Aptitude - Average - Discussion

No of pupil = x.
Avg = Total/no. Of pupil.
1/2=(83-63)/x.
0.5=20/x,
X=40.

It's easy as by simply making equations.

Consider a number of students=N.

As we know the marks of one of the students makes the number of left students= N-1.

So let the average where marks for the student is right to suppose its X.

That makes ( (N-1) +63) /N=X ---------> 1.

NOW CONSIDER WHEN MARKS ARE GIVEN WRONG THAT INCREASES THE AVERAGE BY HALF MARKS.

This means X is increased as X + 0.5.

That makes ( (N-1) +83) /N=X+0.5 --------> 2.

Here, there are two equations, two variables, Now solve for N and we can get our answer.

What if the condition is reverse entry is recorded wrong 39 instead of 63. In this case, the average marks of class reduced by 1/2? Can anyone tell me the answer?

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).

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 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.

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.

@Rakesh.

Thank you.

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.