All ISEE Middle Level Math Resources
Example Questions
Example Question #1 : Sets
What are the next two numbers of this sequence?
The sequence is formed by alternately adding and adding
to each term to get the next term.
and
are the next two numbers.
Example Question #1 : How To Find The Missing Part Of A List
Define two sets as follows:
Which of the following is a subset of ?
Each of the sets listed is a subset of .
Each of the sets listed is a subset of .
We demonstrate that all of the choices are subsets of .
is the intersection of
and
- that is, the set of all elements of both sets. Therefore,
itself is one of the choices; it is a subset of itself. The empty set
is a subset of every set. The other two sets listed comprise only elements from
, making them subsets of
.
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:
Which of the following is an element of the set ?
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:
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:
This eliminates 352 and 336. However,
.
It is also a multiple of 8:
Therefore, .
Example Question #3 : How To Find The Missing Part Of A List
Define two sets as follows:
Which of the following is not an element of the set ?
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:
, so
, so
is the set of all perfect square integers. We can eliminate two additional choices as perfect squares:
, so
, so
All four of the above are therefore elements of .
420, however is in neither set:
, so
and
, so
Therefore, , making this the correct choice.
Example Question #681 : Isee Middle Level (Grades 7 8) Mathematics Achievement
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?
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:
Derreck has 20 ways to fill the ballot.
Example Question #1 : 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?
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
There are 5,040 ways to complete the ballot.
Example Question #2 : 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?
These are four independent events, so by the multiplication principle, the ballot can be filled out ways.
Example Question #3 : 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?
These are three independent events, so by the multiplication principle, the ballot can be filled out ways.
Example Question #4 : 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?
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
Roy can fill out the ballot 3,024 times and have Mike be his first choice.
Example Question #5 : Sets
Find the missing part of the list:
To find the next number in the list, multiply the previous number by .
All ISEE Middle Level Math Resources
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