Section 6.1
Problems 1-4
These problems will be done with both a Venn Diagram and a Table (in parallel) to show you
the advantages of both methods…you will find which is most comfortable for you. But
remember the table can only be used for a 2-circle Venn Diagram…a 3-circle Venn Diagram
would require a cubic table, which is extremely cumbersome…so for 3-circle Venn Diagrams
that’s all you have.
Problem 1
Venn Diagram
Table
Total
Total
“Nineteen items are categorized…” or in other words there are a total of 19 items:
Venn Diagram
Table
Total
Total
19
19
“...5 in A but outside B...” so our 2 circles (or areas are defined as A and B) and the 5 is NOT in
A but is in B:
Venn Diagram
Table
In A
B
A
5
19
In B
Not B
Total
“...4 in B but outside A...” so the 4 is NOT in B but is in A:
Not A Total
5
19
Venn Diagram
Table
In A
B
A
In B
Not B
Total
5
4
19
Not A Total
5
4
19
“...7 outside both A and B.” so the 7 is NOT in B and NOT in A:
Venn Diagram
Table
In A
B
A
In B
Not B
Total
5
4
7
19
4
Not A Total
5
7
19
TECHNICALLY at this point you are done…but some require you completing both the Venn
Diagram and the table, because that is what you will need to do later.
Now to solve each of the above…
For the Venn Diagram, since all areas have a number except one (the intersection of A and B)
then we use the idea that all the areas must add up to the total: 4+5+7+x = 19, so 16+x=19,
therefore the intersection must be 3.
For the Table, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with 5+7 = 12 and 4+7 =11, then in the Totals column and row:
19-11 = 8 and 19-12 = 7; and finally “In A” column gives me 7-4 = 3
Venn Diagram
Table
B
A
4
3
5
7
19
In B
Not B
Total
In A
3
4
7
Not A
5
7
12
Total
8
11
19
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Problem 2
Venn Diagram
Table
Total
Total
“In a universal set of 80 elements…” or in other words there are a total of 80 elements:
Venn Diagram
Table
Total
Total
80
80
“...17 are in C...” so one of our circles is defined as C and the 17 is the sum of the two areas
that make up the circle defined as C:
Venn Diagram
Table
In C
C
Not C Total
17
Total
80
17
80
“...29 are in D...” so the other circle is defined as D and the 29 is the sum of the two areas that
make up the circle defined as D:
Venn Diagram
C
17
Table
In C
D
28
80
In D
Not D
Total
17
Not C Total
28
80
“...39 in CUD.” so the 39 is the total of the union of C and D (which means the total of all the
areas of the two circles equals 39). This also tells us that outside the two circles must be what
is left of the total of 80 (80-39 = 41):
Venn Diagram
Table
In C
D
C
In D
Not D
Total
28
17
41
80
17
Not C Total
28
41
80
TECHNICALLY at this point you are done…but some require you completing both the Venn
Diagram and the table, because that is what you will need to do later.
Now to solve each of the above…
For the Table, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with Totals column and row: 80-28 = 52 and 80-17 = 63; then the
row/column containing the “41”: 52-41 = 11 and 63-41 = 22; and finally “In C” column gives
me: 17-11 = 6
For the Venn Diagram, since we have a union and the totals for each of the two circles, we
must use the equation CUD = C + D - C∩D, which gives us 39 = 17 + 28 - C∩D or
39 = 45 - C∩D, so the intersection is equal to 6. With the intersection equal to 6 then the
other areas must add up to the total…for C: 17 = 6 + x, or x=10; and for D: 28 = 6 + y, or y=22
Venn Diagram
Table
D
C
17
11
6
28
22
41
80
In D
Not D
Total
In C
6
11
17
Not C
22
41
63
Total
28
52
80
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Problem 3
Venn Diagram
Table
Total
Total
“Of 26 elements…” or in other words there are a total of 26 elements:
Venn Diagram
Table
Total
Total
26
26
“...5 are in both R and S...” so the place where you are at both R and S is called the intersection
(the place in R and in S) and that value is 5:
Venn Diagram
Table
R
S
In S
Not S
Total
5
26
In R
5
Not R Total
26
“...9 are in S...” so the circle defined as S contains 9, and 9 is the sum of the two areas that
make up the circle defined as S:
Venn Diagram
R
S
9
Table
5
26
In S
Not S
Total
In R
5
Not R Total
9
26
“...3 are in neither.” so the 3 is the NOT in either circle(so outside the 2 circles):
Venn Diagram
Table
R
S
9
In S
Not S
Total
5
3
26
In R
5
Not R Total
9
3
26
TECHNICALLY at this point you are done…but some require you completing both the Venn
Diagram and the table, because that is what you will need to do later.
Now to solve each of the above…
For the Table, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with Totals column: 26-9 = 17; then the “In S” row: 9-5 = 4; then the
“Not C” column: 4+3 = 7; then the “Not D” row: 17-3 = 14; and finally “In R” column gives me:
14+5 = 19
For the Venn Diagram, since we know one of the two segments of the S circle then the other
must equal 4, so that 5+4 = 9. With 3 of the 4 parts identified, the last part must add up to 26:
26 = 4+5+3+x = 12+x, so x=14.
Venn Diagram
Table
D
C
9
4
5
14
3
26
In S
Not S
Total
In R
5
14
19
Not R
4
3
7
Total
9
17
26
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Problem 4
Venn Diagram
Table
Total
Total
“Of 22 elements…” or in other words there are a total of 22 elements:
Venn Diagram
Table
Total
Total
22
22
“...6 are in V but outside W...” so inside the V circle (which has two parts) but the part that is
NOT in the W circle is where the 6 goes:
Venn Diagram
Table
In V
W
V
In W
Not W
Total
6
22
Not V Total
6
22
“...7 are in W but outside V...” so inside the W circle (which has two parts) but the part that is
NOT in the V circle is where the 7 goes:
Venn Diagram
Table
In V
W
V
6
7
22
In W
Not W
Total
Not V Total
7
6
22
“...8 are in neither W nor V.” so the 8 is the NOT in either circle(so outside the 2 circles):
Venn Diagram
Table
In V
W
V
In W
Not W
Total
7
6
8
22
6
Not V Total
7
8
22
TECHNICALLY at this point you are done…but some require you completing both the Venn
Diagram and the table, because that is what you will need to do later.
Now to solve each of the above…
For the Table, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with “Not V” column: 7+8 = 15; then the “Not W” row: 6+8 = 14;
then the “Total” column and row: 22-14 = 8 and 22-15 = 7; and finally “In V” column gives me:
7-6 = 1
For the Venn Diagram, since we know 3 of the 4 parts identified, the last part must add up to
22: 22 = 6+7+8+x = 21+x, so x=1.
Venn Diagram
Table
W
V
6
1
7
8
22
In V
In W
1
Not W
6
Total
7
Not V
7
8
15
Total
8
14
22
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Three circle Venn Diagrams can use a table as long as it is 3-dimensional (or a cube), but that is
too cumbersome so we just stay with just the Venn Diagram…
Problem 5(a)
G
We know this is a 3-circle Venn Diagram because other
than the Universe (U) there are three other sets (G, H & K).
H
K
Start filling from the center whenever possible, so
looking at G, H and K there is only one element they have in common “l”…so that is placed in
the intersection of the three circles.
G
l
Next we find out what each intersection has…
for example G∩H is where G and H have in
K
H
common…{j, k, l}…so the j and k go in the other
part of the intersection. Likewise, G∩K is {l}, so
there is nothing in the other part of the intersection. And lastly, H∩K is {l, m}, so the m goes in
the other part of the intersection…
G
jk
l
m
K
Next we find what goes in the
H
remainder of any set, by looking at
what is in the set and seeing what has already been displayed…for G, we have j, k and l, but
there needs to also be h and i…so those go in the upper part of the circle…similarly for H, we
have j, k, l and m…which is everything, so nothing needs to be added to the diagram…lastly,
for K, we have l and m, but we need n and p, so those are placed in the right side of K…
G
Lastly, we check to make sure all the elements in
the Universe (U) are in the Venn Diagram…in this
case we are missing r, so that is placed outside the
three circles…
hi
jk
l
m
H
np
K
G
hi
r
jk
l
m
H
<< answer to 5(a)
np
K
Problem 5(b)
G∩K means what elements are in both G and K…in this case: {l}
Problem 5(c)
n(H) means how many elements are in H…so count them up: 4
Problem 5(d)
n(GUH) means how many elements are in both G and H (not counting the duplicates)…so the
union (U) is {h,i,j,k,l,m}, so the number of elements is: 6
Problem 5(e)
n(G∩H) means how many elements are in both G and H (not counting the duplicates)…so the
intersection (∩) is {j,k,l}, so the number of elements is: 3
Problem 5(f)
Hc∩K means what elements are in both Hc (complement of H) and K…first what is NOT in H
(which is the same as Hc): {h,i,n,p,r};
and of these elements only the following are also in K: {n,p}
Problem 5(g)
GU(H∩K) means what elements are in both G and the intersection of H and K…first the
intersection of H and K are all the shared elements: {l,m};
and of these and those in G, the union (bringing all the elements but no duplicates together),
you get: {h,i,j,k,l,m}
Problem 6
A
We know this is a 3-circle Venn Diagram because other
than the Universe (U) there are three other sets (A, B & C).
B
C
Start filling from the center whenever possible, so
n(A∩B∩C) is the intersection and is equal to 1.
A
1
Next we find out what each intersection has…
for example A∩B is where A and B have in
C
B
common (football shaped area); and this must
add up to 5, and since there is a 1 in one of the sections, the other section must equal 4.
Likewise, A∩C equals 6 so the other part of the intersection must be 5. And lastly, B∩C equals
4, so the last piece is 3…
A
4
1
5
3
C
Next we find what goes in the
B
remainder of any set, by looking at
what is in the set and seeing what is left over…for A, it must add up to 12, but there is already
10 in the circle…so the upper part of the circle is 2…similarly for B, it must add up to 12…which
means there is 4 more…and lastly, for C, it must add up to 16, so a 7 is placed in the right side
of C…
A
Lastly, we check to make sure all the elements add
up to the Universe (U), which is 35… in this case
we are short by 9, so 9 is placed outside the
three circles…
2
4
4
B
1
3
5
7
C
9
Problem 7
R
We know this is a 3-circle Venn Diagram because other
than the Universe (U) there are three other sets (R, S & T).
S
T
Start filling from the center whenever possible, but in
this case, it is missing so now there are THREE ways in which to solve this problem:
(1) Guess and correct (almost like trial and error),
(2) use the Union Equation for three circles (which you may or may not have been given yet),
or (3) an elaborate chain of equations, setting the center as “x”. I will show you all three, and
you must decide which is best for you…
First method: GUESS
If you are going to guess I would recommend either the highest or lowest values that the
center could possibly be…the lowest is ZERO (negative values are not allowed in a Venn
Diagram); the highest will be the value of the smallest intersection (in this case: 4), because if
you any value higher than 4 in the center the other part of that intersection would have to be
negative in order for the intersection to be equal to 4. Also the reason for using the end points
is that is you choose another number and it is off by ONE which way will you adjust (one way is
correct and the other will require you to correct again)…this way when you correct it will
always be right the next time. So now let me do this for BOTH zero and four:
With the center being ZERO:
First calculating the other part of each
intersection:
For R∩S, the other part is 7,
For R∩T, the other part is 4, and
For S∩T, the other part is 10
Next complete the value of each circle:
For R, the last part is 8-7-4-0 = -3 (and yes, I
know it is negative, just leave it be),
For S, the last part is 22-7-10-0 = 5, and
For T, the last part is 12-4-10-0 = -2 (we leave
this be too; you will see)
With the center being FOUR:
First calculating the other part of each
intersection:
For R∩S, the other part is 7-4 = 3,
For R∩T, the other part is 4-4 = 0, and
For S∩T, the other part is 10-4 = 6
Next complete the value of each circle:
For R, the last part is 8-3-0-4 = 1,
For S, the last part is 22-3-6-4 = 9, and
For T, the last part is 12-0-6-4 = 2
Since outside the three circles is defined as
n(RUSUT)c…we can add up all the numbers:
7+(-2)+5+(-3)+10+4+7+0 = 28, so a difference
of 4…so the center (in this case) will be 4.
Since outside the three circles is defined as
n(RUSUT)c…we can add up all the numbers:
7+2+9+1+6+0+3+4 = 32, so a difference of
0…so the center (in this case) will be 4. In
most cases it will be something different than
either end point.
Second method: Union Equation
The Union equation for a 3-circle Venn Diagram (or 3 sets) is (using the letters R, S & T):
n(RUSUT) = n(R) + n(S) + n(T) – n(R∩S) – n(R∩T) – n(S∩T) + n(R∩S∩T)
(where n(R∩S∩T) is the center)
You have been given all the information directly for everything but the center, R∩S∩T, and the
union, RUSUT, but you have enough information for the union…because you are given what is
in the “universe” and you are given what is NOT in the union, n(RUSUT)c…so subtracting those
will give you the number in the union: 32 – 7 = 25
So filling in the equation you get: 25 = 8 + 22 + 12 – 7 – 4 - 10 + n(R∩S∩T)
Or 25 = 42 – 21 + n(R∩S∩T)
Or 25 = 21 + n(R∩S∩T)
Or 4 = n(R∩S∩T)
Third method: Chain of Equation
First, if you set the center as “x”, then each of the intersections second piece (remember that
the intersection is the football-shaped part of the diagram with two parts: the center and the
“other piece”) is equal to the value of the intersection minus the center, because the two
pieces must add up to the intersection…so for R∩S, the “other piece” is 7-x and for R∩T, the
“other piece” is 4-x, and for S∩T, the “other piece” is 10-x…so the Venn Diagram looks like:
R
7-x
x
4-x
10-x
S
T
7
Now each of the circles must add up to the value given, so for the circle R, it must add up to 8,
so (7-x) + x +(4-x) + the last piece = 8,
or
11-x + the last piece = 8
or
the last piece = x-3
Similarly for the circle S, it must add up to 22,
so (7-x) + x +(10-x) + the last piece = 22,
or
17-x + the last piece = 22
or
the last piece = x+5
Similarly for the circle T, it must add up to 12,
so (10-x) + x +(4-x) + the last piece = 12,
or
14-x + the last piece = 12
or
the last piece = x-2
So finally, we know that all the pieces must add up to the value of the universe (32), so:
32 = x + (7-x) + (4-x) + (10-x) + (x-3) + (x+5) + (x-2) + 7 or
32 = x + 28 or
4=x
NOW it is up to you how you wish to calculate the center when it is not given to you!
R
4
T
S
7
Next you must make each of the intersections add up to the values given:
R
3
4
0
6
S
T
7
R/S: 3+4 = 7, S/T: 4+6 = 10 and R/T: 4+0 = 4
Finally you must make each circle add up to the values given:
R
1
3
9
4
0
6
S
2
T
7
R: 1+3+4+0 = 8, S: 9+3+4+6 = 22, and T: 2+6+4+0 = 12
Problem 8 (a & b)
This is a 2-circle Venn Diagram and table (knowing that you also have to do a table tells you
that it is a 2-circle Venn Diagram)…but what are the two circles? One of the circles is going to
be either boys or girls (it is your choice), but the other circle will NOT be the other sex, because
if (for example) I choose boys (B) for my circle, then everything in the circle is a boy and
everything outside the circle is not a boy (or a girl)…so that leaves lunches or no lunches for
the other circle (again your choice) for my example I will choose lunches (L)…
Venn Diagram
B
Table
B
L
G
Total
L
No L
Total
“32 students…” or in other words there are a total of 32 students or elements:
Venn Diagram
Table
B
32
L
No L
Total
G
Total
32
“...18 of them boys...” so inside the B circle (which has two parts) adds up to 18:
Venn Diagram
Table
B
L
B
L
No L
Total
18
32
G
18
Total
32
“...7 boys… brought lunches...” so inside the B circle AND also in the L circle is where the 7
goes:
Venn Diagram
Table
L
B
18
7
32
L
No L
Total
B
7
G
18
Total
32
“...11 girls brought lunches...” so the 11 is inside the L circle but outside the B circle:
Venn Diagram
Table
L
B
18
7
11
32
L
No L
Total
B
7
18
G
11
Total
32
Now to solve each of the above…
For the Table, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with “B” column: 18-7 = 11; then the “L” row: 7+11 = 18; then the
“Total” column and row: 32-18 = 14; and finally “G” column gives me: 14-11 = 3
For the Venn Diagram, since V must add up to 18 and one part is 7, then the other part must
be 11, and since we now know 3 of the 4 parts identified, the last part must add up to 32:
32 = 11+7+11+x = 29+x, so x=3.
Venn Diagram
Table
L
B
18
11
7
L
No L
Total
11
3
32
B
7
11
18
G
11
3
14
Total
18
14
32
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Problems 9 & 10
This is a 2-circle Venn Diagram because it says it is…but what are the two circles? One of the
circles is going to be the Morning Sun (S) and the other will be the Evening Bugle (B) … (your
choice on the lettering or designation)…though the problem 9 “suggests” guessing in a Venn
Diagram, you could guess or equation to solve, while problem 10 you use the table (I will show
both in parallel)…
Venn Diagram
B
Table
B
S
No B
Total
S
No S
Total
“Among 60 homes…” or in other words there are a total of 60 homes or elements:
Venn Diagram
B
Table
B
S
60
S
No S
Total
No B
Total
60
“...40 receive at least one of two local papers...” so “at least one: means that 40 of the 60
homes will receive 1 or both of the local papers…it also means that the other 20 homes do not
receive either paper (which is outside both circles):
Venn Diagram
B
Table
B
S
20
S
No S
Total
60
No B
Total
20
60
“Thirty-three homes receive the Sun...” so inside the S circle (the sum of both parts) is 33:
Venn Diagram
B
Table
S
33
20
60
B
S
No S
Total
No B
Total
33
20
60
“...17 receive the Bugle...” so inside the B circle (the sum of both parts) is 17:
Venn Diagram
B
Table
B
S
33
17
20
60
S
No S
Total
No B
Total
33
20
17
60
Now to solve each of the above…(depending on which way you want to address this problem)
For the Table method, you find whatever row or column has two of the three and solve for the
remaining; (you can choose what order you like, it will all work out to be the same answer). So
for this table, I would start with the “Total” column and row: 60-33 = 27 and 60-17 = 43; then
“No B” column: 43-20 = 13; then the “No S” row: 27-20 = 7; then and finally “B” column gives
me: 17-7 = 10
For the Venn Diagram method, you will either guess the center or use the Union Equation:
If you use the Union Equation, which is n(SUB) = n(S) + n(B) – n(S∩B), then the center is S∩B,
and you know S and B; and the union is everything in the two circles, so since we know the
total (60) and what is NOT in the two circles (20) we just subtract: 60-20 = 40
So the equation looks like: 40 = 33 + 17 - n(S∩B) or
40 = 50 - n(S∩B) or
10 = n(S∩B)
If you guess (as “suggested”) I would recommend either the highest or lowest values that the
center could possibly be…the lowest is ZERO (negative values are not allowed in a Venn
Diagram); the highest will be the value of the smallest intersection (in this case: 17), because if
you any value higher than 17 in the center the other part of that intersection would have to be
negative in order for the intersection to be equal to 17. Also the reason for using the end
points is that is you choose another number and it is off by ONE which way will you adjust (one
way is correct and the other will require you to correct again)…this way when you correct it
will always be right the next time. So now let me do this for BOTH zero and 17:
With the center being ZERO:
First calculating the other part of each
intersection:
For S, the other part is 33, and
For B, the other part is 17
Adding up all the elements you get:
33+0+17+20 = 70 , which is 10 off of the total,
so the center is 10 (0+10).
With the center being 17:
First calculating the other part of each
intersection:
For S, the other part is 16, and
For B, the other part is 0
Adding up all the elements you get:
16+17+0+20 = 53 , which is 7 off of the total,
so the center is 10 (17-7).
Now with the center we can solve the other two parts:
33-10 = 23 and 17-10 = 7...giving us the final answer of…
Venn Diagram
Table
S
B
17
7
10
33
23
20
60
S
No S
Total
B
10
7
17
No B
23
20
43
Total
33
27
60
Please note that the numbers within the darker borders of the table are the same in the Venn
Diagram.
Problem 11
L
S
A
We know this is a 3-circle Venn Diagram because
there are three toppings: Strawberries (S),
Liver (L) and Anchovies (A).
Start filling from the center whenever possible, but in
this case, it is missing so now there are THREE ways in which to solve this problem:
(1) Guess and correct (almost like trial and error),
(2) use the Union Equation for three circles (which you may or may not have been given yet),
or (3) an elaborate chain of equations, setting the center as “x”. I will show you all three, and
you must decide which is best for you…
First method: GUESS
If you are going to guess I would recommend either the highest or lowest values that the
center could possibly be…the lowest is ZERO (negative values are not allowed in a Venn
Diagram); the highest will be the value of the smallest intersection (in this case: 4), because if
you any value higher than 4 in the center the other part of that intersection would have to be
negative in order for the intersection to be equal to 4. Also the reason for using the end points
is that is you choose another number and it is off by ONE which way will you adjust (one way is
correct and the other will require you to correct again)…this way when you correct it will
always be right the next time. So now let me do this for BOTH zero and four:
With the center being ZERO:
First calculating the other part of each
intersection:
For S∩L, the other part is 10,
For L∩A, the other part is 4, and
For S∩A, the other part is 5
Next complete the value of each circle:
For L, the last part is 16-10-4-0 = 2,
For S, the last part is 16-10-5-0 = 1, and
For A, the last part is 12-4-5-0 = 3
With the center being FOUR:
First calculating the other part of each
intersection:
For S∩L, the other part is 10-4 = 6,
For A∩L, the other part is 4-4 = 0, and
For S∩A, the other part is 5-4 = 1
Next complete the value of each circle:
For L, the last part is 16-6-0-4 = 6,
For S, the last part is 16-6-1-4 = 5, and
For A, the last part is 12-0-1-4 = 7
Since outside the three circles is defined as
“Three like none of the toppings”…we can add
up all the numbers:
3+3+1+2+5+4+10+0 = 28, so a difference of
3…so the center (in this case) will be 3.
Since outside the three circles is defined as
“Three like none of the toppings”…we can add
up all the numbers:
3+7+5+6+1+0+6+4 = 32, so a difference of
1…so the center (in this case) will be 3 (4-1).
Second method: Union Equation
The Union equation for a 3-circle Venn Diagram (or 3 sets) is (using the letters L, S & A):
n(LUSUA) = n(L) + n(S) + n(A) – n(L∩S) – n(L∩A) – n(S∩A) + n(L∩S∩A)
(where n(L∩S∩A) is the center)
You have been given all the information directly for everything but the center, L∩S∩A, and the
union, LUSUA, but you have enough information for the union…because you are given what is
in the “universe” (31) and you are given what is NOT in the union, “Three like none of the
toppings”…so subtracting those will give you the number in the union: 31 – 3 = 28
So filling in the equation you get: 28 = 16 + 16 + 12 – 10 – 4 - 5 + n(L∩S∩A)
Or 28 = 44 – 19 + n(L∩S∩A)
Or 28 = 25 + n(L∩S∩A)
Or 3 = n(L∩S∩A)
Third method: Chain of Equation
First, if you set the center as “x”, then each of the intersections second piece (remember that
the intersection is the football-shaped part of the diagram with two parts: the center and the
“other piece”) is equal to the value of the intersection minus the center, because the two
pieces must add up to the intersection…so for L∩S, the “other piece” is 10-x and for L∩A, the
“other piece” is 4-x, and for S∩A, the “other piece” is 5-x…so the Venn Diagram looks like:
L
10-x
x
4-x
5-x
S
A
3
Now each of the circles must add up to the value given, so for the circle L, it must add up to 16,
so (10-x) + x +(4-x) + the last piece = 16,
or
14-x + the last piece = 16
or
the last piece = x+2
Similarly for the circle S, it must add up to 16,
so (5-x) + x +(10-x) + the last piece = 16,
or
15-x + the last piece = 16
or
the last piece = x+1
Similarly for the circle A, it must add up to 12,
so (5-x) + x +(4-x) + the last piece = 12,
or
9-x + the last piece = 12
or
the last piece = x+3
So finally, we know that all the pieces must add up to the value of the universe (31), so:
31 = x + (5-x) + (4-x) + (10-x) + (x+2) + (x+1) + (x+3) + 3 or
31 = x + 28 or
3=x
NOW it is up to you how you wish to calculate the center when it is not given to you!
L
3
A
S
3
Next you must make each of the intersections add up to the values given:
L
7
3
1
2
S
A
3
L/S: 3+7 = 10, S/A: 3+2 = 5 and L/A: 3+1 = 4
Finally you must make each circle add up to the values given:
L
5
7
4
3
2
S
1
6
A
3
L: 5+7+3+1 = 16, S: 4+7+3+2 = 16, and A: 6+2+3+1 = 12
Problem 12
Referring to the Venn Diagram above (problem 11)…
Problem 12(a)
“all three” is the intersection of all three circles…3
Problem 12(b)
“exactly two” is where two circles intersect, but not all three…7+1+2 = 10
Problem 12(c)
“at least one” is one or more toppings, so 1 topping OR 2 toppings OR all 3 toppings:
4+2+6+1+5+7+3 = 28
(or you could look at it as if it is everything except none of the toppings):
31-3 = 28
Problem 12(d)
“at most two” is two or less toppings, so 2 toppings OR all 1 topping OR 0 toppings:
3+4+2+6+1+5+7 = 28
(or you could look at it as if it is everything except 3 toppings):
31-3 = 28
Problem 13
B
S
P
We know this is a 3-circle Venn Diagram because
there are three services: Showtime (S),
Bravo (B) and Playboy (P).
Start filling from the center whenever possible, but in
this case, it is missing so now there are THREE ways in which to solve this problem:
(1) Guess and correct (almost like trial and error),
(2) use the Union Equation for three circles (which you may or may not have been given yet),
or (3) an elaborate chain of equations, setting the center as “x”. I will show you all three, and
you must decide which is best for you…
First method: GUESS
If you are going to guess I would recommend either the highest or lowest values that the
center could possibly be…the lowest is ZERO (negative values are not allowed in a Venn
Diagram); the highest will be the value of the smallest intersection (in this case: 4), because if
you any value higher than 4 in the center the other part of that intersection would have to be
negative in order for the intersection to be equal to 4. Also the reason for using the end points
is that is you choose another number and it is off by ONE which way will you adjust (one way is
correct and the other will require you to correct again)…this way when you correct it will
always be right the next time. So now let me do this for BOTH zero and four:
With the center being ZERO:
First calculating the other part of each
intersection:
For S∩B, the other part is 40,
For B∩P, the other part is 4, and
For S∩P, the other part is 10
Next complete the value of each circle:
For B, the last part is 60-40-4-0 = 16,
For S, the last part is 80-40-10-0 = 30, and
For P, the last part is 25-4-10-0 = 11
With the center being FOUR:
First calculating the other part of each
intersection:
For S∩B, the other part is 40-4 = 36,
For B∩P, the other part is 4-4 = 0, and
For S∩P, the other part is 10-4 = 6
Next complete the value of each circle:
For B, the last part is 60-36-0-4 = 20,
For S, the last part is 80-36-6-4 = 34, and
For P, the last part is 25-0-6-4 = 15
Since outside the three circles is defined as
“87 to none of the three”…we can add up all
the numbers:
87+11+30+16+10+4+40+0 = 198, so a
difference of 2…so the center (in this case)
will be 2.
Since outside the three circles is defined as
“87 to none of the three”…we can add up all
the numbers:
87+15+34+20+6+0+36+4 = 202, so a
difference of 2…so the center (in this case)
will be 2 (4-2).
Second method: Union Equation
The Union equation for a 3-circle Venn Diagram (or 3 sets) is (using the letters B, S & P):
n(BUSUP) = n(B) + n(S) + n(P) – n(B∩S) – n(B∩P) – n(S∩P) + n(B∩S∩P)
(where n(B∩S∩P) is the center)
You have been given all the information directly for everything but the center, B∩S∩P, and the
union, BUSUP, but you have enough information for the union…because you are given what is
in the “universe” (200) and you are given what is NOT in the union, “87 to none of the
three”…so subtracting those will give you the number in the union: 200 – 87 = 113
So filling in the equation you get: 113 = 60 + 80 + 25 – 40 – 4 - 10 + n(B∩S∩P)
Or 113 = 165 – 54 + n(B∩S∩P)
Or 113 = 111 + n(B∩S∩P)
Or 2 = n(B∩S∩P)
Third method: Chain of Equation
First, if you set the center as “x”, then each of the intersections second piece (remember that
the intersection is the football-shaped part of the diagram with two parts: the center and the
“other piece”) is equal to the value of the intersection minus the center, because the two
pieces must add up to the intersection…so for B∩S, the “other piece” is 40-x and for B∩P, the
“other piece” is 4-x, and for S∩P, the “other piece” is 10-x…so the Venn Diagram looks like:
B
40-x
x
4-x
10-x
S
P
87
Now each of the circles must add up to the value given, so for the circle B, it must add up to 60,
so (40-x) + x +(4-x) + the last piece = 60,
or
44-x + the last piece = 60
or
the last piece = x+16
Similarly for the circle S, it must add up to 80,
so (40-x) + x +(10-x) + the last piece = 80,
or
50-x + the last piece = 80
or
the last piece = x+30
Similarly for the circle P, it must add up to 25,
so (10-x) + x +(4-x) + the last piece = 25,
or
14-x + the last piece = 25
or
the last piece = x+11
So finally, we know that all the pieces must add up to the value of the universe (200), so:
200 = x + (10-x) + (4-x) + (40-x) + (x+16) + (x+30) + (x+11) + 87 or
200 = x + 198 or
2=x
NOW it is up to you how you wish to calculate the center when it is not given to you!
B
2
P
S
87
Next you must make each of the intersections add up to the values given:
B
37
3
1
7
S
P
3
B/S: 3+37 = 40, S/P: 3+7 = 10 and B/P: 3+1 = 4
Finally you must make each circle add up to the values given:
B
19
37
33
3
7
S
1
14
P
87
B: 19+37+3+1 = 60, S: 33+7+3+37 = 80, and P: 14+7+3+1 = 25
Problem 14
Since the response in this question is 3-pronged then a Venn Diagram will not work since a
circle only represents 2 sides (“opposites”)…but a table will work nicely…starting with
“One hundred voters…”
Yes
No
Undecided
18-25 years old
26-30 years old
Over 30
Total
Total
100
“18-25…9 said “yes” and 12 said “no””
18-25 years old
26-30 years old
Over 30
Total
Yes
9
No
12
Undecided
Total
100
“Of 55 people in the 26-30 age group…”
18-25 years old
26-30 years old
Over 30
Total
Yes
9
No
12
Undecided
Total
55
100
“… 26-30 age group, 42 were against…”
18-25 years old
26-30 years old
Over 30
Total
Yes
9
No
12
42
Undecided
Total
55
100
“Eight people in the over 30…were undecided…”
18-25 years old
26-30 years old
Over 30
Total
Yes
9
No
12
42
Undecided
Total
55
8
100
“In all, 17 people favored… 23 were undecided…”
18-25 years old
26-30 years old
Over 30
Total
Yes
9
No
12
42
Undecided
55
8
23
17
Total
100
“The 18-25 group contained 30…”
Yes
9
No
12
42
Undecided
Total
30
55
18-25 years old
26-30 years old
Over 30
8
Total
17
23
100
So to solve the table you find a row or column that has 3 of the 4 “spaces” filled and calculate
that, until all the spaces are calculated. Let’s start with the “Total” row/column:
100-17-23 = 60 and 100-30-55 = 15; next the “No” column: 60-12-42 = 6; next the “Over 30”
row: 15-8-6 = 1; and then the “18-25 years old” row: 30-9-12 = 9, then the “Undecided”
column: 23-9-8 = 6; and finally the “Yes” column: 17-9-1 = 7…and we are done and ready to
answer the questions…
Yes
9
7
1
17
18-25 years old
26-30 years old
Over 30
Total
No
12
42
6
60
Undecided
9
6
8
23
Total
30
55
15
100
Problem 14 (a)
“…ages under 30…” means the 18-25 AND 26-30 year olds who are “undecided”:
9+6 = 15
Problem 14 (b)
“…over 30 or in favor…” means all those in the “over 30” plus all those in “Yes” but do not
count the intersection twice (this is the wording for the union of these two):
15+17-1 = 31
Problem 14 (c)
“…in favor… or undecided…” means all those in the “Yes” plus all those in “Undecided” (there
is no intersection to worry about (this is the wording for the union of these two):
17+23 = 40
Problem 14 (d)
“…under 31…” means the 18-25 AND 26-30 year olds who said “Yes”:
9+7 = 16
Problem 15
H
S
P
We know this is a 3-circle Venn Diagram because
there are three services: History (H),
Psychology (P) and Sociology (S).
Start filling from the center whenever possible, and we have that with the statement “…5
taking all three…”:
H
5
P
S
Next you must make each of the intersections add up to the values given:
H
15
5
7
3
P
S
H/S: 15+5 = 10, S/P: 3+5 = 8 and H/P: 7+5 = 12
Next you must make each circle add up to the values given:
H
28
15
2
5
7
30
3
P
S
H: 28+15+5+7 = 55, S: 2+15+5+3 = 25, and P: 30+7+5+3 = 45
Finally you must add up all the values and subtract from the “universe” (100) to get the
H
15
2
S
number outside the three circles:
28
5
3
100 – (28+2+30+15+3+7+5) = 100 – 90 = 10
7
30
P
10
Now to answer the questions…
Problem 15 (a)
“…at least one…” means one or more, so 1 subject OR 2 subjects OR 3 subjects, so add all
those with 1, 2 and 3 subjects…which means everything in the 3 circles:
28+2+30+15+3+7+5= 90
OR it means everything except NO subjects…which means everything minus what’s outside
the 3 circles: 100 – 10 = 90
Problem 15 (b)
“…exactly one…” means in a circle but not overlapped (intersected with another):
28+2+30 = 60
Problem 15 (c)
“…at most two…” means two or less, so 2 subjects OR 1 subject OR 0 (NO) subjects, so add all
those with 2, 1 and 0 subjects:
28+2+30+15+3+7+10= 95
OR it means everything except all 3 subjects…which means everything minus what’s in the
center: 100 – 5 = 95
Problem 15 (d)
“…none of the three subjects…” means outside the three circles: 10