A test consists of 4 questions. Four candidates (P, Q, R, S) took the test. Each question has four options (A, B, C, D).
The candidates are divided into two cities: P and Q are from City X; R and S are from City Y.
The scoring scheme is: +2 marks for a correct answer, -1 mark for a wrong answer. All candidates attempted all questions.
The responses of the candidates are:
- P: A, B, C, D
- Q: A, A, A, A
- R: B, B, B, B
- S: A, B, D, C
It is known that the Total Score of City X is 7, and the Total Score of City Y is 1.
Assertion (A): The correct answer for Q3 is C.
Reason (R): The total score of City X is greater than the total score of City Y.
Which of the following is correct?
D
Step-by-Step Solution
Key idea: This is a Demographic Score Reverse-Engineering problem. We must use the city totals to deduce the number of correct answers for each city, and then find the answer key that satisfies these constraints.
Step 1: Understand the Scoring Formula.
Let be the number of correct answers for a candidate.
Score = .
Step 2: Analyze City Totals.
Let be the correct answers for P and Q.
Total Score City X = .
.
Let be the correct answers for R and S.
Total Score City Y = .
.
Step 3: Deduce the Answer Key.
We need to find a key such that the sum of matches for P+Q is 5, and R+S is 3.
Let's analyze each question's contribution to the sums:
- Q1: P=A, Q=A, R=B, S=A. If K1=A, P+Q get 2, R+S get 1. If K1=B, P+Q get 0, R+S get 1.
- Q2: P=B, Q=A, R=B, S=B. If K2=B, P+Q get 1, R+S get 2. If K2=A, P+Q get 1, R+S get 0.
- Q3: P=C, Q=A, R=B, S=D. All different. Any key gives P+Q = 1, R+S = 0 OR P+Q = 0, R+S = 1.
- Q4: P=D, Q=A, R=B, S=C. All different. Any key gives P+Q = 1, R+S = 0 OR P+Q = 0, R+S = 1.
To get , we must maximize the contributions.
Max possible for Q1 is 2 (K1=A).
Max possible for Q2 is 1.
Max possible for Q3 is 1.
Max possible for Q4 is 1.
Total max = 5. So we MUST achieve the maximum for every question!
This forces:
- K1 = A (gives 2)
- Q2 must give 1 (K2=B or A)
- Q3 must give 1 (K3=C or A)
- Q4 must give 1 (K4=D or A)
Now check .
If K1=A, R+S get 1.
If K2=B, R+S get 2. If K2=A, R+S get 0. To reach 3, we MUST have K2=B (gives 2).
If K3=C or A, R+S get 0.
If K4=D or A, R+S get 0.
Total R+S = 1 + 2 + 0 + 0 = 3. Matches perfectly!
So K1=A, K2=B are forced. K3 can be C or A. K4 can be D or A.
Step 4: Evaluate Assertion and Reason.
- Assertion (A): Q3 is C. Since K3 can be C or A, this is NOT necessarily true. (A is false).
- Reason (R): City X score (7) > City Y score (1). This is explicitly given and true. (R is true).
Answer: D