How to find the missing part of a list

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 .

is the intersection of and - the set of all elements appearing in both sets. Thus, an element can be eliminated from by demonstrating either that it is not an element of or 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.

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.