Question

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:

• A. 12
• B. 3
• C. 4
• D. 9
• E. 2

Hint:

When minimizing or maximizing one variable, use the constraints to find the range of possible values.

Answer 1

The Correct Option is B

Step 1: Defining Variables:
Let \(C\) = number of correct answers, \(W\) = number of wrong answers, \(U\) = number of unattempted questions.
Given: \(C + W + U = 50\)
Net score: \(C - \frac{1}{3}W - \frac{1}{6}U = 32\)

Step 2: Eliminating \(U\):
From \(U = 50 - C - W\), substitute:
\[ C - \frac{W}{3} - \frac{50 - C - W}{6} = 32 \]
Multiply by 6:
\[ 6C - 2W - (50 - C - W) = 192 \]
\[ 6C - 2W - 50 + C + W = 192 \]
\[ 7C - W - 50 = 192 \]
\[ 7C - W = 242 \]
\[ W = 7C - 242 \]

Step 3: Applying Constraints:
\(C, W, U \geq 0\) and integers.
\(W \geq 0 \Rightarrow 7C \geq 242 \Rightarrow C \geq 34.57 \Rightarrow C \geq 35\)
\(U = 50 - C - W = 50 - C - (7C - 242) = 50 - C - 7C + 242 = 292 - 8C \geq 0\)
\[ 292 \geq 8C \Rightarrow C \leq 36.5 \Rightarrow C \leq 36 \]
So \(C = 35\) or \(36\).
If \(C = 35\), \(W = 7(35) - 242 = 245 - 242 = 3\).
If \(C = 36\), \(W = 7(36) - 242 = 252 - 242 = 10\).

Step 4: Finding Minimum \(W\):
The possible values of \(W\) are 3 and 10.
The smallest possible \(W\) is 3.