Suraj0184 wrote:
In a test of 50 questions, 3 marks were awarded for a correct answer, 0 for not attempting and -1 for an incorrect answer. In how many ways can the student score 120 marks?
A. 3
B. 4
C. 5
D. 6
E. 7
Source : Wizako
Two ways.
First, brute force, which is the way I would do this question on the actual test, as it requires no formulas to keep track of and there's little chance of making a mistake. There are no bonus points for doing it the other way, so I'll always choose the method that minimizes the risk of making a silly mistake!
We can tell that we need at least 40 correct.
40 correct is 120 points. So we need 0 wrong, which leaves 10 unanswered.
41 correct is 123 points. So we need 3 wrong, which leaves 6 unanswered.
42 correct is 126 points. So we need 6 wrong, which leaves 2 unanswered.
If we get 1 more correct, we don't have enough to get 3 more wrong.
So, there are three ways.
Answer choice A.
Second, equations and stuff.
c = correct, w = wrong
From the fact that there are 50 total questions, we have: c+w <=50
From the fact that we get +3 for each correct answer and -1 for each wrong answer, we have: 3c-w = 120
Solving for w, we get: -w = 120-3c --> w = 3c-120
Substituting for w in the inequality, we get c+3c-120 <= 50 --> 4c <= 170 --> c <= 42.5
So, the number correct is less than or equal to 42, but we know it has to be at least 40. That leaves 40, 41, and 42. That's three ways.
Answer choice A.
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