How to find the missing part of a list - SSAT Middle Level Quantitative

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Question

Define two sets as follows:

$J = \left \{ 5, 15, 25, 35, 45, 55, 65,...\right \}$

$K = \left \{ 9, 18, 27, 36, 45, 54, 63,...\right \}$

Which of the following numbers is an element of $J \cap K$ ?

Answer

$J \cap K$ is the intersection of and - the set of all elements appearing in both sets. Thus, an element can be eliminated from $J \cap K$ by demonstrating either that it is not an element of or that it is not an element of .

is the set of positive integers ending in "5". 513 and 657 are not in , so they are not in $J \cap K$ .

is the set of muliples of 9. We test the three remaining numbers easily by seeing if 9 divides their digit sum:

425 and 565 are not multiples of 9; neither is in , so neither is in $J \cap K$ .

$765 \in J$ and $765 \in K$ , so $765 \in J \cap K$ . This is the correct choice.

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Question

$\left \{ 2,4,6,8,10,x,14,y,18,20\right \}$ What are the two missing values in the following set?

Answer

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

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Question

Which of the following is a subset of the set

$C = \left \{ x | x \textrm{ is a multiple of 3}\right \}$ ?

Answer

For a set to be a subset of , all of its elements must be elements of - that is, all of its elements must be multiples of 3. A set can therefore be proved to not be a subset of by identifying one element not a multiple of 3.

We can do that with four choices:

$\left \{ 9, 15, 36, 84, 88\right \}$ : $88 \div 3 = 29 \textrm{ R }1$

$\left \{ 24, 57, 68, 78, 99\right \}$ : $68 \div 3 = 22 \textrm{ R }2$

$\left \{ 18, 25, 33, 66, 93\right \}$ : $25 \div 3 = 8 \textrm{ R }1$

$\left \{ 33, 42, 54, 87, 91 \right \}$ : $91 \div 3 = 30 \textrm{ R }1$

However, the remaining set, $\left \{ 15, 33, 45, 81, 105 \right \}$ , can be demonstrated to include only multiples of 3:

$15 \div 3 = 5$

$33 \div 3 = 11$

$45 \div 3 = 15$

$81 \div 3 = 27$

$105 \div 3 = 35$

$\left \{ 15, 33, 45, 81, 105 \right \}$ is the correct choice.

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Question

Define two sets as follows:

$J = \left \{ 5, 15, 25, 35, 45, 55, 65,...\right \}$

$K = \left \{ 9, 18, 27, 36, 45, 54, 63,...\right \}$

Which of the following numbers is an element of $J \cap K$ ?

Answer

$J \cap K$ is the intersection of and - the set of all elements appearing in both sets. Thus, an element can be eliminated from $J \cap K$ by demonstrating either that it is not an element of or that it is not an element of .

is the set of positive integers ending in "5". 513 and 657 are not in , so they are not in $J \cap K$ .

is the set of muliples of 9. We test the three remaining numbers easily by seeing if 9 divides their digit sum:

425 and 565 are not multiples of 9; neither is in , so neither is in $J \cap K$ .

$765 \in J$ and $765 \in K$ , so $765 \in J \cap K$ . This is the correct choice.

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Question

Define sets and as follows:

$C = \left \{ x | x \textrm{ is a multiple of 5}\right \}$

$D = \left \{ 345, 456, 574, 600, 724, 855\right \}$

How many elements are in the set $C \cap D$ ?

Answer

The elements of the set $C \cap D$ - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by testing for divisibility by 5 - but this can be accomplished by looking at the last digit. Only 345, 600, and 855 have last digit 5 or 0 so only these three elements are divisible by 5. This makes three the correct answer.

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Question

$\left \{ 2,4,6,8,10,x,14,y,18,20\right \}$ What are the two missing values in the following set?

Answer

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

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Question

Define two sets as follows:

$J = \left \{ 5, 15, 25, 35, 45, 55, 65,...\right \}$

$K = \left \{ 9, 18, 27, 36, 45, 54, 63,...\right \}$

Which of the following numbers is an element of $J \cap K$ ?

Answer

$J \cap K$ is the intersection of and - the set of all elements appearing in both sets. Thus, an element can be eliminated from $J \cap K$ by demonstrating either that it is not an element of or that it is not an element of .

is the set of positive integers ending in "5". 513 and 657 are not in , so they are not in $J \cap K$ .

is the set of muliples of 9. We test the three remaining numbers easily by seeing if 9 divides their digit sum:

425 and 565 are not multiples of 9; neither is in , so neither is in $J \cap K$ .

$765 \in J$ and $765 \in K$ , so $765 \in J \cap K$ . This is the correct choice.

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Question

Define two sets as follows:

$J = \left \{ 5, 15, 25, 35, 45, 55, 65,...\right \}$

$K = \left \{ 9, 18, 27, 36, 45, 54, 63,...\right \}$

Which of the following numbers is an element of $J \cap K$ ?

Answer

$J \cap K$ is the intersection of and - the set of all elements appearing in both sets. Thus, an element can be eliminated from $J \cap K$ by demonstrating either that it is not an element of or that it is not an element of .

is the set of positive integers ending in "5". 513 and 657 are not in , so they are not in $J \cap K$ .

is the set of muliples of 9. We test the three remaining numbers easily by seeing if 9 divides their digit sum:

425 and 565 are not multiples of 9; neither is in , so neither is in $J \cap K$ .

$765 \in J$ and $765 \in K$ , so $765 \in J \cap K$ . This is the correct choice.

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Question

Which of the following is a subset of the set

$C = \left \{ x | x \textrm{ is a multiple of 3}\right \}$ ?

Answer

For a set to be a subset of , all of its elements must be elements of - that is, all of its elements must be multiples of 3. A set can therefore be proved to not be a subset of by identifying one element not a multiple of 3.

We can do that with four choices:

$\left \{ 9, 15, 36, 84, 88\right \}$ : $88 \div 3 = 29 \textrm{ R }1$

$\left \{ 24, 57, 68, 78, 99\right \}$ : $68 \div 3 = 22 \textrm{ R }2$

$\left \{ 18, 25, 33, 66, 93\right \}$ : $25 \div 3 = 8 \textrm{ R }1$

$\left \{ 33, 42, 54, 87, 91 \right \}$ : $91 \div 3 = 30 \textrm{ R }1$

However, the remaining set, $\left \{ 15, 33, 45, 81, 105 \right \}$ , can be demonstrated to include only multiples of 3:

$15 \div 3 = 5$

$33 \div 3 = 11$

$45 \div 3 = 15$

$81 \div 3 = 27$

$105 \div 3 = 35$

$\left \{ 15, 33, 45, 81, 105 \right \}$ is the correct choice.

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Question

Define sets and as follows:

$C = \left \{ x | x \textrm{ is a multiple of 5}\right \}$

$D = \left \{ 345, 456, 574, 600, 724, 855\right \}$

How many elements are in the set $C \cap D$ ?

Answer

The elements of the set $C \cap D$ - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by testing for divisibility by 5 - but this can be accomplished by looking at the last digit. Only 345, 600, and 855 have last digit 5 or 0 so only these three elements are divisible by 5. This makes three the correct answer.

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Question

$\left \{ 2,4,6,8,10,x,14,y,18,20\right \}$ What are the two missing values in the following set?

Answer

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

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Question

How many of the following four numbers are elements of the set

$\left \{ x ; | ; 0.3 < x < 0.4\right \}$ ?

(A) $\frac{1}{3}$

(B) $\frac{2}{5}$

(C) $\frac{3}{8}$

(D) $\frac{5}{13}$

Answer

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

$\frac{1}{3} = 1 \div 3 = 0.333...$

$\frac{2}{5} = 2 \div 5 = 0.4$

$\frac{3}{8} = 0.375$

$\frac{5}{13} = 5 \div 13 = 0.384615...$

All fractions except can be seen to fall between 0.3 and 0.4, exclusive. Three is the correct answer.

Note that $\frac{2}{5}$ is equal to 0.4, so we don't include it. The criterion requires strict inequality.

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Question

Define sets and as follows:

$C = \left \{ x | x \textrm{ is a multiple of 5}\right \}$

$D = \left \{ 345, 456, 574, 600, 724, 855\right \}$

How many elements are in the set $C \cap D$ ?

Answer

The elements of the set $C \cap D$ - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by testing for divisibility by 5 - but this can be accomplished by looking at the last digit. Only 345, 600, and 855 have last digit 5 or 0 so only these three elements are divisible by 5. This makes three the correct answer.

Compare your answer with the correct one above

Question

$\left \{ 2,4,6,8,10,x,14,y,18,20\right \}$ What are the two missing values in the following set?

Answer

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

Compare your answer with the correct one above

Question

How many of the following four numbers are elements of the set

$\left \{ x ; | ; 0.3 < x < 0.4\right \}$ ?

(A) $\frac{1}{3}$

(B) $\frac{2}{5}$

(C) $\frac{3}{8}$

(D) $\frac{5}{13}$

Answer

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

$\frac{1}{3} = 1 \div 3 = 0.333...$

$\frac{2}{5} = 2 \div 5 = 0.4$

$\frac{3}{8} = 0.375$

$\frac{5}{13} = 5 \div 13 = 0.384615...$

All fractions except can be seen to fall between 0.3 and 0.4, exclusive. Three is the correct answer.

Note that $\frac{2}{5}$ is equal to 0.4, so we don't include it. The criterion requires strict inequality.

Compare your answer with the correct one above

Question

Define sets and as follows:

$C = \left \{ x | x \textrm{ is a multiple of 5}\right \}$

$D = \left \{ 345, 456, 574, 600, 724, 855\right \}$

How many elements are in the set $C \cap D$ ?

Answer

The elements of the set $C \cap D$ - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by testing for divisibility by 5 - but this can be accomplished by looking at the last digit. Only 345, 600, and 855 have last digit 5 or 0 so only these three elements are divisible by 5. This makes three the correct answer.

Compare your answer with the correct one above

Question

How many of the following four numbers are elements of the set

$\left \{ x ; | ; 0.3 < x < 0.4\right \}$ ?

(A) $\frac{1}{3}$

(B) $\frac{2}{5}$

(C) $\frac{3}{8}$

(D) $\frac{5}{13}$

Answer

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

$\frac{1}{3} = 1 \div 3 = 0.333...$

$\frac{2}{5} = 2 \div 5 = 0.4$

$\frac{3}{8} = 0.375$

$\frac{5}{13} = 5 \div 13 = 0.384615...$

All fractions except can be seen to fall between 0.3 and 0.4, exclusive. Three is the correct answer.

Note that $\frac{2}{5}$ is equal to 0.4, so we don't include it. The criterion requires strict inequality.

Compare your answer with the correct one above

Question

How many of the following four numbers are elements of the set

$\left \{ x ; | ; 0.3 < x < 0.4\right \}$ ?

(A) $\frac{1}{3}$

(B) $\frac{2}{5}$

(C) $\frac{3}{8}$

(D) $\frac{5}{13}$

Answer

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

$\frac{1}{3} = 1 \div 3 = 0.333...$

$\frac{2}{5} = 2 \div 5 = 0.4$

$\frac{3}{8} = 0.375$

$\frac{5}{13} = 5 \div 13 = 0.384615...$

All fractions except can be seen to fall between 0.3 and 0.4, exclusive. Three is the correct answer.

Note that $\frac{2}{5}$ is equal to 0.4, so we don't include it. The criterion requires strict inequality.

Compare your answer with the correct one above

Question

Define two sets as follows:

$A = \left \{ a, c, d, e, f, h, j, l, p \right \}$

$B = \left \{ c, e, f, h, j, o, p, s, z \right \}$

Which of the following is a subset of $A \cup B$ ?

Answer

$A \cup B$ is the union of and - that is, it is the set of all elements in one set or the other.

$A \cup B = \left \{ a,c,d,e,f,h,j,l,o,p,s,z\right \}$

A set is a subset of $A \cup B$ if and only if every one of its elements is in $A \cup B$ . Three of the listed sets do not meet this criterion:

$b\in \left \{a,b,d,h,l,p,z \right \}$ , $g \in \left \{d,e,f,g,o,l,s \right \}$ , and $r \in \left \{c,d,e,h,o,r,z \right \}$ , but none of those three elements are in $A \cup B$ . All of the elements in $\left \{ c,d,f, h,j,s,z \right \}$ do appear in $A \cup B$ , however, so it is the subset.

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Question

Define sets and as follows:

$C = \left \{ x | x \textrm{ is a multiple of 7}\right \}$

$D = \left \{ 683, 705, 759, 832, 852, 944\right \}$

How many elements are in the set $C \cap D$ ?

Answer

The elements of the set $C \cap D$ - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by dividing each by 7 and noting which divisions yield no remainder:

$D = \left \{ 683, 705, 759, 832, 852, 944\right \}$

$683 \div 7 = 97 \textrm{ R }4$

$705 \div 7 = 100\textrm{ R } 5$

$759 \div 7 = 108\textrm{ R }3$

$832 \div 7 = 118\textrm{ R }6$

$852 \div 7 = 121 \textrm{ R }5$

$944 \div 3 = 134 \textrm{ R }6$

and have no elements in common, so $C \cap D$ has zero elements. This is not one of the choices.

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