Math and Logic Study Guide for the CCAT

Ratios and Proportions

A ratio is a comparison of two quantities, often expressed as a fraction. It represents the relative size or magnitude of one quantity when compared to that of another. A ratio can be expressed in these ways:

\[\frac{a}{b},\;a/b, \; \text{or} \;a\text{:}b\]

A proportion, on the other hand, is an equation stating that two ratios are equal.

A proportion is usually expressed in this form

\[\frac{a}{b} = \frac{c}{d}\]

where \(b\) and \(d\) cannot be equal to \(0\).

Here’s an example of a proportion:

\[\frac{2}{3} = \frac{4}{6}\]

You can check the equality by performing cross multiplication:

\[2 \times 6 = 3 \times 4\] \[12 = 12\]

Recipe Calculations

Recipe calculations involve using proportions to adjust the quantities of ingredients based on the desired yield. This is particularly useful when altering a recipe to make a different amount, serving size, or number of portions. Let’s explore this concept with a simple example.

To make \(20\) cookies, Shawn uses two cups of flour, one cup of sugar, and half a cup of butter. If he wants to adjust the recipe for \(30\) cookies, how many cups each of flour, sugar, and butter will he need?

Solution

We can use the following proportion to figure out the quantities of each of the ingredients:

\[\frac{\text{new amount}}{\text{original amount}} = \frac{\text{desired yield}}{\text{original yield}}\]

First, let’s figure out the amount of flour Shawn needs for \(30\) cookies, letting the unknown be \(x\):

\[\frac{x}{2} = \frac{30}{20}\] \[20x = 2 \times 30\] \[20x = 60\] \[x = 60 \div 20\] \[x = 3\,\text{cups of flour}\]

Now, let’s find the amount of sugar:

\[\frac{x}{1} = \frac{30}{20}\] \[20x = 30\] \[x = 30 \div 20\] \[x = 1.5\]

Written as a fraction, this is:

\[1 \frac{1}{2} \,\text{cups of sugar}\]

Finally, let’s find the amount of butter:

\[\frac{x}{0.5} = \frac{30}{20}\] \[20x = 0.5 \times 30\] \[20x = 15\] \[x = 15 \div 20\] \[x = 0.75\]

Written as a fraction, this is:

\[\frac{3}{4} \,\text{cups of butter}\]

Word Problems

Proportions are useful when solving problems based on real-world scenarios involving the relationship between different quantities.

Let’s look at an example.

Sarah has a mixture of orange juice and water in the ratio \(3\text{:}1\). If she has \(12\) cups of the mixture, how much of it is orange juice?

Solution

Let the unknown quantity be \(x\). Since three parts are orange juice and one part is water, there are a total of \(4\) parts. Now, we can set up the following proportion:

\[\frac{3}{4} = \frac{x}{12}\]

Note that on the left side we have orange juice parts to total parts, and on the right side we have the orange juice cups (\(x\)) to the total number of cups of the original mixture. Let’s solve for \(x\):

\[4 \times x = 3 \times 12\] \[4x = 36\] \[x = 36 \div 4\] \[x = 9\]

Thus, in \(12\) cups of the mixture, \(9\) cups are orange juice.

Basic Algebra

In the expansive realm of mathematics, basic algebra serves as a cornerstone for problem-solving and understanding the relationships between various quantities. Algebra extends our understanding beyond simple arithmetic. Here, variables and constants take center stage, allowing us to express mathematical relationships in a more abstract form.

Algebraic Terms

These are basic terms used in algebra that you should know for the CCAT:

• Variables are symbols that represent unknown or changing values, often denoted by letters like \(x\) or \(y\). In the equation \(3x = 9\), \(x\) is the variable.

• Constants are fixed values that do not change, such as numbers or coefficients, like \(5\) or the \(3\) in \(3x\).

• Expressions are combinations of variables, constants, and mathematical operations, such as \(2y^2\) and \(4x-8\). Expressions do not contain an equal sign (\(=\)) or any other equivalency sign.

Solving Word Problems Using Algebra

Equations are the heart of algebra, providing a means to express relationships and solve for unknowns. An equation does have an \(=\) sign. This is an example of a linear equation:

\[56 = 10x + 2\]

There are two main parts when using equations to solve word problems. First, you build or set up the equation, then you solve it. Each of these parts is broken up into steps, and we will use this word problem to go through them.

Amanda is three times as old as her younger brother. The sum of their ages is \(20\) years. Determine their ages.

Setting up the Equation

Building the equation to solve is the first part. You’ll need to translate textual information into mathematical language to do so. These are the key steps to this part:

Identify the Unknowns

Clearly define the variables representing unknown quantities in the problem. Using the problem above, the question asks for the ages of Amanda and her younger brother, meaning those are our two unknown quantities. We can let Amanda’s age be \(a\) and her younger brother’s age be \(b\).

Translate Statements into Equations

Look for keywords like “is,” “equals,” “more than,” or “less than” that will determine the mathematical operations needed to form an equation.

Since Amanda is three times as old as her younger brother, we know multiplication will be needed and that the younger brother’s age must be multiplied by \(3\) to equal Amanda’s age. Therefore, we can write the following:

\[a = 3b\]

Also, given that the sum of their ages is \(20\), we can write:

\[a + b = 20\]

Record the Equation to Solve

Formulate an equation that accurately represents the given information.

To write the equation to solve, we will need to combine the two equations from above. We can put the first equation (\(a=3b\)) into the second equation (\(a+b=20\)) to get:

\[3b + b = 20\]

Solving the Equation

Solving algebraic equations is the second part in the process, and it follows another step-by-step process that requires precision and logical reasoning. Here’s a general guide:

Isolate the Variable

Isolating the variable means getting a single variable alone on one side of the equation. To do this, you combine like terms and simplify both sides of the equation.

To solve the equation formed above, first, we combine like terms:

\[3b + b = 20\] \[(3+1)b = 20\] \[4b = 20\]

In this form, the variable is isolated on one side of the equal sign, while the number is on the other side.

Perform Inverse Operations

Next, you must undo multiplication or division using their inverses.

Since \(b\) is multiplied by \(4\), to get the variable alone we need to divide both sides by \(4\):

\[\frac{4b}{4} = \frac{20}{4}\] \[b = 5\]

Now we know \(b = 5\). Since \(a = 3b\), we can say this:

\[a = 3 \times 5 = 15\]

Thus, Amanda is \(15\) years old, and her younger brother is \(5\) years old.

Verify the Solution

Once you have a solution, it’s a good idea to verify that your solution is correct. To do this, substitute the obtained value(s) back into the original equations to ensure it works:

\[a = 3b\] \[15 \stackrel{?}{=} 3 \times 5\] \[15 = 15\]

and

\[a + b = 20\] \[15 + 5 \stackrel{?}{=} 20\] \[20 = 20\]

We have found the correct values to satisfy the original word problem.