SAT Math : Data Analysis
Study concepts, example questions & explanations for SAT Math
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Example Questions
Example Question #1 : Intersection
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.
Example Question #2 : How To Find The Intersection Of A Venn Diagram
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 least some of the baseball team are taking statistics.
None of these can be determined.
All members of the baseball team take statistics.
Statistics class is required for the baseball team.
No member of the baseball team takes statistics.
At least some of the baseball team are taking 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.
Example Question #7 : How To Find The Intersection Of A Venn Diagram
Let Set A = 
Find 
Therefore,
Example Question #2 : Intersection
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?
15
10
5
25
30
10
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.
Example Question #21 : Data Analysis
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, 
Example Question #11 : Intersection
Find
A =
B =
The elements that appear in A and B are 4, 5, and 154.
Therefore, the intersection of A and B is 
Example Question #11 : How To Find The Intersection Of A Venn Diagram
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.
Example Question #22 : Venn Diagrams
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 
Example Question #21 : Venn Diagrams
In the above Venn diagram, let the universal set be
How many elements of 
Five
Nine
Eight
Seven
Six
Six
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 
Example Question #1 : Arithmetic Mean
A, B, C, D, and E are integers such that A < B < C < D < E. If B is the average of A and C, and D is the average of C and E, what is the average of B and D?
(2A + C + 2E)/2
(A + E)/4
(A + 2C + E)/4
(A + 2C + E)/2
(A + E)/2
(A + 2C + E)/4
The average of two numbers can be calculated as the sum of those numbers divided by 2. B would thus be calculated as (A + C)/2, and D would be calculated as (C + E)/2. To find the average of those values, you would add them up and divide by 2:
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