Aptitude - Average - Discussion

According to the property of Average, if each quantity is increased by certain valu "X" then the new average is increase by X.

Now in the question, it is given that average has been increased by 1/2 due to the marks entered wrongly. Therefore we can say that the marks of each n every student should increase by 1/2 as per the property.

Let x be no of students.
Therefore (x *1/2) = x/2.
Now, increase value = 83 - 63 = 20.
Therefore x/2 = 20.
x = 40.

Avg = sum * total number.
1 if all are correct then the eqn is,
avg = sum/number.
sum = avg * number----> 1st eqn.

2 if wrong 83 instead of 60 then eqn is,

sum + (83 - 63) = (avg + 0.5) * number.
sum = (avg + 0.5) * number - 20----> 2nd eqn.

let divide 1st eqn by 2nd eqn.
sum/sum = 1.

Avg * number = (avg + 0.5)number - 20.
In solving this avg canceled by avg, So, avg - avg = 0.

0.5 - (20/number) = 0
0.5 * number = 20
Number = 20/0.5
Number =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.

Hai friends,

Let average marks = x.

Total number of students = n, then
total mark = x*n.

According to the question the difference in the marks is = 83-63 = 20.

x*n/n=x.......(Can you remember that)

Here the difference in the marks is 20 and the average is increased by half so that x+1/2.

So, ((x*n)+20)/n = x+1/2.

Cross multiply we get,

2xn+40 = 2xn+n.

From that we get total number of students,

n=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.

Increase/Decrease In(Average) =Increase/Decrease In(sum of entity)//Increase/Decrease In(number of entity).

Therefore =>
Increase/Decrease In(number of entity ) =Increase/Decrease In(sum of entity) //Increase/Decrease In(Average).

In this Example =>
Decrease In(sum of entity) =83-63 =20,
Increase In(Average) =1/2 and,
therefore;
Number of pupil's =20/1/2 ==> 20*2 = 40.

Let x be the sum of marks of all the students except the incorrect mark. and y be the no of students.

Now,

Avg mark of all the students with correct mark =(x+63)/y.
Avg mark of all the students with incorrect mark =(x+83)/y/

Acc to the ques;
(x+83)/y= (x+63)/y+1/2,
(x+83)/y= (2x+126+y)/2y. (taking LCM of denominators)
2(x+83)= 2x +126+y,
2x+166 = 2x+126+y,
y= 40.

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.

Let the total marks be x,

Let the total no of students be x, and let the average marks of all the students be x.

Increase in the total marks = 83-63 = 20.

So, total marks/total students = avg marks.

=> x+40/x = x+1/2.

x gets cancelled on both sides of the equation.

Then the remaining terms are,

=> 20/x = 1/2.

=> x = 40.