Aptitude - Average - Discussion

Here, the Question says A Pupil's which refers to only one student.

While pupils stand for multiple students.


Am I right?

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.