Question

The pupil’s marks were wrongly entered as $83$ instead of $63$. Due to that the average marks of class got increased by half $\left( {\dfrac{1}{2}} \right)$. Find the number of pupils in the class.

Answer 1

First we will assume a variable for number of pupils and total marks of the pupil. Using the average formula, we have to form an equation in the given data. Then add the correct mark and subtract the wrong mark from the total marks to form the equation for the new average mark. Solving the equation, we will get the required answer.

Formula used: ${\text{Average mark = }}\dfrac{{{\text{Total marks scored by all pupils}}}}{{{\text{No}}{\text{. of pupils}}}}$

Complete step-by-step solution:
So here, a pupil’s mark was entered wrongly as $83$ instead of $63$. This mistake made the average increase by half $\left( {\dfrac{1}{2}} \right)$.
First, let us assume ${\text{x}}$ for number of pupils
That particulars pupil’s mark which is mistaken as $83$ is $63$
So let us assume marks of all pupils except for this particular pupil whose mark is mistaken be ${\text{y}}$
$\therefore$ Total marks (mistake) = ${\text{y + 83}}$ and total marks (correct) = ${\text{y + 63}}$
According to the question the mistaken average is half $\left( {\dfrac{1}{2}} \right)$ more than the correct average.
Using average formula we can write it as,
$\Rightarrow \dfrac{{{\text{y + 83}}}}{{\text{x}}}{\text{ = }}\dfrac{{{\text{y + 63}}}}{{\text{x}}}{\text{ + 0}}{\text{.5}}$
Taking LCM on RHSS we get,
$\Rightarrow \dfrac{{{\text{y + 83}}}}{{\text{x}}}{\text{ = }}\dfrac{{{\text{y + 63 + 0}}{\text{.5x}}}}{{\text{x}}}{\text{ }}$
On cancel the denominator term on both sides we get,
$\Rightarrow$ ${\text{y + 83 = y + 63 + 0}}{\text{.5x}}$
Transferring everything except ${\text{0}}{\text{.5x}}$ to the other side, we get
$\Rightarrow$ ${\text{y + 83 - y - 63 = 0}}{\text{.5x}}$
On subtracting we get
$\Rightarrow$ $20 = 0.5{\text{x}}$
Let us divide $0.5$ on both sides we get
$\Rightarrow$ ${\text{x = }}\dfrac{{20}}{{0.5}}$
On dividing we get
$\Rightarrow$ ${\text{x = 40}}$

Therefore the number of pupils in the class is $40$

Note: Here we have alternative method as follows:
First, let us assume ${\text{x}}$ for the number of pupils and let ${\text{a}}$ be the average mark obtained.
Then, Total marks = ${\text{xa}}$
According to the question mistaken average $\left( {{{\text{a}}_1}} \right)$ is half $\left( {\dfrac{1}{2}} \right)$ more than the correct average $\left( {{{\text{a}}_2}} \right)$.
This means, ${{\text{a}}_1}{\text{ = }}{{\text{a}}_2}{\text{ + 0}}{\text{.5}}$
Also, ${{\text{a}}_2} = {\text{ }}\dfrac{{{\text{xa - 83 + 63}}}}{2}$
Since, Total marks =${\text{xa}}$
Now, ${{\text{a}}_1}{\text{ = }}{{\text{a}}_2}{\text{ + 0}}{\text{.5}}$
Putting the values and we get
$\Rightarrow \dfrac{{{\text{xa}}}}{{\text{x}}}{\text{ = }}\dfrac{{{\text{xa - 83 + 63}}}}{{\text{x}}}{\text{ + 0}}{\text{.5}}$
Taking LCM on RHS we get,
$\Rightarrow \dfrac{{{\text{xa }}}}{{\text{x}}}{\text{ = }}\dfrac{{{\text{xa - 83 + 63 + 0}}{\text{.5x}}}}{{\text{x}}}$
On cancel the denominator terms we get,
$\Rightarrow$ ${\text{xa = xa - 20 + 0}}{\text{.5x}}$
Transferring everything except ${\text{0}}{\text{.5x}}$ to the other side, we get
$\Rightarrow$ ${\text{xa - xa + 20 = 0}}{\text{.5x}}$
On cancel the same term with different sign,
$\Rightarrow$ $20 = 0.5{\text{x}}$
Let us divide $0.5$ on both sides we get
$\Rightarrow$ ${\text{x = }}\dfrac{{20}}{{0.5}}$
On dividing we get
$\Rightarrow$ ${\text{x = 40}}$
Hence we get the required answer.