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I came across this question in a Mental Ability Test. The question is as follows —

Anita, a girl, goes to a party. Entry into the party requires a special passcode. In order to know the passcode she listens to the conversation between the guard and two people.

In the first conversation between the guard and the man, the guard says $12$ and the man says $6$ and he is granted entry.

In the next dialogue between the guard and the lady the guard says $6$ and the woman says $3$ and then she is granted entry.

Then Anita thinking she has got the passcode goes to the guard. The guard says $10$ and Anita says $5$ but the guard says "You are wrong!" and denied her entry.

What should Anita have said to gain entry?

I have no idea on how to solve this problem.

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2 Answers 2

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Anita should've said:

3

Because

The responses are the number of letters in each number. 12 = twelve = 6 letters, 6 = six = 3 letters, 10 = ten = 3 letters.

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Anita should've said:

8

Because

The passcode is the number between 0 and 8 that must be added to the guard's number to render a result which is divisible by 9. 12 + 6 = 18; 6 + 3 = 9; 10 + 8 = 18.

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  • $\begingroup$ Well indeed. I can't possibly know that, but I don't think there's any information in the question that rules this out as a possible answer. If this question was really being used in a "mental ability test", I hope it's not on the assumption that there's only one possible correct answer. $\endgroup$
    – atkins
    May 16, 2016 at 10:31
  • $\begingroup$ If you'd like another, how about "multiply the number of digits by three and subtract from the original number". Then the answer is 10 - 6 = 4. $\endgroup$
    – atkins
    May 16, 2016 at 10:31
  • $\begingroup$ (+1)Thanks for your answer...your logic is quite correct but in the MCQ exam,the options included 10,20,6,3 as the valid answers... $\endgroup$
    – Soham
    May 16, 2016 at 10:34
  • $\begingroup$ I'm glad they at least allowed for several possibilities! Now I'll see if I can work out how 10, 20, and 6 can be obtained :) $\endgroup$
    – atkins
    May 16, 2016 at 10:40
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    $\begingroup$ Well, trivially, we can construct a quadratic polynomial for any value: $f(n)=\frac14 n^2-4n+18↦f(6)=3, f(10)=3, f(12)=6$; $f(n)=-\frac18n^2+\frac{11}4n-9↦f(10)=6$; $f(n)=-\frac58n^2+\frac{47}4n-45↦f(10)=10$; and $f(n)=-\frac{15}8n^2+\frac{137}4n-135↦f(10)=20$ (thanks, Wolfram|Alpha).  But,  if  you  want  to  disallow  exponents,  how  about $f(n)=3\times⌈\frac{n+1}{10}⌉↦f(10)=6$; $f(n)=\frac74n-\frac{15}2⌈\frac{n}{10}⌉↦f(10)=10$; and $f(n)=\frac{n}2+15\times(n\bmod 3)↦f(10)=20$. $\endgroup$ May 20, 2016 at 6:57

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