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Three candidates, Amar, Birendra and Chanchal stand for the local election. Opinion polls are conducted and show that fraction $a$ of the voters prefer Amar to Birendra, fraction $b$ prefer Birendra to Chanchal and fraction $c$ prefer Chanchal to Amar. Which of the following is impossible?

1. $(a, b, c) = (0.51, 0.51, 0.51);$
2. $(a, b, c) =(0.61, 0.71, 0.67);$
3. $(a, b, c) = (0.68, 0.68, 0.68);$
4. $(a, b, c) = (0.49, 0.49, 0.49);$
5. None of the above.

recategorized | 807 views

$6$ preference order for voter are possibe:

$ABC.ACB,BCA,BAC,CAB,CBA$ also Given that

$a=ABC+ACB+CAB(A$ prefer over $B) ---(1)$

$b=BCA+BAC+ABC(B$ prefer over $C) ---(2)$

$c=CAB+CBA+BCA(C$ prefer over $A) ---(3)$

Adding $1,2$ and $3$ we get

$a+b+c=2(ABC+BCA+CAB)+ACB+BAC+CBA$

Now we know that $ABC+ACB+BAC+BCA+CAB+CBA=1$ therefore

$[ABC+ACB+BAC+BCA+CAB+CBA]<[2(ABC+BCA+CAB)+ACB+BAC+CBA]<2(ABC+ACB+BAC+BCA+CAB+CBA)$

Hence we can say that value of $a+b+c$ must be between $1$ and $2$

option (c) value greater than $2$ hence correct answer is (c)

by Active (1.6k points)
edited
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Not getting how did you got 2 as an upper bound?
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please explain the upper bound thing ?

+2
We know that (ABC+ACB+BAC+BCA+CAB+CBA)=1 therefore 2 *(ABC+ACB+BAC+BCA+CAB+CBA) = 2.

[ABC+ACB+BAC+BCA+CAB+CBA] < [2(ABC+BCA+CAB)+ACB+BAC+CAB]

< 2(ABC+ACB+BAC+BCA+CAB+CBA)

This can be written as, 1 < (a+b+c) < 2.

It means (a+b+c) must be lies bw 1 and 2, all options follow except option (c).

So,(C) is ans.
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Best explanation for such new question.Thanks, saurav04.

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..
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a+b+c=2(ABC+BCA+CAB)+ACB+BAC+CAB

it should be CBA (last one)
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@Shaik Masthan

Can a + b + c = 2 ?

or is (a + b + c) strictly less than 2?

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on what basis is the equation written . I could not understand . can any one explain it
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@Ruchi Vora

fraction a of the voters prefer Amar to Birendra

It means all the cases where Amar should come before Birendra  i.e. ABC or CAB or ACB (position of C doesn't matter)

so thats why a = ABC+CAB+ACB