[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.

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

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.

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.

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.

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.

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.

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.

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%.

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.

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.

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].

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.

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]

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]

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]

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.

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.

1. 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.
2. There are about 300 million people in the U.S. The military budget is about $2,200 per person.
3. 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.

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.

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.

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.