in Quantitative Aptitude edited by
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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.
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5 Answers

7 votes
7 votes
Best answer

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. 

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4 Comments

@Arjun sir, how did you arrive at the possible values of nc,nw??
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@Arjun Sir,how did you predicted the possbile values?
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it is easy to predict the values as nc must be $\geq$37 because he got 37 marks as correct so he must be attempt 37 correct answer .

so we just need to put value of nc starting from 37 .

and then from the equation we can get nw as well.

for example we are taking nc =37 then,

nw = 9nc - 361 = -28 which is not possible because nw can never be negative.

so we just keep increasing the value and test for which value it comes positive and as well we have to check nc+nw $<$65 .

i hope this clear how we can arrive the possible value of nc and nw.
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Consider equation 361 = 9c−w

                               w – 9c = -361

                               (w-9c) mod 9 = (-361)mod 9

                               w mod 9 = 8

So, w = 9k + 8, k belongs to Z.

PS : Chosen as mod with 9, because it eliminates variable ‘c’
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2 votes
2 votes

Answer is : A 

no. of correct Q = x

no. of wrong Q = y

no. of not attempted Q = z

x+y+z = 65 ........(1)

we need to find proper z

1× x +((-1/4)×y) +((-1/8)×z)= 37 .......(2)

now put value of z = 4 or 5 or 6 or 7  in  both equation until u got integer value of x , y .because you can correct e:g , 4 or 50 question . you can't correct 5.6  question.

put z=4

x+y=61

x-(y/4)-1/2=37

solve these and get x=42.2 ,y=18.8 not integer so  try z=5

put z=5 same as above but not getting integer value

now put z=6

2x+2y=118

8x-2y-6=296

got x=42 ,y = 17 both are integers

so, our answer will be 6 .

 

1 vote
1 vote
  •  65 questions of 1 marks each
  • Student got 37 marks.
  • Remaining marks = 65-37 = 28.

--------------------------------------------------------------------------------------

$\Rightarrow$ 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.

------------------------------------------------------------------------------------------------------------------------------------------------

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.

--------------------------------------------------------------------------------------------------------------------------------------------------------------

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.

--------------------------------------------------------------------------------------------------------------------------------------------------------------

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.

--------------------------------------------------------------------------------------------------------------------------------------------------------------

$\therefore$ Option A is the correct answer.

0 votes
0 votes
Consider, x number of questions answered correctly, y number of questions answered incorrectly. That leaves us with (65 – x – y) questions unattempted.

Total marks(M) = x – (y / 4) – (65 / 8) – (x / 8) – (y / 8) = 37                                                       (1)

We place in the values of z (number of questions unattempted), given in the question to get the equation

x + y = 65 – z                                                                                                                                         (2)

The value of z which leaves us with positive, integral values of x and y, when solved is the correct answer. In this case, it’s z = 6
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