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A student is answering a multiple choice examination with $65$ questions with a marking scheme as follows$:$  $i)$ $1$ marks for each correct answer $,ii)$  $-\frac{1}{4}$ for a wrong answer $,iii)$ $-\frac{1}{8}$ for a question that has not been attempted$.$ If the student gets $37$ marks in the test then the least possible number of questions the student has NOT answered is$:$

1. $6$
2. $5$
3. $7$
4. $4$

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I don't know the exact way to answer this question but from the given options we can try out this possibility of solving it.

Let us consider option $A$

So there are 6 questions which are unattempted and remaining 59 questions were attempted. Now let us consider there are x questions which are answered incorrectly and remaining $59-x$ were answered correctly. So based upon this we can get a linear equation as follows

$\Rightarrow (59-x) \times 1 + x \times \dfrac{-1}{4} + 6 \times \dfrac{-1}{8} = 37$

$\Rightarrow 8 \times (59-x) - 2x -6 = 37 \times 8$

$\Rightarrow 472 -6 -296 = 10x$

$\Rightarrow 170 = 10x$

$\Rightarrow x = 17$.

Hence there are 42 questions answered correctly, 17 incorrectly and 6 unattempted.

Let $n_c$ denote the number of correct answers and $n_w$ denote the number of wrong answers.

Total marks $= 1 \times n_c -\frac{1}{4} \times n_w - \frac{1}{8} \times (65-(n_c+n_w))$

It is given $37 = n_c -\frac{1}{4} \times n_w - \frac{1}{8} \times (65-(n_c+n_w))$

$\implies 37 = n_c -\frac{1}{4} \times n_w-\frac{65}{8} +\frac{n_c}{8}+\frac{n_w}{8}$

$\implies 45 +\frac{1}{8} = n_c +\frac{n_c}{8}-\frac{n_w}{8}$

$\implies 361 = 9n_c-{n_w}$

To minimize the number of unanswered questions we have to maximize $n_c + n_w.$

Possible values of $(n_c, n_w)$ are

• $(41,8)$
• $(42,17)$
• $(43,26) -$ not possible as $43+26 > 65.$

So, maximum value of $n_c + n_w = 42+17 = 59.$

So, minimum number of unanswered questions $= 65-59 = 6.$

Correct Option: A.

by Veteran (416k points)
+1 vote
•  65 questions of 1 marks each
• Student got 37 marks.
• Remaining marks = 65-37 = 28.

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$\Rightarrow$ He has done at least 37 questions correctly.

$\Rightarrow$ He attempted some question incorrectly, left some questions and done some questions correctly in such a way that this remaining 28 marks became 0 and thus he was able to get only 37 marks instead of getting 65 marks.

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Option A

unattempted =6.

$\Rightarrow$  28 -6 = 22 questions remain which are having 22 marks.

$\because$ 1 correct + 4 incorrect question = 0 marks

$\Rightarrow$ 4 correct question + 16 incorrect question = 0 marks

$\Rightarrow$ 20 question resulted in 0 marks and 2 questions remained.

Out of these 2 question

if 1 is correct and 1 is incorrect and also 6 unattempted questions results in  negative marking of -$\frac{6}{8}$

So 1*1 - 1*$\frac{1}{4}$ -$\frac{6}{8}$ =0

Hence option A satisfies.

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Option B

unattempted = 5

$\Rightarrow$  28 -5 = 23 questions remain which are having 23 marks.

$\because$ 1 correct + 4 incorrect question = 0 marks

$\Rightarrow$ 4 correct question + 16 incorrect question = 0 marks

$\Rightarrow$ 20 question resulted in 0 marks and 3 questions remained.

We cant divide these 3 questions such that their sum becomes 0

Hence option B fails.

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Option C

unattempted = 7

No need to check since we already have a value i.e. 7>5 and question is asking for minimum.

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Option D

unattempted = 4

$\Rightarrow$  28 -4 = 24 questions remain which are having 24 marks.

$\because$ 1 correct + 4 incorrect question = 0 marks

$\Rightarrow$ 4 correct question + 16 incorrect question = 0 marks

$\Rightarrow$ 20 question resulted in 0 marks and 4 questions remained.

We cant divide these 4 questions such that their sum becomes 0.

Hence option D fails.

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$\therefore$ Option A is the correct answer.

by Boss (17.5k points)