edited by
2,472 views
4 votes
4 votes

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. $4$
edited by

5 Answers

Best answer
7 votes
7 votes

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. 

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 votes
1 votes
  •  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.

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

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
Answer:

Related questions

4 votes
4 votes
3 answers
3
admin asked May 13, 2019
1,121 views
Consider the series $\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{8}+\frac{1}{9}-\frac{1}{16}+\frac{1}{32}+\frac{1}{27}-\frac{1}{64}+\ldots.$ The sum of the infinite seri...
8 votes
8 votes
3 answers
4
admin asked May 13, 2019
1,378 views
Consider the function $f(x)=\max(7-x,x+3).$ In which range does $f$ take its minimum value$?$ $-6\leq x<-2$ $-2\leq x<2$ $2\leq x<6$$6\leq x<10$