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 2 of 13.
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.
Nehanshu raj badholiya said:
1 decade ago
Rakesh's and habi bbhaijan both are correct so don't get confused.
Ajay said:
1 decade ago
Ans of Savinder question
Let total no of cand-x
Dec in aver- 50-45=5
Total dec in numbers= 5x
then, 5x= 100(90-60)
x=600 ( Ans)
Let total no of cand-x
Dec in aver- 50-45=5
Total dec in numbers= 5x
then, 5x= 100(90-60)
x=600 ( Ans)
(1)
M.V.KRISHNA/PALVONCHA said:
1 decade ago
Hai nik. Your explanation is nice.
Ravi said:
1 decade ago
Thanks habib.
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.
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
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.
Rav said:
1 decade ago
In the question above
is "A pupil's" mean "A student" ?
is "A pupil's" mean "A student" ?
Karthy said:
1 decade ago
The average increase for one person is 0.5
Total increase is 20
Therefore Total no of pupil =20/0.5 = 40.
Total increase is 20
Therefore Total no of pupil =20/0.5 = 40.
(2)
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