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9 years, 2 months ago
a test consists of 120 questions. each correct answer, each wrong answer and each unanswered question in the test carry 1 mark, -1/2 mark and -1/4 mark respectively. Find the maximum no of questions that the candidate could have answered wrongly in the test if he scores 50 marks in the test??? ANS=45.. PLEASE HELP WITH APPROACH
9 years, 2 months ago
Hi Manini,
Let the number of correct answers, number of wrong answers and number of unattempted questions be a, b and c respectively.
=> a + b + c = 120
a - $$\frac{b}{2}$$ - $$\frac{c}{4}$$ = 50
On multiplying the second equation and adding the first equation to it, we get 5a + b = 320.
=> a = 64 + $$\frac{b}{5}$$
Similarly, c = 56 - $$\frac{6b}{5}$$
So, the number of correct answers, number of wrong answers and number of unattempted questions are 64 + $$\frac{b}{5}$$, b and 56 - $$\frac{6b}{5}$$ respectively.
For each of them to be an integer, the value of b must be a multiple of 5.
When the value of b is more than 45, the value of c is negative, which is not possible.
Hence, maximum value of b is 45
1 year, 1 month ago
This type of questions can be solved forming equations. Let's assume the no of wrong questions is W, unattempted is U, and right is R. So, R+W+U = 120 is the first equation (not much useful for this question). Second and the useful equation is 120-(1+1/2)W - (1+1/4)U = 50. Thus 6W+5U=280. Max value of W here is 45. So the required answer is 45.<br><br>I hope this solution make things clear.<br>
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