Understanding arithmetic sets - GMAT Quantitative

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Question

How many functions map from $X=\left \{ a,b \right \}$ to $Y=\left \{ 1,2,3 \right \}$ ?

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Answer

There are three choices for (1, 2, and 3), and similarly there are three choices for (also 1, 2, and 3). Together there are $3\cdot 3=9$ possible functions from to . Remember to multiply, NOT add.

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Question

Let the univeraal set be the set of all positive integers.

Define the sets

$A = \left \{ 1,6,11,16,21,26,...\right\}$ ,

$B = \left \{1, 5, 9, 13,17,21,25,...\right \}$ ,

$C = \left\{1,4,7,10,13,16,19,...\right \}$ .

If the elements in $A \cap B \cap C$ were ordered in ascending order, what would be the fourth element?

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Answer

are the sets of all positive integers that are one greater than a multiple of five, four, and three, respectively. Therefore, for a number to be in all three sets, and subsequently, $A \cap B \cap C$ , the number has to be one greater than a number that is a multiple of five, four, and three. Since , the number has to be one greater than a multiple of 60. The first four numbers that fit this description are 1, 61, 121, and 181, the last of which is the correct choice.

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Question

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?

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Answer

$A \cap (B \cap C)$ 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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Question

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 ?

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Answer

comprises the set of all odd integers except 1; comprises the set of all integers of the form , 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 , a natural number.

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

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

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Question

How many functions map from $X=\left \{ a,b \right \}$ to $Y=\left \{ 1,2,3 \right \}$ ?

Tap to reveal answer

Answer

There are three choices for (1, 2, and 3), and similarly there are three choices for (also 1, 2, and 3). Together there are $3\cdot 3=9$ possible functions from to . Remember to multiply, NOT add.

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Question

Let the univeraal set be the set of all positive integers.

Define the sets

$A = \left \{ 1,6,11,16,21,26,...\right\}$ ,

$B = \left \{1, 5, 9, 13,17,21,25,...\right \}$ ,

$C = \left\{1,4,7,10,13,16,19,...\right \}$ .

If the elements in $A \cap B \cap C$ were ordered in ascending order, what would be the fourth element?

Tap to reveal answer

Answer

are the sets of all positive integers that are one greater than a multiple of five, four, and three, respectively. Therefore, for a number to be in all three sets, and subsequently, $A \cap B \cap C$ , the number has to be one greater than a number that is a multiple of five, four, and three. Since , the number has to be one greater than a multiple of 60. The first four numbers that fit this description are 1, 61, 121, and 181, the last of which is the correct choice.

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Question

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 ?

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Answer

comprises the set of all odd integers except 1; comprises the set of all integers of the form , 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 , a natural number.

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

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

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Question

Let be the set $\{\Gamma ,\Delta ,\Lambda ,\Phi \}$ , and be the set $\{\Pi ,\Lambda ,\Delta ,\Theta ,\Gamma ,\Phi ,\Omega , C^2\}$ .

What are the elements in the set $A \cap B$ ?

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Answer

$A \cap B$ is the set of all elements that are in both and . So in this case the elements that are the same for both sets are $\Delta,\Lambda,\Gamma,\Phi$ , Note that the order that you present the elements in your set doesn't matter. What matters is that you don't exclude any necessary elements, or add any that don't belong.

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Question

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 possibly fall?

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Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

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Question

Given five sets , 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:

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Answer

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

Since all three of 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 .

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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Question

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 possibly fall?

Tap to reveal answer

Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

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Question

Given five sets , 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:

Tap to reveal answer

Answer

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

Since all three of 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 .

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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Question

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 possibly fall?

Tap to reveal answer

Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

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Question

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 possibly fall?

Tap to reveal answer

Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

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Question

Given five sets , 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:

Tap to reveal answer

Answer

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

Since all three of 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 .

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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Question

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 possibly fall?

Tap to reveal answer

Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

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Question

Given five sets , 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:

Tap to reveal answer

Answer

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

Since all three of 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 .

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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Question

Given the set = {2, 3, 4, 5}, what is the value of ?

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Answer

We need to add 3 to every element in .

Then:

$3+A=\left \{ 3+2,3+3,3+4,3+5 \right \}=\left \{ 5,6,7,8 \right \}.$

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Question

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 possibly fall?

Tap to reveal answer

Answer

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 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 to be in Region I. Example:

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

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

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

can fall in either of two different regions.

← Didn't Know|Knew It →

Question

Given five sets , 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:

Tap to reveal answer

Answer

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

Since all three of 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 .

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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