GMAT Math : Understanding arithmetic sets

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #21 : Understanding Arithmetic Sets

Number_sets

Examine the above diagram, which shows a Venn diagram representing the sets of real numbers. 

\(\displaystyle m\) and \(\displaystyle n\) are both numbers in Region V, and they may or may not be equal. In how many of the five regions could the number \(\displaystyle mn\) possibly fall?

Possible Answers:

Three

Two

Four

Five

One

Correct answer:

Four

Explanation:

The numbers in Region V are the irrational numbers, such as \(\displaystyle \sqrt{2}\) and \(\displaystyle \pi\).

Since neither number can be the rational number zero, the product of the two cannot be zero, eliminating the possibility that \(\displaystyle mn = 0\). Region II comprises only this number—only 0 is a whole number but not a natural number—so Region II can be eliminated.

Examples can be produced that would place \(\displaystyle mn\) in any of the other four regions:

Case 1:

\(\displaystyle m = \sqrt{2}, n = \sqrt{2} \Rightarrow mn = \sqrt{2}\cdot \sqrt{2} = 2\),

placing \(\displaystyle mn\) in Region I.

Case 2:

\(\displaystyle m =- \sqrt{2}, n = \sqrt{2} \Rightarrow mn =- \sqrt{2}\cdot \sqrt{2} =- 2\)

placing \(\displaystyle mn\) in Region III (the negative integers).

Case 3:

\(\displaystyle m = \sqrt{\frac{2}{3}}, n = \sqrt{\frac{2}{3}} \Rightarrow mn = \sqrt{\frac{2}{3}}\cdot \sqrt{\frac{2}{3}} = \frac{2}{3}\)

placing \(\displaystyle mn\) in Region IV (the non-integer rational numbers).

Case 4:

\(\displaystyle m = \sqrt{3}, n = \sqrt{2} \Rightarrow mn = \sqrt{3}\cdot \sqrt{2} = \sqrt{6}\)

placing \(\displaystyle mn\) in Region V.

Therefore, \(\displaystyle mn\) can fall in any of four different regions.

Example Question #22 : Understanding Arithmetic Sets

Given five sets \(\displaystyle A,B,C,D,E\), you are given that:

\(\displaystyle A \subseteq B\)

\(\displaystyle B\subseteq C\)

\(\displaystyle D \subseteq B\)

\(\displaystyle E \subseteq D\)

All of the following must be true if \(\displaystyle n \notin B\) except:

Possible Answers:

\(\displaystyle n \notin A\)

\(\displaystyle n \notin D\)

\(\displaystyle n \notin C\)

None of the other choices gives a correct answer.

\(\displaystyle n \notin E\)

Correct answer:

\(\displaystyle n \notin C\)

Explanation:

The subset relation is transitive, so \(\displaystyle E \subseteq D\) and \(\displaystyle D\subseteq B\) together imply that \(\displaystyle E\subseteq B\).

Since all three of \(\displaystyle A, D,E\) are subsets of \(\displaystyle B\), then any element of any of those four sets is an element of \(\displaystyle B\). Contrapositively, any nonelement of \(\displaystyle B\) cannot be an element of any of \(\displaystyle A, D,E\)

However, it is possible for a nonelement of \(\displaystyle B\) to be an element of \(\displaystyle C\) if \(\displaystyle B\subseteq C\). A simple example:

\(\displaystyle B = \{1\}\) and \(\displaystyle C = \{1,2\}\).

\(\displaystyle B\subseteq C\)\(\displaystyle 2 \notin B\), and \(\displaystyle 2 \in C\).

Example Question #23 : Understanding Arithmetic Sets

Given five sets \(\displaystyle A,B,C,D,E\), you are given that:

\(\displaystyle A \subseteq B\)

\(\displaystyle B\subseteq C\)

\(\displaystyle D \subseteq B\)

\(\displaystyle E \subseteq D\)

All of the following must be true if \(\displaystyle n \notin C\) except:

Possible Answers:

\(\displaystyle n \notin D\)

\(\displaystyle n \notin B\)

\(\displaystyle n \notin A\)

\(\displaystyle n \notin E\)

None of the other choices gives a correct answer.

Correct answer:

None of the other choices gives a correct answer.

Explanation:

The subset relation is transitive, so: 

\(\displaystyle A \subseteq B\) and \(\displaystyle B\subseteq C\) together imply that \(\displaystyle A\subseteq C\)

\(\displaystyle D \subseteq B\) and \(\displaystyle B\subseteq C\) together imply that \(\displaystyle D\subseteq C\); and,

\(\displaystyle E \subseteq D\) and \(\displaystyle D\subseteq C\) together imply that \(\displaystyle E\subseteq C\).

Since all four of \(\displaystyle A,B,D,E\) are subsets of \(\displaystyle C\), then any element of any of those four sets is an element of \(\displaystyle C\). Contrapositively, any nonelement of \(\displaystyle C\) cannot be an element of any of \(\displaystyle A,B,D,E\).

Example Question #21 : Sets

Define two sets as follows:

\(\displaystyle A = \{3,5,7,9,11,13,...\}\)

\(\displaystyle B = \{ 1,4,7,10,13,16,... \}\)

 \(\displaystyle N \notin A \cup B\). Which is a possible value of \(\displaystyle N\)

Possible Answers:

\(\displaystyle 150\)

\(\displaystyle 147\)

\(\displaystyle 148\)

\(\displaystyle 151\)

\(\displaystyle 149\)

Correct answer:

\(\displaystyle 150\)

Explanation:

\(\displaystyle A\) comprises the set of all odd integers except 1; \(\displaystyle B\) comprises the set of all integers of the form \(\displaystyle 3M-2\)\(\displaystyle M\) a natural number. Therefore, any number that is not in the union of these two sets must be in neither one.

\(\displaystyle N \notin A\), so \(\displaystyle N\) is even or 1 (although 1 is not a choice). We can eliminate odd choices 147, 149, and 151 immediately. 

\(\displaystyle N \notin B\), so we determine which number cannot be expressed as \(\displaystyle 3M-2\)\(\displaystyle M\) a natural number.

\(\displaystyle 3M-2= 148\)

\(\displaystyle 3M = 150\)

\(\displaystyle M = 50\)

 

\(\displaystyle 3M-2= 150\)

\(\displaystyle 3M = 152\)

\(\displaystyle M = 50 \frac{2}{3}\)

148 is elminated, since it is two less than a multiple of 3. 150 is the correct choice.

Example Question #21 : Sets

Define three sets as follows:

\(\displaystyle A = \{3, 4, 5, 6, 7,8\}\)

\(\displaystyle B =\{2, 4, 6, 7, 9, 10 \}\)

\(\displaystyle C = \{1, 3, 5, 7, 9, 11,... \}\)

How many elements does the set \(\displaystyle A \cup (B \cap C)\) have?

Possible Answers:

\(\displaystyle 1\)

Infinitely many

\(\displaystyle 7\)

\(\displaystyle 8\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle 7\)

Explanation:

The intersection of \(\displaystyle B\) and \(\displaystyle C\) is the set of all elements in \(\displaystyle B\) that are also in \(\displaystyle C\) - namely, \(\displaystyle \{7,9\}\). The union of this set and \(\displaystyle A\) is the set of all elements in one or the other set, or \(\displaystyle \{3, 4, 5, 6, 7,8, 9\}\).

The set has seven elements.

Example Question #1571 : Gmat Quantitative Reasoning

Define three sets as follows:

\(\displaystyle A = \{3, 4, 5, 6, 7, 9, 10\}\)

\(\displaystyle B =\{2, 5, 8, 11, 14,... \}\)

\(\displaystyle C = \{1, 3, 5, 7, 9, 11,... \}\)

How many elements does the set \(\displaystyle A \cap (B \cap C)\) have?

Possible Answers:

None of the other responses gives a correct answer.

\(\displaystyle 1\)

Infinitely many.

\(\displaystyle 7\)

\(\displaystyle 4\)

Correct answer:

\(\displaystyle 1\)

Explanation:

\(\displaystyle A \cap (B \cap C)\) comprises the set of elements common to all three sets. Only 5 fulfills that condition, so the correct choice is 1.

Example Question #21 : Sets

Define three sets as follows:

\(\displaystyle A = \{1,2,3,4,5,6,7,8,9,10,11 \}\)

\(\displaystyle B = \{ 2, 4, 6, 8, 10, 12,... \}\)

\(\displaystyle C = \left \{ 1, 3, 5, 7, 9, 11,...\right \}\)

How many elements does the set \(\displaystyle A \cap (B \cap C)\) have?

Possible Answers:

Infinitely many

\(\displaystyle 11\)

\(\displaystyle 5\)

\(\displaystyle 6\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 0\)

Explanation:

\(\displaystyle A \cap (B \cap C)\) comprises the set of elements common to all three sets. However, since \(\displaystyle B\) is the set of all even integers and \(\displaystyle C\) comprises the set of all odd integers, no element can be common to all three sets. The correct response is 0.

Example Question #28 : Understanding Arithmetic Sets

The Fibonacci sequence is the sequence defined as follows:

\(\displaystyle F_{1} = F_{2} = 1\)

For all integers \(\displaystyle n > 2\)\(\displaystyle F_{n}= F_{n-1}+ F_{n-2}\).

Which of the following expressions is equal to \(\displaystyle F_{100}\) ?

Possible Answers:

\(\displaystyle 3 F_{96} + 2F_{97}\)

\(\displaystyle F_{96} + 2F_{97}\)

\(\displaystyle F_{96} + 3F_{97}\)

\(\displaystyle 2 F_{96} + 3F_{97}\)

\(\displaystyle 2 F_{96} + 2F_{97}\)

Correct answer:

\(\displaystyle 2 F_{96} + 3F_{97}\)

Explanation:

Each term after the second is the sum of the preceding two, ao we can relate the 100th term to the 96th and 97th terms as follows:

\(\displaystyle F_{98}= F_{96}+ F_{97}\)

\(\displaystyle F_{99}= F_{97}+ F_{98}= F_{97}+ F_{96}+ F_{97} = F_{96}+2 F_{97}\)

\(\displaystyle F_{100}= F_{98}+ F_{99}= F_{96}+ F_{97}+ F_{96}+2 F_{97} = 2 F_{96}+ 3 F_{97}\)

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