### All ISEE Middle Level Math Resources

## Example Questions

### Example Question #1 : Sets

What are the next two numbers of this sequence?

**Possible Answers:**

**Correct answer:**

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

Define two sets as follows:

Which of the following is a subset of ?

**Possible Answers:**

Each of the sets listed is a subset of .

**Correct answer:**

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 #3 : 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 ?

**Possible Answers:**

**Correct answer:**

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 #4 : 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 ?

**Possible Answers:**

**Correct answer:**

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

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:**

**Correct answer:**

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 #6 : How To Find The Missing Part Of A List

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:**

**Correct answer:**

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

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:**

**Correct answer:**

These are four independent events, so by the multiplication principle, the ballot can be filled out ways.

### Example Question #8 : How To Find The Missing Part Of A List

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:**

**Correct answer:**

These are three independent events, so by the multiplication principle, the ballot can be filled out ways.

### Example Question #9 : How To Find The Missing Part Of A List

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:**

**Correct answer:**

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 #2 : Sets

Find the missing part of the list:

**Possible Answers:**

**Correct answer:**

To find the next number in the list, multiply the previous number by .

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