SOLUTION: A school teams contain 68 students 33 do field events 40 do track events 23 do swimming 14 do both field and track events 8 do both swimming and field events. If 15 students do

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Question 1209574: A school teams contain 68 students 33 do field events 40 do track events 23 do swimming 14 do both field and track events 8 do both swimming and field events. If 15 students do field event only and 10 do both swimming and track events how many students do
1. Swimming only
2. Track events only
3. All three events
Solve with venn diagram
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
A school teams contain 68 students
33 do field events
40 do track events
23 do swimming
14 do both field and track events
8 do both swimming and field events.
If 15 students do field event only and 10 do both swimming and track events how many students do
(a) Swimming only
(b) Track events only
(c) All three events
Solve with venn diagram
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

You have a universal set of all students U with 68 elements,  and its subsets

    F    33  (field)

    T    40  (track)

    S    23  (swimming)

    FT   14  (field and swimming intersection)

    FS    8  (field and swimming intersection)

    TS   10  (track and swimming intersection)

    Fo    15  (field only)


Field only is the set F minus subset FT minus subset FS.  Therefore, we write this equation for quantities

    |Fo| = |F| - |FT| - |FS| + |FTS|.    (1)


Here FTS is the intersection of F, T and S.  Equation (1) is valid, since when we subtract |FT| and |FS| 
from |F|,  we count the triple intersection twice.


Equation (1) gives us

    15 = 33 - 14 - 8 + |FTS|,  from which we find  |FTS| = -33 + 14 + 8 + 15 = 4.


So,  the triple intersection FTS (all three events) has 4 students.  Question (c) is answered.



For question (a), write similar to equation (1)

    So = |S| - |FS| - |TS| + |FTS|


You know now that |FTS| = 4;  so, you substitute the numbers and get the answer

    So = 23 - 8 - 10 + 4 = 9   (answer to question (a) )



For question (b), write similar to equation (1)

    To = |T| - |FT| - |TS| + |FTS|


Again, you substitute the numbers and get the answer

    To = 40 - 14 - 10 + 4 = 20   (answer to question (b) )

Thus, all questions are answered, so the problem is solved in full,
with complete explanations.

Instead of Venn diagram, you learned how to solve and how to analyze
such problems using basis of elementary set theory.


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Venn diagrams is an activity similar to transferring matches from one matchbox to another.

Doing this way, you will learn transferring matches, but nothing more.

Learning from my solution, you will learn a part (= a piece) of Math and relevant reasoning.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

F = students who do field events
T = students who do track events
S = students who swim

Here's a blank template for a Venn Diagram with 3 overlapped circles.

Letters a through h represent the 8 distinct regions.
They are the subsets of people such that,
a = number who do field events only
b = number who do field and track, but not swim
c = number who do track events only
and so on.

Here's the info from the instructions presented as a numbered list.
  1. There are 68 students total
  2. There are 33 students doing field events
  3. There are 40 students doing track events
  4. There are 23 students swimming
  5. There are 14 students do both field and track events
  6. There are 8 students doing both swimming and field events.
  7. There are 15 students doing field events only
  8. There are 10 doing both swimming and track events
There are 8 bits of info to somehow match with the 8 letters a through h.

I'll refer to those facts as fact (1) through fact (8) in the order presented above.

fact (7) leads immediately to a = 15.
fact (5) tells us that b+e = 14 while fact (6) leads to d+e = 8
Adding these equations straight down gives b+d+2e = 22 and rewrites to b+d = 22-2e

Then turn to fact (2)
a+b+d+e = everyone in circle F = 33 students
a+b+d+e = 33
15+b+d+e = 33 ....... plug in a = 15
15+(22-2e)+e = 33 ....... replace b+d with 22-2e
15+22-e = 33
37-e = 33
-e = 33-37
-e = -4
e = 4
We have determined there are 4 students who performed at all 3 events.

Then,
b+e = 14 becomes b+4 = 14 which solves to b = 10
d+e = 8 becomes d+4 = 8 which solves to d = 4

We have this so far
a = 15
b = 10
d = 4
e = 4

Let's update the Venn Diagram to reflect this.

As a check so far, the numbers in circle F should add to 33 students who did field events.
15+10+4+4 = 33
So far so good.


Now onto fact (8) which gives the equation 4+f = 10. It solves to f = 6.

Once you know f, you can determine g.
This is because all of the values in circle S should add to 23. See fact (4).

4+4+f+g = 23
4+4+6+g = 23
14+g = 23
g = 23-14
g = 9

Similarly, we can determine the value of c when using fact (3).
10+c+4+f = 40
10+c+4+6 = 40
c+20 = 40
c = 40-20
c = 20


At this point we have the following
a = 15
b = 10
c = 20
d = 4
e = 4
f = 6
g = 9
These seven values add to 68
fact (1) says that this is the total number of students, so each student does at least one of the three events mentioned.
This means h = 0.

This is the completed Venn Diagram



Answers:
Only swimming = 9
Only track = 20
All three events = 4

More practice with Venn Diagrams is found here