ORCCA Ratios and Proportions

Section 3.5 Ratios and Proportions

Subsection 3.5.1 Introduction

A ratio is a means of comparing two quantities using division. One common example is a unit price. For example, if a box of cereal costs $\$3.99$ and weighs $21$ ounces then we can write this ratio as:

\begin{equation*} \frac{\$3.99}{21\,\text{oz}} \end{equation*}

If we want to know the unit price (that is, how much each individual ounce costs), then we can divide $\$3.99$ by $21$ ouncesand obtain $\$0.19$ per ounce. These two ratios, $\frac{\$3.99}{21\,\text{oz}}$ and $0.19\frac{\$}{\text{oz}}$ are equivalent, and the equation showing that they are equal is a proportion. In this case, we could write the following proportion:

\begin{equation*} \frac{\$3.99}{21\,\text{oz}}=\frac{\$0.19}{1\,\text{oz}} \end{equation*}

In this section, we will extend this concept and write proportions where one quantity is unknown and solve for that unknown.

Remark 3.5.1.

Sometimes ratios are stated using a colon instead of a fraction. For example, the ratio $\frac{2}{1}$ can be written as $2:1\text{.}$

Example 3.5.2.

Suppose we want to know the total cost for a box of cereal that weighs $18$ ounces, assuming it costs the same per ounce as the $21$-ounce box. Letting $C$ be this unknown cost (in dollars), we could set up the following proportion:

\begin{align*} \frac{\text{cost in dollars}}{\text{weight in oz}}\amp=\frac{\text{cost in dollars}}{\text{weight in oz}}\\ \frac{\$3.99}{21\,\text{oz}}\amp=\frac{\$C}{18\,\text{oz}} \end{align*}

To solve this proportion, we will first note that it will be easier to solve without units:

\begin{equation*} \frac{3.99}{21}=\frac{C}{18} \end{equation*}

Next we want to recognize that each side contains a fraction. Our usual approach for solving this type of equation is to multiply each side by the least common denominator (LCD). In this case, the LCD of $21$ and $18$ is $126\text{.}$ As with many other proportions we solve, it is often easier to just multiply each side by the common denominator of $18\cdot 21\text{,}$ which we know will make each denominator cancel:

\begin{align*} \frac{3.99}{21}\amp=\frac{C}{18}\\ \multiplyleft{18\cdot21}\frac{3.99}{21}\amp=\frac{C}{18}\multiplyright{18\cdot21}\\ 18\cdot\cancelhighlight{21}\frac{3.99}{\cancelhighlight{21}}\amp=\frac{C}{\cancelhighlight{18}}\cdot\cancelhighlight{18}\cdot21\\ 71.82\amp=21C\\ \divideunder{71.82}{21}\amp=\divideunder{21C}{21}\\ C\amp=3.42 \end{align*}

So assuming the cost is proportional to the cost of the $21$-ounce box, the cost for an $18$-ounce box of cereal would be $\$3.42\text{.}$

Subsection 3.5.2 Solving Proportions

Solving proportions uses the process of clearing denominators that we covered in Section 3.3. Because a proportion is exactly one fraction equal to another, we can simplify the process of clearing the denominators simply by multiplying both sides of the equation by both denominators. In other words, we don't specifically need the LCD to clear the denominators.

Example 3.5.3.

Solve $\frac{x}{8} = \frac{15}{12}$ for $x\text{.}$

Instead of finding the LCD of the two fractions, we'll simply multiply both sides of the equation by $8$ and by $12\text{.}$ This will still have the effect of canceling the denominators on both sides of the equation.

\begin{align*} \frac{x}{8} \amp= \frac{15}{12}\\ \highlight{12\cdot8\cdot}\frac{x}{8} \amp= \frac{15}{12}\highlight{\cdot12\cdot8}\\ 12\cdot\cancelhighlight{8}\cdot\frac{x}{\cancelhighlight{8}} \amp= \frac{15}{\cancelhighlight{12}}\cdot\cancelhighlight{12}\cdot8\\ 12\cdot x \amp= 15\cdot 8\\ 12x \amp= 120\\ \divideunder{12x}{12} \amp= \divideunder{120}{12}\\ x \amp= 10 \end{align*}

Our work indicates $10$ is the solution. We can check this as we would for any equation, by substituting $10$ for $x$ and verifying we obtain a true statement:

\begin{align*} \frac{10}{8} \amp\stackrel{?}{=} \frac{15}{12}\\ \frac{5}{4} \amp\stackrel{\checkmark}{=} \frac{5}{4} \end{align*}

Since both fractions reduce to $\frac{5}{4}\text{,}$ we know the solution to the equation $\frac{x}{8} = \frac{15}{12}$ is $10$ and the solution set is $\left\{10\right\}\text{.}$

When solving proportions, we can use the name cross-multiplication to describe the process of what just occurred. Say we have a proportion

\begin{equation*} \frac{a}{b}=\frac{c}{d} \end{equation*}

To remove fractions, we multiply both sides with the common denominator, $bd\text{,}$ and we have:

\begin{align*} \frac{a}{b}\amp=\frac{c}{d}\\ \multiplyleft{bd}\frac{a}{b}\amp=\frac{c}{d}\multiplyright{bd}\\ \cancelhighlight{b}d\cdot\frac{a}{\cancelhighlight{b}}\amp=\frac{c}{\cancelhighlight{d}}\cdot b\cancelhighlight{d}\\ ad\amp=bc \end{align*}

Since $a$ and $d$ are diagonally across the equals sign from each other in $\frac{a}{b}=\frac{c}{d}\text{,}$ as are $b$ and $c\text{,}$ we call this approach cross-multiplication.

\begin{equation*} \text{If }\frac{a}{b}=\frac{c}{d} \text{, then }ad=bc\text{.} \end{equation*}

If we understand cross-multiplication, we are able to rewrite a proportion $\frac{a}{b}=\frac{c}{d}$ in an equivalent form that does not have any fractions, $ad=bc\text{,}$ as our first step of work. If we had used this skill in Example 3.5.3, we would have had:

\begin{align*} \frac{x}{8} \amp= \frac{15}{12}\\ 12\cdot x \amp= 15\cdot 8\\ 12x \amp= 120 \end{align*}

Notice this is the same equation we had in the fifth line of our work in solving Example 3.5.3, but we obtained it without having to contemplate what we need to multiply by to clear the fractions.

We are able to use cross-multiplication when solving proportions, but it is extremely important to note that cross-multiplication only works when we are solving a proportion, an equation that has one ratio or fraction equal to another ratio or fraction. If an equation has anything more than one ratio or fraction on a single side of an equation, we cannot use cross-multiplication. For example, we cannot use cross-multiplication to solve $\frac{3}{4}x-\frac{2}{5} = \frac{9}{4}\text{,}$ unless we first manipulate the equation to have exactly one fraction and nothing else on each side of the equation.

It is also important to be aware of the fact that cross-multiplication is a special version of our general process of clearing fractions: multiplying both sides of an equation by a common denominator of all the fractions in an equation.

Example 3.5.4.

Solve $\frac{t}{5} = \frac{t+2}{3} $ for $t\text{.}$

Explanation

Again this equation is a proportion, so we are able to use cross-multiplication to clear the fractions and solve the equation:

\begin{align*} \frac{t}{5} \amp= \frac{t+2}{3}\\ 3 \cdot t \amp= 5\cdot(t+2) \end{align*}

It is critical that we include the parentheses around $t+2\text{,}$ so that we are multiplying $5$ against the entire numerator.

\begin{align*} 3t \amp= 5(t+2)\\ 3t \amp= 5t+10\\ 3t \highlight{-5t} \amp= 5t+10 \highlight{-5t}\\ -2t \amp= 10\\ \divideunder{-2t}{-2}\amp= \divideunder{10}{-2}\\ t \amp= -5 \end{align*}

We should check that this value $-5$ is actually the solution of the equation:

\begin{align*} \frac{-5}{5} \amp\stackrel{?}{=} \frac{-5+2}{3}\\ -1 \amp\stackrel{?}{=} \frac{-3}{3}\\ -1 \amp\stackrel{\checkmark}{=} -1 \end{align*}

Since we have verified that $-5$ is the solution for $\frac{t}{5} = \frac{t+2}{3}\text{,}$ we know that the solution set is $\{-5\}\text{.}$

Example 3.5.5.

Solve $\frac{r+7}{8} = -\frac{9}{4} $ for $r\text{.}$

Explanation

This proportion is a bit different in the fact that one fraction is negative. The key to working with a negative fraction is to attach the negative sign to either the numerator or denominator, but not both:

\begin{equation*} \frac{-9}{4}=-\frac{9}{4} ~~~\text{ and }~~~ \frac{9}{-4}=-\frac{9}{4}, ~~~\text{ but }~~~ \frac{-9}{-4}=+\frac{9}{4} \end{equation*}

Since we're trying to eliminate the fractions, it will likely make the work a bit easier to attach the negative to the numerator.

We'll work with the equation in the form $\frac{r+7}{8} = \frac{-9}{4} $

\begin{align*} \frac{r+7}{8} \amp= \frac{-9}{4}\\ 4 \cdot (r+7) \amp= 8\cdot(-9)\\ 4r + 28 \amp= -72\\ 4r + 28 \highlight{-28} \amp= -72 \highlight{-28}\\ 4r \amp= -100\\ \divideunder{4r}{4} \amp= \divideunder{-100}{4}\\ r \amp= -25 \end{align*}

We should check that this value $-25$ is actually the solution of the equation:

\begin{align*} \frac{-25+7}{8} \amp\stackrel{?}{=} -\frac{9}{4}\\ \frac{-18}{8} \amp\stackrel{?}{=} -\frac{9}{4}\\ -\frac{9}{4} \amp\stackrel{\checkmark}{=} -\frac{9}{4} \end{align*}

Since we have verified that $-25$ is the solution for $\frac{r+7}{8} = -\frac{9}{4} \text{,}$ we know that the solution set is $\{-25\}\text{.}$

Example 3.5.6.

Solve $\frac{x}{15}=\frac{40}{25}$ for $x\text{.}$

Explanation

To solve this proportion, use cross-multiplication.

\begin{align*} \frac{x}{15}\amp=\frac{40}{25}\\ 25\cdot x\amp=40\cdot15\\ 25x\amp=600\\ \divideunder{25x}{25}\amp=\divideunder{600}{25}\\ x\amp=24 \end{align*}

You can easily verify that this value $24$ is actually the solution of the equation:

\begin{align*} \frac{24}{15}\amp\stackrel{?}{=}\frac{40}{25}\\ \frac{8}{5}\amp\stackrel{\checkmark}{=}\frac{8}{5} \end{align*}

Since we have verified that $24$ is the solution for $\frac{x}{15}=\frac{40}{25}\text{,}$ we know that the solution set is $\{24\}\text{.}$

Example 3.5.7.

Solve $\frac{x-4}{6}=\frac{x+3}{4}$ for $x\text{.}$

Explanation

To solve this proportion, begin by using cross-multiplication.

\begin{align*} \frac{x-4}{6}\amp=\frac{x+3}{4}\\ 4\cdot(x-4)\amp=(x+3)\cdot6\\ 4x-16\amp=6x+18\\ 4x-16\addright{16}\amp=6x+18\addright{16}\\ 4x\amp=6x+34\\ 4x\subtractright{6x}\amp=6x+34\subtractright{6x}\\ -2x\amp=34\\ \divideunder{-2x}{-2}\amp=\divideunder{34}{-2}\\ x\amp=-17 \end{align*}

We can check that this value is correct by substituting it back into the original equation:

\begin{align*} \frac{x-4}{6}\amp=\frac{x+3}{4}\\ \frac{\highlight{-17}-4}{6}\amp\stackrel{?}{=}\frac{\highlight{-17}+3}{4}\\ \frac{-21}{6}\amp\stackrel{?}{=}\frac{-14}{4}\\ \frac{-7}{2}\amp\stackrel{\checkmark}{=}\frac{-7}{2} \end{align*}

Since we have verified that $-17$ is the solution for $\frac{x-4}{6}=\frac{x+3}{4}\text{,}$ we know that the solution set is $\{-17\}\text{.}$

Subsection 3.5.3 Proportionality in Similar Triangles

One really useful example of ratios and proportions involves similar triangles. Two triangles are considered similar if they have the same angles and their side lengths are proportional, as shown in Figure 3.5.8:

Figure 3.5.8. Similar Triangles

In the first triangle in Figure 3.5.8, the ratio of the left side length to the hypotenuse length is $\frac{1\,\text{cm}}{2\,\text{cm}}\text{;}$ in the second triangle, the ratio of the left side length to the hypotenuse length is $\frac{3\,\text{cm}}{6\,\text{cm}}\text{.}$ Since both reduce to $\frac{1}{2}\text{,}$ we can write the following proportion:

\begin{equation*} \frac{1\,\text{cm}}{2\,\text{cm}}=\frac{3\,\text{cm}}{6\,\text{cm}} \end{equation*}

If we extend this concept, we can use it to solve for an unknown side length. Consider the two similar triangles in the next example.

Example 3.5.9.

Figure 3.5.10. Similar Triangles

Since the two triangles are similar, we know that their side lengths should be proportional. To determine the unknown length, we can set up a proportion and solve for $x\text{.}$ This time, as a review, we'll use the least common denominator to clear the fractions:

\begin{align*} \frac{\text{bigger triangle's left side length in cm}}{\text{bigger triangle's bottom side length in cm}}\amp=\frac{\text{smaller triangle's left side length in cm}}{\text{smaller triangle's bottom side length in cm}}\\ \frac{x\,\text{cm}}{6\,\text{cm}}\amp=\frac{3\,\text{cm}}{4\,\text{cm}}\\ \frac{x}{6}\amp=\frac{3}{4}\\ \multiplyleft{12}\frac{x}{6}\amp=\multiplyleft{12}\frac{3}{4}\qquad\text{(12 is the least common denominator)}\\ 2x\amp=9\\ \divideunder{2x}{2}\amp=\divideunder{9}{2}\\ x\amp=\frac{9}{2}\ \text{or}\ 4.5 \end{align*}

The unknown side length is then 4.5 cm.

Remark 3.5.11.

Looking at the triangles in Figure 3.5.8, you may notice that there are many different proportions you could set up, such as:

\begin{align*} \frac{2\,\text{cm}}{1\,\text{cm}}\amp=\frac{6\,\text{cm}}{3\,\text{cm}}\\ \frac{2\,\text{cm}}{6\,\text{cm}}\amp=\frac{1\,\text{cm}}{3\,\text{cm}}\\ \frac{6\,\text{cm}}{2\,\text{cm}}\amp=\frac{3\,\text{cm}}{1\,\text{cm}}\\ \frac{3\sqrt{3}\,\text{cm}}{\sqrt{3}\,\text{cm}}\amp=\frac{3\,\text{cm}}{1\,\text{cm}} \end{align*}

This is often the case when we set up ratios and proportions.

If we take a second look at Figure 3.5.10, there are also several other proportions we could have used to find the value of $x\text{.}$

\begin{align*} \frac{\text{bigger triangle's left side length}}{\text{smaller triangle's left side length}}\amp=\frac{\text{bigger triangle's bottom side length}}{\text{smaller triangle's bottom side length}}\\ \\ \frac{\text{smaller triangle's bottom side length}}{\text{bigger triangle's bottom side length}}\amp=\frac{\text{smaller triangle's left side length}}{\text{bigger triangle's left side length}}\\ \\ \frac{\text{bigger triangle's bottom side length}}{\text{smaller triangle's bottom side length}}\amp=\frac{\text{bigger triangle's left side length} }{\text{smaller triangle's left side length}} \end{align*}

Written as algebraic proportions, these three equations would, respectively, be

\begin{align*} \frac{x\,\text{cm}}{3\,\text{cm}}\amp= \frac{6\,\text{cm}}{4\,\text{cm}}, \amp \frac{4\,\text{cm}}{6\,\text{cm}}\amp= \frac{3\,\text{cm}}{x\,\text{cm}}, \amp \frac{6\,\text{cm}}{4\,\text{cm}}\amp= \frac{x\,\text{cm}}{3\,\text{cm}} \end{align*}

While these are only a few of the possibilities, if we clear the denominators from any properly designed proportion, every one is equivalent to $x=4.5\text{.}$

Subsection 3.5.4 Creating and Solving Proportions

Proportions can be used to solve many real-life applications. The key to using proportions to solve such applications is to first set up a ratio where all values are known. We then set up a second ratio that will be proportional to the first, but has one value in the ratio unknown. Let's look at a few examples.

Example 3.5.12.

Property taxes for a residential property are proportional to the assessed value of the property. Assume that a certain property in a given neighborhood is assessed at $\$234{,}100$ and its annual property taxes are $\$2{,}518.92\text{.}$ What are the annual property taxes for a house that is assessed at $\$287{,}500\text{?}$

Explanation

Let $T$ be the annual property taxes (in dollars) for a property assessed at $\$287{,}500\text{.}$ We can write and solve this proportion:

\begin{align*} \frac{\text{tax}}{\text{property value}}\amp=\frac{\text{tax}}{\text{property value}}\\ \frac{2518.92}{234100}\amp=\frac{T}{287500} \end{align*}

The least common denominator of this proportion is rather large, so we will instead use cross-multiplication to clear the fractions:

\begin{align*} \frac{2518.92}{234100}\amp=\frac{T}{287500}\\ 287500\cdot2518.92 \amp=T\cdot 234100\\ \frac{287500\cdot 2518.92}{234100} \amp=\frac{234100T}{234100}\\ T\amp\approx3093.50 \end{align*}

The property taxes for a property assessed at $\$287,500$ are $\$3,093.50\text{.}$

Example 3.5.13.

Tagging fish is a means of estimating the size of the population of fish in a body of water (such as a lake). A sample of fish is taken, tagged, and then redistributed into the lake. When another sample is taken, the proportion of fish that are tagged out of that sample are assumed to be proportional to the total number of fish tagged out of the entire population of fish in the lake.

\begin{equation*} \frac{\text{number of tagged fish in sample}}{\text{number of fish in sample}}=\frac{\text{number of tagged fish total}}{\text{number of fish total}} \end{equation*}

Assume that $90$ fish are caught and tagged. Once they are redistributed, a sample of $200$ fish is taken. Of these, $7$ are tagged. Estimate how many fish total are in the lake.

Explanation

Let $n$ be the number of fish in the lake. We can set up a proportion for this scenario:

\begin{align*} \frac{7}{200}\amp=\frac{90}{n} \end{align*}

To solve for $n\text{,}$ we'll use cross multiplication:

\begin{align*} \frac{7}{200}\amp=\frac{90}{n}\\ 7n\amp=18000\\ \frac{7n}{7}\amp=\frac{18000}{7}\\ n\amp\approx 2571.4286 \end{align*}

According to this sample, we can estimate that there are about $2,571$ fish in the lake.

Example 3.5.14.

Infant Tylenol contains 160 mg of acetaminophen in each 5 mL of liquid medicine. If Bao's baby is prescribed 60 mg of acetaminophen, how many milliliters of liquid medicine should he give them?

Explanation

Assume Bao should give $q$ milliliters of liquid medicine, and we can set up the following proportion:

\begin{align*} \frac{\text{amount of liquid medicine in mL}}{\text{amount of acetaminophen in mg}}\amp=\frac{\text{amount of liquid medicine in mL}}{\text{amount of acetaminophen in mg}}\\ \frac{5\,\text{mL}}{160\,\text{mg}}\amp=\frac{q\,\text{mL}}{60\,\text{mg}}\\ \frac{5}{160}\amp=\frac{q}{60}\\ 60\cdot 5 \amp=q\cdot 160\\ 300\amp=160q\\ \frac{300}{160}\amp=\frac{160q}{160}\\ q\amp=1.875 \end{align*}

So to give 60 mg of acetaminophen to his baby, Bao should give 1.875 mL of liquid medicine.

Example 3.5.15.

Sarah is an architect and she's making a scale model of a building. The actual building will be 30 ft tall. In the model, the height of the building will be 2 in. How tall should she make the model of a person who is 5 ft6 in tall so that the model is to scale?

Explanation

Let $h$ be the height of the person in Sarah's model, which we'll measure in inches. We'll create a proportion that compares the building and person's heights in the model to their heights in real life:

\begin{align*} \frac{\text{height of model building in inches}}{\text{height of actual building in feet}}\amp=\frac{\text{height of model person in inches}}{\text{height of actual person in feet}}\\ \frac{2\,\text{in}}{30\,\text{ft}}\amp=\frac{h\,\text{in}}{5\,\text{ft}\,6\text{in}} \end{align*}

Before we can just eliminate the units, we'll need to convert 5 ft6 in to feet:

\begin{equation*} \frac{2\,\text{in}}{30\,\text{ft}}=\frac{h\,\text{in}}{5.5\,\text{ft}} \end{equation*}

Now we can remove the units and continue solving:

\begin{align*} \frac{2}{30}\amp=\frac{h}{5.5}\\ 5.5\cdot 2 \amp= h\cdot 30\\ 11\amp=30h\\ \frac{11}{30}\amp=\frac{30h}{30}\\ \frac{11}{30}\amp=h\\ h\amp\approx 0.3667 \end{align*}

Sarah should make the model of a person who is 5 ft6 in tall be $\frac{11}{30}$ inches (about $0.3667$ inches) tall.

Checkpoint 3.5.16.

Exercises 3.5.5 Exercises

Review and Warmup

1.

Reduce the fraction $\displaystyle{ \frac{3}{15} }\text{.}$

2.

Reduce the fraction $\displaystyle{ \frac{2}{12} }\text{.}$

3.

Reduce the fraction $\displaystyle{ \frac{9}{30} }\text{.}$

4.

Reduce the fraction $\displaystyle{ \frac{8}{14} }\text{.}$

5.

Reduce the fraction $\displaystyle{ \frac{84}{105} }\text{.}$

6.

Reduce the fraction $\displaystyle{ \frac{50}{80} }\text{.}$

7.

Reduce the fraction $\displaystyle{ \frac{72}{42} }\text{.}$

8.

Reduce the fraction $\displaystyle{ \frac{112}{98} }\text{.}$

Setting Up Ratios and Proportions

9.

Ibuprofen for infants comes in a liquid form and contains $35$ milligrams of ibuprofen for each $0.875$ milliliters of liquid. If a child is to receive a dose of $50$ milligrams of ibuprofen, how many milliliters of liquid should they be given?

Assume $l$ milliliters of liquid should be given. Write an equation to model this scenario. There is no need to solve it.

10.

Ibuprofen for infants comes in a liquid form and contains $35$ milligrams of ibuprofen for each $0.875$ milliliters of liquid. If a child is to receive a dose of $55$ milligrams of ibuprofen, how many milliliters of liquid should they be given?

Assume $l$ milliliters of liquid should be given. Write an equation to model this scenario. There is no need to solve it.

11.

The property taxes on a $2400$-square-foot house are ${\$3{,}144.00}$ per year. Assuming these taxes are proportional, what are the property taxes on a $1300$-square-foot house?

Assume property taxes on a $1300$-square-foot house is $t$ dollars. Write an equation to model this scenario. There is no need to solve it.

12.

The property taxes on a $1900$-square-foot house are ${\$2{,}717.00}$ per year. Assuming these taxes are proportional, what are the property taxes on a $1200$-square-foot house?

Assume property taxes on a $1200$-square-foot house is $t$ dollars. Write an equation to model this scenario. There is no need to solve it.

Solving Proportions

13.

Solve $\frac{x}{54}=\frac{21}{63}$ for $x\text{.}$

14.

Solve $\frac{x}{24}=\frac{10}{40}$ for $x\text{.}$

15.

Solve $\frac{54}{x}=\frac{27}{6}$ for $x\text{.}$

16.

Solve $\frac{10}{x}=\frac{16}{24}$ for $x\text{.}$

17.

Solve $\frac{x}{2}=\frac{x - 30}{8}$ for $x\text{.}$

18.

Solve $\frac{x}{6}=\frac{x + 6}{9}$ for $x\text{.}$

19.

Solve $\frac{x}{6}=\frac{x + 15}{11}$ for $x\text{.}$

20.

Solve $\frac{x}{7}=\frac{x - 24}{11}$ for $x\text{.}$

21.

Solve $\frac{x + 7}{7}=\frac{x - 11}{10}$ for $x\text{.}$

22.

Solve $\frac{x - 13}{6}=\frac{x + 6}{8}$ for $x\text{.}$

23.

Solve $\frac{x - 9}{7}=\frac{x - 9}{10}$ for $x\text{.}$

24.

Solve $\frac{x - 10}{8}=\frac{x - 12}{12}$ for $x\text{.}$

25.

Solve $\frac{x}{24}=-\frac{21}{12}$ for $x\text{.}$

26.

Solve $\frac{x}{10}=-\frac{21}{15}$ for $x\text{.}$

27.

Solve $\frac{x+3}{8}=-\frac{54}{12}$ for $x\text{.}$

28.

Solve $\frac{x-5}{40}=-\frac{63}{35}$ for $x\text{.}$

Applications

29.

The following two triangles are similar to each other. Find the length of the missing side.

The missing side’s length is

30.

The following two triangles are similar to each other. Find the length of the missing side.

The missing side’s length is

31.

The following two triangles are similar to each other. Find the length of the missing side.

The missing side’s length is

32.

The following two triangles are similar to each other. Find the length of the missing side.

The missing side’s length is

33.

According to a salad recipe, each serving requires $3$ teaspoons of vegetable oil and $21$ teaspoons of vinegar. If $14$ teaspoons of vegetable oil were used, how many teaspoons of vinegar should be used?

If $14$ teaspoons of vegetable oil were used, teaspoons of vinegar should be used.

34.

According to a salad recipe, each serving requires $5$ teaspoons of vegetable oil and $35$ teaspoons of vinegar. If $126$ teaspoons of vinegar were used, how many teaspoons of vegetable oil should be used?

If $126$ teaspoons of vinegar were used, teaspoons of vegetable oil should be used.

35.

Jon makes $\$69$ every six hours he works. How much will he make if he works twenty-six hours this week?

If Jon works twenty-six hours this week, he will make .

36.

Corey makes $\$116$ every eight hours he works. How much will he make if he works thirty-two hours this week?

If Corey works thirty-two hours this week, he will make .

37.

A mutual fund consists of $30$% stock and $70$% bond. In other words, for each $30$ dollars of stock, there are $70$ dollars of bond. For a mutual fund with ${\$2{,}860.00}$ of stock, how many dollars of bond are there?

For a mutual fund with ${\$2{,}860.00}$ of stock, there are approximately of bond.

38.

A mutual fund consists of $39$% stock and $61$% bond. In other words, for each $39$ dollars of stock, there are $61$ dollars of bond. For a mutual fund with ${\$2{,}510.00}$ of bond, how many dollars of stock are there?

For a mutual fund with ${\$2{,}510.00}$ of bond, there are approximately of stock.

39.

Jessica jogs every day. Last month, she jogged $7.5$ hours for a total of $34.5$ miles. At this speed, if Jessica runs $46.5$ hours, how far can she run?

At this speed, Jessica can run in $46.5$ hours.

40.

Gregory jogs every day. Last month, he jogged $17.5$ hours for a total of $85.75$ miles. At this speed, how long would it take Gregory to run $169.05$ miles?

At this speed, Gregory can run ${169.05\ {\rm mi}}$ in .

41.

Randi purchased $4.9$ pounds of apples at the total cost of ${\$21.56}\text{.}$ If he purchases $7.2$ pounds of apples at this store, how much would it cost?

It would cost to purchase $7.2$ pounds of apples.

42.

Connor purchased $2.7$ pounds of apples at the total cost of ${\$13.77}\text{.}$ If the price doesn’t change, how many pounds of apples can Connor purchase with ${\$39.27}\text{?}$

With ${\$39.27}\text{,}$ Connor can purchase of apples.

43.

Gosheven collected a total of $2414$ stamps over the past $17$ years. At this rate, how many stamps would he collect in $25$ years?

At this rate, Gosheven would collect stamps in $25$ years.

44.

Farshad collected a total of $728$ stamps over the past $14$ years. At this rate, how many years would it take he to collect $1508$ stamps?

At this rate, Farshad can collect $1508$ stamps in years.

45.

In a city, the owner of a house valued at $460$ thousand dollars needs to pay ${\$519.80}$ in property tax. At this tax rate, how much property tax should the owner pay if a house is valued at $700$ thousand dollars?

The owner of a $700$-thousand-dollar house should pay in property tax.

46.

In a city, the owner of a house valued at $370$ thousand dollars needs to pay ${\$1{,}024.90}$ in property tax. At this tax rate, if the owner of a house paid ${\$1{,}939.00}$ of property tax, how much is the house worth?

If the owner of a house paid ${\$1{,}939.00}$ of property tax, the house is worth thousand dollars.

47.

To try to determine the health of the Rocky Mountain elk population in the Wenaha Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released $39$ Rocky Mountain elk. A week later, they returned and observed $42$ Rocky Mountain elk, $9$ of which had tags. Approximately how many Rocky Mountain elk are in the Wenaha Wildlife Area?

There are approximately elk in the wildlife area.

48.

To try to determine the health of the black-tailed deer population in the Jewell Meadow Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released $28$ black-tailed deer. A week later, they returned and observed $63$ black-tailed deer, $18$ of which had tags. Approximately how many black-tailed deer are in the Jewell Meadow Wildlife Area?

There are approximately deer in the wildlife area.

49.

A restaurant used ${1308\ {\rm lb}}$ of vegetable oil in $40$ days. At this rate, how many pounds of vegetable oil will be used in $55$ days?

The restaurant will use of vegetable oil in $55$ days.

50.

A restaurant used ${963.2\ {\rm lb}}$ of vegetable oil in $28$ days. At this rate, ${1720\ {\rm lb}}$ of oil will last how many days?

The restaurant will use ${1720\ {\rm lb}}$ of vegetable oil in days.