Average

Assume that the sum of all marks except the wrong entered is a, the total number of pupils is n. Then,
$\frac{a+83}{n}=\frac{a+63}{n}+\frac{1}{2}$ since if we entered 83, the average increased by half.
$\frac{a+83}{n} - \frac{a+63}{n} = \frac{1}{2}$
$\frac{a+83-a-63}{n} = \frac{1}{2}$
$\frac{20}{n} = \frac{1}{2}$
$n = \boxed{40}$

$20=0.5n\\ n=\quad \frac { 20 }{ 0.5 } \\ n=\quad 40$ EASIEST WAY

I JUST DO THIS METHOD AND FIND THE CORRECT ANSWER 83-63/0.5

X:average: N:pupils numbers X+0.5=83/N, X=(83-0.5N)/N X=63/N (83-0.5N)/N=63/N, 83-0.5N=63, 0.5N=20, N=40

[(x+20)/N - x/N] = 0.5, where, x = initial total marks, N = No of pupil & solve for N.. {x will cancelled out}

Let y be the total marks of all the students in the class. Let the number of students be x.

• (y+20)/x - (y/x)= 0.5
• Multiply the equation by x and eliminate y.
• 20 = 0.5x
• x=40

We can do it in a simple way:

let total no. of students = n; then,

n (83) - n (63) = n (0.5) Solve to get n = 40.

20/y=0.5 , where y is no of students.. Implies y= 20/.5= 40

There is difference of 83-63 =20 marks to give avg higher by 0.5 therefore number of pupil in the class = 20/0.5 that is 40 Ans K.K.GARG,India

Let the total students be x & average marks be y. Case 1 : 83/x =y + 0.5 Case 2 : 63/x = y Subtracting Case 2 from Case 1, we get value of x i.e. 40