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(B) Let $x_i$ be the marks obtained in the $i$-th question,where $x_i \in \{3, -2, 0\}$.We need to find the number of ways such that $\sum_{i=1}^{5} x

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let $x_i$ be the marks obtained in the $i$-th question,where $x_i \in \{3, -2, 0\}$.We need to find the number of ways such that $\sum_{i=1}^{5} x_i = 5$.Let $n_1$ be the number of correct answers,$n_2$ be the number of wrong answers,and $n_3$ be the number of unattempted questions.We have $n_1 + n_2 + n_3 = 5$ and $3n_1 - 2n_2 + 0n_3 = 5$.From the second equation,$3n_1 - 2n_2 = 5$. Testing integer values for $n_1$ and $n_2$:If $n_1 = 1$,$3 - 2n_2 = 5 \implies n_2 = -1$ (not possible).If $n_1 = 3$,$9 - 2n_2 = 5 \implies 2n_2 = 4 \implies n_2 = 2$.Then $n_3 = 5 - (3 + 2) = 0$.So,the student must have $3$ correct answers and $2$ wrong answers.For each wrong answer,there are $2$ incorrect choices available.The number of ways to choose which questions are correct is $\binom{5}{3} = 10$.The number of ways to choose the incorrect options for the $2$ wrong answers is $2^2 = 4$.Total number of ways = $10 	imes 4 = 40$.

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