How to find the missing part of a list - SSAT Middle Level Quantitative

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 that is equal to 0.4, so we don't include it. The criterion requires strict inequality.

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 that is equal to 0.4, so we don't include it. The criterion requires strict inequality.

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 that is equal to 0.4, so we don't include it. The criterion requires strict inequality.

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

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.

The elements of the set - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by 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.

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

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.

The elements of the set - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by 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.

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

The elements of the set - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by 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.

The pattern for the set is all even numbers up until . The missing even number between and would be and the missing number between and would be .

The elements of the set - that is, the intersection of and - are exactly those in both sets. We can test each of the six elements in for inclusion in set by 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.

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.

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.

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.

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

A set is a subset of if 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 in do appear in , however, so it is the subset.

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

and have no elements in common, so has zero elements. This is not one of the choices.

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.