Sets - ISEE Middle Level Math

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Question

What are the next two numbers of this sequence?

Answer

The sequence is formed by alternately adding and adding to each term to get the next term.

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

and are the next two numbers.

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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 $A \cap B$ ?

Answer

We demonstrate that all of the choices are subsets of $A \cap B$ .

$A \cap B$ is the intersection of and - that is, the set of all elements of both sets. Therefore,

$A \cap B = \left { c,e,f,h,j,p\right }$

$\left { c,e,f,h,j,p\right }$ itself is one of the choices; it is a subset of itself. The empty set $\varnothing$ is a subset of every set. The other two sets listed comprise only elements from $A \cap B$ , making them subsets of $A \cap B$ .

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Question

Let the universal set be the set of all positive integers. Also, define two sets as follows:

$A = \left { 8, 16, 24, 32, 40, 48, 56...\right }$

$B = \left { 1, 4, 9, 16, 25, 36, 49,...\right }$

Which of the following is an element of the set $A \cap B$ ?

Answer

We are looking for an element that is in the intersection of and - in other words, we are looking for an element that appears in both sets.

is the set of all multiples of 8. We can eliminate two choices as not being in by demonstrating that dividing each by 8 yields a remainder:

$484 \div 8 = 60 \textrm{ R }4$

$900 \div 8 = 112 \textrm{ R }4$

is the set of all perfect square integers. We can eliminate two additional choices as not being perfect squares by showing that each is between two consecutive perfect squares:

$18^{2}= 324< 352 < 361 = 19^{2}$

$18^{2}= 324< 336 < 361 = 19^{2}$

This eliminates 352 and 336. However,

$256 = 16 ^{2}$ .

It is also a multiple of 8:

$256 \div 8 = 32$

Therefore, $256 \in A \cap B$ .

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Question

Define two sets as follows:

$A = \left { 8, 16, 24, 32, 40, 48, 56...\right }$

$B = \left { 1, 4, 9, 16, 25, 36, 49,...\right }$

Which of the following is not an element of the set $A \cup B$ ?

Answer

$A \cup B$ is the union of and , the set of all elements that appear in either set_._ Therefore, we are looking to eliminate the elements in and those in to find the element in neither set.

is the set of all multiples of 8. We can eliminate two choices as mulitples of 8:

$272 \div 8 = 34$ , so $272 \in A$

$344\div 8 = 43$ , so $344 \in A$

is the set of all perfect square integers. We can eliminate two additional choices as perfect squares:

$\sqrt{225}=15$ , so $225 \in B$

$\sqrt{441}=21$ , so $441 \in B$

All four of the above are therefore elements of $A \cup B$ .

420, however is in neither set:

$420 \div 8 = 52 \textrm{ R } 4$ , so

and

$20 = \sqrt{400}<\sqrt{420} < \sqrt{441} = 21$ , so

Therefore, $420 otin A \cup B$ , making this the correct choice.

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Question

Find the missing part of the list:

$\small \small 1, 3, 9, 27,...,243,729$

Answer

To find the next number in the list, multiply the previous number by $\small 3$ .

$27\times3=81$

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Question

Seven students are running for student council; each member of the student body will vote for three. Derreck does not want to vote for Anne, whom he does not like. How many ways can he cast a ballot so as not to include Anne among his choices?

Answer

Derreck is choosing three students from a field of six (seven minus Anne) without respect to order, making this a combination. He has ways to choose. This is:

$C(6,3)= \frac{6!}{(6-3)! ; 3! } = \frac{6!}{3! ; 3! } = \frac{720 }{6 \times 6}= 20$

Derreck has 20 ways to fill the ballot.

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Question

Ten students are running for Senior Class President. Each member of the student body will choose four candidates, and mark them 1-4 in order of preference.

How many ways are there to fill out the ballot?

Answer

Four candidates are being selected from ten, with order being important; this means that we are looking for the number of permutations of four chosen from a set of ten. This is

$P (10,4) = \frac{10!}{(10-4)!}= \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5,040$

There are 5,040 ways to complete the ballot.

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Question

Ten students are running for Senior Class President. Each member of the student body will choose five candidates, and mark them 1-5 in order of preference.

Roy wants Mike to win. How many ways can Roy fill out the ballot so that Mike is his first choice?

Answer

Since Mike is already chosen, Roy is in essence choosing four candidates from nine, with order being important. This is a permutation of four elements out of nine. The number of these is

$P (9,4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3,024$

Roy can fill out the ballot 3,024 times and have Mike be his first choice.

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Question

The junior class elections have four students running for President, five running for Vice-President, four running for Secretary-Treasurer, and seven running for Student Council Representative. How many ways can a student fill out a ballot?

Answer

These are four independent events, so by the multiplication principle, the ballot can be filled out $4 \times 5 \times 4\times 7 = 560$ ways.

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Question

The sophomore class elections have six students running for President, five running for Vice-President, and six running for Secretary-Treasurer. How many ways can a student fill out a ballot if he is allowed to select one name per office?

Answer

These are three independent events, so by the multiplication principle, the ballot can be filled out $6 \times 5 \times 6 = 180$ ways.

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Question

What is the value of y in the pattern below?

$\frac{2}{3}, \frac{4}{6}, \frac{6}{9}, \frac{y}{12}$

Answer

What that the fractions in this pattern have in common is that they are all the equivalent of $\frac{2}{3}$ .

The value of y should be a number that is the equivalent of $\frac{2}{3}$ when divided by 12.

Given that $\frac{1}{3}$ of 12 is 4, $\frac{2}{3}$ of 12 would be equal to 8, the correct answer.

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Question

Mary is making a very long necklace with a variety of beads. The beads are white, blue, and black, and she strings them on the necklace, in that order. What color is the 213th bead?

Answer

A number is divisible by 3 when the sum of its digits is divisble by 3. The sum of the digits of 213 equals 6, which is evenly divisible by 3.

Therefore, because 213 is a number that is evenly divisble by 3, the 213th bead is going to be the third color that Mary uses, which is black.

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Question

What is the value of in the sequence below?

Answer

In this sequence, every subsequent number is equal to one third of the preceding number:

$108\div3=36$

$36\div3=12$

$12\div3=4$

Given that $4\div3 =\frac{4}{3}$ , that is the correct answer.

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Question

The sum of three consecutive odd numbers is 81. What is the largest number?

Answer

In order to solve this problem, it is best to work backwards by "plugging in" the answer choices to see which one yields a correct answer.

If 29 is the largest of the three odd consecutive numbers, then that means that the numbers being added together would be 25, 27, and 29.

Given that , is the correct answer.

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Question

Find the next number that should appear in the set below:

$\frac{1}{2}, \frac{1}{4},\frac{1}{8}, \frac{1}{16}$

Answer

In this set, each subsequent fraction is half the size of the preceding fraction; (the denominator is doubled for each successive fraction, but the numerator stays the same). Given that the last fraction in the set is $\frac{1}{16}$ , it follows that the subsequent fraction will be $\frac{1}{32}$ .

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Question

If every number in set appears in set , which consists of multiples of , which of the following could describe set ?

Answer

If every number that appears in set also appears in set , that means that set must be broader than set .

Any number that is a multiple of 16 will also be a multiple of 8 (characteristic of set ); therefore, if set consists of multiples of 16, set will include all those numbers.

Therefore, Set can consist of multiples of 16.

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Question

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

$\left { x ; | ; 0.6 < x < 0.8\right }$ ?

(A) $\frac{3}{5}$

(B) $\frac{5}{7}$

(C) $\frac{7}{9}$

(D) $\frac{9}{11}$

Answer

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

$\frac{3}{5} = 3 \div 5 = 0.6$

$\frac{5}{7} = 5\div 7 = 0.714285...$

$\frac{7}{9}= 7 \div 9 = 0.777...$

$\frac{9}{11} =9 \div 11 = 0.8181...$

Of the four, $\frac{5}{7}$ and $\frac{7}{9}$ fall between 0.6 and 0.8 exclusive. The correct response is "two"

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Question

Which of the following is a subset of the set

$C = \left { x | x \textrm{ is a multiple of 4}\right }$ ?

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 4. A set can therefore be proved to not be a subset of by identifying one element not a multiple of 4.

We can do that with all four given sets:

$\left { 24, 36, 42, 84, 88\right }$ : $42 \div 4 = 10 \textrm{ R }2$

$\left { 12, 40, 66, 92, 100\right }$ : $66 \div 4 = 16 \textrm{ R }2$

$\left { 18, 28, 44, 68, 76\right }$ : $18 \div 4 = 4 \textrm{ R }2$

$\left { 32, 52, 64, 74, 92\right }$ : $74\div 4 = 18 \textrm{ R }2$

The correct response is therefore "None of the other responses are correct."

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Question

Define sets and as follows:

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

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

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 3 and noting which divisions yield no remainder:

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

$683 \div 3 = 227 \textrm{ R }2$

$705 \div 3 = 235$

$759 \div 3 = 253$

$832 \div 3 = 277 \textrm{ R }1$

$852 \div 3 = 284$

$944 \div 3 = 314 \textrm{ R }2$

and have three elements in common, so $C \cap D$ has that many elements.

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Question

Define $K = \left { x ; | ; x \textrm{ is a multiple of 5 }\right }$

How many of the four sets listed are subsets of the set ?

(A) $\left { 4385, 8930, 2980, 5385, 8725 \right }$

(B) $\left { 8465, 6675, 7300, 9230, 7665\right }$

(C) $\left { 4925, 7655, 5580, 9340, 8755\right }$

(D) $\left { 9965, 8450, 4980, 6640, 5875\right }$

Answer

For a set to be a subset of , all of its elements must also be elements of - that is, all of its elements must be multiples of 5. An integer is a multple 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. Every number in every set ends in 5 or 0, so every number in every set is a multiple of 5. This makes all four sets subsets of .

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