Venn Diagrams - SAT Math
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Fifty 6th graders were asked what their favorite school subjects were. Three students like math, science and English. Five students liked math and science. Seven students liked math and English. Eight people liked science and English. Twenty students liked science. Twenty-eight students liked English. Fourteen students liked math. How many students didn’t like any of these classes?
Fifty 6th graders were asked what their favorite school subjects were. Three students like math, science and English. Five students liked math and science. Seven students liked math and English. Eight people liked science and English. Twenty students liked science. Twenty-eight students liked English. Fourteen students liked math. How many students didn’t like any of these classes?
Draw a Venn diagram with three subsets: Math, Science, and English. Start in the center with students that like all three subjects. Next, look at students that liked two subjects. Be sure to subtract out the ones already counted in the middle. Then, look at the students that only like one subject. Be sure to subtract out the students already accounted for. Once all of the subsets are filled, look at those students who don’t like any of these subjects. To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50.
M = math
S = science
E = English
M∩S∩E = 3
M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY
M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY
S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY
M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY
S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY
E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY
Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students
So, those students who don’t like any of these subjects are 50 – 45 = 5 students
Draw a Venn diagram with three subsets: Math, Science, and English. Start in the center with students that like all three subjects. Next, look at students that liked two subjects. Be sure to subtract out the ones already counted in the middle. Then, look at the students that only like one subject. Be sure to subtract out the students already accounted for. Once all of the subsets are filled, look at those students who don’t like any of these subjects. To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50.
M = math
S = science
E = English
M∩S∩E = 3
M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY
M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY
S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY
M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY
S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY
E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY
Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students
So, those students who don’t like any of these subjects are 50 – 45 = 5 students
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Set A contains the positive even integers less than 14. Set B contains the positive multiples of three less than 20. What is the intersection of the two sets?
Set A contains the positive even integers less than 14. Set B contains the positive multiples of three less than 20. What is the intersection of the two sets?
A = {2, 4, 6, 8, 10, 12}
B = {3, 6, 9, 12, 15, 18}
The intersection of a set means that the elements are in both sets: A∩B = {6, 12}
A = {2, 4, 6, 8, 10, 12}
B = {3, 6, 9, 12, 15, 18}
The intersection of a set means that the elements are in both sets: A∩B = {6, 12}
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There are 75 juniors at a high school. 15 of the students are enrolled in Physics and 40 students are enrolled in Chemistry. 30 students are not enrolled in either Physics or Chemistry. How many students are enrolled in both Physics and Chemistry?
There are 75 juniors at a high school. 15 of the students are enrolled in Physics and 40 students are enrolled in Chemistry. 30 students are not enrolled in either Physics or Chemistry. How many students are enrolled in both Physics and Chemistry?
First, subtract the students that are in neither class; 75 – 30 = 45 students.
Thus, 45 students are enrolled in Chemistry, Physics, or both. Of these 45 students, we know 40 are in Chemistry, so that leaves 5 students who are enrolled in Physics only; with 15 total students in Physics, that means 10 must be in Chemistry as well. So 10 students are in both Physics and Chemistry.
First, subtract the students that are in neither class; 75 – 30 = 45 students.
Thus, 45 students are enrolled in Chemistry, Physics, or both. Of these 45 students, we know 40 are in Chemistry, so that leaves 5 students who are enrolled in Physics only; with 15 total students in Physics, that means 10 must be in Chemistry as well. So 10 students are in both Physics and Chemistry.
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100 students are in the 10th grade class. 30 are swimmers, 40 are runners, and 20 are swimmers and runners. What is the probability that a student is a swimmer OR a runner?
100 students are in the 10th grade class. 30 are swimmers, 40 are runners, and 20 are swimmers and runners. What is the probability that a student is a swimmer OR a runner?
The formula for intersection is P(a or b) = P(a) + P(b) – P(a and b).
Now, 30 students out of 100 swim, so P(swim) = 30/100 = 3/10.
40 students run out of 100, so P(run) = 40/100 = 4/10. Notice how we are keeping 10 as the common denominator even though we could simplify this further. Keeping all of the fractions similar will make the addition and subtraction easier later on.
Finally, 20 students swim AND run, so P(swim AND run) = 20/100 = 2/10. (Again, we keep this as 2/10 instead of 1/5 so that we can combine the three fractions more easily.)
P(swim OR run) = P(swim) + P(run) – P(swim and run)
= 3/10 + 4/10 – 2/10 = 5/10 = 1/2.
The formula for intersection is P(a or b) = P(a) + P(b) – P(a and b).
Now, 30 students out of 100 swim, so P(swim) = 30/100 = 3/10.
40 students run out of 100, so P(run) = 40/100 = 4/10. Notice how we are keeping 10 as the common denominator even though we could simplify this further. Keeping all of the fractions similar will make the addition and subtraction easier later on.
Finally, 20 students swim AND run, so P(swim AND run) = 20/100 = 2/10. (Again, we keep this as 2/10 instead of 1/5 so that we can combine the three fractions more easily.)
P(swim OR run) = P(swim) + P(run) – P(swim and run)
= 3/10 + 4/10 – 2/10 = 5/10 = 1/2.
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and 
.
Find 
.
Find
The intersection of two sets contains every element that is present in both sets, so 
is the correct answer.
The intersection of two sets contains every element that is present in both sets, so
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We have two sports clubs offered to a class of 100 students. 70 students joined the basketball club, 40 students joined the swimming club, and 10 students joined neither. How many students joined both the swimming club and the basketball club?
We have two sports clubs offered to a class of 100 students. 70 students joined the basketball club, 40 students joined the swimming club, and 10 students joined neither. How many students joined both the swimming club and the basketball club?
The idea is to draw a Venn Diagram and find the intersection. We have one circle of 70 and another with 40. When we add the two circles plus the 10 students who joined neither, we should get 100 students. However, when adding the two circles, we are adding the intersections twice, therefore we need to subtract the intersection once.
We get 
, which means the intersection is 20.
The idea is to draw a Venn Diagram and find the intersection. We have one circle of 70 and another with 40. When we add the two circles plus the 10 students who joined neither, we should get 100 students. However, when adding the two circles, we are adding the intersections twice, therefore we need to subtract the intersection once.
We get
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Students at a local high school are given the option to take one gym class, one music class or one of each. Out of 100 students, 60 say that they are currently taking a gym class and 70 say that they are taking a music class. How many students are taking both?
Students at a local high school are given the option to take one gym class, one music class or one of each. Out of 100 students, 60 say that they are currently taking a gym class and 70 say that they are taking a music class. How many students are taking both?
This problem can be solved two ways, with a formula or with reason.
Using the formula, the intersection of the Venn diagram for which classes students take is:
By using reason, it is clear that 60 + 70 is greater than 100 by 30. It is assumed that this extra 30 students come from students who were counted twice because they took both classes.
This problem can be solved two ways, with a formula or with reason.
Using the formula, the intersection of the Venn diagram for which classes students take is:
By using reason, it is clear that 60 + 70 is greater than 100 by 30. It is assumed that this extra 30 students come from students who were counted twice because they took both classes.
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High school freshmen can take Biology, Chemistry, or both. If 
freshmen take Biology, 
freshmen take Chemistry, and there are 
freshmen in total. How many freshmen take both Biology and Chemistry?
High school freshmen can take Biology, Chemistry, or both. If
If 
students are enrolled in sciences, but there are only 
students then we must find how many overlap in the subjects they take.
To do this we can subtract 
from 
.
Therefore, 
of those "enrollments" must be doubles.
Those 
students take both Chemistry and Biology.
If
To do this we can subtract
Therefore,
Those
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At a certain college, some members of the baseball team are seniors and all seniors are in statistics class. Which statement is must be true?
At a certain college, some members of the baseball team are seniors and all seniors are in statistics class. Which statement is must be true?
The statement says all seniors take statistics so if you are a senior you are in statistics automatically. It also said some baseball team members are seniors which means at least some teammates must be in statistics.
The statement says all seniors take statistics so if you are a senior you are in statistics automatically. It also said some baseball team members are seniors which means at least some teammates must be in statistics.
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Let Set A = 
and Set B =
.
Find 
.
Let Set A =
Find
represents the intersection of the two sets. In other words, we want all the elements that appear in both sets. The elements that appear in both sets are 2, 4, 6, and 10.
Therefore, 
Therefore,
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What is the intersection of the Venn Diagram shown below?
What is the intersection of the Venn Diagram shown below?
The intersection of the Venn Diagram is only the numbers in both circles.
The section in the middle contains the answer set.
Thus the intersection is, 
.
The intersection of the Venn Diagram is only the numbers in both circles.
The section in the middle contains the answer set.
Thus the intersection is,
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What is the intersection of the sets A and B?
A = 
B = 
What is the intersection of the sets A and B?
A =
B =
The intersection of 2 sets A and B is the set of the items that are included in both sets. The items that appear in both A and B are 4 and 35.
Therefore, the intersection of A and B is 
.
The intersection of 2 sets A and B is the set of the items that are included in both sets. The items that appear in both A and B are 4 and 35.
Therefore, the intersection of A and B is
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Find 
A = 
B = 
Find
A =
B =
represents the intersection of the two sets A and B. The intersection of A and B is the set of elements that are contained in both A and B.
The elements that appear in A and B are 4, 5, and 154.
Therefore, the intersection of A and B is 
.
The elements that appear in A and B are 4, 5, and 154.
Therefore, the intersection of A and B is
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Farmer John has one hundred plots of land. Sixty plots grow corn. Forty plots grow carrots. These numbers take into account that some of the plots grow both corn and carrots.
How many plots grow both corn and carrots?
Farmer John has one hundred plots of land. Sixty plots grow corn. Forty plots grow carrots. These numbers take into account that some of the plots grow both corn and carrots.
How many plots grow both corn and carrots?
To find how many plots grow both carrots and corn, we subtract 
.
That means that fifty plots grow corn only, while thirty plots grow carrots only.
To find how many plots grow both carrots and corn, we subtract
That means that fifty plots grow corn only, while thirty plots grow carrots only.
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In the above Venn diagram, let the universal set be
yields a remainder of 1 when divided by 4 
yields a remainder of 1 when divided by 3 
How many elements of 
would be placed in the shaded portion of the above diagram?
In the above Venn diagram, let the universal set be
How many elements of
The shaded portion of the Venn diagram is 
- the set of all elements in 
but not 
.
The following elements of 
yield a remainder of 1 when divided by 4, and therefore comprise set 
:
If 3 is subtracted from each, what results are the elements whose division by 4 yields a remainder of 1, and thus, elements of 
.
The following elements of yield a remainder of 1 when divided by 4, and therefore comprise set 
:
The elements in 
that are also elements in 
are 1, 13, 25, and 37 - four elements out of ten. Therefore, the set 
comprises six elements.
The shaded portion of the Venn diagram is
The following elements of
If 3 is subtracted from each, what results are the elements whose division by 4 yields a remainder of 1, and thus, elements of
The following elements of
The elements in
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Doug has a cow farm. Some of Doug's cows are used for milk, some are used for reproduction and some are used for both. If he has a total of 40 cows and 10 are used only for milk and 3 are used for both milk and reproduction, then how many cows are used for reproduction?
Doug has a cow farm. Some of Doug's cows are used for milk, some are used for reproduction and some are used for both. If he has a total of 40 cows and 10 are used only for milk and 3 are used for both milk and reproduction, then how many cows are used for reproduction?
Since we know that only 10 cows are for milk only we must subtract this number from the total amount of cows to get our answer: 40 – 10 = 30 cows. The cows that do both are still used for reproduction, so the correct answer is 30 cows.
Since we know that only 10 cows are for milk only we must subtract this number from the total amount of cows to get our answer: 40 – 10 = 30 cows. The cows that do both are still used for reproduction, so the correct answer is 30 cows.
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All students have to take at least one math class and one language class. Twenty students take calculus, and thirty students take statistics. Fifteen students take Spanish and twenty-five take French. If there are thirty-five students total, what is the maximum number of students taking both two math classes and two language classes.
All students have to take at least one math class and one language class. Twenty students take calculus, and thirty students take statistics. Fifteen students take Spanish and twenty-five take French. If there are thirty-five students total, what is the maximum number of students taking both two math classes and two language classes.
Totalling the number in math there are 50 students on the rosters of all the math classes. With 35 total students this means that there are 15 students taking 2 math classes. For the language classes there are 40 students on the roster, showing that 5 students are taking 2 language classes. The maximum number of students taking two math classes and two language classes is only as great as the smallest number taking a double math or language class, which is 5 students (limited by the language doubles).
Totalling the number in math there are 50 students on the rosters of all the math classes. With 35 total students this means that there are 15 students taking 2 math classes. For the language classes there are 40 students on the roster, showing that 5 students are taking 2 language classes. The maximum number of students taking two math classes and two language classes is only as great as the smallest number taking a double math or language class, which is 5 students (limited by the language doubles).
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Sets P, Q, and R consist of the positive factors of 48, 90, and 56, respectively. If set T = P U (Q ∩ R), which of the following does NOT belong to T?
Sets P, Q, and R consist of the positive factors of 48, 90, and 56, respectively. If set T = P U (Q ∩ R), which of the following does NOT belong to T?
First, let's find the factors of 48, which will give us all of the elements in P. In order to find the factors of 48, list the pairs of numbers whose product is 48.
The pairs are as follows:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8
Therefore the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
Now we can write P = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
Next, we need to find the factors of 90.
Again list the pairs:
1 and 90; 2 and 45; 3 and 30; 5 and 18; 6 and 15; 9 and 10
Then the factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Thus, Q = {1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90}.
Next find the factors of 56:
1 and 56; 2 and 28; 4 and 14; 7 and 8
Set R = {1, 2, 4, 7, 8, 14, 28, 56}
Now, we need to find set T, which is P U (Q ∩ R).
We have to start inside the parantheses with Q ∩ R. The intersection of two sets consists of all of the elements that the two sets have in common. The only elements that Q and R have in common are 1 and 2.
Q ∩ R = {1, 2}
Lastly, we must find P U (Q ∩ R).
The union of two sets consists of any element that is in either of the two sets. Thus, the union of P and Q ∩ R will consist of the elements that are either in P or in Q ∩ R. The following elements are in either P or Q ∩ R:
{1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Therefore, T = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
The problem asks us to determine which choice does NOT belong to T. The number 28 doesn't belong to T.
The answer is 28.
First, let's find the factors of 48, which will give us all of the elements in P. In order to find the factors of 48, list the pairs of numbers whose product is 48.
The pairs are as follows:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8
Therefore the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
Now we can write P = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
Next, we need to find the factors of 90.
Again list the pairs:
1 and 90; 2 and 45; 3 and 30; 5 and 18; 6 and 15; 9 and 10
Then the factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Thus, Q = {1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90}.
Next find the factors of 56:
1 and 56; 2 and 28; 4 and 14; 7 and 8
Set R = {1, 2, 4, 7, 8, 14, 28, 56}
Now, we need to find set T, which is P U (Q ∩ R).
We have to start inside the parantheses with Q ∩ R. The intersection of two sets consists of all of the elements that the two sets have in common. The only elements that Q and R have in common are 1 and 2.
Q ∩ R = {1, 2}
Lastly, we must find P U (Q ∩ R).
The union of two sets consists of any element that is in either of the two sets. Thus, the union of P and Q ∩ R will consist of the elements that are either in P or in Q ∩ R. The following elements are in either P or Q ∩ R:
{1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Therefore, T = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
The problem asks us to determine which choice does NOT belong to T. The number 28 doesn't belong to T.
The answer is 28.
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The class of 2034 at Make Believe High School graduated 50 students. 13 students studied only math. 35 students studied English. 30 students studied only 2 subjects. Only 4 students studied writing, it was the third subject for all of them. How many students did not study anything?
The class of 2034 at Make Believe High School graduated 50 students. 13 students studied only math. 35 students studied English. 30 students studied only 2 subjects. Only 4 students studied writing, it was the third subject for all of them. How many students did not study anything?
The answer is 2. Taking away the 4 writing students, the 13 math-only students, and the remaining 31 English students, we have 2 students remaining.
The answer is 2. Taking away the 4 writing students, the 13 math-only students, and the remaining 31 English students, we have 2 students remaining.
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In a group of 83 gym members, 51 are taking kickboxing and 25 are taking yoga. Of the students taking kickboxing or yoga, 11 are taking both classes. How many members are not taking either course?
In a group of 83 gym members, 51 are taking kickboxing and 25 are taking yoga. Of the students taking kickboxing or yoga, 11 are taking both classes. How many members are not taking either course?
If 11 people are taking both courses, this means 51-11 or 40 are taking kickboxing only and 25-11 or 14 are taking yoga only. The number of people taking at least one course, therefore, is 40 + 14 + 11 = 65. The 83 members minus the 65 that are taking courses leaves 18 who are not taking any courses.
If 11 people are taking both courses, this means 51-11 or 40 are taking kickboxing only and 25-11 or 14 are taking yoga only. The number of people taking at least one course, therefore, is 40 + 14 + 11 = 65. The 83 members minus the 65 that are taking courses leaves 18 who are not taking any courses.
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