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:
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:
123 comments Page 10 of 13.
Sanchit said:
6 years ago
Take sum as S and number if people as N.
Avg + (1/2) = (s + 83)/n----> (1)
Now, Avg = (s+63)/n ----> (2)
Just substitute in place of avg
[(s + 63)/n] + 1/2 = [(s + 83)/n].
1/2 = [(s + 83)/n] - [(s + 63)/n].
1/2 = (83-63)/n,
n. = 2 x 20.
n = 40.
Avg + (1/2) = (s + 83)/n----> (1)
Now, Avg = (s+63)/n ----> (2)
Just substitute in place of avg
[(s + 63)/n] + 1/2 = [(s + 83)/n].
1/2 = [(s + 83)/n] - [(s + 63)/n].
1/2 = (83-63)/n,
n. = 2 x 20.
n = 40.
Prashanth.A said:
6 years ago
Let a Total number of students = x.
Form an equation of averages from given statement,
(Total marks/x)+((83-63)/x)=(total marks/x)+ (1/2).
Solve for x.
i.e x = 40.
Form an equation of averages from given statement,
(Total marks/x)+((83-63)/x)=(total marks/x)+ (1/2).
Solve for x.
i.e x = 40.
Karan S.Bisht said:
6 years ago
Let suppose x is the total students.
Average increases due to error is = x.5.
Total error is 83-63 which is 20,
x+20=x.5,
0.5x=20,
X=40 answer.
Average increases due to error is = x.5.
Total error is 83-63 which is 20,
x+20=x.5,
0.5x=20,
X=40 answer.
Rudrendra said:
6 years ago
@Nik.
Well explained, Thanks.
Well explained, Thanks.
Kabir said:
6 years ago
Increased by half is old + 1/2 not (1.5 * old).
Parth said:
6 years ago
s= other student.
n = number of student.
avg (when its 83) = avg(when its 63)+1/2.
(s+83)/n = (s+63)/n + 1/2,
(s+83)/n = [(s+63) + 0.5n]/n,
s+83 = s+63+0.5n,
83 = 63+0.5n,
0.5n = 83-63,
n = 20/0.5.
n=20.
n = number of student.
avg (when its 83) = avg(when its 63)+1/2.
(s+83)/n = (s+63)/n + 1/2,
(s+83)/n = [(s+63) + 0.5n]/n,
s+83 = s+63+0.5n,
83 = 63+0.5n,
0.5n = 83-63,
n = 20/0.5.
n=20.
Saravanan said:
6 years ago
Let x be the cumulative marks excepting the wrongly entered mark.
Let y be the no of pupils,
(x+83)/y = ((x+63)/y) + 0.5.
x+83 = x+63+0.5y,
20 = 0.5y,
y = 40.
Let y be the no of pupils,
(x+83)/y = ((x+63)/y) + 0.5.
x+83 = x+63+0.5y,
20 = 0.5y,
y = 40.
Sneha Agarwal said:
6 years ago
Simply:
x = num of pupils.
y = total average.
Therefore,
Total marks of class=xy
Total marks of class due to wrong entry = xy-63 + 83.
New average of class due to wrong entry = (xy-63 + 83)/x
Also, (according to question) new average = y+0.5 (or y+1/2)
Thus, we get:
(xy-63+83)/x = y + 0.5,
x = 40.
x = num of pupils.
y = total average.
Therefore,
Total marks of class=xy
Total marks of class due to wrong entry = xy-63 + 83.
New average of class due to wrong entry = (xy-63 + 83)/x
Also, (according to question) new average = y+0.5 (or y+1/2)
Thus, we get:
(xy-63+83)/x = y + 0.5,
x = 40.
(1)
Prithika said:
6 years ago
I am not understanding this, please explain it.
Chetan shahi said:
6 years ago
1/2 means 0.5.
Because of difference of 20 this avg 0.5 is increased.
So distributing 0.5 equally to get total 20.
We divide 20/0.5 will give no.of student.
i.e. 20/0.5 = 40 student.
Because of difference of 20 this avg 0.5 is increased.
So distributing 0.5 equally to get total 20.
We divide 20/0.5 will give no.of student.
i.e. 20/0.5 = 40 student.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers