In order to solve this question, we need to use both the equation and the table. We are looking for the corresponding value for . We can plug into the in our equation to solve for .

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 sets and , 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.

In order to solve this question, we need to use both the equation and the table. We are looking for the corresponding value for . We can plug into the in our equation to solve for .

In this sequence, every subsequent number is equal to one third of the preceding number:

Given that , that is the correct answer.

In order to solve this question, we need to use both the equation and the table. We are looking for the corresponding value for . We can plug into the in our equation to solve for .

In order to solve this question, we need to use both the equation and the table. We are looking for the corresponding value for . We can plug into the in our equation to solve for .

In this set, each subsequent fraction is half the size of the preceding fraction; (the denominator is doubled for each successive fraction, but the numerator stays the same). Given that the last fraction in the set is , it follows that the subsequent fraction will be .

In order to solve this question, we need to use both the equation and the table. We are looking for the corresponding value for . We can plug into the in our equation to solve for .

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 of by 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.

Add 23 to the first element to get the second element. The increment decreases by two with each successive entry. The pattern can be seen below:

To get the element in the circle, add 15:

, the correct response.

This is confirmed by adding 13;

, the next number in the sequence.