One of the more advanced concepts the GRE tests is that of Overlapping Sets. For many, these problems take a significant amount of time because the default position is to use a table or Venn Diagram of overlapping circles to comprehensively address all aspects of the problem graphically. While this approach can be effective, it often leads to solving for more values than are required to answer the question. In this post, we'll look at some common grouping scenarios to see how we can use simple logic and direct equations to more efficiently solve these types of problems.
Standard Two-Group EquationIn the Official Guide to the GRE, ETS uses some very technical terminology to define the components of overlapping sets, but it all boils down to this equation:
Total = Group 1 + Group 2 - Overlap + NeitherNow, let's define these values:
- Total = The sum of the terms
- Group 1 = Those terms included in the first subgroup
- Group 2 = Those terms included in the second subgroup
- Overlap = Those terms included in each of the two subgroups
- Neither = Those terms included in neither of the two identified subgroups but in the overall group
Let's now understand why the equation works as it does by using an example of 100 total students.
The total is technically known as "union" and it is all of the terms, so in this case it would be the 100 students.
Next, we could define Group 1 as the students out of the 100 taking geometry and give that a value of 70.
Then, we could define Group 2 as the students out of the 100 taking chemistry and give that a value of 60.
Now, the overlap, or "intersection" is those students in each of the two groups, which we could give a value of 45.
This idea of the overlap is the key to the equation. Conceptually, if Pascal takes both geometry and chemistry, then Pascal has been counted twice in the inventory of 100 students. However, Pascal should only be accounted for once in the total number of students, so we would need to subtract him (and any other students in each of the two independent groups) once from the total count because of this double counting.
Lastly, we have the neither group, represented by
n who have not been accounted for in either of the two subgroups. Using the group equation and our already defined values for Group 1, Group 2, and the overlap we can solve for those in neither group:
100 = 70 + 60 - 45 + n100 = 85 + n15 = Neither
Finding for Group X Only with Two GroupsIn addition to solving for any of the discrete values comprising the group equation total, the GRE could also ask for the number of terms in exactly one of the two groups. This too has a simple algebraic approach:
Group X Only = Group X - OverlapOnce again let's consider Pascal as he relates to those students taking geometry, but not chemistry. Conceptually, if Pascal takes both geometry and chemistry, then we would not want to include him in the group of students taking only geometry, so we would need to subtract him (and any other students in each of the two independent groups) from the geometry group to determine how many students take only geometry as
Group 1 Only = 70 - 45 = 25Standard Three-Group EquationBuilding on the logic we've already established for two group overlapping sets, we can extrapolate to three groups as:
Total = Group 1 + Group 2 + Group 3 - Exactly Two Groups - 2(All Three Groups) + NoneNow, let's again define these values:
- Total = The sum of the terms
- Group 1 = Those terms included in the first subgroup
- Group 2 = Those terms included in the second subgroup
- Group 3 = Those terms included in the third subgroup
- Exactly Two Groups = Those terms included in exactly two of the three subgroups
- All Three Groups = Those terms included in each of the three subgroups
- None = Those terms included in none of the three identified subgroups but in the overall group
Let's once more understand why the equation works as it does by using an example of 100 total students.
Again, define Group 1 as the students out of the 100 taking geometry and give that a value of 70.
Then, define Group 2 as the students out of the 100 taking chemistry and give that a value of 60.
Next, define Group 3 as the students out of the 100 taking literature and give that a value of 50.
Now, the overlap of those students in exactly two groups, in any combination, could once again have a value of 45.
Lastly, those students in each of the three groups could be 20.
Once again, the idea of overlap is the key to the equation. Conceptually, if Pascal takes both geometry and chemistry or geometry and literature or chemistry and literature, then Pascal would be counted twice in the inventory of 100 students and must be subtracted once. However, if Pascal were in each of the three groups, then he would be counted thrice, when he should only be accounted for once in the total number of students, so we would need to subtract him (and any other students in each of the three independent groups) twice from the total count because of this triple counting.
Still, we would then have the none group, represented by
n who have not been accounted for in any of the three subgroups. Using the three-group equation and our already defined values for Group 1, Group 2, Group 3, exactly two groups, and all three groups, we can solve for those in the none group:
100 = 70 + 60 + 50 - 45 - 2(20) + n100 = 95 + n5 = None
By following these three basic equations, you should be able to accurately, and most importantly, efficiently address grouping questions in the GRE quantitative section!
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