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The average marks per student in a class of $30$ students were $45$. On rechecking it was found that marks had been entered wrongly in two cases. After correction these marks were increased by $24$ and $34$ in the two cases. The correct average marks per student are

$(1) 75 ;\ \ (2) 60 ;\ \ (3) 56 ;\ \ (4) 47.$

On the first look this looked very easy but when done I was left guessing I did not understand the question properly.

The average marks per student in a class of $30$ students were $45$ when the counting was flawed. So total marks in that case was $30\times 45=1350$ . Suppose for student $A$ and $B$ marks had been entered wrongly as $'a'$ and $'b'.$ After rechecking the marks were increased by $24$ and $34$. So , now $A$ has $a+24$ and $B$ has $b+34$. So total marks now is $1350+24+34=1408$. Then the correct average is $1408/30 =46.93$ which is not same as any of the options.

So has there been a mistake in printing or my understanding $?$

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    $\begingroup$ They probably just rounded off 46.93 to 47. So, correct answer is (4) 47. $\endgroup$ – SilverSlash Jan 8 '16 at 15:10
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Your understanding looks correct to me.

I'm guessing that either there was a typo (it's supposed to be $26$ and $34$ or $24$ and $36$) or you're expected to round your answer to the nearest integer for some reason.

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I agree with both your understanding and solution of the problem. I would assume that you are expected to round to then nearest integer.

In general, for an average of $A$ of $n$ elements, if x scores are increased by $x_1, x_2, x_3...x_x$, then the new average, $A'$, is $A \cdot n+x_1+x_2+x_3+\cdots x_x$. Note that $x_1$ is the change in score for the first element, $x_2$ is the change in score for the second element, and so on.

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