manini dhar

CAT Preparation Community 10y 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

4

Praneeth Madhunanthu 10y 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

K Raj Kishan 10y ago

Incomplete question

kajol Sarda 6y ago

Plz help with approach

Kratos 2y 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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