How to solve following question? In an upcoming election, 15% of married voters will vote for Candidate A, while the rest will vote for Candidate B; 80% of unmarried voters will vote for Candidate A, while the rest will vote for Candidate B. Which of the following represents the lowest percentage from all voters combined (married and unmarried) that must be unmarried (not married) in order for Candidate A to win the election?

The lowest percentage from all voters combined that must be unmarried in order for Candidate A to win the election is 53.85%.

Step-by-step explanation:

Proportion married:

x are married

1 - x are unmarried.

Will vote for candidate A:

15% of x

80% of 1 - x. So

Candidate A wins:

If his proportion is more than 50%, that is:

Highest percentage of married is 46.15%, so:

The lowest percentage of unmarried is:

100 - 46.15 = 53.85%.

The lowest percentage from all voters combined that must be unmarried in order for Candidate A to win the election is 53.85%.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.