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ISEE Upper Level Math : Sets

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Example Questions

Example Question #687 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Define set $A = \left \{ 1, 2, 4, 8, 10, a, g, j, r, w \right \}$ . Which of the following sets could we define to be set so that $A \cap B = \varnothing$ ?

Possible Answers:

$B= \left \{ 3,5, 6,8,14,d,f,s,t,x\right \}$

$B = \left \{ 3, 5, 6, 7, 9, e, g, t, u, x\right \}$

$B = \left \{ 5,6,7,11,17, f,h,i,k,r\right \}$

$B = \left \{ 3, 6, 7, 9, 12, b, e, h, s, t\right \}$

$B =\left \{ 3, 5, 7, 10, 13, e, f, h, k, y \right \}$

Correct answer:

$B = \left \{ 3, 6, 7, 9, 12, b, e, h, s, t\right \}$

Explanation:

For $A \cap B = \varnothing$ to be a true statement, sets and cannot have any elements in common. We can eliminate four of these choices of by noting that in each, there is one element (underlined) also in :

$\left \{ 3,5, 6,\underline{8},14,d,f,s,t,x\right \}$

$\left \{ 3, 5, 7, \underline{10}, 13, e, f, h, k, y \right \}$

$\left \{ 3, 5, 6, 7, 9, e, \underline{g}, t, u, x\right \}$

$\left \{ 5,6,7,11,17, f,h,i,k,\underline{r}\right \}$

However, $B = \left \{ 3, 6, 7, 9, 12, b, e, h, s, t\right \}$ shares no elements with $A = \left \{ 1, 2, 4, 8, 10, a, g, j, r, w \right \}$ , so their intersection is $\varnothing$ by definition. This is the correct choice.

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Example Question #688 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Given a set $A = \left \{ 1,3, 6, 7, 8\right \}$ , which of the following sets could we assign to such that $A \cup B = \left \{ 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right \}$ ?

Possible Answers:

$\left \{ 2, 9,5, 4, 10, 8\right \}$

None of the choices answer the question correctly.

$\left \{ 6,5,4, 9, 10\right \}$

$\left \{ 2, 4, 9, 10 \right \}$

$\left \{1, 2, 4, 5, 6, 9, 10, 12\right \}$

Correct answer:

$\left \{ 2, 9,5, 4, 10, 8\right \}$

Explanation:

$A \cup B$ is the union of the sets, i.e. the set of all elements in , , or both.

For $A \cup B = \left \{ 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right \}$ , must include all elements in $\left \{ 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right \}$ that are not in , or $\left \{ 2, 4, 5, 9, 10\right \}$ . This allows us to eliminate $\left \{ 2, 4, 9, 10 \right \}$ , which excludes 5, and $\left \{ 6,5,4, 9, 10\right \}$ , which excludes 2.

Also, cannot include any element not in $\left \{ 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right \}$ . This allows us to eliminate $\left \{1, 2, 4, 5, 6, 9, 10, 12\right \}$ , which includes 12.

This leaves $\left \{ 2, 9,5, 4, 10, 8\right \}$ .

If $B = \left \{ 2, 9,5, 4, 10, 8\right \}$ , then

$A \cup B =\left \{ 1,3, 6, 7, 8\right \} \cup \left \{ 2, 9,5, 4, 10, 8\right \}$

$= \left \{ 1,2, 3, 4, 5, 6, 7, 8, 9, 10\right \}$ .

This is the correct choice.

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Example Question #689 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Examine the sequence:

$5, 6, 8, 11, 15, 20, 26, \square, \bigcirc,...$

What number goes into the circle?

Possible Answers:

Correct answer:

Explanation:

Each element is obtained by adding a number to the previous one; the number added increases by 1 each time:

5 + 1 = 6

6 + 2 = 8

8 + 3 = 11

11 + 4 = 15

15 + 5 = 20

20 + 6 = 26

26 + 7 = 33 - this number replaces the square

33 + 8 = 41 - this number replaces the circle

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

A pair of fair dice are tossed. What is the probability that the product of the numbers of the faces is greater than or equal to ?

Possible Answers:

$\frac {5}{18}$

$\frac {1}{3}$

$\frac {1}{4}$

$\frac {1}{6}$

$\frac {2}{9}$

Correct answer:

$\frac {2}{9}$

Explanation:

Out of a possible thirty-six rolls, the following result in a product of twenty or greater:

$\left\{ (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) \right\}$

This is eight out of thirty-six, making the probability

$\frac{8}{36} = \frac {2}{9}$ .

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Example Question #1 : Sets

Ten students are running for student council; each member of the student body will choose four.

Two of the candidates are Kevin's brothers, Mickey and Steve. Kevin wants to vote for one, but not both, of his brothers. How many ways can Kevin fill out his ballot so that he can vote for exactly one of his brothers?

Possible Answers:

672

224

336

112

Correct answer:

112

Explanation:

Kevin will choose three students from the eight who are not his brothers, and will do so without respect to order. This is the number of combinations of three from eight:

$C(8,3) = \frac{8!}{(8-3)!\; 3!} = \frac{8!}{5!\; 3!} = \frac{40,320}{120 \times 6} = 56$

Kevin will also choose one of his two brothers. By the multiplication principle, Kevin has $56 \times 2 = 112$ ways to fill out the ballot.

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

Define $f (x) = \frac{1}{x ^{2}+ 9 }$

What is the natural domain of ?

Possible Answers:

$(3 , \infty)$

$(-\infty , 3) \cup (3 , \infty)$

$(-\infty ,3)$

$(-\infty , \infty)$

Correct answer:

$(-\infty , \infty)$

Explanation:

The only possible restriction of the domain here is the denominator $x ^{2}+ 9$ , which cannot be equal to 0. We can find any such values of as follows:

$x ^{2}+ 9 = 0$

$x ^{2}+ 9- 9 = 0 - 9$

$x ^{2} = - 9$

This has no real solution, so the domain is the set of all real numbers, $(-\infty , \infty)$ .

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Example Question #5 : How To Find The Missing Part Of A List

Nine students are running for student council; each member of the student body will choose four.

The candidates include two of Eileen's sisters, Maureen and Colleen. Eileen wants to vote for one or both of them. How many ways can Eileen fill out her ballot so that one or both of her sisters is among her choices?

Possible Answers:

154

112

133

Correct answer:

133

Explanation:

If Eileen wants to choose both sisters, then she will choose two out of the other seven candidates. This is the number of combinations of two out of seven:

$C (7,2) = \frac{7!}{(7-2)!\; 2!} = \frac{7!}{5!\; 2!} = \frac{5,040}{120 \times 2} = 21$

If Eileen wants to choose one sister, then she will choose three out of the other eight candidates. This is the number of combinations of three out of eight:

$C (8,3) = \frac{8!}{(8-3)!\; 3!} = \frac{8!}{5!\; 3!} = \frac{40,320}{120 \times 6} = 56$

Since there are also two ways Eileen can choose one sister, by the multiplication principle, she can fill out the ballot $56\times 2 = 112$ ways that include one sister.

Add these two:

112 + 21 = 133

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

Define $f (x) = \frac{1}{x ^{2}+ 9 }$

What is the range of ?

Possible Answers:

$\left ( \frac{1}{9}, \infty \right )$

$\left ( 0, \frac{1}{9}\right ]$

$\left ( 0, \frac{1}{9}\right )$

$\left [\frac{1}{9}, \infty \right )$

Correct answer:

$\left ( 0, \frac{1}{9}\right ]$

Explanation:

$x ^{2} \geq 0$ , so

$x ^{2} + 9 \geq 0+ 9$ , or

$x ^{2} + 9 \geq 9$

Since $x ^{2} + 9 \geq 9$ , then

$0 < \frac{1 }{x ^{2} + 9} \leq \frac{1}{9}$

The range is therefore

$\left ( 0, \frac{1}{9}\right ]$

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Example Question #7 : How To Find The Missing Part Of A List

The Department of Motor Vehicles wants to make all of the state's license numbers conform to two rules:

Rule 1: The number must comprise two letters (A-Z) followed by four numerals (0-9).

Rule 2: Numerals can be repeated but not letters.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

$25 ^{2} \times 10 ^{4}$

$25 ^{2} \times P (10,4 )$

$26 \times 25 \times P (10,4 )$

$26 \times 25 \times 10 ^{4}$

Correct answer:

$26 \times 25 \times 10 ^{4}$

Explanation:

The first letter can be any one of the 26 letters, but after this choice is made, since the letter cannot be repeated, there are 25 choices for the second letter. Each of the next four characters can each be one of the 10 numerals, with repetition allowed. By the multiplication principle, the number of possible license numbers is

$26 \times 25 \times 10 \times 10 \times 10 \times 10 = 26\times 25 \times 10 ^{4}$

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Example Question #6 : Sets

The Department of Motor Vehicles wants to make all of the state's license numbers conform to two rules:

Rule 1: The number must comprise two letters (A-Z) followed by four numerals (0-9).

Rule 2: Repeats are allowed, but neither of the letters can be an "O" or an "I".

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

$24 \times 23 \times C (10,4 )$

$24 ^{2} \times C (10,4 )$

$24 \times 23 \times 10 ^{4}$

$24 ^{2} \times 10 ^{4}$

Correct answer:

$24 ^{2} \times 10 ^{4}$

Explanation:

Repeats are allowed, so the first character can be one of the 24 allowed letters (26 minus the "I" and the "O"), as can the second; each of the next four characters can each be one of the 10 numerals. By the multiplication principle, the number of possible license plate numbers is:

$24 \times 24 \times 10 \times 10 \times 10 \times 10 = 24^{2} \times 10 ^{4}$

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Sets

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #687 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #688 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #689 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Example Question #1 : Sets

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

Example Question #5 : How To Find The Missing Part Of A List

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

Example Question #7 : How To Find The Missing Part Of A List

Example Question #6 : Sets

All ISEE Upper Level Math Resources

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