Aptitude - Average - Discussion

Let pupils be x.

Given avg is increased by 1/2 after changing 83 to 63.
So, avg2 - avg1 = 1/2,
(sum2/x) - (sum1/x) =1/2,
(sum2 - sum1)/x = 1/2,
(83-63)/x=1/2,
20/x=1/2.
x = 40.

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

My calculation:
1) The mistake is: 83-63 = 20.
2) Sum total after mistake = (sum + 20).
3) Avg after mistake = (avg + 0.5).

Therefore:
avg = sum/N.
(avg + 0.5) = (sum + 20)/N,
(sum/N) + 0.5 = (sum + 20)/N,
N(sum/N + 0.5) = (sum + 20),
sum + 0.5N = (sum + 20),
0.5N = 20 --> N = 40.

Assume number of pupils is x.
avg1 = 63/x --->eq (1)
avg2 = 83/x --->eq (2)
avg2 = avg1 + 0.5 ---> eq (3)

Substituting in eq (2) by eq (3).
(avg1) +0.5 = (83/x),
(63/x) +0.5 = (83/x),
(83/x) -(63/x) = 0.5,
(83-63)/x = 0.5,
20/x = 0.5,
x = 20/0.5 = 40.

Good Explanation, thanks all.

Let considered no students be - n.
and average be - a.
And total marks be - x.

So, if correction was applied then average marks obtained will be;
a= x/n i.e n=x/a ------------> eqn(1).

Now correction is not applied then average will increase by 0.5.
And marks will also increase by 20 (83-63).

Then average will be;
a+0.5=x+20/n i.e n=x+20/a+0.5 --------> eqn(2)
From eqn 1 and 2,
x/a=x+20/a+0.5,

By doing little aligabra
You get x=40a --------------------------> eqn(3)
Substitution eqn 3 in 1.

n = 40a/a
Then, n = 40.

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

While pupils stand for multiple students.


Am I right?

0.5 is a portion of (the sum of student's marks/no. of students). Therefore, if no of students is 40, then the sum of marks is 40, which is not true.

Please correct me, if I am wrong.

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.