# GMAT / GRE Mathematics Tutorial Series: Set Theory

Set theory has notations and symbols that are quite unique. In this tutorial, we look at a few solved examples to understand how set theory works and its applications.

Definition

A set, in simple terms, is a collection of objects.

It is usually represented within flower braces.

For example:
Set of natural numbers           = {1,2,3,…..}
Set of whole numbers             = {0,1,2,3,…..}

Each object in a set is called an element.

The set that contains all the elements of a given collection is called the universal set and is represented by the symbol ‘µ’, pronounced as ‘mu’.

For two sets A and B,

• n(AᴜB) is the number of elements present in either of the sets A or B.
• n(A∩B) is the number of elements present in both the sets A and B.
• n(AᴜB) = n(A) + (n(B) – n(A∩B)

For three sets A, B, and C,

• n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)

Consider the following example:

Question: In a class of 100 students, 35 like science and 45 like math. 10 like both. How many like either of them and how many like neither?

Solution:

Total number of students, n(µ) = 100

Number of science students, n(S) = 35

Number of math students, n(M) = 45

Number of students who like both, n(M∩S) = 10

Number of students who like either of them,

n(MᴜS) = n(M) + n(S) – n(M∩S)

→ 45+35-10 = 70

Number of students who like neither = n(µ) – n(MᴜS) = 100 – 70 = 30

The easiest way to solve problems based on GMAT exam set theory questions is by using Venn diagrams, as shown below. A pictorial representation of data can convey a lot of information. Using a Venn diagram, problems can be solved at a much faster rate, especially in the case of problems that have more than two categories.

Take a look at these solved examples.

Problem 1: There are 30 students in a class. 8 of these students are learning both English and French. A total of 18 students are learning English. If every student is learning at least one language, what is the total number of students learning French?

Solution:

The Venn diagram for this problem can be represented as follows Each student is learning at least one language. This means that there are no students who fall into the category ‘neither’.

So in this case, n(EᴜF) = n(µ).

According to the problem, a total of 18 students are learning English. This, however, DOES NOT mean that the 18 are learning ONLY English. It is important to remember that only when the word ‘only’ is mentioned in a problem, should we consider it to be so.

Now, 18 are learning English and 8 are learning both. This means that 18 – 8 = 10 are learning ONLY English.

n(µ) = 30, n(E) = 10

n(EᴜF) = n(E) + n(F) – n(E∩F)

30 = 18+ n(F) – 8

n(F) = 20

Therefore, total number of students learning French = 20.

Note: The question was only about the total number of students learning of French and not about those learning ONLY French, for which the answer is 12.

The below figure is the Venn diagram for this problem. Problem 2: Among a group of graduates, 50 played cricket, 50 played hockey and 40 played volleyball. 15 played both cricket and hockey, 20 played both hockey and volleyball, 15 played cricket and volleyball and 10 played all three. If every graduate played at least one game, find the number of graduates and how many played only cricket, only hockey and only volleyball?

Solution:

n(C) = 50, n(H) = 50, n(V) = 40

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

No. of graduates who played at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Total number of graduates = 100.

Let a denote the number of people who played cricket and volleyball only.
Let b denote the number of people who played cricket and hockey only.
Let c denote the number of people who played hockey and volleyball only.
Let d denote the number of people who played all three games.

Accordingly, d = n (CnHnV) = 10

Now, n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Therefore, a = 15 – 10 = 5 [cricket and volleyball only]

b = 15 – 10 = 5 [cricket and hockey only]

c = 20 – 10 = 10 [hockey and volleyball only]

No. of graduates who played only cricket = n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

No. of graduates who played only hockey = n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

No. of graduates who played only volley ball = n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15 Alternatively, we can solve it faster with the help of a Venn diagram.

The Venn diagram for the given information can be represented as the following. Subtracting the values in the intersections from the individual values gives us the number of graduates who played only one game.

Set Theory Quiz: Practize problems

1) In a town, there were 115 people whose proofs of identity were being verified. Some had a passport, some had voter id and some had both. If 65 had a passport and 30 had both, how many had voter id only and not passport?

A.30
B. 50
C. 80
D. None of the above

Explanation

Let us draw the Venn diagram for the given information. n(PᴜV) = n(P) + n(V) – n(P∩V)

115 = 65+n(V) – 30

n(V) = 80

People with only voter id = 80-30 = 50

2) Among a group of children, 40% liked red, 30% liked blue and 30% liked green. 7% liked both red and green, 5% liked both red and blue, 10% liked both green and blue. If 86% of them liked at least one color, what percentage of children liked all three?

A.10
B. 6
C. 8
D. None