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Ratio Problem Solving

Here we will learn about ratio problem solving, including how to set up and solve problems. We will also look at real life ratio problems.

There are also ratio problem solving worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is ratio problem solving?

Ratio problem solving is a collection of word problems that link together aspects of ratio and proportion into more real life questions. This requires you to be able to take key information from a question and use your knowledge of ratios (and other areas of the curriculum) to solve the problem.

A ratio is a relationship between two or more quantities . They are usually written in the form a:b where a and b are two quantities. When problem solving with a ratio, the key facts that you need to know are,

• What is the ratio involved?
• What order are the quantities in the ratio?
• What is the total amount / what is the part of the total amount known?
• What are you trying to calculate ?

As with all problem solving, there is not one unique method to solve a problem. However, this does not mean that there aren’t similarities between different problems that we can use to help us find an answer. 

The key to any problem solving is being able to draw from prior knowledge and use the correct piece of information to allow you to get to the next step and then the solution.

Let’s look at a couple of methods we can use when given certain pieces of information.

When solving ratio problems it is very important that you are able to use ratios. This includes being able to use ratio notation. 

For example, Charlie and David share some sweets in the ratio of 3:5. This means that for every 3 sweets Charlie gets, David receives 5 sweets.

Charlie and David share 40 sweets, how many sweets do they each get?

We use the ratio to divide 40 sweets into 8 equal parts. 

Then we multiply each part of the ratio by 5.

3 x 5:5 x 5 = 15:25

This means that Charlie will get 15 sweets and David will get 25 sweets.

• Dividing ratios

Step-by-step guide: Dividing ratios (coming soon)

Ratios and fractions (proportion problems)

We also need to consider problems involving fractions. These are usually proportion questions where we are stating the proportion of the total amount as a fraction.

Simplifying and equivalent ratios

• Simplifying ratios

Equivalent ratios

Units and conversions ratio questions

Units and conversions are usually equivalent ratio problems (see above).

• If £1:\$1.37 and we wanted to convert £10 into dollars, we would multiply both sides of the ratio by 10 to get £10 is equivalent to \$13.70.
• The scale on a map is 1:25,000. I measure 12cm on the map. How far is this in real life, in kilometres? After multiplying both parts of the ratio by 12 you must then convert 12 \times 25000=300000 \ cm to km by dividing the solution by 100 \ 000 to get 3km.

Notice that for all three of these examples, the units are important. For example if we write the mapping example as the ratio 4cm:1km, this means that 4cm on the map is 1km in real life.

Top tip: if you are converting units, always write the units in your ratio.

Usually with ratio problem solving questions, the problems are quite wordy . They can involve missing values , calculating ratios , graphs , equivalent fractions , negative numbers , decimals and percentages .

Highlight the important pieces of information from the question, know what you are trying to find or calculate , and use the steps above to help you start practising how to solve problems involving ratios.

How to do ratio problem solving

In order to solve problems including ratios:

Identify key information within the question.

Know what you are trying to calculate.

Use prior knowledge to structure a solution.

Explain how to do ratio problem solving

Ratio problem solving worksheet

Get your free ratio problem solving worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on ratio

Ratio problem solving is part of our series of lessons to support revision on ratio . You may find it helpful to start with the main ratio lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• How to work out ratio  
• Ratio to fraction
• Ratio scale
• Ratio to percentage

Ratio problem solving examples

Example 1: part:part ratio.

Within a school, the number of students who have school dinners to packed lunches is 5:7. If 465 students have a school dinner, how many students have a packed lunch?

Within a school, the number of students who have school dinners to packed lunches is \bf{5:7.} If \bf{465} students have a school dinner , how many students have a packed lunch ?

Here we can see that the ratio is 5:7 where the first part of the ratio represents school dinners (S) and the second part of the ratio represents packed lunches (P).

We could write this as

Where the letter above each part of the ratio links to the question.

We know that 465 students have school dinner.

2 Know what you are trying to calculate.

From the question, we need to calculate the number of students that have a packed lunch, so we can now write a ratio below the ratio 5:7 that shows that we have 465 students who have school dinners, and p students who have a packed lunch.

We need to find the value of p.

3 Use prior knowledge to structure a solution.

We are looking for an equivalent ratio to 5:7. So we need to calculate the multiplier. We do this by dividing the known values on the same side of the ratio by each other.

So the value of p is equal to 7 \times 93=651.

There are 651 students that have a packed lunch.

Example 2: unit conversions

The table below shows the currency conversions on one day.

Use the table above to convert £520 (GBP) to Euros € (EUR).

Use the table above to convert \bf{£520} (GBP) to Euros \bf{€} (EUR).

The two values in the table that are important are GBP and EUR. Writing this as a ratio, we can state

We know that we have £520.

We need to convert GBP to EUR and so we are looking for an equivalent ratio with GBP = £520 and EUR = E.

To get from 1 to 520, we multiply by 520 and so to calculate the number of Euros for £520, we need to multiply 1.17 by 520.

1.17 \times 520=608.4

So £520 = €608.40.

Example 3: writing a ratio 1:n

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the 500ml of concentrated plant food must be diluted into 2l of water. Express the ratio of plant food to water respectively in the ratio 1:n.

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the \bf{500ml} of concentrated plant food must be diluted into \bf{2l} of water . Express the ratio of plant food to water respectively as a ratio in the form 1:n.

Using the information in the question, we can now state the ratio of plant food to water as 500ml:2l. As we can convert litres into millilitres, we could convert 2l into millilitres by multiplying it by 1000.

2l = 2000ml

So we can also express the ratio as 500:2000 which will help us in later steps.

We want to simplify the ratio 500:2000 into the form 1:n.

We need to find an equivalent ratio where the first part of the ratio is equal to 1. We can only do this by dividing both parts of the ratio by 500 (as 500 \div 500=1 ).

So the ratio of plant food to water in the form 1:n is 1:4.

Example 4: forming and solving an equation

Three siblings, Josh, Kieran and Luke, receive pocket money per week proportional to their age. Kieran is 3 years older than Josh. Luke is twice Josh’s age. If Josh receives £8 pocket money, how much money do the three siblings receive in total?

Three siblings, Josh, Kieran and Luke, receive pocket money per week proportional to their ages. Kieran is \bf{3} years older than Josh . Luke is twice Josh’s age. If Luke receives \bf{£8} pocket money, how much money do the three siblings receive in total ?

We can represent the ages of the three siblings as a ratio. Taking Josh as x years old, Kieran would therefore be x+3 years old, and Luke would be 2x years old. As a ratio, we have

We also know that Luke receives £8.

We want to calculate the total amount of pocket money for the three siblings.

We need to find the value of x first. As Luke receives £8, we can state the equation 2x=8 and so x=4.

Now we know the value of x, we can substitute this value into the other parts of the ratio to obtain how much money the siblings each receive.

The total amount of pocket money is therefore 4+7+8=£19.

Example 5: simplifying ratios

Below is a bar chart showing the results for the colours of counters in a bag.

Express this data as a ratio in its simplest form.

From the bar chart, we can read the frequencies to create the ratio.

We need to simplify this ratio.

To simplify a ratio, we need to find the highest common factor of all the parts of the ratio. By listing the factors of each number, you can quickly see that the highest common factor is 2.

\begin{aligned} &12 = 1, {\color{red} 2}, 3, 4, 6, 12 \\\\ &16 = 1, {\color{red} 2}, 4, 8, 16 \\\\ &10 = 1, {\color{red} 2}, 5, 10 \end{aligned}

HCF (12,16,10) = 2

Dividing all the parts of the ratio by 2 , we get

Our solution is 6:8:5 .

Example 6: combining two ratios

Glass is made from silica, lime and soda. The ratio of silica to lime is 15:2. The ratio of silica to soda is 5:1. State the ratio of silica:lime:soda.

Glass is made from silica, lime and soda. The ratio of silica to lime is \bf{15:2.} The ratio of silica to soda is \bf{5:1.} State the ratio of silica:lime:soda .

We know the two ratios

We are trying to find the ratio of all 3 components: silica, lime and soda.

Using equivalent ratios we can say that the ratio of silica:soda is equivalent to 15:3 by multiplying the ratio by 3.

We now have the same amount of silica in both ratios and so we can now combine them to get the ratio 15:2:3.

Example 7: using bar modelling

India and Beau share some popcorn in the ratio of 5:2. If India has 75g more popcorn than Beau, what was the original quantity?

India and Beau share some popcorn in the ratio of \bf{5:2.} If India has \bf{75g} more popcorn than Beau , what was the original quantity?

We know that the initial ratio is 5:2 and that India has three more parts than Beau.

We want to find the original quantity.

Drawing a bar model of this problem, we have

Where India has 5 equal shares, and Beau has 2 equal shares.

Each share is the same value and so if we can find out this value, we can then find the total quantity.

From the question, India’s share is 75g more than Beau’s share so we can write this on the bar model.

We can find the value of one share by working out 75 \div 3=25g.

We can fill in each share to be 25g.

Adding up each share, we get

India = 5 \times 25=125g

Beau = 2 \times 25=50g

The total amount of popcorn was 125+50=175g.

Common misconceptions

• Mixing units

Make sure that all the units in the ratio are the same. For example, in example 6 , all the units in the ratio were in millilitres. We did not mix ml and l in the ratio.

• Ratio written in the wrong order

For example the number of dogs to cats is given as the ratio 12:13 but the solution is written as 13:12.

• Ratios and fractions confusion

Take care when writing ratios as fractions and vice-versa. Most ratios we come across are part:part. The ratio here of red:yellow is 1:2. So the fraction which is red is \frac{1}{3} (not \frac{1}{2} ).

• Counting the number of parts in the ratio, not the total number of shares

For example, the ratio 5:4 has 9 shares, and 2 parts. This is because the ratio contains 2 numbers but the sum of these parts (the number of shares) is 5+4=9. You need to find the value per share, so you need to use the 9 shares in your next line of working.

• Ratios of the form \bf{1:n}

The assumption can be incorrectly made that n must be greater than 1 , but n can be any number, including a decimal.

Practice ratio problem solving questions

1. An online shop sells board games and computer games. The ratio of board games to the total number of games sold in one month is 3:8. What is the ratio of board games to computer games?

8-3=5 computer games sold for every 3 board games.

2. The volume of gas is directly proportional to the temperature (in degrees Kelvin). A balloon contains 2.75l of gas and has a temperature of 18^{\circ}K. What is the volume of gas if the temperature increases to 45^{\circ}K?

3. The ratio of prime numbers to non-prime numbers from 1-200 is 45:155. Express this as a ratio in the form 1:n.

4. The angles in a triangle are written as the ratio x:2x:3x. Calculate the size of each angle.

5. A clothing company has a sale on tops, dresses and shoes. \frac{1}{3} of sales were for tops, \frac{1}{5} of sales were for dresses, and the rest were for shoes. Write a ratio of tops to dresses to shoes sold in its simplest form.

6. During one month, the weather was recorded into 3 categories: sunshine, cloud and rain. The ratio of sunshine to cloud was 2:3 and the ratio of cloud to rain was 9:11. State the ratio that compares sunshine:cloud:rain for the month.

Ratio problem solving GCSE questions

1. One mole of water weighs 18 grams and contains 6.02 \times 10^{23} water molecules.

Write this in the form 1gram:n where n represents the number of water molecules in standard form.

2. A plank of wood is sawn into three pieces in the ratio 3:2:5. The first piece is 36cm shorter than the third piece.

Calculate the length of the plank of wood.

5-3=2 \ parts = 36cm so 1 \ part = 18cm

3. (a) Jenny is x years old. Sally is 4 years older than Jenny. Kim is twice Jenny’s age. Write their ages in a ratio J:S:K.

(b) Sally is 16 years younger than Kim. Calculate the sum of their ages.

Learning checklist

You have now learned how to:

• Relate the language of ratios and the associated calculations to the arithmetic of fractions and to linear functions
• Develop their mathematical knowledge, in part through solving problems and evaluating the outcomes, including multi-step problems
• Make and use connections between different parts of mathematics to solve problems

The next lessons are

• Compound measures
• Best buy maths

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Solving Ratio Problems

• We add the parts of the ratio to find the total number of parts.
• There are 2 + 3 = 5 parts in the ratio in total.
• To find the value of one part we divide the total amount by the total number of parts.
• 50 ÷ 5 = 10.
• We multiply the ratio by the value of each part.
• 2:3 multiplied by 10 gives us 20:30.
• The 50 counters are shared into 20 counters to 30 counters.

• 2 + 3 = 5 and so there are 5 parts in the ratio in total.
• We divide by this total number of parts to find the value of each part.
• We multiply the original ratio by the value of each part.
• We have 20:30.

• Sharing in a Ratio: Part 1

Ratio Problems: Worksheets and Answers

How to Solve Ratio Problems

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Ratio Math Problems - Three Term Ratios

In these lessons, we will learn how to solve ratio word problems that involve three terms.

Related Pages Two-Term Ratio Word Problems More Ratio Word Problems Math Word Problems More Algebra Lessons

Ratio problems are word problems that use ratios to relate the different items in the question.

Ratio problems: Three-term Ratios

Example 1: A special cereal mixture contains rice, wheat and corn in the ratio of 2:3:5. If a bag of the mixture contains 3 pounds of rice, how much corn does it contain?

Step 2: Solve the equation: Cross Multiply

2 × x = 3 × 5 2x = 15

Answer: The mixture contains 7.5 pounds of corn.

Example 2: Clothing store A sells T-shirts in only three colors: red, blue and green. The colors are in the ratio of 3 to 4 to 5. If the store has 20 blue T-shirts, how many T-shirts does it have altogether?

Solution: Step 1: Assign variables: Let x = red shirts y = green shirts

Step 2: Solve the equation: Cross Multiply both equations 3 × 20 = x × 4 60 = 4x x = 15

5 × 20 = y × 4 100 = 4y y = 25

The total number of shirts would be 15 + 25 + 20 = 60

Answer: There are 60 shirts.

How to solve Ratio Word Problems with three terms?

Example: A piece of string that is 63 inches long is cut into 3 parts such that the lengths of the parts of the string are in the ratio of 5 to 6 to 10. Find the lengths of the 3 parts.

How to solve Two Term and Three Term Ratio Problems?

A Ratio compares two things that have the same units A Part to Part Ratio compares one thing to another thing A Part to Total (whole) Ratio compares one thing to the total number

Example: In a class of 30 students, there are 18 girls and 12 boys. What is the ratio of boys to girls? What is the ratio of girls to boys? What is the ratio of girls to total?

We can have a three term ratio of red to blue to green marbles.

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1.4: Proportions

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In the previous section, we learned that a ratio is a comparison of two quantities. However, many of the problems we solved involved comparing multiple ratios, and often required finding an equality between two ratios. These sorts of problems are most easily solved using proportions , which are the subject of this section.

In this section, you will learn to:

• Recognize and set up proportion problems
• Apply the Cross Multiplication and Division undoes Multiplication methods to solve proportion problems
• Use the Compare to the Whole method to solve problems involving proportions

Proportions: Definition and Basic Methods

Let's recall our starting example from the previous section, in which we had a recipe that called for \(2\) cups of flour and \(1\) cup of sugar. We asked the question: if we wanted to make a larger version of the same recipe using \(3\) cups of sugar, how much flour should we use?

We know now that ratios can be expressed as fractions, and whenever ratios are equivalent, the fractional representations of those ratios are equal. So, let's call the number of cups of flour \(x\) and set up the following equality:

\[\frac{2 \text{ cups of flour}}{1 \text{ cup of sugar}} = \frac{x \text{ cups of flour}}{3 \text{ cups of sugar}}\]

We need to figure out the value of \(x\) that makes this equation true. That is, what number can be substituted in place of \(x\) so that these fractions truly are equal? In this case, the easiest way is to guess-and-check, which is a perfectly valid solution method with small numbers. If you just try some small numbers, you can find that \(x = 6\) is the solution, because \(\frac{6}{3}\) reduces to \(\frac{2}{1}\), and thus \(\frac{6}{3}\) and \(\frac{2}{1}\)are equivalent ratios.

In this chapter, we'll learn how to solve problems like this generally, including those that can't be solved using guess-and-check. Our primary tool will be a proportion.

Definition: Proportion

A proportion is an equality between ratios.

This means that, in a strictly mathematical sense, a proportion is an equation. For example,

\[\frac{5}{2} = \frac{40}{16}\]

is a proportion, because it contains two ratios that are equal to one another. You may have heard the word proportion used in other ways, and that's because the word "proportion" has a different mathematical meaning than its typical English usage. That doesn't mean either usage is wrong; rather, it is dependent upon context. In this book, we will use the word proportion to mean any equation that looks like this:

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

where \(a, b, c\), and \(d\) will usually be numbers or variables.

The reasons we care about proportions is that they give us a way to find an unknown part of one of the ratios involved . Recall the following example, which we saw in the Ratios section:

Example \(\PageIndex{1}\)

On the Western Oregon University website , the total enrollment is listed as \(3752\) students, and the student-faculty ratio is listed as \(13 \colon 1\). You want to know how many faculty there are at WOU. How might you find this out, and how do you explain your answer?

In solving this problem before, we set up two ratios

\[3752 \colon x \quad \text{and} \quad 13 \colon 1 \]

Why did we do this? Well, it turns out that all proportion problems can be solved using a method from algebra known as cross multiplication . While this text mostly stays away from algebra, this procedure is essential. The good news is that it works the same way every time, and it's not very complicated.

Cross Multiplication

If you have a proportion of the form:

then "cross multiplication" refers to rewriting the equation in the following equivalent way:

\[c \times b= a \times d\]

In other words, we are "crossing" from \(a\) to \(d\) in the \(\searrow\) direction and from \(c\) to \(b\) in the \(\swarrow\) direction.

A quick mathematical note: what we're really doing is multiplying both sides by \(b\) and \(d\), and then canceling common factors -- but calling it cross multiplication seems to make it easier for students to understand and remember.

Let's practice cross multiplying in our example. We had the proportion:

\[\frac{3752}{x} = \frac{13}{1}\]

Cross multiplying this gives us the following equation:

\[13 \times x = 3752 \times 1\]

What good did that do? Well, note that we can simplify a little bit. It's typical to omit the \(\times\) when a variable is multiplied by a number, so we can rewrite \(13x\) for \(13 \times x\). We also know that \(3752 \times 1 = 3752\). So our equation becomes

\[13x = 3752\]

Now what? We've eliminated the fractions, but we can't yet say what \(x\) is. In order to find \(x\), we need one more algebra procedure, which we will call Division undoes Multiplication.

Division undoes Multiplication

Given an equation of the form \[Ax = B\] where \(A\) and \(B\) are numbers, we can find the value of \(x\) by dividing both sides by \(A\). That is, \[x = \frac{B}{A} = B \div A\]

Once again, we are using properties of fractions here: mathematically, we are dividing both sides by \(A\) and then reducing:

\[\begin{align*} Ax & = B\\ \frac{Ax}{A} & = \frac{B}{A} \\ \frac{\cancel{A}x}{\cancel{A}} & = \frac{B}{A} \\ x & = \frac{B}{A} \\ \end{align*} \]

But this is another procedure used so frequently that it's worth giving it a name.

Back to our example: we had the equation \[13x = 3752\]

We now have a tool to find \(x\) -- the fact that Division undoes Multiplication ! Using this procedure, we have \[x = \frac{3752}{13} = 3752 \div 13 \approx 288.6\]

This is the same answer we found before, but we used a slightly different method. And keep in mind that, just as the previous section, we would need to round this answer to \(289\) faculty to make sense in context. That said; the main point is now we now have a fool-proof way to solve this type of equation! 

While this may seem more complicated at first, you'll find that the following sequence of steps will always work to solve proportions:

Solving Proportions

• Set up the proportion with exactly one unknown value, called \(x\).
• Apply the Cross Multiplication.
• Apply Division undoes Multiplication.

We will get lots of practice with this procedure in the exercises for this section. Once you practice with the procedures above, you'll find that it's not too bad. The hardest part is often the first step — setting up the proportion correctly. That's the part that depends on reading the question very carefully! In general, the way to set up a proportion involves keeping track of units. Let's see an example to understand.

Example \(\PageIndex{2}\)

In an office supply store, \(8\) markers cost a total of \(\$12.00\). Assuming all markers are equally priced, how much would 6 markers cost?

This is a problem that is suitable to be solved using proportions because the markers are all equally priced, meaning that the ratio of total cost : number of markers purchased will be the same, no matter how many markers are purchased. That means we can set up the following proportion:

\[\frac{12 \text{ dollars}}{8 \text{ markers}} = \frac{x \text{ dollars}}{6 \text{ markers}}\]

Notice how, in the equation above, we are labeling the units of all quantities involved. Moreover, the units on each side match: dollars are on top, markers are on bottom, and the corresponding quantities are grouped on each side of the equation -- \(\$12\) for \(8\) markers, and \(\$x\) for 6 markers. Labeling your units in this way will help you avoid mistakes with units!

Now that we've gotten our proportion set up correctly, we can rewrite it without labels: \[\frac{12}{8} = \frac{x}{6}\]

From here, we'll follow the last two steps: cross multiply, and then use division to find \(x\). Using Cross Multiplication, we have \[x \times 8 = 12 \times 6 \]

On the left, we can rewrite \(x \times 8\) as \(8 x\), since multiplication can always switch orders. Then we can simplify to get \[8x = 72\]

Now we can use Division undoes Multiplication to get

\[x = \frac{72}{8} = 72 \div 8 = 9\]

Therefore, \(x = 9\). Now, we want to make sure our answer actually means something. What are the units on \(x\)? Well, if we look back at our original proportion,

we see that \(x\) is a number of dollars. Thus, we can say that \(x = \$9\), which means that 6 markers will cost \(\$9\).

You may be thinking: there is a much faster way to do that! And that may be true for you. Once again, the point is not to mimic a particular method for problem solving here — these notes will show some good ways of solving a problem, but they cannot cover every good solution. They are intended to highlight themes and strategies that will work for many types of situations. Other ways you may have solved the problem above include:

• Calculate the cost per marker to be \(\$1.50\), and multiply that number by 6 markers to get \(\$9\).
• Calculate that \(6\) is \(\frac{3}{4}\) of \(8\), so the cost of \(6\) markers would be \(\frac{3}{4}\) the cost of \(8\) markers, and \(\frac{3}{4}\) of \(\$12\) is \(\$9\).
• Set up a different initial proportion, such as \(\frac{12 \text{ dollars}}{x \text{ dollars}} = \frac{8 \text{ markers}}{6 \text{ markers}}\) or \(\frac{8 \text{ markers}}{12 \text{ dollars}} = \frac{6 \text{ markers}}{x \text{ dollars}}\) and then solved that proportion.

What's amazing about the last point above is that both of those proportions — which were different than the method used in the solution above — still give the same answer! This shows that there are many  different ways of approaching the same problem. All you need to do is find the one that works for you, and be able to explain your work.

Comparing to the Whole

Sometimes a problem involving proportions will be less straightforward. For example, consider the following:

Example \(\PageIndex{3}\)

In a rainforest in Panama, the ratio of two-toed sloths to three-toed sloths is \(10 \colon 3\). There are \(741\) total sloths in the rainforest. How many of them are two-toed?

In the problem above, we are given one ratio that compares the quantity of two-toed versus three-toed sloths. However, we are not given any information about the actual numbers of either two- or three-toed sloths. We simply know the comparison between them. Instead, we are just given the total number of sloths, but no actual breakdown into how many fall into each category. How are we supposed to find the number of two-toed sloths from just this information? We can't readily write down a proportion like we were able to in the previous example, because the units would be wrong; we need to compare like quantities. This situation calls for one more procedure.

Compare to the Whole

Assume there are two quantities, \(x\) and \(y\), neither of which you know. However, you know two things about them

• The total \(x + y\) (the total number of both quantities)
• The ratio of quantity \(x\) to quantity \(y\) is \(a \colon b\)

Then you can use the Compare to the Whole method. This says that, to find quantity \(x\), you use the proportion \[\frac{a}{a+b} = \frac{x}{x+y}\] and then find \(x\). Note: you already know \(x+y\), since it is the total number of both quantities.

Let's see how this procedure can be applied to the sloth example.

Example \(\PageIndex{3}\) Revisited

In this question, our two quantities \(x\) and \(y\) are the number of two- and three-toed sloths, respectively. We are asked to find \(x\), the number of two-toed sloths. Our known ratio is \(10 \colon 3\), so using the notation of the Compare to the Whole method, we have \(a = 10\) and \(b = 3\), and \(a +b = 13\). We also know that the total number of sloths is 741, so \(x + y = 741\). So we'll set up the following proportion -- pay close attention to the labels!

\[\frac{10 \text{ two-toed sloths}}{13 \text{ total sloths}} = \frac{x \text{ two-toed sloths}}{741 \text{ total sloths}}\]

On the righthand side of the proportion above, the ratio \(\frac{x}{741}\) represents the actual number of sloths, in which there are \(x\) two-toed sloths out of a total of \(741\) total sloths.

On the lefthand side, the ratio \(\frac{10}{13}\) represents an imaginary "smaller but proportional rainforest," in which there are only \(10\) two-toed and \(3\) three-toed sloths, for a total of \(13\) sloths in our imaginary smaller rainforest.

Proportionality says that these proportions must be equal, but since we don't know the breakdown of the total number of sloths, we must compare to the whole , which means we must compare the total number of sloths on each side. We get a total number of \(13\) on the left by computing \(10 + 3\), and on the right, we know the total to be \(741\).

Once we have that proportion, we can simply solve it using our processes from the previous section. From the proportion \[\frac{10}{13} = \frac{x}{741}\]

we use Cross Multiplication to obtain \[13x = 7410\]

and then use Division undoes Multiplication to get \[x = \frac{7410}{13} = 570\]

Looking back, we see that \(x\) represents the number of two-toed sloths. Therefore, there are \(570\) two-toed sloths in the rainforest.

That's the best way to think about the Compare to the Whole method -- the ratio you are given represents a "smaller version" of the situation described, and to find the total quantity in the smaller version, you simply add the two parts together. Then compare that to the actual total quantity using a proportion. If the problem statement contains words like "total," "whole," or "all together," it's likely that you'll need to use the Compare to the Whole method . However, as always, the most important thing is to read the problem and think critically about what it's asking!

P.S. for this section: You may notice that some of the algebra is becoming less explicit as we see more and more examples. If you are confused about why an algebra or arithmetic step is true, try looking for a similar problem earlier in this book — there is likely an explanation there. If you can't find one, or are still confused, you should ask your instructor or email the author of this book at [email protected] .

When you are completing these exercises, make sure to show supporting work. 

• You can walk 2 miles in 36 minutes. How long will it take you to walk 5 miles? Give you answer as a number of hours plus a number of minutes (that is, you would express 70 minutes as "1 hour and 10 minutes"). Remember that there are 60 minutes in an hour!
• You can mow 1/3 of an acre of lawn in 90 minutes. How long would it take you to mow 2 acres of lawn? Give your answer as a number of hours.
• When brewing an amber ale (a type of beer), recipes typically call for an 8:2 ratio of pale malts to crystal malts (these are types of grain in the beer). If you are brewing a 10 gallon batch of amber ale, you need a total of 22 pounds of malt. How many pounds of each type of malt (pale and crystal) should you buy? Make sure to indicate both answers clearly, and do not round them -- decimals are fine. [Hint: they should add up to 22 pounds!]
• How long of a shadow does a 6 foot tall person cast?
• If a shadow of a tree is 20 feet long, how tall is the tree?
• The ratio of registered Democrats to registered Republicans is 47 : 52 in Polk County. There is a total of 8920 registered Democrat and Republican voters. How many of them are Democrats?

"More than 200,000 sea turtles nest on or near Raine, a tiny 80-acre curl of sand along the northern edge of the  Great Barrier Reef , the portion hardest hit by warming waters. The other portion of that sea turtle population nests further from the equator, near Brisbane, where temperature increases have not been as dramatic.

What Allen and Jensen discovered was significant. Older turtles that had emerged from their eggs 30 or 40 years earlier were also mostly female, but only by a 6 to 1 ratio. But younger turtles for at least the last 20 years had been  more than 99 percent female . And as evidence that rising temperatures were responsible, female turtles from the cooler sands near Brisbane currently still only outnumber males 2 to 1.

Six weeks after Allen and Jensen published their results,  another study  from Florida looking at loggerheads revealed that temperature is just one factor. If sands are moist and cool, they produce more males. If sands are hot and dry, hatchlings are more female.

But new research in the last year also offered rays of hope." 

• What ratios can you find above? Write them down, stating explicitly what they are comparing.
• What two factors does this article assert affect the sex of sea turtles? List them.
• Given the information in the article, if a randomly selected group of 120 turtles from Brisbane have their sex examined, how many do you expect to be female? Show your work.
• Write a 2-4 sentence reaction to the article excerpt above, and make sure to answer the following question: do you feel that that ratios in the article are presented in a way that makes sense? If not, how else could you present this same information?

[*Note: you can access the article for free if you enter your email when prompted; however, you do NOT need to access the article answer this question.] 

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Use the generator to make customized ratio worksheets. Experiment with the options to see what their effect is.

Primary Grade Challenge Math by Edward Zaccaro

A good book on problem solving with very varied word problems and strategies on how to solve problems. Includes chapters on: Sequences, Problem-solving, Money, Percents, Algebraic Thinking, Negative Numbers, Logic, Ratios, Probability, Measurements, Fractions, Division. Each chapter’s questions are broken down into four levels: easy, somewhat challenging, challenging, and very challenging.

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Proportion word problems

It is very important to notice that if the ratio on the left is a ratio of number of liters of water to number of lemons, you have to do the same ratio on the right before you set them equal. 

More interesting proportion word problems

Check this site if you want to solve more proportion word problems.

Ratio word problems

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Course: 7th grade   >   Unit 1

• Worked example: Solving proportions

Solving proportions

• Writing proportions example
• Writing proportions
• Proportion word problem: cookies
• Proportion word problem: hot dogs
• Proportion word problems

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍