Aptitude - Average - Discussion

s= other student.
n = number of student.

avg (when its 83) = avg(when its 63)+1/2.
(s+83)/n = (s+63)/n + 1/2,
(s+83)/n = [(s+63) + 0.5n]/n,
s+83 = s+63+0.5n,
83 = 63+0.5n,
0.5n = 83-63,
n = 20/0.5.
n=20.

Let x be the cumulative marks excepting the wrongly entered mark.
Let y be the no of pupils,
(x+83)/y = ((x+63)/y) + 0.5.
x+83 = x+63+0.5y,
20 = 0.5y,
y = 40.

I am not understanding this, please explain it.

1/2 means 0.5.

Because of difference of 20 this avg 0.5 is increased.
So distributing 0.5 equally to get total 20.

We divide 20/0.5 will give no.of student.
i.e. 20/0.5 = 40 student.

Avg = total marks/No. Of pupil.
1/2 = 20/x,
X = 40.

Average = total marks/number of pupils.

Average=x+63/number of pupils -------------> (1)
1.5 average=x+83/number of pupils ---------> (2)
eqn(2)-eqn(1).
0.5 = 20/number of pupils.
Number of pupils =20/0.5,
The number of pupils = 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.

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.

@Rakesh.

Thank you.

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.