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ISEE Middle Level Math : Sets

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

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Example Questions

Example Question #7 : Sets

What are the next two numbers of this sequence?

10, 14, 19, 23, 28, 32, 37...

Possible Answers:

41, 47

41, 46

41, 45

42, 46

42, 47

Correct answer:

41, 46

Explanation:

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

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 $A \cap B$ ?

Possible Answers:

$\left \{ h,c,e,p\right \}$

$\left \{ c,e,f,h,j,p\right \}$

Each of the sets listed is a subset of $A \cap B$ .

$\left \{ f\right \}$

$\varnothing$

Correct answer:

Each of the sets listed is a subset of $A \cap B$ .

Explanation:

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

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 $A \cap B$ ?

Possible Answers:

900

484

352

256

336

Correct answer:

256

Explanation:

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

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 $A \cup B$ ?

Possible Answers:

441

225

420

272

344

Correct answer:

420

Explanation:

$A \cup B$ 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 $420 \notin A$

and

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

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

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Example Question #1 : Sets

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?

Possible Answers:

120

Correct answer:

Explanation:

Derreck is choosing three students from a field of six (seven minus Anne) without respect to order, making this a combination. He has C(6,3) 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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Example Question #2 : Sets

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?

Possible Answers:

210

151,200

10,000

5,040

Correct answer:

5,040

Explanation:

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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Example Question #3 : Sets

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?

Possible Answers:

560

450

630

500

Correct answer:

560

Explanation:

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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Example Question #4 : Sets

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?

Possible Answers:

240

180

120

Correct answer:

180

Explanation:

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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Example Question #5 : Sets

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?

Possible Answers:

210

3,024

252

5,040

Correct answer:

3,024

Explanation:

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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Example Question #6 : Sets

Find the missing part of the list:

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

Possible Answers:

$\small 81$

$\small 45$

$\small 36$

$\small 135$

Correct answer:

$\small 81$

Explanation:

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

$27\times3=81$

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ISEE Middle Level Math : Sets

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

All ISEE Middle Level Math Resources

Example Questions

Example Question #7 : Sets

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

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

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

Example Question #1 : Sets

Example Question #2 : Sets

Example Question #3 : Sets

Example Question #4 : Sets

Example Question #5 : Sets

Example Question #6 : Sets

All ISEE Middle Level Math Resources

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