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 4 of 13.
Adi said:
1 decade ago
Let total number be x.
Then it is given that average for 63 is A.
i.e, sum/x = A.
When it is wrongly entered as 83 then.
(sum-63+83) /x = A + 1/2.
Then (sum- 20) /x =A + 1/2.
(sum/x) - (20/x) =A+1/2.
Now pick from above that sum/x=A.
Now A- (20/x) =A + 1/2.
Thus, it brings x=40.
Then it is given that average for 63 is A.
i.e, sum/x = A.
When it is wrongly entered as 83 then.
(sum-63+83) /x = A + 1/2.
Then (sum- 20) /x =A + 1/2.
(sum/x) - (20/x) =A+1/2.
Now pick from above that sum/x=A.
Now A- (20/x) =A + 1/2.
Thus, it brings x=40.
Saji said:
1 decade ago
let x be the no.of students.
When it is 83, the avg is increased by 1/2.
So we get,
x(avg+1/2) = 83 --(1).
Also given that 63 is the correct mark so we get an equ like,
x(avg) = 63 --(2).
Now (1)-(2)we get,
x(avg)+x(1/2)-x(avg) = 83-63.
Therefore, x/2 = 20.
x = 40.
When it is 83, the avg is increased by 1/2.
So we get,
x(avg+1/2) = 83 --(1).
Also given that 63 is the correct mark so we get an equ like,
x(avg) = 63 --(2).
Now (1)-(2)we get,
x(avg)+x(1/2)-x(avg) = 83-63.
Therefore, x/2 = 20.
x = 40.
Rakesh said:
1 decade ago
If mark=63,then total mark(mark1+mark2+---+63) =(average mark*x)-(1)
but in the qst given that enterd mark is 83 instead of 63,then avg is increased by 1/2(half),
then total mark(mark1+mark2+----+83)=(avg mark+1/2)*x -(2)
eq(2)-eq(1)
83-63=1/2*x;
x=20*2=40
but in the qst given that enterd mark is 83 instead of 63,then avg is increased by 1/2(half),
then total mark(mark1+mark2+----+83)=(avg mark+1/2)*x -(2)
eq(2)-eq(1)
83-63=1/2*x;
x=20*2=40
Sachin Bende said:
1 decade ago
@ swathy
rakesh's logic is absolutely correct..
explanation:
=(avg mark+1/2)*x - (avg mark*x)
taking "x" common we get,
=x(avg mark + 1/2 - avg mark)
cancelling avg mark
hence,we get
= x(1/2)= 1/2*x
therefore,
83-63=1/2*x
x=20*2=40
ok got it.
rakesh's logic is absolutely correct..
explanation:
=(avg mark+1/2)*x - (avg mark*x)
taking "x" common we get,
=x(avg mark + 1/2 - avg mark)
cancelling avg mark
hence,we get
= x(1/2)= 1/2*x
therefore,
83-63=1/2*x
x=20*2=40
ok got it.
Mohammed Aijaz said:
1 decade ago
Hi friends.
Let us consider p be the no. of pupil, x be the average.
Case1: original marks 63/p=x ----- average of case 1.
Case2: new marks 83/p=x+1/2 ----average of case 2.
2-1 => (83/p-63/p)=x+1/2-x
=> 20/p=1/2
=> p=40.
Let us consider p be the no. of pupil, x be the average.
Case1: original marks 63/p=x ----- average of case 1.
Case2: new marks 83/p=x+1/2 ----average of case 2.
2-1 => (83/p-63/p)=x+1/2-x
=> 20/p=1/2
=> p=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.
Eliwa Mohamed Eliwa Salman said:
7 months ago
Assume number of pupils is x.
avg1 = 63/x --->eq (1)
avg2 = 83/x --->eq (2)
avg2 = avg1 + 0.5 ---> eq (3)
Substituting in eq (2) by eq (3).
(avg1) +0.5 = (83/x),
(63/x) +0.5 = (83/x),
(83/x) -(63/x) = 0.5,
(83-63)/x = 0.5,
20/x = 0.5,
x = 20/0.5 = 40.
avg1 = 63/x --->eq (1)
avg2 = 83/x --->eq (2)
avg2 = avg1 + 0.5 ---> eq (3)
Substituting in eq (2) by eq (3).
(avg1) +0.5 = (83/x),
(63/x) +0.5 = (83/x),
(83/x) -(63/x) = 0.5,
(83-63)/x = 0.5,
20/x = 0.5,
x = 20/0.5 = 40.
(7)
Sanchit said:
7 years ago
Take sum as S and number if people as N.
Avg + (1/2) = (s + 83)/n----> (1)
Now, Avg = (s+63)/n ----> (2)
Just substitute in place of avg
[(s + 63)/n] + 1/2 = [(s + 83)/n].
1/2 = [(s + 83)/n] - [(s + 63)/n].
1/2 = (83-63)/n,
n. = 2 x 20.
n = 40.
Avg + (1/2) = (s + 83)/n----> (1)
Now, Avg = (s+63)/n ----> (2)
Just substitute in place of avg
[(s + 63)/n] + 1/2 = [(s + 83)/n].
1/2 = [(s + 83)/n] - [(s + 63)/n].
1/2 = (83-63)/n,
n. = 2 x 20.
n = 40.
Savinder said:
1 decade ago
In an exam, the average was found to be 50 marks. After detecting computer errors, the marks of 100 candidates had to be changed from 90 to 60 each and the average came down to 45. The total number of candidates who took the exam was:.
Please solve.
Please solve.
Sahithi said:
1 decade ago
Short cut method:
Here given avg incresed 1/2 means 0.5
Taken 83 instead of 63 the differnce of avg incresed 83-63=20
Therefore how many 1/2 s in 20 ?
for detailed
1 means 1/2+1/2 contain 2 halfs
20 contain 20*2=40 halfs
Therefore n=40 pupils.
Here given avg incresed 1/2 means 0.5
Taken 83 instead of 63 the differnce of avg incresed 83-63=20
Therefore how many 1/2 s in 20 ?
for detailed
1 means 1/2+1/2 contain 2 halfs
20 contain 20*2=40 halfs
Therefore n=40 pupils.
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