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.
Dinesh said:
1 decade ago
Its very simple given difference of avg is half hence,
(83/x)-(63/x) = 0.5.
Hence by solve x = 40.
(83/x)-(63/x) = 0.5.
Hence by solve x = 40.
Danny Blac said:
1 decade ago
@Midhun J.
You are right. I can so much relate with your answer.
You are right. I can so much relate with your answer.
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.
Pritisha said:
1 decade ago
Let the total marks be x,
Let the total no of students be x, and let the average marks of all the students be x.
Increase in the total marks = 83-63 = 20.
So, total marks/total students = avg marks.
=> x+40/x = x+1/2.
x gets cancelled on both sides of the equation.
Then the remaining terms are,
=> 20/x = 1/2.
=> x = 40.
Let the total no of students be x, and let the average marks of all the students be x.
Increase in the total marks = 83-63 = 20.
So, total marks/total students = avg marks.
=> x+40/x = x+1/2.
x gets cancelled on both sides of the equation.
Then the remaining terms are,
=> 20/x = 1/2.
=> x = 40.
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.
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).
Vignesh BS said:
1 decade ago
Correct mark = 63.
Wrongly entered = 83.
The average marks for the class got increased = 1/2 => 0.5.
A = For single pupil extra mark = 1*1/2 = 0.5.
B = Extra added mark = 83 - 63 => 20.
The number of pupils in the class "X"= B/A.
Therefore x = 20/0.5; 0.5 =1/2.
x = 20/(1/2).
x = 20* 2.
x = 40.
The number of pupils in the class is 40.
Wrongly entered = 83.
The average marks for the class got increased = 1/2 => 0.5.
A = For single pupil extra mark = 1*1/2 = 0.5.
B = Extra added mark = 83 - 63 => 20.
The number of pupils in the class "X"= B/A.
Therefore x = 20/0.5; 0.5 =1/2.
x = 20/(1/2).
x = 20* 2.
x = 40.
The number of pupils in the class is 40.
Shahin said:
1 decade ago
Very simple.
Let no. of. pupil be x.
New mark (wrongly entered) = 83.
Replaced mark (correct mark) = 63.
Increased avg = 0.5(1/2).
Formula: New mark=Replaced mark+(Number of pupil*inc avg).
We can apply this formula for this type of question.
Now,
83 = 63+(x*0.5).
= 83-63 = x*0.5.
= 20 = x*0.5.
= 20/(1/2) = x.
= 20*2 = x.
= 40 = x.
Number of pupils in the class is 40.
Hope it clear your doubts.
Let no. of. pupil be x.
New mark (wrongly entered) = 83.
Replaced mark (correct mark) = 63.
Increased avg = 0.5(1/2).
Formula: New mark=Replaced mark+(Number of pupil*inc avg).
We can apply this formula for this type of question.
Now,
83 = 63+(x*0.5).
= 83-63 = x*0.5.
= 20 = x*0.5.
= 20/(1/2) = x.
= 20*2 = x.
= 40 = x.
Number of pupils in the class is 40.
Hope it clear your doubts.
San_B said:
1 decade ago
Let N be the total number of students.
Total mark increase => 83-63 = 20. (because Old Sum-New Sum of marks will be affected only due to change of 83 to 63) __ (1).
Now, average marks are increased by 1/2.
It means each student's mark is increased by 1/2. (Average is equally applicable to all the students).
Hence, total mark increased => 1/2*N __(2).
From (1) and (2) ,
20 = (1/2)*N.
N = 40.
Total mark increase => 83-63 = 20. (because Old Sum-New Sum of marks will be affected only due to change of 83 to 63) __ (1).
Now, average marks are increased by 1/2.
It means each student's mark is increased by 1/2. (Average is equally applicable to all the students).
Hence, total mark increased => 1/2*N __(2).
From (1) and (2) ,
20 = (1/2)*N.
N = 40.
NITISH GULERIA said:
10 years ago
First we have average formula = Sum of all the marks/No of students.
Then for this question let be assume average as = A.
And no of students = X.
Now A.T.Q = A = 63 +mark1+. /X. equation no. 1 means average formula.
No second he say that when marks increase by 83 then again A = 83+ mark1+. /x.
But he said that average increased by 0.5. So A+0.5 = 83 + marks1+. /x. equation no 2.
Now equate these equations you will find exact answer.
Then for this question let be assume average as = A.
And no of students = X.
Now A.T.Q = A = 63 +mark1+. /X. equation no. 1 means average formula.
No second he say that when marks increase by 83 then again A = 83+ mark1+. /x.
But he said that average increased by 0.5. So A+0.5 = 83 + marks1+. /x. equation no 2.
Now equate these equations you will find exact answer.
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