All HSPT Math Resources
Example Questions
Example Question #51 : How To Do Other Word Problems
Define the universal set to be the set of all people. Let be the set of people who know how to play spades, and be the set of people who know how to play hearts.
Beyonce knows how to play hearts, but she doesn't know how to play spades. Aaliyah knows how to play spades, but she doesn't know how to play hearts. Which of the ladies falls in the set ?
Aaliyah, but not Beyonce
Neither Beyonce nor Aaliyah
Beyonce, but not Aaliyah
Both Beyonce and Aaliyah
Neither Beyonce nor Aaliyah
By DeMorgan's Laws,
That is, the complement of the union of the sets is the intersection of the complements of the sets. Therefore, for a person to be in this set, it is necessary for the person to be in the complements of both and - that is, the person cannot be in either set. Any person who can play spades or hearts must be excluded, and since both Beyonce and Aaliyah can play at least one game, both are excluded.
Example Question #52 : How To Do Other Word Problems
Let the universal set be the set of positive integers. Let be the set of prime numbers and be the set of odd numbers. Which of the following is in the set ?
is the intersection of two sets. One is the complement of , which is the set of all elements not in ; these are the numbers that are not prime. The other is , the set of all odd numbers. Therefore, the correct choice must be odd and not prime.
2 and 32 are eliminated for being even. Of the two odd choices, 17 is a prime number, having only 1 and 17 as factors. 39 has more than two factors (3 is a factor, since 39 divided by 3 yields quotient 13), so it is not prime. 39 is the correct choice.
Example Question #81 : Word Problems
A number of balls are placed in a box, each havng a letter on it. Each vowel is represented by three balls; each consonant, two. A ball is then drawn at random. What are the odds in favor of, or against, the ball having a vowel on it?
Note: for purposes of this problem, "Y" is considered a consonant.
3 to 2 in favor
14 to 5 against
63 to 10 in favor
21 to 5 against
14 to 5 against
There are five vowels, each represented by three balls; the number of balls with vowels is
.
There are twenty-one consonants, each represented by two balls; the number of balls with consonants is
.
The odds against drawing a vowel are therefore
, or 14 to 5.
Example Question #54 : How To Do Other Word Problems
A number of balls are placed in a box, each havng a letter on it. Each vowel is represented by balls; each consonant, one ball. A ball is then drawn at random. In terms of , give the probability that the ball will have an "E" on it.
Note: for purposes of this problem, "Y" is considered a consonant.
There are five vowels, each represented by balls, so there will be balls with vowels; each of the twenty-one consonants will be represented by one ball, so the total number of balls is . Since there are balls with an "E" on them, the probability of drawing one of these balls is .
Example Question #51 : How To Do Other Word Problems
A number of balls are placed in a box, each with an integer from 1 to 10 inclusive. Each prime number is represented by three balls; each other number, two balls. Give the odds against drawing a ball with a "1".
12 to 1
25 to 3
22 to 3
11 to 1
11 to 1
In the range of 1 to 10, the prime numbers - the numbers with exactly two factors, one and itself - are 2, 3, 5, and 7. Therefore, there are six nonprime numbers (including 1, which, having only one factor, 1 itself, is considered to not be a prime number) represented by two balls each, and four prime numbers represented by three balls each. Therefore, the box will contain
balls.
Since 2 of the balls have a "1" on them - and 22 do not - the odds against drawing a ball with a "1" are
, or 11 to 1.
Example Question #53 : How To Do Other Word Problems
A number of balls are placed in a box, each with an integer from 1 to 10 inclusive. Each prime number is represented by balls; each other number, balls. If a ball is drawn at random, in terms of and , give the probability that the number on the ball will be a "2".
In the range of 1 to 10, the prime numbers - the numbers with exactly two factors, one and itself - are 2, 3, 5, and 7. Therefore, there are six nonprime numbers (including 1, which, having only one factor, 1 itself, is considered to not be a prime number) represented by balls each, and four prime numbers represented by balls each. Therefore, the box will contain
balls. Since 2 is prime, balls will have a "2" on them, making the probability of drawing one
.
Example Question #51 : How To Do Other Word Problems
A number of balls are placed in a box, each havng printed on it either a letter or an integer from "1" to "10" inclusive. Each of the integers is represented by three balls; each vowel, by two balls; each consonant, by one ball.
What is the probability that a ball drawn at random will have either a "1" or an "A" printed on it?
Note: for purposes of this problem, "Y" is considered a consonant.
Each of the ten integers is represented by three balls; each of the five vowels, two balls; each of the twenty-one consonants, one ball. Therefore, the total number of balls in the box is
.
Of these balls, three have a "1" on them; since "A" is a vowel, two have an "A" on them. Therefore, the probability of drawing a ball with either is .
Example Question #534 : Hspt Mathematics
Let the universal set be the set of all people. Let , , and be the sets of people who like Shania Twain, Kelly Clarkson, and Avril Lavigne, respectively.
Let represent Roshonda. How would you rewrite the statement:
"Roshonda dislikes one or both of Avril Lavigne and Kelly Clarkson, but she likes Shania Twain"
in set notation?
Roshonda likes Shania Twain, putting her in set .
She dislikes one or both of Avril Lavigne and Kelly Clarkson. The set of people who like Avril Lavigne comprise ; the set of people who dislike Avril Lavigne comprise the complement of this set, . Similarly, the set of people who dislike Kelly Clarkson is . She falls in either or both of these sets, putting her in their union, .
Since she is in both and , she is in their intersection,
.
The correct choice is
Example Question #535 : Hspt Mathematics
Define the universal set to be the set of all people.
Let be the set of all people who like The Allman Brothers Band, be the set of all people who like The Beach Boys, and be the set of all people who like Crosby, Stills, and Nash.
John likes The Allman Brothers Band and The Beach Boys, but not Crosby, Stills, and Nash.
Mary likes The Allman Brothers Band and Crosby, Stills, and Nash, but not The Beach Boys.
Who would be in the set ?
John, but not Mary
Both John and Mary
Mary, but not John
Neither John nor Mary
John, but not Mary
For a person to fall in the set - that is, the intesection of and - (s)he must fall in both and .
To fall in , the person is in either , , or both; in oher words, the person must like either The Allman Brothers Band or The Beach Boys, or both.
To fall in , the person must fall outside ; in other words, the person must not like Crosby, Stills, and Nash.
John and Mary, both of whom like The Allman Brothers Band, fall in . However, Mary likes Crosby, Stills, and Nash, so she does not fall in . John does not like Crosby, Stills, and Nash, so he does fall in . Only John falls in the set .
Example Question #53 : How To Do Other Word Problems
Mr. and Mr. Smith have invited six guests to dinner. The above show their kitchen table with the locations of its eight seats.
How many ways can the eight participants be seated if the Smiths are to occupy seats 1 and 5?
First, there are two ways to seat the Smiths; Mr. Smith at seat 1 and his wife at seat 5, and vice versa.
Second, the number of ways to seat the remaining six guests is the number of ways to permute 6 from 6:
Multiply: