In a multiple-choice test, each question is followed by n options.

(1) The test is designed in such a way that if a candidate appearing for the test, decides to mark the options for every question in a random manner, the candidate would expect to receive ‘0’ marks in the test. - Insufficient since we do not know the weightage of correct and incorrect answers

(2) For every correct answer, a candidate is awarded 1 mark and for every wrong answer, there are −1/3 marks assigned - Insufficient - now we only the weightage of correct and incorrect answers - marking system

We know the test taker selects options at random and total score when test takers marks answer options in random manner is 0
1 + (n-1)* -1/3 = 0
=>n/3 = 4/3
=> n =4
Answer C

1. We cannot infer n since we do not know the marking system.

2. We know the marking system but cannot infer n since we do not know about total marks earned by the candidate.

Combining both the statements, if the candidate randomly marks the options and earns 0 marks in the test,
since one option is correct and rest n-1 options are wrong-

1 + (n-1)*(-1/3) = 0

Sufficient to calculate n.

Hence ans C.

i dont understand.
how do everyone get the formula -- 1 + (n-1)*(-1/3) = 0
does that mean the candidate get 1 correct, and the other one wrong?
the question is asking how many options for the each question, how do you all get this formula?

Hello pclawong,

Let me try to answer your question.

Let's say, the candidate guesses the 1st question correctly and gets 1 mark. For each question, only one option is correct and rest of the options are wrong.

If there are n options for each question to choose from, he gets 1 for selecting the right option and (n-1)*(-1/3) for choosing the other (n-1) wrong options for the following questions.

So, our equation becomes 1 + (n-1)*(-1/3) = 0. Sufficient to calculate n.

Hence, both A and B are needed to answer the question correctly.

Doubts?

Does that mean this person can select all the options?
so he can at least get 1 point?
does that mean he select all the option for each question?
thank you so much.
it just very confusing to me.
the question states that he ramdomly choose an option but didn't say he chooses all the options

Does that mean this person can select all the options? No.
does that mean he select all the option for each question? Also no.

Each question has n options, out of which only one is correct. He chooses only one option per question, which could be either correct or incorrect.

Let's say, he chooses the right option for the first question. His score is 1.
For the next question, he chooses one out of those (n-1) wrong options. He gets -1/3. So his total score now is 1-(1/3) = 2/3.
He continues to do this till his score is 0.

So, the equation is 1+(n-1) * (-1/3) = 0

Statements 1 and 2 are clearly insufficient on their own.

combining,

Let number of correct answers be \(a\)
Let number of incorrect answers be \(b\)

Total marks in random manner = \(a(1) + b(-1/3) = 0\)
=> \(b = 3a\)

=> For every correct question answered, 3 questions will be incorrectly answered
=> for this to happen, there has to be 4 options, so that out of 4 questions attended, 3 will be incorrect and 1 will be correct

so answer is 4 options (C)

Nobody told us that every question has only 1 correct answer.

I agree that (1)+(2) gives us the ratio between correct and wrong answers to be 1:3, but the number of correct answers could be 1, 2, 3, ... hence n could be 4, 8, 12, ...

Hence (E) is the correct answer.

I followed a method different from the ones discussed above. Although I got the right answer, I am not sure if the approach is correct.

Statements 1 & 2 individually are not sufficient.

Combining statements 1 & 2 we get,
i) Total score will be 0 if the + and - marks are balanced out. And for that the correct and incorrect options selected should balance out accordingly.
ii) For correct option(C), 1 mark is rewarded and for incorrect option(IC), -1/3 is deducted.

So for the total marks to be zero, the count of C and IC options selected should give us, IC = 3*C.

We can say that the probability of selecting a correct option is \(\frac{C}{T}\), where T is total options.
This can also be written as, \(\frac{C}{T} = \frac{1}{n}\) where n is the total options for each question. There will always be 1 correct option.

\(\frac{C}{T} = \frac{1}{n}\)

\(\frac{C}{(C+IC)}=\frac{1}{n}\)

\(nC = C + IC\)
\(nC = C + 3C\)
\(nC = 4C\)
\(n = 4\)


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HI chetan2u
Why can't we say that n=8,12,16....
If n=4 then ratio of correct and incorrect is 1:3
But it could easily be 2:6,.....
Can you please delve deeper into why it can't be 8,12,16

Thank you
Regards

Hi,
A ratio of 2:6 would mean two correct answers and 6 wrong answers.
But can a question have two right answers. May not be.
So each question has 4 options where 1 is correct and other three wrong.
Eight options with one correct would make ratio 1:7

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