Section 2.3 Equations and Inequalities with Fractions
Imagine an equation with fractions present in the coefficients or terms, for example:
\begin{equation*}
\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}
\end{equation*}
We could solve this using the same steps from Section 1. We’d have to do some fraction subtraction, followed by some fraction division. In other words, we’d use fraction arithmetic, which is a bit more likely to lead to human error than whole number arithmetic.
An alternative is to multiply each side of the equation by a helpful number that “clears” the denominator(s). In the above example, the helpful number would be \(12\text{:}\)
\begin{align*}
\frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\
\multiplyleft{12}\left(\frac{1}{4}x+\frac{2}{3}\right)\amp=\multiplyleft{12}\left(\frac{1}{6}\right)\\
3x+8\amp=2
\end{align*}
Somehow multiplying by \(12\) cleared all of the fractions from the equation. We could solve this equation easily now. This section covers this technique, which generally makes it easier to solve linear equations and inequalities when fractions are present.
Subsection 2.3.1 Eliminating Denominators
Example 2.3.2.
Deshawn planted a sapling in his yard that was 4 ft tall. The tree will grow \(\frac{2}{3}\) of a foot every year. How many years will it take for his tree to become 10 ft tall?
We can use a table to help understand the rate of growth and then write a formula that models the tree’s growth:
| Years Passed | Tree’s Height (ft) |
|---|---|
| \(0\) | \(4\) |
| \(1\) | \(4+\frac{2}{3}\) |
| \(2\) | \(4+\frac{2}{3}\cdot2\) |
| \(3\) | \(4+\frac{2}{3}\cdot3\) |
| \(\vdots\) | \(\vdots\) |
| \(y\) | \(4+\frac{2}{3}y\) |
We find that \(y\) years since the tree was planted, the tree’s height is \(4+\frac{2}{3}y\) feet. To find when Deshawn’s tree will be \(10\) feet tall, we set up the equation
\begin{equation*}
4+\frac{2}{3}y=10
\end{equation*}
Note that the equation has a fraction present as a coefficient.
To solve the equation, note that the fraction’s denominator is \(3\text{.}\) As the very first step, multiplying by \(3\) on each side of the equation will leave us with no fractions.
\begin{align*}
4+\frac{2}{3}y\amp=10\\
\multiplyleft{3}\left(4+\frac{2}{3}y\right)\amp=\multiplyleft{3}10\\
3\cdot4+3\cdot\frac{2}{3}y\amp=30\\
12+2y\amp=30\amp\text{(it's smooth sailing from here)}\\
2y\amp=18\\
y\amp=9
\end{align*}
Let’s check the solution \(9\) to be sure:
\begin{align*}
4+\frac{2}{3}y\amp=10\\
4+\frac{2}{3}(\substitute{9})\amp\wonder{=}10\\
4+6\amp\confirm{=}10
\end{align*}
In summary, it will take \(9\) years for Deshawn’s tree to reach \(10\) feet tall. The point of this example was to demonstrate how clearing denominators can make an equation relatively easier to solve.
Checkpoint 2.3.3.
In a science lab, a container had \(21\) ounces of water at 9:00 AM. Water has been evaporating at the rate of \(3\) ounces every \(5\) minutes. When will there be \(8\) ounces of water left?
Explanation.
Since the container has been losing \(3\) oz of water every \(5\) minutes, it loses \(\frac{3}{5}\) oz every minute. At \(t\) minutes past 9:00 AM, the container will have lost \(\frac{3}{5}t\) oz of water. Since the initial amount was \(21\) oz, the amount of water in the container (in oz) can be modeled by \(21-\frac{3}{5}t\text{.}\)
So to work out when there would be only \(8\) oz of water left, we write the following equation. Note there is a denominator of \(5\text{,}\) which we can clear.
\begin{equation*}
\begin{aligned}
21-\frac{3}{5}t\amp=8\\
\multiplyleft{5}\left(21-\frac{3}{5}t\right)\amp=\multiplyleft{5}8\\
5\cdot21-5\cdot\frac{3}{5}t\amp=40\\
105-3t\amp=40
\end{aligned}
\end{equation*}
From here it is straightforward to use steps from Section 1 to solve the equation. You can try that on your own and you should find that it will take between \(21\) and \(22\) minutes for the container to only have \(8\) ounces of water left.
The previous examples only had one fraction in the equation. What happens if there are more?
Example 2.3.4.
Solve for \(x\) in \(\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{.}\)
Explanation.
Once again, we have an equation with fractions present. This time there are multiple denominators: \(4\text{,}\) \(3\text{,}\) and \(6\text{.}\) Is there one number we could use to clear all three denominators? The “Least Common Multiple” of the denominators (abbreviated as LCM or LCD) would work. Anything that is a multiple of all three of these denominators would work, and using the smallest such number is just a good habit. In this case, the LCD is \(12\text{.}\) So we will multiply each side of the equation by \(12\text{:}\)
\begin{align*}
\frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\
\multiplyleft{12}\left(\frac{1}{4}x+\frac{2}{3}\right)\amp=\multiplyleft{12}\frac{1}{6}\\
12\cdot\left(\frac{1}{4}x\right)+12\cdot\left(\frac{2}{3}\right)\amp=12\cdot\frac{1}{6}\\
3x+8\amp=2\\
3x\amp=-6\\
x\amp=-2
\end{align*}
Checking the solution \(-2\text{:}\)
\begin{align*}
\frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\
\frac{1}{4}(\substitute{-2})+\frac{2}{3}\amp\wonder{=}\frac{1}{6}\\
-\frac{2}{4}+\frac{2}{3}\amp\wonder{=}\frac{1}{6}\\
-\frac{6}{12}+\frac{8}{12}\amp\wonder{=}\frac{1}{6}\\
\frac{2}{12}\amp\confirm{=}\frac{1}{6}
\end{align*}
The solution is therefore \(-2\) and the solution set is \(\{-2\}\text{.}\)
Checkpoint 2.3.5.
Solve for \(z\) in \(-\frac{2}{5}z-\frac{3}{2}=-\frac{1}{2}z+\frac{4}{5}\text{.}\)
Explanation.
With fractions present, we decide to identify the LCD. The denominators are \(2\) and \(5\text{,}\) so the LCD is \(10\text{.}\) And so our first algebra step is to multiply each side of the equation by \(10\text{:}\)
\begin{equation*}
\begin{aligned}
-\frac{2}{5}z-\frac{3}{2}\amp=-\frac{1}{2}z+\frac{4}{5}\\
\multiplyleft{10}\left(-\frac{2}{5}z-\frac{3}{2}\right)\amp=\multiplyleft{10}\left(-\frac{1}{2}z+\frac{4}{5}\right)\\
10\left(-\frac{2}{5}z\right)-10\left(\frac{3}{2}\right)\amp=10\left(-\frac{1}{2}z\right)+10\left(\frac{4}{5}\right)\\
-4z-15\amp=-5z+8\\
z-15\amp=8\\
z\amp=23
\end{aligned}
\end{equation*}
Checking the solution \(23\text{:}\)
\begin{equation*}
\begin{aligned}
-\frac{2}{5}z-\frac{3}{2}\amp=-\frac{1}{2}z+\frac{4}{5}\\
-\frac{2}{5}(\substitute{23})-\frac{3}{2}\amp\wonder{=}-\frac{1}{2}(\substitute{23})+\frac{4}{5}\\
-\frac{46}{5}-\frac{3}{2}\amp\wonder{=}-\frac{23}{2}+\frac{4}{5}\\
-\frac{92}{10}-\frac{15}{10}\amp\wonder{=}-\frac{115}{10}+\frac{8}{10}\\
-\frac{107}{10}\amp\confirm{=}-\frac{107}{10}
\end{aligned}
\end{equation*}
So the solution is \(23\) and so the solution set is \(\{23\}\text{.}\)
Example 2.3.6.
Solve for \(a\) in the equation \(\frac{2}{3}(a+1)+5=\frac{1}{3}\text{.}\)
Explanation.
There are two fractions appearing here, and both have the same denominator \(3\text{.}\) We can multiply by \(3\) on each side and that alone will clear the denominators.
\begin{align*}
\frac{2}{3}(a+1)+5\amp=\frac{1}{3}\\
\multiplyleft{3}\left(\frac{2}{3}(a+1)+5\right)\amp=\multiplyleft{3}\frac{1}{3}\\
3\cdot\frac{2}{3}(a+1)+3\cdot5\amp=1\\
2(a+1)+15\amp=1\\
2a+2+15\amp=1\\
2a+17\amp=1\\
2a\amp=-16\\
a\amp=-8
\end{align*}
Checking the solution \(-8\text{:}\)
\begin{align*}
\frac{2}{3}(a+1)+5\amp=\frac{1}{3}\\
\frac{2}{3}(\substitute{-8}+1)+5\amp\wonder{=}\frac{1}{3}\\
\frac{2}{3}(-7)+5\amp\wonder{=}\frac{1}{3}\\
-\frac{14}{3}+\frac{15}{3}\amp\confirm{=}\frac{1}{3}
\end{align*}
The solution is therefore \(-8\) and the solution set is \(\{-8\}\text{.}\)
Example 2.3.7.
Solve for \(b\) in the equation \(\frac{2b+1}{3}=\frac{2}{5}\text{.}\)
Explanation.
The structure of the equation is a little different from previous examples, but we can still see that there are denominators \(3\) and \(5\text{,}\) and that their LCM is \(15\text{.}\) Multiplying each side of the equation by \(15\) will help.
\begin{align*}
\frac{2b+1}{3}\amp=\frac{2}{5}\\
\multiplyleft{15}\frac{2b+1}{3}\amp=\multiplyleft{15}\frac{2}{5}\\
5(2b+1)\amp=6\\
10b+5\amp=6\\
10b\amp=1\\
b\amp=\frac{1}{10}
\end{align*}
Checking the solution \(\frac{1}{10}\text{:}\)
\begin{align*}
\frac{2\left(\substitute{\frac{1}{10}}\right)+1}{3}\amp\wonder{=}\frac{2}{5}\\
\frac{\frac{1}{5}+1}{3}\amp\wonder{=}\frac{2}{5}\\
\frac{\frac{6}{5}}{3}\amp\wonder{=}\frac{2}{5}\\
\frac{6}{5}\cdot \frac{1}{3}\amp\confirm{=}\frac{2}{5}
\end{align*}
The solution is \(\frac{1}{10}\) and the solution set is \(\left\{\frac{1}{10}\right\}\text{.}\)
Subsection 2.3.2 Proportional Equations
Proportional equations are used to solve many real-life applications where two quantities vary together. For example if you have more liquid Tylenol in a medicine cup, then you have more milligrams of the drug itself dissolved in that liquid. The two quantities (volume of liquid and mass of the drug) vary together.
Example 2.3.8.
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 a \(21\)-ounce box. Letting \(C\) be this unknown cost (in dollars), we could set up a proportional equation. Whether we are dealing with the smaller box or the larger box, the ratio of cost to weight is supposed to be the same. So below on the left, we enter details for the larger box. And on the right, for the smaller box.
\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*}
We can rewrite this equation without the units:
\begin{equation*}
\frac{3.99}{21}=\frac{C}{18}
\end{equation*}
And we have an equation with fractions present. We want to multiply each side by the LCD, but this time the denominators are larger than previous examples. Can we still find the LCD? We could! But it is maybe not worth the time and effort. Having the least common multiple is a nice thing, but any common multiple of denominators will work out for our needs. A simple way to get a common multiple is just to multiply the numbers together. So we choose to multiply each side by \(18\cdot 21\) instead of whatever the actual LCD is.
\begin{align*}
\frac{3.99}{21}\amp=\frac{C}{18}\\
\multiplyleft{18\cdot21}\frac{3.99}{21}\amp=\multiplyleft{18\cdot21}\frac{3.99}{21}\frac{C}{18}\\
18\cdot\cancelhighlight{21}\cdot\frac{3.99}{\cancelhighlight{21}}\amp=\cancelhighlight{18}\cdot21\cdot\frac{C}{\cancelhighlight{18}}\\
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 \(\$3.42\text{.}\)
Example 2.3.9.
Property taxes for a residential property are proportional to the assessed value of the property. A certain home is assessed at \(\$234{,}100\) and its annual property taxes are \(\$2{,}518.92\text{.}\) What are the annual property taxes for the house next door 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 multiply each side by \(234100\) and \(287500\) and simplify from there:
\begin{align*}
\frac{2518.92}{234100}\amp=\frac{T}{287500}\\
\multiplyleft{234100\cdot287500}\cdot\frac{2518.92}{234100}\amp=\multiplyleft{234100\cdot287500}\cdot\frac{T}{287500}\\
287500\cdot2518.92 \amp=234100T \\
\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 \(\$3{,}093.50\text{.}\)
Checkpoint 2.3.10.
Infant Tylenol contains 160 mg of acetaminophen in each 5 mL of liquid. If Bao’s baby is prescribed 60 mg of acetaminophen, how many milliliters of liquid should he give the baby?
Explanation.
Assume Bao should give \(q\) milliliters of liquid medicine, so we can set up the following proportion:
\begin{equation*}
\begin{aligned}
\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}\\
\multiplyleft{160\cdot 60}\cdot\frac{5}{160}\amp=\multiplyleft{160\cdot 60}\cdot\frac{q}{60}\\
300\amp=160q\\
\divideunder{300}{160}\amp=\divideunder{160q}{160}\\
q\amp=1.875
\end{aligned}
\end{equation*}
So to give 60 mg of acetaminophen to his baby, Bao should measure out 1.875 mL of the liquid medication.
Example 2.3.11.
Tagging fish is a means of estimating the size of the population of fish in a lake. A sample of fish is taken, tagged, and then redistributed into the lake. Later when another sample is taken, some of those fish will have tags. The number of tagged fish are assumed to be proportional to the total number of fish. We can look at that relationship from the perspective of the entire lake, or just the second sample, and we get two ratios that should be proportional.
\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 \(n\text{,}\) which is in a denominator, we’ll need to multiply each side by both \(200\) and \(n\text{:}\)
\begin{align*}
\frac{7}{200}\amp=\frac{90}{n}\\
\multiplyleft{200\cdot n}\cdot\frac{7}{200}\amp=\multiplyleft{200\cdot n}\cdot\frac{90}{n}\\
{\cancelhighlight{200}\cdot n}\cdot\frac{7}{\cancelhighlight{200}}\amp=200\cdot\cancelhighlight{n}\cdot\frac{90}{\cancelhighlight{n}}\\
7n\amp=18000\\
\divideunder{7n}{7}\amp=\divideunder{18000}{7}\\
n\amp\approx 2571
\end{align*}
According to this sample, we can estimate that there are about \(2571\) fish in the lake.
Subsection 2.3.3 Solving Inequalities with Fractions
The technique of clearing denominators also applies to inequalities, not just equations.
Example 2.3.12.
Solve for \(x\) in the inequality \(\frac{3}{4}x-2\gt\frac{4}{5}x\text{.}\) Write the solution set in both set-builder notation and interval notation.
Explanation.
The LCM of the denominators is \(20\text{,}\) so we start out multiplying each side of the inequality by \(20\text{.}\)
\begin{align*}
\frac{3}{4}x-2\amp\gt\frac{4}{5}x\\
\multiplyleft{20}\left(\frac{3}{4}x-2\right)\amp\gt\multiplyleft{20}\frac{4}{5}x\\
20\cdot\frac{3}{4}x-20\cdot2\amp\gt16x\\
15x-40\amp\gt16x\\
-40\amp\gt x\\
x\amp\lt-40
\end{align*}
The solution set in set-builder notation is \(\{x\mid x\lt-40\}\text{.}\) And in interval notation, it’s \((-\infty,-40)\text{.}\)
Checkpoint 2.3.13.
Solve for \(x\) in the inequality \(\frac{4}{7}-\frac{4}{3}x\le\frac{2}{3}\text{.}\) Write the solution set in both set-builder notation and interval notation.
Explanation.
The LCM of the denominators is \(21\text{,}\) so we start out by multiplying each side of the inequality by \(21\text{.}\)
\begin{equation*}
\begin{aligned}
\frac{4}{7}-\frac{4}{3}x\amp\le\frac{2}{3}\\
\multiplyleft{21}\left(\frac{4}{7}-\frac{4}{3}x\right)\amp\le\multiplyleft{21}\left(\frac{2}{3}\right)\\
21\left(\frac{4}{7}\right)-21\left(\frac{4}{3}x\right)\amp\le21\left(\frac{2}{3}\right)\\
12-28x\amp\le14\\
-28x\amp\le2\\
\divideunder{-28x}{-28}\amp\secondhighlight{{}\ge{}}\divideunder{2}{-28}\\
x\amp\ge-\frac{1}{14}
\end{aligned}
\end{equation*}
Note that when we divided each side of the inequality by \(-28\text{,}\) the inequality symbol reversed direction. The solution set in set-builder notation is \(\left\{x\mid x\ge-\frac{1}{14}\right\}\text{.}\) The solution set in interval notation is \(\left[-\frac{1}{14},\infty\right)\text{.}\)
Reading Questions 2.3.4 Reading Questions
1.
What does LCD stand for? And using different words, explain what it is.
2.
When you clear denominators from an equation like \(\frac{2}{3}x+5=\frac{2}{7}\text{,}\) you will multiply by \(21\text{.}\) What are the two things that you multiply by \(21\text{?}\) Hint: it is not the two fractions.
3.
What is a proportional equation?
Exercises 2.3.5 Exercises
Review and Warmup
Exercise Group.
Find the Least Common Multiple of the given integers.
1.
\(4\) and \(11\)
2.
\(5\) and \(9\)
3.
\(45\) and \(14\)
4.
\(36\) and \(42\)
5.
\(30\text{,}\) \(42\text{,}\) and \(35\)
6.
\(21\text{,}\) \(56\text{,}\) and \(24\)
Skills Practice
Linear Equations with Fractions.
Solve the given equation. Practice clearing the denominator(s) as a first step.
7.
\({11x+{\frac{10}{3}}} = {7}\)
8.
\({2x+{\frac{3}{2}}} = {5}\)
9.
\({3x+11} = {{\frac{7}{12}}}\)
10.
\({4x+8} = {{\frac{1}{5}}}\)
11.
\({{\frac{1}{5}}x-1} = {{\frac{12}{5}}}\)
12.
\({{\frac{11}{7}}x+9} = {{\frac{1}{7}}}\)
13.
\({{\frac{4}{5}}x+{\frac{3}{7}}} = {-8}\)
14.
\({{\frac{1}{6}}x - {\frac{5}{7}}} = {8}\)
15.
\({{\frac{1}{25}}x+8} = {{\frac{7}{20}}}\)
16.
\({{\frac{3}{20}}x+8} = {{\frac{7}{24}}}\)
17.
\({{\frac{1}{15}}x+{\frac{7}{10}}} = {{\frac{5}{6}}}\)
18.
\({{\frac{6}{35}}x+{\frac{3}{10}}} = {{\frac{1}{14}}}\)
19.
\({{\frac{7}{4}}x-8} = {-2x - {\frac{3}{4}}}\)
20.
\({{\frac{1}{5}}x - {\frac{6}{5}}} = {11x+8}\)
21.
\({7x - {\frac{4}{3}}} = {-{\frac{6}{7}}x-12}\)
22.
\({{\frac{11}{12}}x - {\frac{2}{7}}} = {7x-5}\)
23.
\({{\frac{5}{21}}x-8} = {4x - {\frac{3}{35}}}\)
24.
\({6x - {\frac{3}{20}}} = {-x+{\frac{7}{25}}}\)
25.
\({{\frac{3}{20}}x+{\frac{5}{12}}} = {{\frac{8}{15}}x-5}\)
26.
\({8x+{\frac{7}{6}}} = {-{\frac{7}{10}}x+{\frac{2}{15}}}\)
Linear Inequalities with Fractions.
Solve the given inequality. Practice clearing the denominator(s) as a first step.
27.
\({3x+{\frac{1}{11}}} \leq {8}\)
28.
\({4x+5} \gt {{\frac{7}{4}}}\)
29.
\({{\frac{11}{5}}x+12} \geq {{\frac{8}{5}}}\)
30.
\({{\frac{4}{5}}x+{\frac{4}{3}}} \lt {5}\)
31.
\({{\frac{7}{12}}x+5} \gt {{\frac{4}{15}}}\)
32.
\({{\frac{5}{28}}x+{\frac{1}{20}}} \lt {{\frac{2}{35}}}\)
33.
\({6x - {\frac{3}{10}}} \lt {-9x+{\frac{7}{10}}}\)
34.
\({{\frac{12}{11}}x+{\frac{3}{7}}} \gt {4x+1}\)
35.
\({{\frac{7}{18}}x-1} \leq {-5x+{\frac{7}{12}}}\)
36.
\({{\frac{7}{6}}x+5} \gt {-{\frac{4}{15}}x - {\frac{3}{10}}}\)
Applications
37.
Danica is running on a track. We start the clock as she passes the 11 m mark, and she is running 31 m every 6 s. How long does it take for Danica to reach the 90 m mark?
38.
Everett filled the gas tank in his car to \({17\ {\rm gal}}\text{.}\) When the tank reaches \({1\ {\rm gal}}\text{,}\) the low gas light will come on. On average, Everett’s car uses \({{\frac{4}{91}}\ {\rm gal}}\) per mile driven. How many miles will Everett’s car be able to drive before the low gas light comes on?
39.
You planted a young tree in front of your house, and it was \(5\) feet tall. Ever since, it has been growing by \({{\frac{5}{12}}\ {\rm ft}}\) each year. How many years will it take for the tree to grow to be 10 feet tall?
40.
One of the tires on your car looks a little flat. You measure its air pressure and are alarmed to see it so low at \({23\ {\rm psi}}\text{.}\) You have a portable device that can pump air into the tire increasing the pressure at a rate of \({{\frac{5}{2}}\ {\textstyle\frac{\rm\mathstrut psi}{\rm\mathstrut min}}}\text{.}\) How long will it take to fill the tire to the manual’s recommended pressure of \({32\ {\rm psi}}\text{?}\)
41.
In one college math course, the final grade (out of \(100\) points) is calculated by adding together one-half of the homework average, one-sixth of the midterm score, and one-third of the final exam score.
Makena has 95 for her homework average and 90 for her midterm score. Her goal is to finish the course with a final grade of 90. What must she score on the final exam?
42.
In one college math course, the final grade (out of \(100\) points) is calculated by adding together one-third of the homework average, one-sixth of the class project score, one-fourth of the midterm score, and one-fourth of the final exam score.
Orlando has 71 for his homework average, 87 for his class project score, and 80 for his midterm score. His goal is to finish the course with a final grade of 80. What must he score on the final exam?
43.
An old cookbook has a recipe for soup that uses \(20\) g of flour for thickener. The recipe makes \(1\) gal of soup, but you are adjusting it to make only \(2.5\) gal. How much flour should you use?
44.
An internet recipe for \(3\) tbsp of garam masala calls for \(2.75\) tbsp of ground nutmeg. You want to use up the last of your ground nutmeg supply, which is \(2.25\) tbsp. Assuming you have enough of all of the other ingredients, how much garam masala will you make?
45.
Autumn and Ezra went to the weekend market to buy apples in bulk for a school event, canning lots of applesauce. Autumn bought 63 lb of apples and paid $62.37. Ezra has bagged their apples and has 87 lb of apples ready. How much will it Ezra’s bag of apples cost?
46.
Dangelo is an architect and he is making a scale model of a building. The actual building will be \(80\) ft tall. In the model, the height of the building will be \(2\) in tall and there will be a little model person standing next to the building. How tall should Dangelo make the model of that person who is meant to represent a \(5\) ft \(6\) in tall person?
47.
To estimate the health of the Rocky Mountain elk population in the Wenaha Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released \(33\) Rocky Mountain elk. A month later, they returned and observed \(43\) Rocky Mountain elk, \(11\) of which had tags. Approximately how many Rocky Mountain elk are in the Wenaha Wildlife Area?
48.
To estimate 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 \(89\) black-tailed deer. A month later, they returned and observed \(77\) black-tailed deer, \(12\) of which had tags. Approximately how many black-tailed deer are in the Jewell Meadow Wildlife Area?
Challenge
49.
The ratio of girls to boys in a preschool is 4 to 7. If there are 66 kids in the school, how many girls are there in the preschool?
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