Intersection - SAT Math
Card 0 of 105
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
Compare your answer with the correct one above
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}
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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,
Compare your answer with the correct one above
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,
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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}
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above