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 6 of 13.
Tran said:
3 years ago
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.
Please correct me, if I am wrong.
(6)
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.
Ajay Makala said:
2 years ago
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.
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.
(114)
Yamuna said:
1 decade ago
Let x be the no. of pupil, y is the avg.
63/x = y,
83/x = y+1/2,
By dividing these 2 equations you will get y value. Substitute y in 1st equation, you get x as 40.
Simple.
63/x = y,
83/x = y+1/2,
By dividing these 2 equations you will get y value. Substitute y in 1st equation, you get x as 40.
Simple.
Tasleem said:
1 decade ago
Average of the total as it is increased by 20 can be written as:
= s+20/n.
= s+20/n = s/n*1/2 (as the original avg is increased by 1/2).
= 2(s+20) = s.
Hence s = 40.
= s+20/n.
= s+20/n = s/n*1/2 (as the original avg is increased by 1/2).
= 2(s+20) = s.
Hence s = 40.
Deekshith.D said:
8 years ago
1st condition : 83/n =av,
Here av is average.
2nd one is: 63/n = 1/2 +av,
So by solving these 2 eq. We have,
1/2 = 83/n - 63/n.
Solving this we will get n=2*20=40.
Here av is average.
2nd one is: 63/n = 1/2 +av,
So by solving these 2 eq. We have,
1/2 = 83/n - 63/n.
Solving this we will get n=2*20=40.
Prashanth.A said:
7 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.
Rida said:
3 years ago
What if the condition is reverse entry is recorded wrong 39 instead of 63. In this case, the average marks of class reduced by 1/2? Can anyone tell me the answer?
(2)
Anonymous said:
1 decade ago
Before..
Av = Sum/num ------(1).
After..
Av+1/2 = (Sum+20)/num -------(2).
Putting value of (1) in (2).
Sum/num + 1/2 = Sum/num + 20/num.
num = 40.
Av = Sum/num ------(1).
After..
Av+1/2 = (Sum+20)/num -------(2).
Putting value of (1) in (2).
Sum/num + 1/2 = Sum/num + 20/num.
num = 40.
Bitu said:
9 years ago
Let the number of pupils = x.
Avg of mistaken marks = 83/x,
Avg of correct marks = 63/x,
By question;
83/x = 63/x + 1/2,
83/x - 63/x = 1/2,
20/x = 1/2,
x = 40.
Avg of mistaken marks = 83/x,
Avg of correct marks = 63/x,
By question;
83/x = 63/x + 1/2,
83/x - 63/x = 1/2,
20/x = 1/2,
x = 40.
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