GATE CS 2015 Set 1 Q.6 (GA Q.6): Data Interpretation and Graphs

The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.

Q.No.	Marks	Answered Correctly	Answered Wrongly	Not Attempted
1	2	21	17	6
2	3	15	27	2
3	1	11	29	4
4	2	23	18	3
5	5	31	12	1

What is the average of the marks obtained by the class in the examination?

$2.290$

$2.970$

$6.795$

$8.795$

Official answer and solution

Official answer: C
From: Official GATE 2015 answer key

Solution

Compute the total marks obtained by all students and divide by the total number of students in the class.

From each row of the table, the total number of students in the class is constant:

For Question 1: $21 + 17 + 6 = 44$
For Question 2: $15 + 27 + 2 = 44$
For Question 3: $11 + 29 + 4 = 44$
For Question 4: $23 + 18 + 3 = 44$
For Question 5: $31 + 12 + 1 = 44$

Thus, total number of students $N = 44$ .

Because there is no negative or partial marking, marks are awarded only for correct answers:

Q1: $21 \times 2 = 42$
Q2: $15 \times 3 = 45$
Q3: $11 \times 1 = 11$
Q4: $23 \times 2 = 46$
Q5: $31 \times 5 = 155$

Sum of all marks scored by the entire class:

\text{Total Marks} = 42 + 45 + 11 + 46 + 155 = 299.

The class average is:

\text{Average} = \frac{299}{44} = 6.79545... \approx 6.795.

This matches option C.

Final answer

Common trap

Dividing the total marks by 5 (the number of questions) instead of 44 (the total number of students in the class).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 6 of 65, printed as GA Q.6
Section: General Aptitude

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