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.

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 .

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 .

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 .

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 .