Aptitude - Average - Discussion

See the difference 83 and 63, which means extra 20 marks they have given. It increased by 1/2 means we shoud divide 20 by 1/2 which is nothing but 40.

Thanks @Ravi.

Let x = total no of students, y=Avg of students when 63 is taken.

Sum of students when 63 is taken=xy.
So, (xy-63+83) /x=y+0.5.
(xy-20) =xy-0.5x.
x = 40.

When 63 is entered the let avg is x.

x = (sum of the marks of rest of students+63)/n.
Let the sum of the marks of rest of the students=y.
Therefore, x=(y+63)/n.

When 83 is entered avg is increased by 1/2 or 0.5.
So, x+0.5= (y+83)/n.
Equating both the terms we get;
0.5=20/n.
Therefore, n=40.

I am not getting it. Please someone explain to me briefly.

Please explain me, I am not getting this.

Thank you @Nik.

Avg = (sum of observations)/(no. of observations).

Given,
Avg = (x+63)/n ------> (1)
Avg + 1/2 = (x+83)/n ------> (2) (Avg increased by half).

Substituting eqn 1 in eqn 2,
(x+63)/n + 1/2 = (x+83)/n.
(x+83)/n - (x+63)/n = 1/2.
n = 40.

Take average except for the person who's marks wrongly entered = x.
LET, total no of students = N.
AND, initial average =A.

So, CORRECT AVERAGE should be ----> (x+63)/N = A { EQ. 1}
and, WRONG AVERAGE calculated is----> (x+83)/N = A+(1/2) {EQ. 2} --->(i.e increase of the average marks by 1/2).


NOW, put the value of A from EQ.1 in EQ.2 and solve for N.
i.e, (x+83)/N=(x+63)/N + 1/2,
(x+83-x-63)/N=1/2,
20/N=1/2,
N = 40.

Let, pupil=x and real avrge=63.

xy+20=(y+1/2)x.