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.MarksAnswered CorrectlyAnswered WronglyNot Attempted
1221176
2315272
3223183

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

Choose one option
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.

  1. Determine the total number of students in the class (NN):
    For any question, the sum of students who answered correctly, answered wrongly, and did not attempt equals the total class size:

    • From Question 1: 21+17+6=4421 + 17 + 6 = 44 students.
    • From Question 2: 15+27+2=4415 + 27 + 2 = 44 students.
    • From Question 3: 23+18+3=4423 + 18 + 3 = 44 students.
      Thus, N=44N = 44 students in the class.
  2. 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): 21×2=4221 \times 2 = 42 marks.
    • Question 2 (3 marks each): 15×3=4515 \times 3 = 45 marks.
    • Question 3 (2 marks each): 23×2=4623 \times 2 = 46 marks.
    Total Marks=42+45+46=133 marks.\text{Total Marks} = 42 + 45 + 46 = 133\text{ marks}.
  3. Calculate the average marks per student:

    Average=Total MarksN=13344≈3.0227.\text{Average} = \frac{\text{Total Marks}}{N} = \frac{133}{44} \approx 3.0227.

Rounding to two decimal places gives 3.023.02, which matches option C.

Final answer

C

Common trap

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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