is the union ofand- that is, it is the set of all elements in one set or the other.

A set is a subset ofif and only if every one of its elements is in. Three of the listed sets do not meet this criterion:

,, and, but none of those three elements are in. All of the elements indo appear in, however, so it is the subset.

is the intersection ofand- the set of all elements appearing inbothsets. Thus, an element can be eliminated fromby demonstrating either that it is not an element ofor that it is not an element of.

is the set of positive integers ending in "5". 513 and 657 are not in, so they are not in.

is the set of muliples of 9. We test the three remaining numbers easily by seeing if 9 divides their digit sum:

425 and 565 are not multiples of 9; neither is in, so neither is in.

and, so. This is the correct choice.

The set is composed of consecutive squares.

We can see thatwill b equal to

Therefore, 36 is the correct answer.

The elements of the set- that is, the intersection ofand- are exactly those in both sets. We can test each of the six elements infor inclusion in setby dividing each by 7 and noting which divisions yield no remainder:

andhave no elements in common, sohas zero elements. This is not one of the choices.

For a set to be a subset of, all of its elements must be elements of- that is, all of its elements must be multiples of 3. A set can therefore be proved to not be a subset ofby identifying one element not a multiple of 3.

We can do that with four choices:

:

:

:

:

However, the remaining set,, can be demonstrated to include only multiples of 3:

is the correct choice.

The elements of the set- that is, the intersection ofand- are exactly those in both sets. We can test each of the six elements infor inclusion in setby testing for divisibility by 5 - but this can be accomplished by looking at the last digit. Only 345, 600, and 855 have last digit 5 or 0 so only these three elements are divisible by 5. This makes three the correct answer.

We show that none of the four listed sets can be a subset of the primes by identifying one composite number in each - that is, by proving that there is at least one factor not equal to 1 or itself:

, so 25 has 5 as a factor, and 25 is not prime.

, so 9 has 3 as a factor, and 9 is not prime.

, so 21 has 3 and 7 as factors, and 21 is not prime.

,所以21 3和9因素,和27不整洁的e.

Since each set has at least one element that is not a prime, each has at least one element not in, and none of the sets are subsets of.

By dividing the numerator of each fraction by its denominator, each fraction can be rewritten as its decimal equivalent:

All fractions except can be seen to fall between 0.3 and 0.4, exclusive. Three is the correct answer.

Note thatisequalto 0.4, so we don't include it. The criterion requiresstrictinequality.

For a set to be a subset of, all of its elements must also be elements of- that is, all of its elements must be multiples of 5. An integer is a multiple of 5 if and only if its last digit is 5 or 0, so all we have to do is examine the last digit of each number in all four sets.

In the setsand, every element ends in a 5 or a 0, so all elements of both sets are in; both sets are subsets of.

However,includes one element that does not end in either 5 or 0, namely 8934, so 8934 is not an element in; subsequently, this set is not a subset of. Similarly,is not a subset of, since it includes 7472, which ends in neither 0 nor 5.

The correct answer is therefore two.

The sequence is formed by alternately addingand addingto each term to get the next term.

andare the next two numbers.