the question,
Required average = $$\frac{\left(40\times\ 68+130-112\right)}{40}=\frac{2738}{40}=68.45$$
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