A test has 50 questions. A student scores 1 mark for a correct answer, - 1/3 for a wrong answer, and -1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than :

Let Correct Answers, Wrong Answers and Unattempted be C, W and N respectively. 

Total questions: C+W+N=50

Score: $$C-\dfrac{W}{3}-\dfrac{N}{6}=32$$

Multiply the score equation by 6 to eliminate the fraction: 6C - 2W - N = 192

Substituting N with N=50−C−W from equation 1:

6C − 2W − (50 − C − W) = 192

6C − 2W − 50 + C + W = 192

7C − W − 50 = 192

7C − W = 242

W = 7C − 242

We want minimum W, so minimise C, but ensure: $$W\ge\ 0,\ N\ge\ 0$$

Condition 1: W ≥ 0

7C − 242 ≥ 0 ⇒ C ≥ 34.57

So C ≥ 35

Condition 2: N ≥ 0 

N = 50 − C − W

Substitute W: N = 50 − C − (7C − 242) = 292 − 8C ≥ 0

C ≤ 36.5 ⇒ C ≤ 36

So, C = 35 or 36

If C = 35:  W = 7(35) − 242 = 245 − 242 = 3

If C = 36: W = 252 − 242 = 10

Hence, the minimum possible wrong answers = 3