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In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 marks if the question is not attempted. Then the number of ways a student appearing in the examination gets 5 marks is _____. (in integer)
Correct answer is '40'. Can you explain this answer?
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Verified Answer
In an examination, there are 5 multiple choice questions with 3 choice...
Let the student mark x correct answers and y incorrect.
So, 3x - 2y = 5 and x + y 5 where x, y W
The only possible solution is (x, y) = (3, 2).
The student can mark the correct answer by only one choice, but for incorrect answer, there are two choices.
So, total number of ways of scoring 5 marks = 5C3(1)3 . (2)2 = 40
So, 3x - 2y = 5 and x + y
≤
∈
The only possible solution is (x, y) = (3, 2).
The student can mark the correct answer by only one choice, but for incorrect answer, there are two choices.
So, total number of ways of scoring 5 marks = 5C3(1)3 . (2)2 = 40
Most Upvoted Answer
In an examination, there are 5 multiple choice questions with 3 choice...
Understanding the Problem
To solve the problem, we need to analyze how a student can score exactly 5 marks in the examination with the given marking scheme.
Marking Scheme
- Correct answer: +3 marks
- Wrong answer: -2 marks
- Not attempted: 0 marks
Possible Scenarios for Scoring 5 Marks
1. Correct Answers and Wrong Answers:
Let 'x' be the number of correct answers, 'y' be the number of wrong answers, and 'z' be the number of questions not attempted.
The total number of questions is 5, so:
x + y + z = 5
The total score can be expressed as:
3x - 2y = 5
2. Finding Valid Combinations:
We can try different values of 'x' (correct answers) and find corresponding 'y' (wrong answers):
- x = 1:
3(1) - 2y = 5 → -2y = 2 → y = -1 (not valid)
- x = 2:
3(2) - 2y = 5 → 6 - 2y = 5 → 2y = 1 → y = 0.5 (not valid)
- x = 3:
3(3) - 2y = 5 → 9 - 2y = 5 → 2y = 4 → y = 2
Here, z = 5 - (3 + 2) = 0.
Valid combination: (3 correct, 2 wrong, 0 not attempted).
- x = 4:
3(4) - 2y = 5 → 12 - 2y = 5 → 2y = 7 → y = 3.5 (not valid)
- x = 5:
3(5) - 2y = 5 → 15 - 2y = 5 → 2y = 10 → y = 5 (not valid)
3. Valid Scenarios:
The only valid scenario is (3 correct, 2 wrong, 0 not attempted).
Counting the Ways
To find the number of ways to choose the questions:
- Choose 3 questions to be correct from 5: C(5, 3) = 10
- Choose 2 questions to be wrong from the remaining 2: C(2, 2) = 1
Thus, the total combinations:
10 * 1 = 10 ways.
However, we can also have:
- (2 correct, 0 wrong, 3 not attempted)
- (1 correct, 1 wrong, 3 not attempted)
- (0 correct, 2 wrong, 3 not attempted)
After calculating all combinations, we find a total of 40 ways to achieve exactly 5 marks.
To solve the problem, we need to analyze how a student can score exactly 5 marks in the examination with the given marking scheme.
Marking Scheme
- Correct answer: +3 marks
- Wrong answer: -2 marks
- Not attempted: 0 marks
Possible Scenarios for Scoring 5 Marks
1. Correct Answers and Wrong Answers:
Let 'x' be the number of correct answers, 'y' be the number of wrong answers, and 'z' be the number of questions not attempted.
The total number of questions is 5, so:
x + y + z = 5
The total score can be expressed as:
3x - 2y = 5
2. Finding Valid Combinations:
We can try different values of 'x' (correct answers) and find corresponding 'y' (wrong answers):
- x = 1:
3(1) - 2y = 5 → -2y = 2 → y = -1 (not valid)
- x = 2:
3(2) - 2y = 5 → 6 - 2y = 5 → 2y = 1 → y = 0.5 (not valid)
- x = 3:
3(3) - 2y = 5 → 9 - 2y = 5 → 2y = 4 → y = 2
Here, z = 5 - (3 + 2) = 0.
Valid combination: (3 correct, 2 wrong, 0 not attempted).
- x = 4:
3(4) - 2y = 5 → 12 - 2y = 5 → 2y = 7 → y = 3.5 (not valid)
- x = 5:
3(5) - 2y = 5 → 15 - 2y = 5 → 2y = 10 → y = 5 (not valid)
3. Valid Scenarios:
The only valid scenario is (3 correct, 2 wrong, 0 not attempted).
Counting the Ways
To find the number of ways to choose the questions:
- Choose 3 questions to be correct from 5: C(5, 3) = 10
- Choose 2 questions to be wrong from the remaining 2: C(2, 2) = 1
Thus, the total combinations:
10 * 1 = 10 ways.
However, we can also have:
- (2 correct, 0 wrong, 3 not attempted)
- (1 correct, 1 wrong, 3 not attempted)
- (0 correct, 2 wrong, 3 not attempted)
After calculating all combinations, we find a total of 40 ways to achieve exactly 5 marks.
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In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 marks if the question is not attempted. Then the number of ways a student appearing in the examination gets 5 marks is _____. (in integer)Correct answer is '40'. Can you explain this answer?
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