Sets - ISEE Upper Level Quantitative Reasoning

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

This is ten equally probable rolls out of thirty-six, resulting in a probability of

.

Since , (b) is the greater quantity.

This question can be most easily answered by matching each element in the set in (a) with the next consecutive integer, which is in the set in (b):

...

Every element in the second set has a match, but there is an unmatched element in the first set. Therefore (a) is the greater quantity.

The easiest way to answer this is to try to match each element in the first set to one the second set as follows:

...

In other words, each element in the set in (a) is paired with the element in the set in (b) that is its double. Since there is a one-to-one correspondence, the two sets are of equal aize, and (a) and (b) are equal quantities.

The easiest way to answer this is to try to match each element in the first set to one in the second set as follows:

...

Since there is a one-to-one correspondence between the elements of the two sets, (a) and (b) are equal.

For the roll to be unfavorable to the event that at least one of the dice is 5 or 6, both dice would have to be 1, 2, 3, or 4. There are ways out of 36 that this can happen, so there are ways for either or both of the two dice to be 5 or 6. Since half of 36 is 18, the probability of this event is greater than .

The quantities are equal. This can be proved as follows:

The sum of the integers from to is

.

(b) is twice this:

This is the same value as (a), the sum of the even integers from to .

The common difference of the sequence is , so the next two terms of the sequence are:

215 is the fourth term. This makes (a) greater.

The common ratio of the geometric sequence is

The next four terms of the sequence are:

- the fourth term

- the sixth term

, so the fourth term, which is (a), is greater

The common ratio of the sequence is

The next two terms of the sequence can be found as follows:

This is the fourth term, which is greater than 30.

The common difference of the sequence is ; then tenth element is

,

which is greater than 70.

The three removed jacks must either comprise two red cards and one black card, or one red card and two black cards. Since the two red queens are also removed, the net result is either the removal of four red cards and one black card, or three red cards and two black cards. Either way, more black cards than red cards are left in the altered deck, and (b) must be greater.

The common difference of the sequence is

.

The fifth number in the sequence is

.

This makes (b) greater.

The common ratio of the sequence is

.

The next three terms of the sequence are:

, the fifth term.

, making (a) greater.

The common ratio is , which means that the terms will alternate in sign. The first term is negative; therefore, all of the odd-numbered terms will be negative and all of the even-numbered terms will be positive. This makes the one thousandth term greater than the nine hundred ninety-ninth.

(a) The arithmetic sequence has common difference ; its tenth term is .

(b) By beginning with and taking the sum of the previous two terms to get each successive term, we can generate the Fibonacci sequence:

Therefore the two are equal.

By beginning with and taking the sum of the previous two terms to get each successive term, we can generate the Fibonacci sequence:

The seventh through tenth terms, added together, yield a sum of

,

which is less than the twelfth term, .

The number of ways Samantha can choose two Democrats out of the six is the number of combinations of 2 chosen from a set of 6:

This is also the number of ways Samantha can choose two Republicans out of the six.

These are two independent events, so we multiply them to get the number of total ways Samantha can vote with these specifications:

Therefore (b) is greater.

The terms are generated by alternately subtracting two and adding two. The next four terms are:

Therefore (b) is greater.

(a) There are 30 ways to fill the office of President. Afterward, there will be 29 ways to fill the office of Vice-President, 28 ways to fill the office of Secretary, and 27 ways to fill the office of Treasurer. Multiply:

(b)

This makes (b) greater.