Holiday dinners at June’s house are always a festive occasion. One particular holiday, June has 16 guests. June offers the guests apple pie and blueberry pie. While any guest can have slices of both pies, June never serves a particlar guest more than one slice of each pie and each guest is allowed to decline any slice of pie they don’t care for.

June serves a total of 10 slices of apple pie and a total of 11 slices of blueberry pie.

June serves 8 guests slices of both pies.

How many guests had no pie at all?

a. 3

b.2

c.0

a. 3

Step-by-step explanation:

We solve this question gathering the information given in the question.

June serves 8 guests slices of both pies.

This is soon going to be used

June serves a total of 10 slices of apple pie

This includes the 8 guests which got served both pieces. So 10 - 8 = 2 guests were served only apple pie.

11 slices of blueberry pie.

By the same logic as above, 11 - 8 = 3 guests were served only blueberry pie.

So

8 + 2 + 3 = 13 guests were served at least one pie.

16 - 13 = 3 guests had no pie at all.

answer: 7

step-by-step explanation:

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.