Aptitude - Average - Discussion

The Correct average increased by half.

Let suppose average marks A.
Number of students be X.

If the marks entered correctly i.e. 63 then.

A = (Total_1) / X -----------> (1).

Total_1= A * X -------------> (2).

If the marks entered is wrong, ie 83 then,

A+ (1/2) = (Total_2) /X -----------> (3).
Total_2 = [A + (1/2)] * X-------------> (4).
Total_2 = A * X + X/2 -----------> (5).

And the difference between Total_1 and total_2 is (83 - 63 = 20).
Therefore Eq (5) - (2).

A * X + X/2 - A * X = 20.
X/2 = 20.
X = 40.

Once read the question, it's stated that. Instead of 83, it's wrongly entered as 63.

Now. As we know sum/num=avg ------->1 (when mark entered is correct).
83 - 63 = 20. This is the new difference when the mark is entered wrongly.

So, sum+20/num=avg+avg (1/2) ------->2 (when mark entered is the wrong Avg is increased by 1/2).

Sum + 20/num = avg (1 + 1/2).
Sum + 20/num = avg (3/2).

Substituting 1 in 2;

Sum + 20/num = sum/num (3/2).
Sum + 20 = 3/2sum.
Sum = 40.

I too have that same doubt @Maria Ninan.

Can anyone clarify it?

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

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.

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

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.

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.

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.