Aptitude - Average - Discussion

Let, average marks of class = x & also let, total no. of pupil in the class = A.

When the mark is 63. Then, the total marks of all pupil = Ax.

When, 83 is entered instead of 63,

Then avg marks of the class = A(x+1/2).

Then, A(x+1/2)-Ax = 83-63.

=> A/2 = 20.

=> A = 40. Option(C).

Correct mark = 63.
Wrongly entered = 83.

The average marks for the class got increased = 1/2 => 0.5.

A = For single pupil extra mark = 1*1/2 = 0.5.

B = Extra added mark = 83 - 63 => 20.

The number of pupils in the class "X"= B/A.

Therefore x = 20/0.5; 0.5 =1/2.

x = 20/(1/2).
x = 20* 2.
x = 40.

The number of pupils in the class is 40.

Very simple.

Let no. of. pupil be x.

New mark (wrongly entered) = 83.
Replaced mark (correct mark) = 63.

Increased avg = 0.5(1/2).

Formula: New mark=Replaced mark+(Number of pupil*inc avg).

We can apply this formula for this type of question.

Now,

83 = 63+(x*0.5).
= 83-63 = x*0.5.

= 20 = x*0.5.
= 20/(1/2) = x.
= 20*2 = x.
= 40 = x.

Number of pupils in the class is 40.

Hope it clear your doubts.

Let N be the total number of students.

Total mark increase => 83-63 = 20. (because Old Sum-New Sum of marks will be affected only due to change of 83 to 63) __ (1).

Now, average marks are increased by 1/2.

It means each student's mark is increased by 1/2. (Average is equally applicable to all the students).

Hence, total mark increased => 1/2*N __(2).

From (1) and (2) ,

20 = (1/2)*N.
N = 40.

First we have average formula = Sum of all the marks/No of students.

Then for this question let be assume average as = A.

And no of students = X.

Now A.T.Q = A = 63 +mark1+. /X. equation no. 1 means average formula.

No second he say that when marks increase by 83 then again A = 83+ mark1+. /x.

But he said that average increased by 0.5. So A+0.5 = 83 + marks1+. /x. equation no 2.

Now equate these equations you will find exact answer.

Its simple,

Let us assume that the total number of pupils in class be x+1,

Now Let the sum of the marks of x students (whose marks have been entered correctly be S) , and let original average be A, then (S+63)/(x+1) = A; Equation 1.

And (S+83)/(x+1) = (A+1/2); Equation 2.

Divide Equation 1 by Equation 2;

(S+63)/(S+83) = (2*A)/(2*A+1) ;

Solving it, we get: S+63=40*A; Equation 3.

Now, put value of Equation 3 in Equation 1, we get:

(40*A)/(x+1) = A;

Solve it, we get: x+1=40.

And we have assumed the total students to be x+1. Hence total students = x+1 = 40.

83-63/(1/2) = 2*20 = 40.

Let as take, m is actual total marks of students and p is number of the pupil.

Difference between average = 1/2.
m/p - (m+20)/p = 1/2.
p = 40.

But the question says increased by half, so shouldn't it mean the incorrect average was increased by half of the correct average?

Thanks in advance

I too have that same doubt @Maria Ninan.

Can anyone clarify it?