In this explainer, we will learn how to use our understanding of ratios to solve word problems.

A ratio is a way of comparing two or more values. More specifically, a ratio compares part to part. For instance, consider a box of chocolates that contains 8 dark chocolates and 6 milk chocolates. The ratio of dark chocolates to milk chocolates is represented as an ordered pair of numbers written as follows:

Similarly, the ratio of milk chocolates to dark chocolates is

In each case, the ratio was reduced to its *simplest form* by dividing both parts by their greatest
common factor. It should be clear that since we can divide by a common number, we can also
*scale up* the ratio by multiplying each part by a common number. This technique can be
helpful in comparing two or more ratios.

In our first example, we will demonstrate how this process works in more detail.

### Example 1: Finding a Ratio given Two Other Ratios

The ratio of the height of Amer to that of Karim is and the ratio of the height of Karim to that of Bassem is . Find the ratio of the height of Amer to that of Bassem.

### Answer

Ratios can be manipulated to create equivalents by multiplying or dividing each part by some constant number. In order to find the ratio of the height of Amer to that of Bassem, we will need to find a way to use the height of the person they have in common, that is, Karim. In fact, we will make the term that represents Karim’s height equal in both ratios by finding their least common multiple.

In the first ratio, the height of Amer to that of Karim is . In the second ratio, the height of Karim to that of Bassem is . Hence, we need to find the least common multiple of 2 and 1:

To make the term that represents Karim’s height equal in both ratios, we multiply the first ratio by 2:

Since the ratio of the height of Amer to the height of Karim to the height of Bassem is , we can now find the ratio of the height of Amer to that of Bassem by using only those parts.

The ratio is .

In our previous example, we saw how we can scale ratios by multiplying or dividing by some constant to allow us to compare a pair of ratios. We can also equivalently express these relationships using fraction notation, where the ratio of Amer’s height to Bassem’s height is , although it is much more common to see the fractional representation as a ratio of one part to a whole. In our second example, we will demonstrate how to use this representation.

### Example 2: Finding the Measurement of the Angles of a Triangle given the Ratio between Them

The ratio among the measures of the angles of a triangle is . Find the measure of the greatest angle of this triangle.

### Answer

Remember, the interior angles in a triangle sum to . To find the measure of the greatest angle of a triangle where the ratio between the angles is , we can follow one of a number of methods.

Arguably, the most efficient method is to convert the ratio to a fraction of one part of the whole. The greatest angle is the angle represented by 19 parts, and there are a total of parts in the ratio. Hence, the greatest angle is of :

This method is equivalent to first calculating the measure of 1 part of the ratio by sharing between 45 parts and then using this calculation to find the measure of 19 parts.

In our next example, we will once again demonstrate how to share an amount into a ratio, this time using an alternative, but equivalent, method.

### Example 3: Sharing a Quantity into a Three-Term Ratio

A school has 588 LE to spend on decorations. The money will be divided between classes , , and in the ratio . How much money does each class get to spend on decorations?

### Answer

To share an amount into a ratio, we can think about the ratio as dividing a whole into a number of parts. For the ratio , the total number of parts is given by

Hence, to share 588 LE between the classes, we will begin by sharing it equally into 12 parts:

Each part in the ratio is equal to 49 LE. With this fact, we can now calculate the amount of money that each class gets.

Since the number of parts class gets is 2, class gets

Similarly, class gets 4 parts:

Finally, class gets 6 parts:

Observe that the sum of these values is, as expected, 588 LE.

So, we have that , , and .

In our fourth example, we will demonstrate how to share a quantity given a relationship between parts of the ratio.

### Example 4: Finding Two Values given Their Ratio and Their Difference

Given that , and , calculate the value of .

### Answer

To find the value of , we will use the information about the relationship between and . We are given that . Observe that, in the ratio, the difference between the number of parts represented by and the number of parts represented by is part.

Hence, 1 part 49.

The ratio tells us that is given by 3 parts, so

### Example 5: Solving a Ratio Problem Involving Sharing a Quantity given a Relationship between the Shares

Fares, Ramy, and Seif share some money in the ratio . Twice Fares’s share exceeds Ramy’s share by 12 LE. How much money do Fares, Ramy, and Seif each have?

### Answer

We will begin by using the information that twice Fares’s share exceeds Ramy’s share by 12 LE. In the ratio, the amount of money that Fares receives is given by 4 parts, while the amount of money that Ramy receives is given by 5 parts.

Twice Fares’s share is parts, so twice Fares’s share exceeds Ramy’s share by

Since 3 parts are equal to 12 LE, 1 part is .

Hence, we will multiply each part of the ratio by 4:

Fares has 16 LE, Ramy has 20 LE, and Seif has 32 LE.

We now have the skills required to allow us to compare ratios, share a quantity into a given ratio, and use information about the difference between two parts of a ratio to solve problems. There will also be times where listing multiples of the ratio can help us solve problems. This may not feel like a particularly efficient method, but it can help us avoid having to manipulate complicated algebraic expressions. In our next example, we will look at such a process.

### Example 6: Finding a New Ratio after a Number Is Subtracted from Each Term of the Ratio

The ratio between two integers is . If you subtract 33 from each, then the ratio becomes . What are the two integers?

### Answer

While not entirely necessary, it can help us to begin by defining the two integers as and . We are given , so let’s list the first few possible options for these two integers by multiplying each by 1, 2, 3, and so on.

2 | 5 |

4 | 10 |

6 | 15 |

8 | 20 |

10 | 25 |

12 | 30 |

14 | 35 |

We are told that the ratio becomes when 33 is subtracted from each number, so we will now subtract 33 from each value in the table.

2 | 5 | ||

4 | 10 | ||

6 | 15 | ||

8 | 20 | ||

10 | 25 | ||

12 | 30 | ||

14 | 35 | 2 |

Finally, we locate the pair of numbers in these two columns that are now in the ratio :

Since this ratio results from the two starting numbers and , the integers are 6 and 15.

In the previous example, we saw how we can use our understanding of the equivalence of ratios to solve problems. It is worth noting that we could have also solved this particular problem algebraically. While this can make things more complicated, it also increases the efficiency of the solution.

To demonstrate this, we use the fact that the ratio can be scaled up by multiplying each part by any number. We assign a variable such that the ratio can be written as

Subtracting 33 from each gives

We can now create a single equation in by recognizing that this is equivalent to ; hence,

Solving for ,

Since the initial ratio was , we substitute into each expression to find that the integers are 6 and 15, as expected.

Let’s recap the key concepts from this explainer.

### Key Points

- A ratio is a way of comparing two or more values.
- Ratios are written as an ordered group of numbers.
- A ratio is reduced to its simplest form by dividing both parts by the greatest common factor.
- Ratios can also be scaled up by multiplying each part by a common number.
- Listing multiples of a ratio can help us solve problems and avoid using complicated algebraic equations, although solving problems algebraically can lead to a more efficient solution.