GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #1571 : Problem Solving Questions

Number_sets

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

and are both numbers in Region I; also, $m \ne n$ . In how many of the five regions could the number m - n possibly fall?

Possible Answers:

Three

Five

Two

Four

One

Correct answer:

Two

Explanation:

The numbers in Region I are exactly the natural numbers $\left \{ 1, 2, 3, 4, 5,...\right \}$ . All natural numbers are integers, and the integers are closed under subtraction, so m - n cannot fall in Region IV or Region V. Also, since it is given in the problem that $m \ne n$ , it follows that $m - n \ne 0$ , so the difference cannot be in Region II (the only whole number that is not a natural number is ).

It is possible for m - n to be in Region I. Example:

$m = 2, n = 1 \Rightarrow m - n = 1$

It is possible for m - n to be in Region III (the integers that are not whole numbers, or the negative integers). Example:

$m = 1, n = 2 \Rightarrow m - n= -1$

m - n can fall in either of two different regions.

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Example Question #16 : Sets

Number_sets

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

and are both numbers in Region I; also, $m \ne n$ . In how many of the five regions could the number $m \div n$ possibly fall?

Possible Answers:

Three

Five

Two

Four

One

Correct answer:

Two

Explanation:

As natural numbers, and are also rational numbers; since the set of rational numbers is closed under division, and neither nor is equal to zero (zero not being a natural number), $m \div n$ is rational and cannot fall in Region V. Regions III (negative integers) and II (zero only) can be eliminated, since both and are positive. This leaves Regions I and IV.

Examples can be produced that would place $m \div n$ in Region I:

$m = 2, n = 1 \Rightarrow m \div n = 2$

Examples can be produced that would place $m \div n$ in Region IV (the rational numbers that are not integers):

$m = 1, n = 2 \Rightarrow m \div n = \frac{1}{2}$

$m \div n$ can fall in either of two different regions.

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Example Question #21 : Understanding Arithmetic Sets

Number_sets

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

and are both numbers in Region V, and they may or may not be equal. In how many of the five regions could the number possibly fall?

Possible Answers:

Five

Two

Four

Three

One

Correct answer:

Four

Explanation:

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

Since neither number can be the rational number zero, the product of the two cannot be zero, eliminating the possibility that 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 in any of the other four regions:

Case 1:

$m = \sqrt{2}, n = \sqrt{2} \Rightarrow mn = \sqrt{2}\cdot \sqrt{2} = 2$ ,

placing in Region I.

Case 2:

$m =- \sqrt{2}, n = \sqrt{2} \Rightarrow mn =- \sqrt{2}\cdot \sqrt{2} =- 2$

placing in Region III (the negative integers).

Case 3:

$m = \sqrt{\frac{2}{3}}, n = \sqrt{\frac{2}{3}} \Rightarrow mn = \sqrt{\frac{2}{3}}\cdot \sqrt{\frac{2}{3}} = \frac{2}{3}$

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

Case 4:

$m = \sqrt{3}, n = \sqrt{2} \Rightarrow mn = \sqrt{3}\cdot \sqrt{2} = \sqrt{6}$

placing in Region V.

Therefore, can fall in any of four different regions.

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Example Question #21 : Understanding Arithmetic Sets

Given five sets A,B,C,D,E , you are given that:

$A \subseteq B$

$B\subseteq C$

$D \subseteq B$

$E \subseteq D$

All of the following must be true if $n \notin B$ except:

Possible Answers:

$n \notin A$

None of the other choices gives a correct answer.

$n \notin C$

$n \notin D$

$n \notin E$

Correct answer:

$n \notin C$

Explanation:

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

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

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

$B = \{1\}$ and $C = \{1,2\}$ .

$B\subseteq C$ , $2 \notin B$ , and $2 \in C$ .

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Example Question #21 : Arithmetic

Given five sets A,B,C,D,E , you are given that:

$A \subseteq B$

$B\subseteq C$

$D \subseteq B$

$E \subseteq D$

All of the following must be true if $n \notin C$ except:

Possible Answers:

$n \notin B$

None of the other choices gives a correct answer.

$n \notin D$

$n \notin A$

$n \notin E$

Correct answer:

None of the other choices gives a correct answer.

Explanation:

The subset relation is transitive, so:

$A \subseteq B$ and $B\subseteq C$ together imply that $A\subseteq C$ ;

$D \subseteq B$ and $B\subseteq C$ together imply that $D\subseteq C$ ; and,

$E \subseteq D$ and $D\subseteq C$ together imply that $E\subseteq C$ .

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

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Example Question #22 : Understanding Arithmetic Sets

Define two sets as follows:

$A = \{3,5,7,9,11,13,...\}$

$B = \{ 1,4,7,10,13,16,... \}$

$N \notin A \cup B$ . Which is a possible value of ?

Possible Answers:

148

147

149

151

150

Correct answer:

150

Explanation:

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

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

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

3M-2= 148

3M = 150

M = 50

3M-2= 150

3M = 152

$M = 50 \frac{2}{3}$

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

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Example Question #22 : Arithmetic

Define three sets as follows:

$A = \{3, 4, 5, 6, 7,8\}$

$B =\{2, 4, 6, 7, 9, 10 \}$

$C = \{1, 3, 5, 7, 9, 11,... \}$

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

Possible Answers:

Infinitely many

Correct answer:

Explanation:

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

The set has seven elements.

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Example Question #26 : Sets

Define three sets as follows:

$A = \{3, 4, 5, 6, 7, 9, 10\}$

$B =\{2, 5, 8, 11, 14,... \}$

$C = \{1, 3, 5, 7, 9, 11,... \}$

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

Possible Answers:

None of the other responses gives a correct answer.

Infinitely many.

Correct answer:

Explanation:

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

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Example Question #23 : Arithmetic

Define three sets as follows:

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

$B = \{ 2, 4, 6, 8, 10, 12,... \}$

$C = \left \{ 1, 3, 5, 7, 9, 11,...\right \}$

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

Possible Answers:

Infinitely many

Correct answer:

Explanation:

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

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Example Question #21 : Understanding Arithmetic Sets

The Fibonacci sequence is the sequence defined as follows:

$F_{1} = F_{2} = 1$

For all integers n > 2 , $F_{n}= F_{n-1}+ F_{n-2}$ .

Which of the following expressions is equal to $F_{100}$ ?

Possible Answers:

$F_{96} + 3F_{97}$

$2 F_{96} + 2F_{97}$

$F_{96} + 2F_{97}$

$3 F_{96} + 2F_{97}$

$2 F_{96} + 3F_{97}$

Correct answer:

$2 F_{96} + 3F_{97}$

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:

$F_{98}= F_{96}+ F_{97}$

$F_{99}= F_{97}+ F_{98}= F_{97}+ F_{96}+ F_{97} = F_{96}+2 F_{97}$

$F_{100}= F_{98}+ F_{99}= F_{96}+ F_{97}+ F_{96}+2 F_{97} = 2 F_{96}+ 3 F_{97}$

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1571 : Problem Solving Questions

Example Question #16 : Sets

Example Question #21 : Understanding Arithmetic Sets

Example Question #21 : Understanding Arithmetic Sets

Example Question #21 : Arithmetic

Example Question #22 : Understanding Arithmetic Sets

Example Question #22 : Arithmetic

Example Question #26 : Sets

Example Question #23 : Arithmetic

Example Question #21 : Understanding Arithmetic Sets

All GMAT Math Resources

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