Question
The table below has question-wise data on the performance of students in an examination. The marks for each question are also listed. There is no negative or partial marking in the examination.
| Q No. | Marks | Answered Correctly | Answered Wrongly | Not Attempted |
|---|---|---|---|---|
| 1 | 2 | 21 | 17 | 6 |
| 2 | 3 | 15 | 27 | 2 |
| 3 | 2 | 23 | 18 | 3 |
What is the average of the marks obtained by the class in the examination?
Official answer and solution
- Official answer
- C
- From
- Official GATE 2014 answer key
Solution
Compute total marks scored by all students across all questions and divide by the total number of students in the class.
-
Determine the total number of students in the class ():
For any question, the sum of students who answered correctly, answered wrongly, and did not attempt equals the total class size:- From Question 1: students.
- From Question 2: students.
- From Question 3: students.
Thus, students in the class.
-
Calculate the total marks scored by all students:
Marks are awarded only for correct answers (no negative or partial marking):- Question 1 (2 marks each): marks.
- Question 2 (3 marks each): marks.
- Question 3 (2 marks each): marks.
-
Calculate the average marks per student:
Rounding to two decimal places gives , which matches option C.
C
Dividing total marks by the total number of responses (44 x 3 = 132) instead of the total number of students (44).
The answer is the one in the official GATE answer key. The solution is this site's own explanation.
Source
- Paper
- GATE 2014 Computer Science and Information Technology (CS) Session 3
- Question
- 5 of 65, printed as GA Q.5
- Section
- General Aptitude
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