1.3 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.

Example 3

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

Show solution

We can set up a proportion by setting equal two $\frac{map inches}{real miles}$ rates, and introducing a variable, $x$ , to represent the unknown quantity – the mile distance between the cities.

$\begin{matrix} \frac{\frac{1}{2} map inch}{3 miles} = \frac{2 \frac{1}{4} map inches}{x miles} & Multiply both sides by x and rewriting the mixed number \\ \frac{\frac{1}{2}}{3} \cdot x = \frac{9}{4} & Multiply both sides by 3 \\ \frac{1}{2} x = \frac{27}{4} & Multiply both sides by 2 (or divide by \frac{1}{2}) \\ x = \frac{27}{2} = 13 \frac{1}{2} miles \end{matrix}$

Many proportion problems can also be solved using dimensional analysis, the process of multiplying a quantity by rates to change the units.

Example 4

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

Show solution

We could certainly answer this question using a proportion: $$ \frac{300 miles}{15 gallons} = \frac{x miles}{40 gallons} $$ .

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:

$40 gallons \cdot \frac{20 miles}{gallon} = \frac{40 gallons}{1} \cdot \frac{20 miles}{gallon} = 800 miles$

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:

$50 miles \cdot \frac{lgallon}{20 miles} = \frac{50 miles}{1} \cdot \frac{1 gallon}{20 miles} = \frac{50 gallons}{20} = 2.5 gallons$

Dimensional analysis can also be used to do unit conversions. Here are some unit conversions for reference.

Note

Unit Conversions

Length

$\begin{matrix} 1 foot (ft) = 12 inches (in) & 1 yard (yd) = 3 feet (ft) \\ 1 mile = 5, 280 feet \\ 1000 millimeters m m = 1 meter (m) & 100 centimeters (cm) = 1 meter \\ 1000 meters (m) = 1 kilometer (km) & 2.54 centimeters (cm) = 1 inch \end{matrix}$

Weight and Mass

$\begin{matrix} 1 pound (lb) = 16 ounces (oz) & 1 ton = 2000 pounds \\ 1000 milligrams (mg) = 1 gram (g) & 1000 grams = 1 kilogram (kg) \\ 1 kilogram = 2.2 pounds (on earth) \end{matrix}$

Capacity

$\begin{matrix} 1 cup = 8 {fluid ounces (fl oz)}^{*} & 1 pint = 2 cups \\ 1 quart = 2 pints = 4 cups & 1 gallon = 4 quarts = 16 cups \\ 1000 milliliters (ml) = 1 liter (L) \end{matrix}$

^*Fluid ounces are a capacity measurement for liquids. 1 fluid ounce ≈ 1 ounce (weight) for water only.

Example 5

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

from fractions import Fraction

# Dimensional analysis, done the way the book does it: multiply by rates that
# equal 1, and cross off units until only the unit you want is left.
def convert(amount, unit, chain):
    """chain: list of (top_amount, top_unit, bottom_amount, bottom_unit)"""
    value, top, bottom = Fraction(str(amount)), [unit], []
    print("   start:%-28s %14s  %s" % ("", str(value), unit))
    for ta, tu, ba, bu in chain:
        value = value * Fraction(str(ta)) / Fraction(str(ba))
        top.append(tu)
        bottom.append(bu)
        cancelled = [u for u in list(top) if u in bottom]      # cross off matching units
        for u in cancelled:
            top.remove(u)
            bottom.remove(u)
        units = "*".join(top) + ("/" + "*".join(bottom) if bottom else "")
        print("   x %-32s %14s  %-14s (cancelled %s)"
              % ("%s %s / %s %s" % (ta, tu, ba, bu), str(value), units,
                 ", ".join(cancelled) or "nothing"))
    return value, top, bottom

# The book's bicycle: 15 miles per hour, how many feet in 20 seconds?
# CHANGE the 20, the 15, or add/remove links in the chain and re-run.
value, top, bottom = convert(20, "second", [
    (1, "minute", 60, "second"),
    (1, "hour",   60, "minute"),
    (15, "mile",   1, "hour"),      # <-- the bicycle's speed
    (5280, "foot", 1, "mile"),
])
print()
print("answer: %s = %.1f %s" % (value, float(value), "*".join(top)))
print()

# Try it Now 4 from the book: 18 inches of wire from a 1000 ft, 19.8 lb spool.
value, top, bottom = convert(18, "inch", [
    (1, "foot", 12, "inch"),
    (19.8, "pound", 1000, "foot"),
    (16, "ounce", 1, "pound"),
])
print("answer: %.3f %s" % (float(value), "*".join(top)))
# If a unit never cancels, you chained the rate upside down -- flip it and re-run.

Show 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:

$\begin{matrix} 20 seconds \cdot \frac{1 minute}{60 seconds} \cdot \frac{1 hour}{60 minutes} = \frac{1}{180} hour & Now we can multiply by the 15 miles/hr \\ \frac{1}{180} hour \cdot \frac{15 miles}{Ihour} = \frac{1}{12} mile & Now we can convert to feet \\ \frac{1}{12} mile \cdot \frac{5280 feet}{1 mile} = 440 feet \end{matrix}$

We could have also done this entire calculation in one long set of products:

$20 seconds \cdot \frac{1 minute}{60 seconds} \cdot \frac{1 hour}{60 minutes} \cdot \frac{15 miles}{1 hour} \cdot \frac{5280 feet}{1 mile} = 440 feet$

Your Turn

Try it Now 4

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?

Answer

$18 inches \cdot \frac{1 foot}{12 inches} \cdot \frac{19.8 pounds}{1000 feet} \cdot \frac{16 ounces}{1 pound} \approx 0.475 ounces$

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 6

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?

Show 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:

$\frac{100 tiles}{100 {f t}^{2}} = \frac{n tiles}{400 {f t}^{2}}$

Other quantities just don’t scale proportionally at all.

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 9

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

# A total and a rate can point in opposite directions. Book data (2011).
# ADD a country of your own: "name": (kilowatt-hours per year, population)
countries = {
    "China":  (4_693_000_000_000, 1_344_130_000),
    "Japan":    (859_700_000_000,   127_817_277),
    "Iceland":    (17_000_000_000,       320_000),   # <-- try deleting this line
}

print("%-10s %22s %15s %16s" % ("country", "total KWH/yr", "population", "KWH per person"))
print("-" * 66)
rows = []
for name, (kwh, pop) in countries.items():
    per_capita = kwh / pop
    rows.append((name, kwh, pop, per_capita))
    print("%-10s %22s %15s %16.1f" % (name, "{:,}".format(kwh), "{:,}".format(pop), per_capita))
print("-" * 66)

by_total = sorted(rows, key=lambda r: -r[1])
by_rate  = sorted(rows, key=lambda r: -r[3])
print("ranked by TOTAL use : " + " > ".join(r[0] for r in by_total))
print("ranked by PER PERSON: " + " > ".join(r[0] for r in by_rate))
print()

a, b = "China", "Japan"      # <-- CHANGE these two names to compare any pair
ka, pa = countries[a]
kb, pb = countries[b]
print("%s uses %.1f times the electricity of %s overall," % (a, ka / kb, b))
print("but %s uses %.1f times as much PER PERSON as %s."
      % (b, (kb / pb) / (ka / pa), a))
print("Same two numbers, opposite headline. The denominator decides.")

To address this question, we will first need data. From the CIA[1] 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[2], 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.

Show solution

Computing the consumption per capita for each country:

China: $\frac{4,693,000,000,000 K W H}{1,344,130,000 people} \approx 3491.5$ KWH per person

Japan: $\frac{859,700,000,000 K W H}{127,817,277 people} \approx 6726$ 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.

[1] www.cia.gov/library/publicat.../2042rank.html

[2] http://data.worldbank.org/indicator/SP.POP.TOT

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.