Aptitude - Average - Discussion

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.

Let us consider p be the no. of pupil, x be the average.

Case1: Original marks 63/p=x -----> average of case 1.
Case2: New marks 83/p=x+1/2 ----> average of case 2.

2-1 => (83/p-63/p) = x+1/2-x.
=> 20/p = 1/2,
=> p = 40.

Let us consider p be the no. of pupil, x be the average.

Case1: Original marks 63/p=x -----> average of case 1.
Case2: New marks 83/p=x+1/2 ----> average of case 2.

2-1 => (83/p-63/p) = x+1/2-x.
=> 20/p = 1/2,
=> p = 40.

Here the simple solution;

A pupil's marks -> one student marks entered wrong.
Here 's1' means student mark.

Let (S1+S2+...63) = S total marks, 63 is the one student marks. So we don't know about remain students marks.thats why I take it as S1,S2.
Like this;
Marks entered wrong here that's 83.

(S1+S2+....83) = (S1+S2+....63) + 20 we can write like this isn't it.
(S1+S2+....83) = S +20 we can also write this.

Avg = total marks/total students.

Avg = S/X -------> (1) when correct marks(63).
Avg + 1/2 = S+20/X ----> (2) when 83 marks.

Substitute (1) in (2).

S/X + 1/2 = (S+20)/X.
( 2(S) +X ) / 2X = ( S+20 )/X

Now cross multiple.
X ( 2(S) + X ) = 2X ( S + 20).
Now cancel one 'x' from both sides.

2(S) + X = 2 ( S + 20 ) => 2S + X = 2S + 40.

Now, 2S will cancel.
Then, X = 40.

Hope it helps.

Let's take the actual avg marks = X and the number of pupil as Y.

Given, 63 is the actual mark obtained by Y pupil.
Which gives us, X = 63/Y (This should have been the actual avg)
So, XY = 63 ---------> (1).

But instead, due to the error, it was replaced to 83 which increased the actual avg by half
So, the new eqn for the error should be,

X + 1/2 = 83/ Y (Since the number of pupil remains the same)
Solve: (2X+1)/2 = 83/Y.
2X+1 = 166/Y.
2XY+Y = 166 -------> (2)
(1) in (2).
2(63)+Y = 166.
Y = 40 (Number of pupil).