SSAT Middle Level Math : How to find the missing part of a list

Study concepts, example questions & explanations for SSAT Middle Level Math

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Example Questions

Example Question #1 : How To Find The Missing Part Of A List

Define two sets as follows:

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

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

Which of the following is a subset of \(\displaystyle A \cup B\) ?

 

Possible Answers:

\(\displaystyle \left \{a,b,d,h,l,p,z \right \}\)

\(\displaystyle \left \{d,e,f,g,o,l,s \right \}\)

\(\displaystyle \left \{ c,d,f, h,j,s,z \right \}\)

\(\displaystyle \left \{c,d,e,h,o,r,z \right \}\)

Each of the sets listed is a subset of \(\displaystyle A \cup B\).

Correct answer:

\(\displaystyle \left \{ c,d,f, h,j,s,z \right \}\)

Explanation:

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

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

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

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

Example Question #2 : How To Find The Missing Part Of A List

Define two sets as follows:

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

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

Which of the following numbers is an element of \(\displaystyle J \cap K\) ? 

Possible Answers:

\(\displaystyle 657\)

\(\displaystyle 565\)

\(\displaystyle 513\)

\(\displaystyle 765\)

\(\displaystyle 425\)

Correct answer:

\(\displaystyle 765\)

Explanation:

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

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

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

\(\displaystyle 425: 4 + 2 + 5 = 11\)

\(\displaystyle 565: 5 + 6 + 5 = 16\)

\(\displaystyle 765: 7 + 6 + 5 = 18\)

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

\(\displaystyle 765 \in J\) and \(\displaystyle 765 \in K\), so   \(\displaystyle 765 \in J \cap K\). This is the correct choice.

Example Question #3 : How To Find The Missing Part Of A List

Complete the set by determining the value of \(\displaystyle x\).

\(\displaystyle 1, 4, 9, 16, 25, x\)

Possible Answers:

\(\displaystyle 45\)

\(\displaystyle 42\)

\(\displaystyle 34\)

\(\displaystyle 36\)

\(\displaystyle 49\)

Correct answer:

\(\displaystyle 36\)

Explanation:

The set is composed of consecutive squares.

\(\displaystyle 1, 4, 9, 16, 25, x\)

\(\displaystyle 1^2, 2^2, 3^2, 4^2, 5^2, x\)

We can see that \(\displaystyle x\) will b equal to \(\displaystyle 6^2\)

\(\displaystyle x=6^2=36\)

Therefore, 36 is the correct answer. 

Example Question #2 : Sets

Define sets \(\displaystyle C\) and \(\displaystyle D\) as follows:

\(\displaystyle C = \left \{ x | x \textrm{ is a multiple of 7}\right \}\) 

\(\displaystyle D = \left \{ 683, 705, 759, 832, 852, 944\right \}\)

How many elements are in the set \(\displaystyle C \cap D\) ?

Possible Answers:

The correct answer is not given among the other responses.

One

Two

Four

Three

Correct answer:

The correct answer is not given among the other responses.

Explanation:

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

\(\displaystyle D = \left \{ 683, 705, 759, 832, 852, 944\right \}\)

\(\displaystyle 683 \div 7 = 97 \textrm{ R }4\)

\(\displaystyle 705 \div 7 = 100\textrm{ R } 5\)

\(\displaystyle 759 \div 7 = 108\textrm{ R }3\)

\(\displaystyle 832 \div 7 = 118\textrm{ R }6\)

\(\displaystyle 852 \div 7 = 121 \textrm{ R }5\)

\(\displaystyle 944 \div 3 = 134 \textrm{ R }6\)

 

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

Example Question #3 : Sets

Which of the following is a subset of the set

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

Possible Answers:

\(\displaystyle \left \{ 33, 42, 54, 87, 91 \right \}\)

\(\displaystyle \left \{ 9, 15, 36, 84, 88\right \}\)

\(\displaystyle \left \{ 15, 33, 45, 81, 105 \right \}\)

\(\displaystyle \left \{ 18, 25, 33, 66, 93\right \}\)

\(\displaystyle \left \{ 24, 57, 68, 78, 99\right \}\)

Correct answer:

\(\displaystyle \left \{ 15, 33, 45, 81, 105 \right \}\)

Explanation:

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

We can do that with four choices:

\(\displaystyle \left \{ 9, 15, 36, 84, 88\right \}\)\(\displaystyle 88 \div 3 = 29 \textrm{ R }1\)

\(\displaystyle \left \{ 24, 57, 68, 78, 99\right \}\)\(\displaystyle 68 \div 3 = 22 \textrm{ R }2\)

\(\displaystyle \left \{ 18, 25, 33, 66, 93\right \}\)\(\displaystyle 25 \div 3 = 8 \textrm{ R }1\)

\(\displaystyle \left \{ 33, 42, 54, 87, 91 \right \}\)\(\displaystyle 91 \div 3 = 30 \textrm{ R }1\)

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

\(\displaystyle 15 \div 3 = 5\)

\(\displaystyle 33 \div 3 = 11\)

\(\displaystyle 45 \div 3 = 15\)

\(\displaystyle 81 \div 3 = 27\)

\(\displaystyle 105 \div 3 = 35\)

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

Example Question #2 : How To Find The Missing Part Of A List

Define sets \(\displaystyle C\) and \(\displaystyle D\) as follows:

\(\displaystyle C = \left \{ x | x \textrm{ is a multiple of 5}\right \}\) 

\(\displaystyle D = \left \{ 345, 456, 574, 600, 724, 855\right \}\)

How many elements are in the set \(\displaystyle C \cap D\) ?

Possible Answers:

None

The correct answer is not given among the other responses.

Three

One

Two

Correct answer:

Three

Explanation:

The elements of the set \(\displaystyle C \cap D\) - that is, the intersection of \(\displaystyle C\) and \(\displaystyle D\) - are exactly those in both sets. We can test each of the six elements in \(\displaystyle D\) for inclusion in set \(\displaystyle C\) 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.

Example Question #3 : How To Find The Missing Part Of A List

Which of the following is a subset of the set

\(\displaystyle A = \left \{ x | x \textrm{ is a prime number}\right \} ?\) ?

Possible Answers:

\(\displaystyle \left \{ 5, 11, 17, 25\right \}\)

\(\displaystyle \left \{ 2, 13, 27, 41\right \}\)

The correct answer is not among the answer choices.

\(\displaystyle \left \{ 9, 17, 19, 31\right \}\)

\(\displaystyle \left \{ 3, 21, 23, 37\right \}\)

Correct answer:

The correct answer is not among the answer choices.

Explanation:

We show that none of the four listed sets can be a subset of the primes by identifying one composite number in each - that is, by proving that there is at least one factor not equal to 1 or itself:

 

\(\displaystyle \left \{ 5, 11, 17, 25\right \}\)

\(\displaystyle 25 = 5 \times 5\), so 25 has 5 as a factor, and 25 is not prime.

 

\(\displaystyle \left \{ 9, 17, 19, 31\right \}\)

\(\displaystyle 9 = 3 \times 3\), so 9 has 3 as a factor, and 9 is not prime.

 

\(\displaystyle \left \{ 3, 21, 23, 37\right \}\)

\(\displaystyle 21 = 3 \times 7\), so 21 has 3 and 7 as factors, and 21 is not prime.

 

\(\displaystyle \left \{ 2, 13, 27, 41\right \}\)

\(\displaystyle 27 = 3 \times 9\), so 21 has 3 and 9 as factors, and 27 is not prime.

 

Since each set has at least one element that is not a prime, each has at least one element not in \(\displaystyle A\), and none of the sets are subsets of \(\displaystyle A\).

Example Question #4 : How To Find The Missing Part Of A List

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

\(\displaystyle \left \{ x \; | \; 0.3 < x < 0.4\right \}\) ?

(A) \(\displaystyle \frac{1}{3}\)

(B) \(\displaystyle \frac{2}{5}\)

(C) \(\displaystyle \frac{3}{8}\)

(D) \(\displaystyle \frac{5}{13}\)

Possible Answers:

Two

None

Four

Three

One

Correct answer:

Three

Explanation:

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

\(\displaystyle \frac{1}{3} = 1 \div 3 = 0.333...\)

\(\displaystyle \frac{2}{5} = 2 \div 5 = 0.4\)

\(\displaystyle \frac{3}{8} = 0.375\)

\(\displaystyle \frac{5}{13} = 5 \div 13 = 0.384615...\)

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

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

Example Question #581 : Ssat Middle Level Quantitative (Math)

Define \(\displaystyle K = \left \{ x \; | \; x \textrm{ is a multiple of 5 }\right \}\).

How many of the four sets listed are subsets of the set \(\displaystyle K\)?

(A) \(\displaystyle \left \{ 7335, 8934, 1970, 5585, 8625 \right \}\)

(B) \(\displaystyle \left \{ 8655, 4135, 7472, 9530, 2265\right \}\)

(C) \(\displaystyle \left \{ 5725, 7695, 3640, 9375, 8405\right \}\)

(D) \(\displaystyle \left \{ 9730, 8390, 6380, 5980, 3975\right \}\)

Possible Answers:

Two

One

None

Four

Three

Correct answer:

Two

Explanation:

For a set to be a subset of \(\displaystyle K\), all of its elements must also be elements of \(\displaystyle K\) - that is, all of its elements must be multiples of 5. An integer is a multiple of 5 if and only if its last digit is 5 or 0, so all we have to do is examine the last digit of each number in all four sets. 

In the sets \(\displaystyle \left \{ 9730, 8390, 6380, 5980, 3975\right \}\) and \(\displaystyle \left \{ 5725, 7695, 3640, 9375, 8405\right \}\), every element ends in a 5 or a 0, so all elements of both sets are in \(\displaystyle K\); both sets are subsets of \(\displaystyle K\)

However, \(\displaystyle \left \{ 7335, 8934, 1970, 5585, 8625 \right \}\) includes one element that does not end in either 5 or 0, namely 8934, so 8934 is not an element in \(\displaystyle K\); subsequently, this set is not a subset of \(\displaystyle K\). Similarly, \(\displaystyle \left \{ 8655, 4135, 7472, 9530, 2265\right \}\) is not a subset of \(\displaystyle K\), since it includes 7472, which ends in neither 0 nor 5.

The correct answer is therefore two.

 

Example Question #1 : How To Find The Missing Part Of A List

What are the next two numbers of this sequence?

\(\displaystyle 10, 14, 19, 23, 28, 32, 37...\)

Possible Answers:

\(\displaystyle 41, 46\)

\(\displaystyle 41, 45\)

\(\displaystyle 41, 47\)

\(\displaystyle 42, 46\)

\(\displaystyle 42, 47\)

Correct answer:

\(\displaystyle 41, 46\)

Explanation:

The sequence is formed by alternately adding \(\displaystyle 4\) and adding \(\displaystyle 5\) to each term to get the next term.

\(\displaystyle \begin{matrix} 10 + 4 = 14\\ 14 + 5 = 19\\ 19+4= 23\\ 23+5=28\\ 28+4 = 32\\ 32+5=37\\ 37+4=41\\ 41+5=46 \end{matrix}\)

\(\displaystyle 41\) and \(\displaystyle 46\) are the next two numbers.

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