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.
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.
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.
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.
Deepak kumar said:
8 years ago
Let x is average marks when total marks are 63.
Replacement,
83 - 63 = 20.
Then, x + 20 = x\2.
X = 40.
Replacement,
83 - 63 = 20.
Then, x + 20 = x\2.
X = 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.
MAHESH BHOI said:
9 years ago
LET, X=TOTAL MARKS AND N=NUMBER OF STUDENT.
AS GIVEN INCREASE AVG+1/2 WHEN WRONGLY ADDED.
AVG + 1/2 = X/N.........1
LET CORRECT IT,
AVG = X - 63 + 83/N........2
FROM 1 AND 2.
X/N-1/2 = X/N-20/N,
1/2 = 20/N,
N = 40.
AS GIVEN INCREASE AVG+1/2 WHEN WRONGLY ADDED.
AVG + 1/2 = X/N.........1
LET CORRECT IT,
AVG = X - 63 + 83/N........2
FROM 1 AND 2.
X/N-1/2 = X/N-20/N,
1/2 = 20/N,
N = 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.
Sumaira said:
9 years ago
Thanks for your explanation @Nik.
Akash soni said:
9 years ago
Avg = sum * total number.
1 if all are correct then the eqn is,
avg = sum/number.
sum = avg * number----> 1st eqn.
2 if wrong 83 instead of 60 then eqn is,
sum + (83 - 63) = (avg + 0.5) * number.
sum = (avg + 0.5) * number - 20----> 2nd eqn.
let divide 1st eqn by 2nd eqn.
sum/sum = 1.
Avg * number = (avg + 0.5)number - 20.
In solving this avg canceled by avg, So, avg - avg = 0.
0.5 - (20/number) = 0
0.5 * number = 20
Number = 20/0.5
Number =40
1 if all are correct then the eqn is,
avg = sum/number.
sum = avg * number----> 1st eqn.
2 if wrong 83 instead of 60 then eqn is,
sum + (83 - 63) = (avg + 0.5) * number.
sum = (avg + 0.5) * number - 20----> 2nd eqn.
let divide 1st eqn by 2nd eqn.
sum/sum = 1.
Avg * number = (avg + 0.5)number - 20.
In solving this avg canceled by avg, So, avg - avg = 0.
0.5 - (20/number) = 0
0.5 * number = 20
Number = 20/0.5
Number =40
Thiru said:
9 years ago
(83 - 63) = 20,
X * 20 = 1/2,
= 40.
X * 20 = 1/2,
= 40.
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