Aptitude - Average - Discussion

Simple logic.

Let no.of candidate = x.
Their average marks =y.
Generate an equation,
xy+83-63=x(y+1/2).
Solving we get y=40.

The simplest logic is;

If by increasing 20 marks(83-63) in the class the average increases by it's half = x+x/2.
Average of total marks = x
Average of total marks + 20 = x + x/2
Thus 20 = x/2;
=> x = 40.

1st condition : 83/n =av,
Here av is average.
2nd one is: 63/n = 1/2 +av,
So by solving these 2 eq. We have,
1/2 = 83/n - 63/n.
Solving this we will get n=2*20=40.

@Utsav Sinha.

RHS should be;
X (y + 1/2y). Because average is increasing by half that means avg + 1/2 of avg.

@Shubham.

In your case, You calculated Average Marks Not num of the pupil.

@Deekshit.

It's INSUFFICIENT DATA.

63/n = 1/2 +av is incorrect . the correct one should be;
63/n = 1/2av + av : because av (average) is increased by half which means avg was changed to av + av/2 . And Most of the folks have done thia mistake only. Please ignore this question, there is Insufficient data to solve this.

Let we take n be the number of students, x be the average of the students, so nx be the sum of observations,

(nx+20)/n=x+0.5,
nx+20=nx+0.5n,
0.5n=20, n=40.

@Nik.

Thanks for your explanation.

Let, pupil=x and real avrge=63.

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

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.