Aptitude - Average - Discussion

Good session, thanks.

Thank you all.

Thank you for explaining the answer.

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.

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.

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.

Good Explanation, thanks all.

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.

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.

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

While pupils stand for multiple students.


Am I right?