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.
Anriudh Swaminathan said:
8 years ago
P1+P2+......+ let it be = x.
Let Y be the number of students.
Let Z be the correct avg marks.
(i) due to the mistake, the equation is:-
(x+83)/y =z + 0.5.
(ii) now without mistake:-
(x+63)/y = z.
Solve equation (i) and (ii).
20/y = 0.5.
Therefore Y = 40.
Let Y be the number of students.
Let Z be the correct avg marks.
(i) due to the mistake, the equation is:-
(x+83)/y =z + 0.5.
(ii) now without mistake:-
(x+63)/y = z.
Solve equation (i) and (ii).
20/y = 0.5.
Therefore Y = 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.
Chinmay said:
8 years ago
Total marks=x,
Number of pupils=y,
Average=x/y.
But,
(x+y)/20=x/y+1/2,
(x+20)/y=(2x+y)/2y,
Therefore y=40.
Number of pupils=y,
Average=x/y.
But,
(x+y)/20=x/y+1/2,
(x+20)/y=(2x+y)/2y,
Therefore y=40.
MGM said:
8 years ago
How about this?
(83- 63)/ ? = 1/2
that is 20/ (1/2) = ?
Ans 40.
(83- 63)/ ? = 1/2
that is 20/ (1/2) = ?
Ans 40.
Sharma said:
8 years ago
@Vishal explain it please.
7x/4y +2x/9y = 1/1.
we get 2:3.
7x/4y +2x/9y = 1/1.
we get 2:3.
Utsav Sinha said:
8 years ago
Simple logic.
Let no.of candidate = x.
Their average marks =y.
Generate an equation,
xy+83-63=x(y+1/2).
Solving we get y=40.
Let no.of candidate = x.
Their average marks =y.
Generate an equation,
xy+83-63=x(y+1/2).
Solving we get y=40.
Shubham Singh said:
8 years ago
The simplest logic is;
If by increasing 20 marks(83-63) in the class the average increases by it's half = x+x/2.
Average of total marks = x
Average of total marks + 20 = x + x/2
Thus 20 = x/2;
=> x = 40.
If by increasing 20 marks(83-63) in the class the average increases by it's half = x+x/2.
Average of total marks = x
Average of total marks + 20 = x + x/2
Thus 20 = x/2;
=> x = 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.
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.
Abhishek Ojha said:
8 years ago
@Shubham.
In your case, You calculated Average Marks Not num of the pupil.
In your case, You calculated Average Marks Not num of the pupil.
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