Bunuel wrote:
In a multiple-choice test, each question is followed by n options. What is the value of n?
(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.
(2) For every correct answer, a candidate is awarded 1 mark and for every wrong answer, there are −1/3 marks assigned.
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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