Chapter 1: General Problem Solving
Learning Outcomes
- Given the part and the whole, write a percent
- Calculate both relative and absolute change of a quantity
- Calculate tax on a purchase
In the 2004 vice-presidential debates, Democratic contender John Edwards claimed that US forces have suffered “90% of the coalition casualties” in Iraq. Incumbent Vice President Dick Cheney disputed this, saying that in fact Iraqi security forces and coalition allies “have taken almost 50 percent” of the casualties.[1]
Who was correct? How can we make sense of these numbers?
In this section, we will show how the idea of percent is used to describe parts of a whole. Percents are prevalent in the media we consume regularly, making it imperative that you understand what they mean and where they come from.
We will also show you how to compare different quantities using proportions. Proportions can help us understand how things change or relate to each other.
Percents
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].
Percent
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.
For a refresher on basic percentage rules, using the examples on this page, view the following video.
Examples
In a survey, 243 out of 400 people state that they like dogs. What percent is this?
Solution:
[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
Write each as a percent:
- [latex]\displaystyle\frac{1}{4}[/latex]
- 0.02
- 2.35
Solutions:
- [latex]\displaystyle\frac{1}{4}=0.25[/latex] = 25%
- 0.02 = 2%
- 2.35 = 235%
TRY IT NOW
Throughout this text, you will be given opportunities to answer questions and know immediately whether you answered correctly. To answer the question below, do the calculation on a separate piece of paper and enter your answer in the box. Click on the submit button , and if you are correct, a green box will appear around your answer. If you are incorrect, a red box will appear. You can click on “Try Another Version of This Question” as many times as you like. Practice all you want!
Example
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?
Solution:
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
Example
The value of a car dropped from $7400 to $6800 over the last year. What percent decrease is this?
Solution:
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.
For a deeper dive on absolute and relative change, using the examples on this page, view the following video.
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.[2]
Example
There are about 75 QFC supermarkets in the United States. Albertsons has about 215 stores. Compare the size of the two companies.
Solution:
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
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?
Solution:
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.
A video walk-through of this example can be seen here.
Consideration of the base of percentages is explored in this video, using the examples on this page.
TRY IT NOW
Example
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.”
- Is the “decreased by 17” number a useful comparison?
- Considering the last sentence, can we conclude that the number of “dropout factories” was originally 34?
Solutions:
- 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.
- 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
Let’s return to the example at the top of this page. In the 2004 vice-presidential debates, Democratic candidate John Edwards 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?
Solution:
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.
A detailed explanation of these examples can be viewed here.
Think About It
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
A politician’s support increases from 40% of voters to 50% of voters. Describe the change.
Solution:
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
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.
Solution:
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 that player made 200(0.40) = 80 2-point shots and 100(0.30) = 30 3-point shots. Overall, they player made 110 shots out of 300, for a [latex]\displaystyle\frac{110}{300}=0.367=36.7\%[/latex] overall field goal percentage.
For more information about these cautionary tales using percentages, view the following.
Proportions and Rates
If you wanted to power the city of Lincoln, Nebraska using wind power, how many wind turbines would you need to install? Questions like these can be answered using rates and proportions.
Rates
A rate is the ratio (fraction) of two quantities.
A unit rate is a rate with a denominator of one.
Example
Your car can drive 300 miles on a tank of 15 gallons. Express this as a rate.
Solution:
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.[/hidden-answer]
Proportion Equation
A proportion equation is an equation showing the equivalence of two rates or ratios.
For an overview on rates and proportions, using the examples on this page, view the following video.
Example
Solve the proportion [latex]\displaystyle\frac{5}{3}=\frac{x}{6}[/latex] for the unknown value x.
Solution:
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
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
Solution:
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 x 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
Your car can drive 300 miles on a tank of 15 gallons. How far can it drive on 40 gallons?
Solution:
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]\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]
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]\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]
A worked example of this last question can be found in the following video.
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.
You have likely encountered distance, rate, and time problems in the past. This is likely because they are easy to visualize and most of us have experienced them first hand. In our next example, we will solve distance, rate and time problems that will require us to change the units that the distance or time is measured in.
Example
A bicycle is traveling at 15 miles per hour. How many feet will it cover in 20 seconds?
Solution:
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]\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]
Now we can multiply by the 15 miles/hr
[latex]\displaystyle\frac{1}{180}\text{ hour}\cdot\frac{15\text{ miles}}{1\text{ hour}}=\frac{1}{12}\text{ mile}[/latex]
Now we can convert to feet
[latex]\displaystyle\frac{1}{12}\text{ mile}\cdot\frac{5280\text{ feet}}{1\text{ mile}}=440\text{ feet}[/latex]
We could have also done this entire calculation in one long set of products:
[latex]\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]
View the following video to see this problem worked through.
Example
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?
Solution:
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]\displaystyle\frac{100\text{ tiles}}{100\text{ft}^2}=\frac{n\text{ tiles}}{400\text{ft}^2}[/latex]
Other quantities just don’t scale proportionally at all.
Example
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?
Solution:
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.
Matters of scale in this example and the previous one are explained in more detail here.
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.
Examples
The 2010 U.S. military budget was $683.7 billion. To gain perspective on how much money this is, answer the following questions.
- What would the salary of each of the 1.4 million Walmart employees in the US be if the military budget were distributed evenly amongst them?
- If you distributed the military budget of 2010 evenly amongst the 300 million people who live in the US, how much money would you give to each person?
- If you converted the US budget into $100 bills, how long would it take you to count it out – assume it takes one second to count one $100 bill.
Solutions:
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
Compare the electricity consumption per capita in China to the rate in Japan.
Solution:
To address this question, we will first need data. From the CIA[3] 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,[4] 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.
Working with large numbers is examined in more detail in this video.
A Bit of Geometry
Geometric shapes, as well as area and volumes, can often be important in problem solving.
Let’s start things off with an example, rather than trying to explain geometric concepts to you.
Example
You are curious how tall a tree is, but don’t have any way to climb it. Describe a method for determining the height.
Solution:
There are several approaches we could take. We will 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 case 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]\displaystyle\frac{6}{1.5}=\frac{h}{15}[/latex]
If we multiply each side by 15, h, or the height of the tree, is approximately 60 feet tall.
It may be helpful to recall some formulas for areas and volumes of a few basic shapes:
| Rectangular Box | Cylinder |
| Volume: [latex]L\times{W}\times{H}[/latex] | Volume: [latex]\pi{r^2}h[/latex] |
In our next two examples, we will combine the ideas we have explored about ratios with the geometry of some basic shapes to answer questions. In the first example, we will predict how much dough will be needed for a pizza that is 16 inches in diameter given that we know how much dough it takes for a pizza with a diameter of 12 inches. The second example uses the volume of a cylinder to determine the number of calories in a marshmallow.
Examples
If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?
Solution:
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]\displaystyle\frac{10\text{ ounces}}{113\text{in}^2}=\frac{x\text{ ounces}}{201\text{in}^2}[/latex]
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.
The following video illustrates how to solve this problem.
Example
A company makes regular and jumbo marshmallows. The regular marshmallow has 25 calories. How many calories will the jumbo marshmallow have?
Solution:
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]\displaystyle{118.8}\text{ units}^3\cdot\frac{25\text{ calories}}{33.7\text{ units}^3}=88.1\text{ calories}[/latex]
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.
For more about the marshmallow example, watch this video.
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?
Solution:
The original sandbox has volume [latex]64\text{ft}^3[/latex]. The smaller sandbox has volume [latex]24\text{ft}^3[/latex].
[latex]\displaystyle\frac{48\text{bags}}{64\text{ft}^3}=\frac{x\text{ bags}}{24\text{ft}^3}[/latex] results in x = 18 bags.
Mary (from the application that started this topic), decides to use what she knows about the height of the roof to measure the height of her second daughter. If her second daughter casts a shadow that is 1.5 feet long when she is 13.5 feet from the house, what is the height of the second daughter? Draw an accurate diagram and use similar triangles to solve.
Solution:
2.5 ft
In the next section, we will explore the process of combining different types of information to answer questions.
Attributions
This chapter contains material taken from Math in Society (on OpenTextBookStore) by David Lippman, and is used under a CC Attribution-Share Alike 3.0 United States (CC BY-SA 3.0 US) license.
This chapter contains material taken from of Math for the Liberal Arts (on Lumen Learning) by Lumen Learning, and is used under a CC BY: Attribution license.