No of different ways of marks can be obtained in MCQ?












0












$begingroup$


No of different possible marks we can get if 'a' marks will be added for the correct answer and 'b' marks will be deducted if the answer happened to be wrong. An unattempted question is awarded 0 marks.
The number of questions will be n.



For example:
For n=2 and a=1 and b=1 is -2,-1,0,1,2.
I have read this link.
Possible marks in a MCQ test






combinatorics contest-math







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The answer will depend on the number of questions on the test, which you have left unstated.
    $endgroup$
    – N. F. Taussig
    Dec 15 '18 at 12:10






  • 1




    $begingroup$
    Number of questions will be n.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:11










  • $begingroup$
    Knock Knock.....
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:27










  • $begingroup$
    @SagarSharma do you mean the no of different marks or the possible marks?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 12:52










  • $begingroup$
    @SagarSharma Yes for?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 13:00
















0












$begingroup$


No of different possible marks we can get if 'a' marks will be added for the correct answer and 'b' marks will be deducted if the answer happened to be wrong. An unattempted question is awarded 0 marks.
The number of questions will be n.



For example:
For n=2 and a=1 and b=1 is -2,-1,0,1,2.
I have read this link.
Possible marks in a MCQ test






combinatorics contest-math







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The answer will depend on the number of questions on the test, which you have left unstated.
    $endgroup$
    – N. F. Taussig
    Dec 15 '18 at 12:10






  • 1




    $begingroup$
    Number of questions will be n.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:11










  • $begingroup$
    Knock Knock.....
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:27










  • $begingroup$
    @SagarSharma do you mean the no of different marks or the possible marks?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 12:52










  • $begingroup$
    @SagarSharma Yes for?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 13:00














0












0








0





$begingroup$


No of different possible marks we can get if 'a' marks will be added for the correct answer and 'b' marks will be deducted if the answer happened to be wrong. An unattempted question is awarded 0 marks.
The number of questions will be n.



For example:
For n=2 and a=1 and b=1 is -2,-1,0,1,2.
I have read this link.
Possible marks in a MCQ test






combinatorics contest-math







share|cite|improve this question











$endgroup$




No of different possible marks we can get if 'a' marks will be added for the correct answer and 'b' marks will be deducted if the answer happened to be wrong. An unattempted question is awarded 0 marks.
The number of questions will be n.



For example:
For n=2 and a=1 and b=1 is -2,-1,0,1,2.
I have read this link.
Possible marks in a MCQ test






combinatorics contest-math




combinatorics contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 13:02







Sagar Sharma

















asked Dec 15 '18 at 12:07









Sagar SharmaSagar Sharma

143




143








  • 2




    $begingroup$
    The answer will depend on the number of questions on the test, which you have left unstated.
    $endgroup$
    – N. F. Taussig
    Dec 15 '18 at 12:10






  • 1




    $begingroup$
    Number of questions will be n.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:11










  • $begingroup$
    Knock Knock.....
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:27










  • $begingroup$
    @SagarSharma do you mean the no of different marks or the possible marks?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 12:52










  • $begingroup$
    @SagarSharma Yes for?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 13:00














  • 2




    $begingroup$
    The answer will depend on the number of questions on the test, which you have left unstated.
    $endgroup$
    – N. F. Taussig
    Dec 15 '18 at 12:10






  • 1




    $begingroup$
    Number of questions will be n.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:11










  • $begingroup$
    Knock Knock.....
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 12:27










  • $begingroup$
    @SagarSharma do you mean the no of different marks or the possible marks?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 12:52










  • $begingroup$
    @SagarSharma Yes for?
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 13:00








2




2




$begingroup$
The answer will depend on the number of questions on the test, which you have left unstated.
$endgroup$
– N. F. Taussig
Dec 15 '18 at 12:10




$begingroup$
The answer will depend on the number of questions on the test, which you have left unstated.
$endgroup$
– N. F. Taussig
Dec 15 '18 at 12:10




1




1




$begingroup$
Number of questions will be n.
$endgroup$
– Sagar Sharma
Dec 15 '18 at 12:11




$begingroup$
Number of questions will be n.
$endgroup$
– Sagar Sharma
Dec 15 '18 at 12:11












$begingroup$
Knock Knock.....
$endgroup$
– Sagar Sharma
Dec 15 '18 at 12:27




$begingroup$
Knock Knock.....
$endgroup$
– Sagar Sharma
Dec 15 '18 at 12:27












$begingroup$
@SagarSharma do you mean the no of different marks or the possible marks?
$endgroup$
– Ankit Kumar
Dec 15 '18 at 12:52




$begingroup$
@SagarSharma do you mean the no of different marks or the possible marks?
$endgroup$
– Ankit Kumar
Dec 15 '18 at 12:52












$begingroup$
@SagarSharma Yes for?
$endgroup$
– Ankit Kumar
Dec 15 '18 at 13:00




$begingroup$
@SagarSharma Yes for?
$endgroup$
– Ankit Kumar
Dec 15 '18 at 13:00










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.



$$f(x)=f(y)$$$$implies x=y$$or$$a+b=0$$



Case $1$: $x=y implies injection implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.



The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 13:54










  • $begingroup$
    Oh. I'm sorry. I forgot about the unattempted ones
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 14:01











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1 Answer
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1 Answer
1






active

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active

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0












$begingroup$

Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.



$$f(x)=f(y)$$$$implies x=y$$or$$a+b=0$$



Case $1$: $x=y implies injection implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.



The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 13:54










  • $begingroup$
    Oh. I'm sorry. I forgot about the unattempted ones
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 14:01
















0












$begingroup$

Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.



$$f(x)=f(y)$$$$implies x=y$$or$$a+b=0$$



Case $1$: $x=y implies injection implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.



The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 13:54










  • $begingroup$
    Oh. I'm sorry. I forgot about the unattempted ones
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 14:01














0












0








0





$begingroup$

Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.



$$f(x)=f(y)$$$$implies x=y$$or$$a+b=0$$



Case $1$: $x=y implies injection implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.



The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!






share|cite|improve this answer









$endgroup$



Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.



$$f(x)=f(y)$$$$implies x=y$$or$$a+b=0$$



Case $1$: $x=y implies injection implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.



The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 13:15









Ankit KumarAnkit Kumar

1,479221




1,479221












  • $begingroup$
    5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 13:54










  • $begingroup$
    Oh. I'm sorry. I forgot about the unattempted ones
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 14:01


















  • $begingroup$
    5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
    $endgroup$
    – Sagar Sharma
    Dec 15 '18 at 13:54










  • $begingroup$
    Oh. I'm sorry. I forgot about the unattempted ones
    $endgroup$
    – Ankit Kumar
    Dec 15 '18 at 14:01
















$begingroup$
5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
$endgroup$
– Sagar Sharma
Dec 15 '18 at 13:54




$begingroup$
5 possible marks is coming as follows:(1)2 correct (2) 1 correct and 1 unattempted (3) 2 unattempted (4)1 wrong and 1 unattempted (5)2 wrong.
$endgroup$
– Sagar Sharma
Dec 15 '18 at 13:54












$begingroup$
Oh. I'm sorry. I forgot about the unattempted ones
$endgroup$
– Ankit Kumar
Dec 15 '18 at 14:01




$begingroup$
Oh. I'm sorry. I forgot about the unattempted ones
$endgroup$
– Ankit Kumar
Dec 15 '18 at 14:01


















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