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

The sequence is formed by alternately adding and adding to each term to get the next term.

and are the next two numbers.

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.

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 question asks that you first determine the pattern, then use said pattern to find A and B, which can then be substituted into the equation . The pattern is to add 5 to the previous number to achieve the next number, which results in

and

Putting A and B into the equation gives:

The other answers are wrong for the following reasons:

286 is , not

48 is

63 is

624 is a result of having , a pattern addition mistake

To answer this question, the pattern must first be found. There is no common interval between the numbers, however it can be seen that the next number is the sum of the number plus the number before. A good way to address a tricky pattern is to write the difference of the two numbers above the number, as seen below where the bold numbers are the orginal numbers and the nonbolded numbers are the differences.

0 1 1 2 3

1 1 2 3 5 8

Using this pattern the value of A will be

The first stop to determining what the tile will be is to determine how many repeats of the pattern will occur before hand. There are colors in the pattern, so divide by . This results in with a remainder of . This means that there will be repeats of the pattern and at the tile, it will be at the second spot in the pattern, which is Blue. So, the tile is Blue.

What that the fractions in this pattern have in common is that they are all the equivalent of .

The value of y should be a number that is the equivalent of when divided by 12.

Given that of 12 is 4, of 12 would be equal to 8, the correct answer.

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

Given that , that is the correct answer.

The set is composed of consecutive squares.

We can see that will b equal to

Therefore, 36 is the correct answer.

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 .

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 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.

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.

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.

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.

, so 21 has 3 and 9 as factors, and 27 is not prime.

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 .

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.

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 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 .