Easy but I don't understand! Sets, Venn diagrams

Question

In the figure ξ is the set of all students in Secondary One. (The figure shows a Venn diagram with two overlapping circles. There is only 1 intersection, labeled y. The left circle is represents G and the right circle represents C. circle G, without the intersection is part x and circle C without the intersection is part z. the letter w is outside the Venn diagram but within the box.)

G= {students who joined uniformed groups} and 

C= {students who join clubs and societies},

 

The letters w, x, y, and z in the figure, represent the number of students in each subset. Given that n(ξ)= 200, n(G)= 120, and n(C)= 90,

 

a) find the smallest possible value of y

 

b) find the largest possible value of w 

 

c) find the value of x and z if w=30

 

 

 

i need step by step answers for both parts! Thanks in advance. It is quite a simple question but I dont understand how to do both parts. Can't the smallest possible value of y be a zero?

Answer 1

Sorry, not circles.
= = = = = = = = = = = = = = = = = =
=        w                                            =
=                                                       =
=    G* * * * * *                                 =
=    *      * * * * * * * * * C                =
=    *      *  y   *               *                =
=    * x   *       *               *                =
=    * * * * * * *     z        *                =
=            *                       *                =
=            * * * * * * * * * *                =
=                                                       =
= = = = = = = = = = = = = = = = = =

Given that n(ξ)= 200, n(G)= 120, and n(C)= 90
    n(G) = 120 = x + y      [120 students who joined uniformed groups]

    n(C) = 90 = y + z [90 students who join clubs and societies]

    There were y students who joined both uniformed groups and clubs/societies. Note that y appears in each equation.

n(ξ)= 200 = w + x + y + z       [all students in Secondary One] [This is the "universe"]

The math:
   n(ξ)= w + (x + y) + (z + y) - y = 200    [rearrange; include y that cancels]
   w + n(G) + n(C) - y = 200        [ah, we know those]
   w + 120 + 90 - y = 200

   w - y = -10
   y - w = 10

So, if w=0 [no uninvolved Secondary One students], y=10 [This the minimum (smallest possible) value for y]. Note (to answer your question): y cannot equal 0 because that means that w = -10.

We also have:

    x + y = 120
     so, if x=0 then y = 120 thus y <= 120
 and
    y + z = 90
     so, if z=0 then y = 90 thus y <= 90

Use y <=90   (because they both must be true)
     y-10 <= 80
         w <= 80     (because w-y=-10 means w=y-10)

The largest (maximum value) for w is 80.
    So, it could be that w=80, x=30, y = 90, and z=0.
    Other values of x, y, or z make w smaller.

Answer 2

Let's goof around with this situation for a bit.

 

For a moment, let's assume that all students are required to participate in G or C or both.  That means that w = 0.  W stands for the students that are a part of the 200 but not participating in either.  So:  if w = 0, then x + y + z = 200.

  

If you add 120 + 90 you get 210.  But there are only 200 girls.  So 210-200 = 10 students must be in both.

 

So, since there are 120 students in Groups, that includes x, the number of students in ONLY G and y, the number of students who are in both.  x + 10 = 120.  There are 110 students in only G.

 

There are 90 in C.  z + 10 = 90.  80 students are only in C.

 

Double checking 110 + 10 + 80 = 200.  yep. 

______________________________________

 

What if there are 5 students who do not participate.  w = 5.

 

x + y + z + w = 200.  Since we said w = 5, x + y + z + 5 = 200.

 

Go through the same gyrations we did in the first part, except instead of 200, use 195.

 

Does this help?