Venn Diagrams - PSAT Math

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

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}

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.

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 intersection of two sets contains every element that is present in both sets, so is the correct answer.

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 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.

This question is asking you to find the intersection of Set A and Set B. The intersection of two sets contains only the things which are already in both Set A and Set B. Because the numbers 5 and 9 are the only numbers that are shared in common between Set A and Set B, the intersection consists of only those two numbers. : {5, 9}

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.

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.

You can construct a Venn diagran in which one circle represents art (A), the other represents gym (B), the region of overlap is designated (C), and the number of students not present in either circle is designated (S).

First, it is given that S=450.

50% of students take art and 50% take gym and the total number of students is 1250. 50% of 1250 is 625. Thus, A+C=625 and B+C=625.

Setting them equal to each other we get A+C=B+C.

Subtract C from both sides to get A=B; so the same number of students take art only and gym only.

Now 1250-450= A+B+C=800.

Since A=B we can use substitution to get 2A+C=800.

Finally you can solve the system of equations using the method of your choice (substitution or elimination) to solve the system with A+C=625 and 2A+C=800.

For substitution, we solve for C in the first equation to get C=625-A. Then we substitute this value into the second equation to get 2A+ (625-A)=800. Solve for A to get A=175, so B=175 also. Since A+B+C = 800, C=450.

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.

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).

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.

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.

means intersection and means union

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {1, 3, 5}

A' = {2, 4, 6, 7, 8, 9, 10}

B = {2, 4, 6}

B' = {1, 3, 5, 7, 8, 9, 10}

= {2, 4, 6}

= {1, 3, 5}

= {1, 2, 3, 4, 5, 6}

We can draw a Venn diagram to see these two sets of students.

We need to find the overlap between these two sets. To find that, add up the total number of students who are taking history and the total number of students who are taking calculus.

Notice that we have more students this way than the total number who were polled. That is because the students who are taking history AND calculus have been double counted. Subtract the total number of students polled to find out how many students were counted twice.

The symbol stands for the union between two sets. Therefore, means the set of all numbers that are in either A or B. Looking at our choices, the only number that isn't in either A, B, or both is 23.

We can draw a Venn diagram of these students.

Drawn this way, there are more students on the Venn diagram than we have.

This is because some of the students play both sports and should be in the overlap on the Venn diagram. To find the number of students in the overlap, subtract the total number of students given from the number on the diagram.

This represents the number of students who were counted twice, or the number in the overlap.

We can redraw the correct Venn diagram with this number.

The notation stands for "A union C," which refers to everything that is in either set or set .

When we add the numbers together, we get: