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Study Guides > Mathematics for the Liberal Arts

Problem Solving

In previous math courses, you’ve no doubt run into the infamous “word problems.” Unfortunately, these problems rarely resemble the type of problems we actually encounter in everyday life. In math books, you usually are told exactly which formula or procedure to use, and are given exactly the information you need to answer the question. In real life, problem solving requires identifying an appropriate formula or procedure, and determining what information you will need (and won’t need) to answer the question. In this chapter, we will review several basic but powerful algebraic ideas: percents, rates, and proportions. We will then focus on the problem solving process, and explore how to use these ideas to solve problems where we don’t have perfect information.


In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq. Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties.[footnote]http://www.factcheck.org/cheney_edwards_mangle_facts.html[/footnote] Who is correct? How can we make sense of these numbers? Percent literally means “per 100,” or “parts per hundred.” When we write 40%, this is equivalent to the fraction [latex]\displaystyle\frac{40}{100}\\[/latex] or the decimal 0.40. Notice that 80 out of 200 and 10 out of 25 are also 40%, since [latex]\displaystyle\frac{80}{200}=\frac{10}{25}=\frac{40}{100}\\[/latex].

Example 1

243 people out of 400 state that they like dogs. What percent is this?


[latex]\displaystyle\frac{243}{400}=0.6075=\frac{60.75}{100}\\[/latex]. This is 60.75%. Notice that the percent can be found from the equivalent decimal by moving the decimal point two places to the right.

Example 2

Write each as a percent:
  1. [latex]\displaystyle\frac{1}{4}\\[/latex]
  2. 0.02
  3. 2.35


  1. [latex]\displaystyle\frac{1}{4}=0.25\\[/latex] = 25%
  2. 0.02 = 2%
  3. 2.35 = 235%


If we have a part that is some percent of a whole, then [latex]\displaystyle\text{percent}=\frac{\text{part}}{\text{whole}}\\[/latex], or equivalently,[latex]\text{part}\cdot\text{whole}=\text{percent}\\[/latex]. To do the calculations, we write the percent as a decimal.

Example 3

The sales tax in a town is 9.4%. How much tax will you pay on a $140 purchase?


Here, $140 is the whole, and we want to find 9.4% of $140. We start by writing the percent as a decimal by moving the decimal point two places to the left (which is equivalent to dividing by 100). We can then compute: tax = 0.094(140) = $13.16 in tax.

Example 4

In the news, you hear “tuition is expected to increase by 7% next year.” If tuition this year was $1200 per quarter, what will it be next year?


The tuition next year will be the current tuition plus an additional 7%, so it will be 107% of this year’s tuition: $1200(1.07) = $1284. Alternatively, we could have first calculated 7% of $1200: $1200(0.07) = $84. Notice this is not the expected tuition for next year (we could only wish). Instead, this is the expected increase, so to calculate the expected tuition, we’ll need to add this change to the previous year’s tuition: $1200 + $84 = $1284.

Try It Now

A TV originally priced at $799 is on sale for 30% off. There is then a 9.2% sales tax. Find the price after including the discount and sales tax.

Example 5

The value of a car dropped from $7400 to $6800 over the last year. What percent decrease is this?


To compute the percent change, we first need to find the dollar value change: $6800 – $7400 = –$600. Often we will take the absolute value of this amount, which is called the absolute change: |–600| = 600. Since we are computing the decrease relative to the starting value, we compute this percent out of $7400: [latex]\displaystyle\frac{600}{7400}=0.081=[/latex] 8.1% decrease. This is called a relative change.

Absolute and Relative Change

Given two quantities,

Absolute change =[latex]\displaystyle|\text{ending quantity}-\text{starting quantity}|[/latex]

Relative change: [latex]\displaystyle\frac{\text{absolute change}}{\text{starting quantity}}[/latex]

Absolute change has the same units as the original quantity. Relative change gives a percent change. The starting quantity is called the base of the percent change.
The base of a percent is very important. For example, while Nixon was president, it was argued that marijuana was a “gateway” drug, claiming that 80% of marijuana smokers went on to use harder drugs like cocaine. The problem is, this isn’t true. The true claim is that 80% of harder drug users first smoked marijuana. The difference is one of base: 80% of marijuana smokers using hard drugs, vs. 80% of hard drug users having smoked marijuana. These numbers are not equivalent. As it turns out, only one in 2,400 marijuana users actually go on to use harder drugs.[footnote]http://tvtropes.org/pmwiki/pmwiki.php/Main/LiesDamnedLiesAndStatistics[/footnote]

Example 6

There are about 75 QFC supermarkets in the United States. Albertsons has about 215 stores. Compare the size of the two companies.


When we make comparisons, we must ask first whether an absolute or relative comparison. The absolute difference is 215 – 75 = 140. From this, we could say “Albertsons has 140 more stores than QFC.” However, if you wrote this in an article or paper, that number does not mean much. The relative difference may be more meaningful. There are two different relative changes we could calculate, depending on which store we use as the base:

Using QFC as the base, [latex]\displaystyle\frac{140}{75}=1.867\\[/latex].

This tells us Albertsons is 186.7% larger than QFC.

Using Albertsons as the base,[latex]\displaystyle\frac{140}{215}=0.651\\[/latex].

This tells us QFC is 65.1% smaller than Albertsons.

Notice both of these are showing percent differences. We could also calculate the size of Albertsons relative to QFC:[latex]\displaystyle\frac{215}{75}=2.867\\[/latex], which tells us Albertsons is 2.867 times the size of QFC. Likewise, we could calculate the size of QFC relative to Albertsons:[latex]\displaystyle\frac{75}{215}=0.349\\[/latex], which tells us that QFC is 34.9% of the size of Albertsons.

Example 7

Suppose a stock drops in value by 60% one week, then increases in value the next week by 75%. Is the value higher or lower than where it started?


To answer this question, suppose the value started at $100. After one week, the value dropped by 60%: $100 – $100(0.60) = $100 – $60 = $40. In the next week, notice that base of the percent has changed to the new value, $40. Computing the 75% increase: $40 + $40(0.75) = $40 + $30 = $70. In the end, the stock is still $30 lower, or [latex]\displaystyle\frac{\$30}{100}[/latex] = 30% lower, valued than it started.

Try It Now

The US federal debt at the end of 2001 was $5.77 trillion, and grew to $6.20 trillion by the end of 2002. At the end of 2005 it was $7.91 trillion, and grew to $8.45 trillion by the end of 2006.[footnote]http://www.whitehouse.gov/sites/default/files/omb/budget/fy2013/assets/hist07z1.xls[/footnote] Calculate the absolute and relative increase for 2001–2002 and 2005–2006. Which year saw a larger increase in federal debt?

Example 8

A Seattle Times article on high school graduation rates reported “The number of schools graduating 60 percent or fewer students in four years—sometimes referred to as “dropout factories”—decreased by 17 during that time period. The number of kids attending schools with such low graduation rates was cut in half.”
  1. Is the “decrease by 17” number a useful comparison?
  2. Considering the last sentence, can we conclude that the number of “dropout factories” was originally 34?


  1. This number is hard to evaluate, since we have no basis for judging whether this is a larger or small change. If the number of “dropout factories” dropped from 20 to 3, that’d be a very significant change, but if the number dropped from 217 to 200, that’d be less of an improvement.
  2. The last sentence provides relative change, which helps put the first sentence in perspective. We can estimate that the number of “dropout factories” was probably previously around 34. However, it’s possible that students simply moved schools rather than the school improving, so that estimate might not be fully accurate.

Example 9

In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq. Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties. Who is correct?


Without more information, it is hard for us to judge who is correct, but we can easily conclude that these two percents are talking about different things, so one does not necessarily contradict the other. Edward’s claim was a percent with coalition forces as the base of the percent, while Cheney’s claim was a percent with both coalition and Iraqi security forces as the base of the percent. It turns out both statistics are in fact fairly accurate.

Try It Now

In the 2012 presidential elections, one candidate argued that “the president’s plan will cut $716 billion from Medicare, leading to fewer services for seniors,” while the other candidate rebuts that “our plan does not cut current spending and actually expands benefits for seniors, while implementing cost saving measures.” Are these claims in conflict, in agreement, or not comparable because they’re talking about different things?
We’ll wrap up our review of percents with a couple cautions. First, when talking about a change of quantities that are already measured in percents, we have to be careful in how we describe the change.

Example 10

A politician’s support increases from 40% of voters to 50% of voters. Describe the change.


We could describe this using an absolute change: [latex]|50\%-40\%|=10\%[/latex]. Notice that since the original quantities were percents, this change also has the units of percent. In this case, it is best to describe this as an increase of 10 percentage points. In contrast, we could compute the percent change:[latex]\displaystyle\frac{10\%}{40\%}=0.25=25\%[/latex] increase. This is the relative change, and we’d say the politician’s support has increased by 25%.
Lastly, a caution against averaging percents.

Example 11

A basketball player scores on 40% of 2-point field goal attempts, and on 30% of 3-point of field goal attempts. Find the player’s overall field goal percentage.


It is very tempting to average these values, and claim the overall average is 35%, but this is likely not correct, since most players make many more 2-point attempts than 3-point attempts. We don’t actually have enough information to answer the question. Suppose the player attempted 200 2-point field goals and 100 3-point field goals. Then they made 200(0.40) = 80 2-point shots and 100(0.30) = 30 3-point shots. Overall, they made 110 shots out of 300, for a [latex]\displaystyle\frac{110}{300}=0.367=36.7\%\\[/latex] overall field goal percentage.

Proportions and Rates

If you wanted to power the city of Seattle using wind power, how many windmills would you need to install? Questions like these can be answered using rates and proportions.


A rate is the ratio (fraction) of two quantities. A unit rate is a rate with a denominator of one.

Example 12

Your car can drive 300 miles on a tank of 15 gallons. Express this as a rate.


Expressed as a rate, [latex]\displaystyle\frac{300\text{ miles}}{15\text{ gallons}}\\[/latex]. We can divide to find a unit rate:[latex]\displaystyle\frac{20\text{ miles}}{1\text{ gallon}}\\[/latex], which we could also write as[latex]\displaystyle{20}\frac{\text{miles}}{\text{gallon}}\\[/latex], or just 20 miles per gallon.

Proportion Equation

A proportion equation is an equation showing the equivalence of two rates or ratios.

Example 13

Solve the proportion [latex]\displaystyle\frac{5}{3}=\frac{x}{6}\\[/latex] for the unknown value x.


This proportion is asking us to find a fraction with denominator 6 that is equivalent to the fraction[latex]\displaystyle\frac{5}{3}\\[/latex]. We can solve this by multiplying both sides of the equation by 6, giving [latex]\displaystyle{x}=\frac{5}{3}\cdot6=10\\[/latex].

Example 14

A map scale indicates that ½ inch on the map corresponds with 3 real miles. How many miles apart are two cities that are [latex]\displaystyle{2}\frac{1}{4}\\[/latex] inches apart on the map?


We can set up a proportion by setting equal two [latex]\displaystyle\frac{\text{map inches}}{\text{real miles}}\\[/latex] rates, and introducing a variable, x, to represent the unknown quantity—the mile distance between the cities.
[latex]\displaystyle\frac{\frac{1}{2}\text{map inch}}{3\text{ miles}}=\frac{2\frac{1}{4}\text{map inches}}{x\text{ miles}}\\[/latex] Multiply both sides by and rewriting the mixed number
[latex]\displaystyle\frac{\frac{1}{2}}{3}\cdot{x}=\frac{9}{4}\\[/latex] Multiply both sides by 3
[latex]\displaystyle\frac{1}{2}x=\frac{27}{4}\\[/latex] Multiply both sides by 2 (or divide by ½)
[latex]\displaystyle{x}=\frac{27}{2}=13\frac{1}{2}\text{ miles}\\[/latex]
Many proportion problems can also be solved using dimensional analysis, the process of multiplying a quantity by rates to change the units.

Example 15

Your car can drive 300 miles on a tank of 15 gallons. How far can it drive on 40 gallons?


We could certainly answer this question using a proportion: [latex]\displaystyle\frac{300\text{ miles}}{15\text{ gallons}}=\frac{x\text{ miles}}{40\text{ gallons}}\\[/latex]. However, we earlier found that 300 miles on 15 gallons gives a rate of 20 miles per gallon. If we multiply the given 40 gallon quantity by this rate, the gallons unit “cancels” and we’re left with a number of miles: [latex-display]\displaystyle40\text{ gallons}\cdot\frac{20\text{ miles}}{\text{gallon}}=\frac{40\text{ gallons}}{1}\cdot\frac{20\text{ miles}}{\text{gallons}}=800\text{ miles}\\[/latex-display] Notice if instead we were asked “how many gallons are needed to drive 50 miles?” we could answer this question by inverting the 20 mile per gallon rate so that the miles unit cancels and we’re left with gallons: [latex-display]\displaystyle{50}\text{ miles}\cdot\frac{1\text{ gallon}}{20\text{ miles}}=\frac{50\text{ miles}}{1}\cdot\frac{1\text{ gallon}}{20\text{ miles}}=\frac{50\text{ gallons}}{20}=2.5\text{ gallons}\\[/latex-display]
Dimensional analysis can also be used to do unit conversions. Here are some unit conversions for reference.

Unit Conversions


1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft)
1 mile = 5,280 feet
1000 millimeters (mm) = 1 meter (m) 100 centimeters (cm) = 1 meter
1000 meters (m) = 1 kilometer (km) 2.54 centimeters (cm) = 1 inch

Weight and Mass

1 pound (lb) = 16 ounces (oz) 1 ton = 2000 pounds
1000 milligrams (mg) = 1 gram (g) 1000 grams = 1kilogram (kg)
1 kilogram = 2.2 pounds (on earth)


1 cup = 8 fluid ounces (fl oz)[footnote]Fluid ounces are a capacity measurement for liquids. 1 fluid ounce ≈ 1 ounce (weight) for water only.[/footnote] 1 pint = 2 cups
1 quart = 2 pints = 4 cups 1 gallon = 4 quarts = 16 cups
1000 milliliters (ml) = 1 liter (L)

Example 16

A bicycle is traveling at 15 miles per hour. How many feet will it cover in 20 seconds?


To answer this question, we need to convert 20 seconds into feet. If we know the speed of the bicycle in feet per second, this question would be simpler. Since we don’t, we will need to do additional unit conversions. We will need to know that 5280 ft = 1 mile. We might start by converting the 20 seconds into hours: [latex-display]\displaystyle{20}\text{ seconds}\cdot\frac{1\text{ minute}}{60\text{ seconds}}\cdot\frac{1\text{ hour}}{60\text{ minutes}}=\frac{1}{180}\text{ hour}\\[/latex-display] Now we can multiply by the 15 miles/hr [latex-display]\displaystyle\frac{1}{180}\text{ hour}\cdot\frac{15\text{ miles}}{1\text{ hour}}=\frac{1}{12}\text{ mile}\\[/latex-display] Now we can convert to feet [latex-display]\displaystyle\frac{1}{12}\text{ mile}\cdot\frac{5280\text{ feet}}{1\text{ mile}}=440\text{ feet}\\[/latex-display] We could have also done this entire calculation in one long set of products: [latex-display]\displaystyle20\text{ seconds}\cdot\frac{1\text{ minute}}{60\text{ seconds}}\cdot\frac{1\text{ hour}}{60\text{ minutes}}=\frac{15\text{ miles}}{1\text{ miles}}=\frac{5280\text{ feet}}{1\text{ mile}}=\frac{1}{180}\text{ hour}\\[/latex-display]

Try It Now

A 1000 foot spool of bare 12-gauge copper wire weighs 19.8 pounds. How much will 18 inches of the wire weigh, in ounces?
Notice that with the miles per gallon example, if we double the miles driven, we double the gas used. Likewise, with the map distance example, if the map distance doubles, the real-life distance doubles. This is a key feature of proportional relationships, and one we must confirm before assuming two things are related proportionally.

Example 17

Suppose you’re tiling the floor of a 10 ft by 10 ft room, and find that 100 tiles will be needed. How many tiles will be needed to tile the floor of a 20 ft by 20 ft room?


In this case, while the width the room has doubled, the area has quadrupled. Since the number of tiles needed corresponds with the area of the floor, not the width, 400 tiles will be needed. We could find this using a proportion based on the areas of the rooms: [latex-display]\displaystyle\frac{100\text{ tiles}}{100\text{ft}^2}=\frac{n\text{ tiles}}{400\text{ft}^2}\\[/latex-display]
Other quantities just don’t scale proportionally at all.

Example 18

Suppose a small company spends $1000 on an advertising campaign, and gains 100 new customers from it. How many new customers should they expect if they spend $10,000?


While it is tempting to say that they will gain 1000 new customers, it is likely that additional advertising will be less effective than the initial advertising. For example, if the company is a hot tub store, there are likely only a fixed number of people interested in buying a hot tub, so there might not even be 1000 people in the town who would be potential customers.
Sometimes when working with rates, proportions, and percents, the process can be made more challenging by the magnitude of the numbers involved. Sometimes, large numbers are just difficult to comprehend.

Example 19

Compare the 2010 U.S. military budget of $683.7 billion to other quantities.


Here we have a very large number, about $683,700,000,000 written out. Of course, imagining a billion dollars is very difficult, so it can help to compare it to other quantities. If that amount of money was used to pay the salaries of the 1.4 million Walmart employees in the U.S., each would earn over $488,000. There are about 300 million people in the U.S. The military budget is about $2,200 per person. If you were to put $683.7 billion in $100 bills, and count out 1 per second, it would take 216 years to finish counting it.

Example 20

Compare the electricity consumption per capita in China to the rate in Japan.


To address this question, we will first need data. From the CIA[footnote]https://www.cia.gov/library/publications/the-world-factbook/rankorder/2042rank.html[/footnote] website we can find the electricity consumption in 2011 for China was 4,693,000,000,000 KWH (kilowatt-hours), or 4.693 trillion KWH, while the consumption for Japan was 859,700,000,000, or 859.7 billion KWH. To find the rate per capita (per person), we will also need the population of the two countries.   From the World Bank,[footnote]http://data.worldbank.org/indicator/SP.POP.TOTL[/footnote] we can find the population of China is 1,344,130,000, or 1.344 billion, and the population of Japan is 127,817,277, or 127.8 million. Computing the consumption per capita for each country:

China: [latex]\displaystyle\frac{4,693,000,000,000\text{KWH}}{1,344,130,000\text{ people}}\\[/latex] ≈ 3491.5 KWH per person

Japan: [latex]\displaystyle\frac{859,700,000,000\text{KWH}}{127,817,277\text{ people}}\\[/latex] ≈ 6726 KWH per person

While China uses more than 5 times the electricity of Japan overall, because the population of Japan is so much smaller, it turns out Japan uses almost twice the electricity per person compared to China.


Geometric shapes, as well as area and volumes, can often be important in problem solving.

Example 21

You are curious how tall a tree is, but don’t have any way to climb it. Describe a method for determining the height.


There are several approaches we could take. We’ll use one based on triangles, which requires that it’s a sunny day. Suppose the tree is casting a shadow, say 15 ft long. I can then have a friend help me measure my own shadow. Suppose I am 6 ft tall, and cast a 1.5 ft shadow. Since the triangle formed by the tree and its shadow has the same angles as the triangle formed by me and my shadow, these triangles are called similar triangles and their sides will scale proportionally. In other words, the ratio of height to width will be the same in both triangles. Using this, we can find the height of the tree, which we’ll denote by h: [latex-display]\displaystyle\frac{6\text{ft tall}}{1.5\text{ft shadow}}=\frac{h\text{ft tall}}{15\text{ft shadow}}\\[/latex-display] Multiplying both sides by 15, we get h = 60. The tree is about 60 ft tall.
It may be helpful to recall some formulas for areas and volumes of a few basic shapes.



Fig1_1_2 Area: L · W Perimeter: 2L + 2W


Fig1_1_1 Radius: r Area: πr2 Circumference: 2πr


Rectangular Box

Fig1_1_3 Volume: L·W·H


Fig1_1_4 Volume: πr2H

Example 22

If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?


To answer this question, we need to consider how the weight of the dough will scale. The weight will be based on the volume of the dough. However, since both pizzas will be about the same thickness, the weight will scale with the area of the top of the pizza. We can find the area of each pizza using the formula for area of a circle, [latex]A=\pi{r}^2\\[/latex]:

A 12" pizza has radius 6 inches, so the area will be [latex]\pi6^2\\[/latex] = about 113 square inches.

A 16" pizza has radius 8 inches, so the area will be [latex]\pi8^2\\[/latex] = about 201 square inches.

Notice that if both pizzas were 1 inch thick, the volumes would be 113 in3 and 201 in3 respectively, which are at the same ratio as the areas. As mentioned earlier, since the thickness is the same for both pizzas, we can safely ignore it. We can now set up a proportion to find the weight of the dough for a 16" pizza: [latex-display]\displaystyle\frac{10\text{ ounces}}{113\text{in}^2}=\frac{x\text{ ounces}}{201\text{in}^2}\\[/latex-display] Multiply both sides by 201 [latex]\displaystyle{x}=201\cdot\frac{10}{113}\\[/latex] = about 17.8 ounces of dough for a 16" pizza. It is interesting to note that while the diameter is [latex]\displaystyle\frac{16}{12}\\[/latex] = 1.33 times larger, the dough required, which scales with area, is 1.332 = 1.78 times larger.

Example 23

A company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have?


We would expect the calories to scale with volume. Since the marshmallows have cylindrical shapes, we can use that formula to find the volume. From the grid in the image, we can estimate the radius and height of each marshmallow. The regular marshmallow appears to have a diameter of about 3.5 units, giving a radius of 1.75 units, and a height of about 3.5 units. The volume is about π(1.75)2(3.5) = 33.7 units3. The jumbo marshmallow appears to have a diameter of about 5.5 units, giving a radius of 2.75 units, and a height of about 5 units. The volume is about π(2.75)2(5) = 118.8 units3. We could now set up a proportion, or use rates. The regular marshmallow has 25 calories for 33.7 cubic units of volume. The jumbo marshmallow will have: [latex-display]\displaystyle{118.8}\text{ units}^3\cdot\frac{25\text{ calories}}{33.7\text{ units}^3}=88.1\text{ calories}\\[/latex-display] It is interesting to note that while the diameter and height are about 1.5 times larger for the jumbo marshmallow, the volume and calories are about 1.53 = 3.375 times larger.

Try It Now

A website says that you’ll need 48 fifty-pound bags of sand to fill a sandbox that measure 8ft by 8ft by 1ft. How many bags would you need for a sandbox 6ft by 4ft by 1ft?

Problem Solving and Estimating

Finally, we will bring together the mathematical tools we’ve reviewed, and use them to approach more complex problems. In many problems, it is tempting to take the given information, plug it into whatever formulas you have handy, and hope that the result is what you were supposed to find. Chances are, this approach has served you well in other math classes. This approach does not work well with real life problems. Instead, problem solving is best approached by first starting at the end: identifying exactly what you are looking for. From there, you then work backwards, asking “what information and procedures will I need to find this?” Very few interesting questions can be answered in one mathematical step; often times you will need to chain together a solution pathway, a series of steps that will allow you to answer the question.

Problem Solving Process

  1. Identify the question you’re trying to answer.
  2. Work backwards, identifying the information you will need and the relationships you will use to answer that question.
  3. Continue working backwards, creating a solution pathway.
  4. If you are missing necessary information, look it up or estimate it. If you have unnecessary information, ignore it.
  5. Solve the problem, following your solution pathway.
In most problems we work, we will be approximating a solution, because we will not have perfect information. We will begin with a few examples where we will be able to approximate the solution using basic knowledge from our lives.

Example 24

How many times does your heart beat in a year?


This question is asking for the rate of heart beats per year. Since a year is a long time to measure heart beats for, if we knew the rate of heart beats per minute, we could scale that quantity up to a year. So the information we need to answer this question is heart beats per minute. This is something you can easily measure by counting your pulse while watching a clock for a minute. Suppose you count 80 beats in a minute. To convert this beats per year: [latex-display]\displaystyle\frac{80\text{ beats}}{1\text{ minute}}\cdot\frac{60\text{ minutes}}{1\text{ hour}}\cdot\frac{24\text{ hours}}{1\text{ day}}\cdot\frac{365\text{ days}}{1\text{ year}}=42,048,000\text{ beats per year}\\[/latex-display]

Example 25

How thick is a single sheet of paper? How much does it weigh?


While you might have a sheet of paper handy, trying to measure it would be tricky. Instead we might imagine a stack of paper, and then scale the thickness and weight to a single sheet. If you’ve ever bought paper for a printer or copier, you probably bought a ream, which contains 500 sheets. We could estimate that a ream of paper is about 2 inches thick and weighs about 5 pounds. Scaling these down, [latex-display]\displaystyle\frac{2\text{ inches}}{\text{ream}}\cdot\frac{1\text{ ream}}{500\text{ pages}}=0.004\text{ inches per sheet}\\[/latex-display] [latex-display]\displaystyle\frac{5\text{ pounds}}{\text{ream}}\cdot\frac{1\text{ ream}}{500\text{ pages}}=0.01\text{ pounds per sheet, or }=0.16\text{ ounces per sheet.}\\[/latex-display]

Example 26

A recipe for zucchini muffins states that it yields 12 muffins, with 250 calories per muffin. You instead decide to make mini-muffins, and the recipe yields 20 muffins. If you eat 4, how many calories will you consume?


There are several possible solution pathways to answer this question. We will explore one. To answer the question of how many calories 4 mini-muffins will contain, we would want to know the number of calories in each mini-muffin. To find the calories in each mini-muffin, we could first find the total calories for the entire recipe, then divide it by the number of mini-muffins produced. To find the total calories for the recipe, we could multiply the calories per standard muffin by the number per muffin. Notice that this produces a multi-step solution pathway. It is often easier to solve a problem in small steps, rather than trying to find a way to jump directly from the given information to the solution. We can now execute our plan: [latex-display]\displaystyle{12}\text{ muffins}\cdot\frac{250\text{ calories}}{\text{muffin}}=3000\text{ calories for the whole recipe}\\[/latex-display] [latex-display]\displaystyle\frac{3000\text{ calories}}{20\text{ mini-muffins}}=\text{ gives }150\text{ calories per mini-muffin}\\[/latex-display] [latex-display]\displaystyle4\text{ mini-muffins}\cdot\frac{150\text{ calories}}{\text{mini-muffin}}=\text{totals }600\text{ calories consumed.}\\[/latex-display]

Example 27

You need to replace the boards on your deck. About how much will the materials cost?


There are two approaches we could take to this problem: 1) estimate the number of boards we will need and find the cost per board, or 2) estimate the area of the deck and find the approximate cost per square foot for deck boards. We will take the latter approach. For this solution pathway, we will be able to answer the question if we know the cost per square foot for decking boards and the square footage of the deck. To find the cost per square foot for decking boards, we could compute the area of a single board, and divide it into the cost for that board. We can compute the square footage of the deck using geometric formulas. So first we need information: the dimensions of the deck, and the cost and dimensions of a single deck board. Suppose that measuring the deck, it is rectangular, measuring 16 ft by 24 ft, for a total area of 384 ft2. From a visit to the local home store, you find that an 8 foot by 4 inch cedar deck board costs about $7.50. The area of this board, doing the necessary conversion from inches to feet, is: [latex-display]\displaystyle{8}\text{ feet}\cdot4\text{ inches}\cdot\frac{1\text{ foot}}{12\text{ inches}}=2.667\text{ft}^2{.}\\[/latex] The cost per square foot is then [latex]\displaystyle\frac{\$7.50}{2.667\text{ft}^2}=\$2.8125\text{ per ft}^2{.}\\[/latex-display] This will allow us to estimate the material cost for the whole 384 ft2 deck [latex-display]\displaystyle\$384\text{ft}^2\cdot\frac{\$2.8125}{\text{ft}^2}=\$1080\text{ total cost.}\\[/latex-display] Of course, this cost estimate assumes that there is no waste, which is rarely the case. It is common to add at least 10% to the cost estimate to account for waste.

Example 28

Is it worth buying a Hyundai Sonata hybrid instead the regular Hyundai Sonata?


To make this decision, we must first decide what our basis for comparison will be. For the purposes of this example, we’ll focus on fuel and purchase costs, but environmental impacts and maintenance costs are other factors a buyer might consider. It might be interesting to compare the cost of gas to run both cars for a year. To determine this, we will need to know the miles per gallon both cars get, as well as the number of miles we expect to drive in a year. From that information, we can find the number of gallons required from a year. Using the price of gas per gallon, we can find the running cost. From Hyundai’s website, the 2013 Sonata will get 24 miles per gallon (mpg) in the city, and 35 mpg on the highway. The hybrid will get 35 mpg in the city, and 40 mpg on the highway. An average driver drives about 12,000 miles a year. Suppose that you expect to drive about 75% of that in the city, so 9,000 city miles a year, and 3,000 highway miles a year. We can then find the number of gallons each car would require for the year. Sonata: [latex]\displaystyle{9000}\text{ city miles}\cdot\frac{1\text{ gallon}}{24\text{ city miles}}+3000\text{ highway miles}\cdot\frac{1\text{ gallon}}{35\text{ highway miles}}=460.7\text{ gallons}[/latex] Hybrid: [latex]\displaystyle{9000}\text{ city miles}\cdot\frac{1\text{ gallon}}{35\text{ city miles}}+3000\text{ highway miles}\cdot\frac{1\text{ gallon}}{40\text{ highway miles}}=332.1\text{ gallons}[/latex] If gas in your area averages about $3.50 per gallon, we can use that to find the running cost: Sonata: [latex]\displaystyle{460.7}\text{ gallons}\cdot\frac{\$3.50}{\text{gallon}}=\$1612.45\\[/latex] Hybrid: [latex]\displaystyle{332.1}\text{ gallons}\cdot\frac{\$3.50}{\text{gallon}}=\$1162.35\\[/latex] The hybrid will save $450.10 a year. The gas costs for the hybrid are about [latex]\displaystyle\frac{\$450.10}{\$1612.45}\\[/latex] = 0.279 = 27.9% lower than the costs for the standard Sonata. While both the absolute and relative comparisons are useful here, they still make it hard to answer the original question, since “is it worth it” implies there is some tradeoff for the gas savings. Indeed, the hybrid Sonata costs about $25,850, compared to the base model for the regular Sonata, at $20,895. To better answer the “is it worth it” question, we might explore how long it will take the gas savings to make up for the additional initial cost. The hybrid costs $4965 more. With gas savings of $451.10 a year, it will take about 11 years for the gas savings to make up for the higher initial costs. We can conclude that if you expect to own the car 11 years, the hybrid is indeed worth it. If you plan to own the car for less than 11 years, it may still be worth it, since the resale value of the hybrid may be higher, or for other non-monetary reasons. This is a case where math can help guide your decision, but it can’t make it for you.

Try It Now

If traveling from Seattle, WA to Spokane WA for a three-day conference, does it make more sense to drive or fly?

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