The Gateway to Computer Science Excellence
+14 votes
807 views

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.
in Mathematical Logic by Boss (30.8k points)
recategorized by | 807 views

1 Answer

+25 votes
Best answer

$6$ 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 by
0
Not getting how did you got 2 as an upper bound?
0

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.
0

Best explanation for such new question.Thanks, saurav04.

0
..
0
Some one edit the answer

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

it should be CBA (last one)
0
0

@Shaik Masthan

Can a + b + c = 2 ?

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

0
on what basis is the equation written . I could not understand . can any one explain it
0

@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

Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,292 answers
198,224 comments
104,909 users