All HSPT Math Resources
Example Questions
Example Question #91 : 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".
25 to 3
12 to 1
11 to 1
22 to 3
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 #92 : 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:
Example Question #61 : How To Do Other Word Problems
Fill in the box with a single digit to form a number that is divisible by 24:
.
The number is not divisible by 24 regardless of the digit chosen.
The number is not divisible by 24 regardless of the digit chosen.
A whole number divisible by 24 is also divisible by any factor of 24, including 8.
A whole number is divisible by 8 if and only if its last three digits form a number which itself is divisible by 8. We can try each of the three digits:
None of the choices yields a number divisible by 8. Since 8 cannot be a factor of this number, neither can 24.
Example Question #538 : Hspt Mathematics
Fill in the square with a single digit to form a number that is divisible by 36:
A number divisible by 36 must be divisible by any factor of 36; two such factors are 4 and 9, so we can test for divisibliity by both.
The number is divisible by 4 if and only if its last two digits form a number divisible by 4. 32, 52, 72, and 92 are all divisible by 4:
Therefore, all of the choices make the number divisible by 4. Therefore, we see which one makes the number divisible by 9, which happens if and only if the sum of the digits is a multiple of 9. If we let be the missing digit, then the digit sum is
.
We can test each digit by substituting for :
Since 18 is the only digit sum divisible by 9, 3 is the correct digit to fill in the box.
Example Question #92 : Word Problems
Define the universal set to be the set of all people.
Let be the set of all people who own canoes, and be the set of all people who own motorboats.
Donald owns a canoe and a motorboat. Ed owns a canoe, but not a motorboat. Who belongs in the set ?
Neither Donald nor Ed
Ed, but not Donald
Both Donald and Ed
Donald, but not Ed
Donald, but not Ed
is the complement of , making the set of all persons not in - that is, the set of all people who do not own a canoe. is the set of all persons who own a motorboat. To be in , the union of the sets, a person must be in one or the other set, or both, or, equivalently, the person must either not own a canoe or must own a motorboat, or both. Donald is in by virtue of owning a motorboat. Ed, however, owns a canoe, but does not own a motorboat, so he does not meet either qualification.
Example Question #540 : Hspt Mathematics
Mr. and Mrs. Gregory are hosting a party with their six children. The above shows their two tables along with their seats.
How many ways can the eight persons be seated so that Mr. Gregory and Mrs. Gregory are at different tables?
First, the table at which Mr. Gregory is seated must be decided - there are two alternatives. Mrs. Gregory will be at the other table.
Then it must be decided which of the four seats Mr. Gregory will occupy at his table, and which of the four seats Mrs. Gregory will occupy at hers.
Finally, it must be decided how the six children will be arranged in the remaining six seats - this can happen in
ways.
Apply the multiplication principle; there will be
possible arrangements meeting the specifications.