Aptitude - Average - Discussion

Discussion Forum : Average - General Questions (Q.No. 15)
15.
A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:
10
20
40
73
Answer: Option
Explanation:

Let there be x pupils in the class.

Total increase in marks = x x 1 = x
2 2

x = (83 - 63)    x = 20      x= 40.
2 2

Discussion:
125 comments Page 13 of 13.

K Chandan Achary said:   5 years ago
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.

S.Rohini said:   5 years ago
@Rakesh.

Thank you.

Akash Ghadage said:   5 years ago
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.

Sarika Jain said:   12 months ago
Thank you all.

Usman Khan said:   12 months ago
Good session, thanks.


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