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 3 of 13.
Phani said:
1 decade ago
Hi. I have come up with one more simple solution:
1st case:
Lets say avg is X when marks are 63. So X = 63/n.
2nd case:
When marks were noted as 83, the avg is increased by 0.5 so,
X+0.5 = 83/n.
Now if you subtract 2nd case - 1st case then (83-63) /n = X+0.5-X.
So, 20/n = 0.5 and n = 20/0.5 so n = 40.
Have a great day dudes. !
1st case:
Lets say avg is X when marks are 63. So X = 63/n.
2nd case:
When marks were noted as 83, the avg is increased by 0.5 so,
X+0.5 = 83/n.
Now if you subtract 2nd case - 1st case then (83-63) /n = X+0.5-X.
So, 20/n = 0.5 and n = 20/0.5 so n = 40.
Have a great day dudes. !
Yuvraj.. said:
1 decade ago
@priya:
Increase in avg. marks of class * no. of students = increase in marks of a student---->eqn 1
Now; increase in avg marks of class = 1/2= 1.5
and increase in marks of a student = 83-63= 20
therefore
from eqn 1
no. of students = increase in marks of a student/ increase in avg marks of class
So,
No. of students=20/1.5= 40
Increase in avg. marks of class * no. of students = increase in marks of a student---->eqn 1
Now; increase in avg marks of class = 1/2= 1.5
and increase in marks of a student = 83-63= 20
therefore
from eqn 1
no. of students = increase in marks of a student/ increase in avg marks of class
So,
No. of students=20/1.5= 40
Habib said:
1 decade ago
Frnz....dnt get confused boy that 1/2 factor !!!!
see,
(total marks)/(no. of pupils)= avg
OR
total marks = avg X no. of pupils
Given, increase is of 1/2 , no. of pupils=x
total marks increased : 83-63 = 20
Hence, for 1/2 mark avg increase, we use
20 = (1/2) * x OR x=40
see,
(total marks)/(no. of pupils)= avg
OR
total marks = avg X no. of pupils
Given, increase is of 1/2 , no. of pupils=x
total marks increased : 83-63 = 20
Hence, for 1/2 mark avg increase, we use
20 = (1/2) * x OR x=40
Abhishek Ojha said:
8 years ago
@Deekshit.
It's INSUFFICIENT DATA.
63/n = 1/2 +av is incorrect . the correct one should be;
63/n = 1/2av + av : because av (average) is increased by half which means avg was changed to av + av/2 . And Most of the folks have done thia mistake only. Please ignore this question, there is Insufficient data to solve this.
It's INSUFFICIENT DATA.
63/n = 1/2 +av is incorrect . the correct one should be;
63/n = 1/2av + av : because av (average) is increased by half which means avg was changed to av + av/2 . And Most of the folks have done thia mistake only. Please ignore this question, there is Insufficient data to solve this.
Priya said:
1 decade ago
Please trace where I went wrong.
Let n be total number of students
When marks was entered as 63 -----(x+63)/n=1 . . eqn1
When enterd as 83--------(x+83)/n=1.5. . . .eqn2
eqn1 --->n=x+63
eqn2---->n=(x+83)/1.5
equatin both x+63=(x+83)/1.5
1.5x+94.5=x+83
.5x=11.5
x=23
bak sub in eqn1
23+63=n
n=86
Please help me.
Let n be total number of students
When marks was entered as 63 -----(x+63)/n=1 . . eqn1
When enterd as 83--------(x+83)/n=1.5. . . .eqn2
eqn1 --->n=x+63
eqn2---->n=(x+83)/1.5
equatin both x+63=(x+83)/1.5
1.5x+94.5=x+83
.5x=11.5
x=23
bak sub in eqn1
23+63=n
n=86
Please help me.
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)
Rituparna Sengupta said:
1 decade ago
Let, average marks of class = x & also let, total no. of pupil in the class = A.
When the mark is 63. Then, the total marks of all pupil = Ax.
When, 83 is entered instead of 63,
Then avg marks of the class = A(x+1/2).
Then, A(x+1/2)-Ax = 83-63.
=> A/2 = 20.
=> A = 40. Option(C).
When the mark is 63. Then, the total marks of all pupil = Ax.
When, 83 is entered instead of 63,
Then avg marks of the class = A(x+1/2).
Then, A(x+1/2)-Ax = 83-63.
=> A/2 = 20.
=> A = 40. Option(C).
Bharat Gupta said:
7 years ago
When 63 is entered the let avg is x.
x = (sum of the marks of rest of students+63)/n.
Let the sum of the marks of rest of the students=y.
Therefore, x=(y+63)/n.
When 83 is entered avg is increased by 1/2 or 0.5.
So, x+0.5= (y+83)/n.
Equating both the terms we get;
0.5=20/n.
Therefore, n=40.
x = (sum of the marks of rest of students+63)/n.
Let the sum of the marks of rest of the students=y.
Therefore, x=(y+63)/n.
When 83 is entered avg is increased by 1/2 or 0.5.
So, x+0.5= (y+83)/n.
Equating both the terms we get;
0.5=20/n.
Therefore, n=40.
Natasha said:
7 years ago
Let 'x' be the avg marks and 'y' be the no of pupils due to entering 83 instead of 63 the avg mark 'x' became 'x + 0.5'. Please note that the avg was increased by 0.5 = 1/2 and not by x/2. Since the difference between 83 answer 63 yields 20 which is divided among y pupils.
y/2 =20 gives y=40.
y/2 =20 gives y=40.
PaniniLuncher said:
8 months ago
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.
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.
(23)
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