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 7 of 13.
Nelson said:
7 years ago
Let we take n be the number of students, x be the average of the students, so nx be the sum of observations,
(nx+20)/n=x+0.5,
nx+20=nx+0.5n,
0.5n=20, n=40.
(nx+20)/n=x+0.5,
nx+20=nx+0.5n,
0.5n=20, n=40.
Siddhesh said:
8 years ago
Let x is average marks when total marks are 63.
Let n=total no.of pupils
then, 63/n =x --<1>
83/n = x + 0.5.
From 1, 83/n=63/n+0.5,
20/n = 0.5.
n = 40.
Let n=total no.of pupils
then, 63/n =x --<1>
83/n = x + 0.5.
From 1, 83/n=63/n+0.5,
20/n = 0.5.
n = 40.
Prashant ekal said:
8 years ago
Let n = no of pupils.
X = avg marks.
Now,
For 63 marks avg is,
63/n = X ----> (1).
For 83 marks avg is
83/n = X+0.5 ----> (2)
Solve 1 and 2.
N = 40.
X = avg marks.
Now,
For 63 marks avg is,
63/n = X ----> (1).
For 83 marks avg is
83/n = X+0.5 ----> (2)
Solve 1 and 2.
N = 40.
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.
Arushi Mathur said:
7 years ago
Let x = total no of students, y=Avg of students when 63 is taken.
Sum of students when 63 is taken=xy.
So, (xy-63+83) /x=y+0.5.
(xy-20) =xy-0.5x.
x = 40.
Sum of students when 63 is taken=xy.
So, (xy-63+83) /x=y+0.5.
(xy-20) =xy-0.5x.
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.
Ashwini said:
7 years ago
See the difference 83 and 63, which means extra 20 marks they have given. It increased by 1/2 means we shoud divide 20 by 1/2 which is nothing but 40.
Adnan said:
1 decade ago
Let avg=y, no. of pupils=x, sum of marks=s
Case 1) y= s/x -------------(i)
Case 2) y + 1/2= (s+20)/x ----------(ii)
Solving eq. i & ii gives,x=40
Case 1) y= s/x -------------(i)
Case 2) y + 1/2= (s+20)/x ----------(ii)
Solving eq. i & ii gives,x=40
Maria Ninan said:
9 years ago
But the question says increased by half, so shouldn't it mean the incorrect average was increased by half of the correct average?
Thanks in advance
Thanks in advance
Deepak said:
7 years ago
Let avg.+1/2 = y+83/x- ------> eq1.
Y=remaining student's marks.
X =total no of students.
Now avg.= y+63/x-----> eq2
Eq2-eq1.
83-63/x=1/2.
X=40.
Y=remaining student's marks.
X =total no of students.
Now avg.= y+63/x-----> eq2
Eq2-eq1.
83-63/x=1/2.
X=40.
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