- 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.**