DISCOVERY PROJECT

Real Numbers in the Real World

When we use real numbers in our everyday life, the numbers almost always are associated with a unit. For example, if you count the number of students in your class, you may find out that there are $30$ people—in this case the units are people. If you measure your height you may find out that you are $70$ inches tall—in this case the units are inches. You may also state your height in terms of other units; so you may say that you are $178$ centimeters tall. Of course your height is the same, but the units in which you choose to measure your height may be different. So how did we convert a height of $70$ inches to the equivalent height in centimeters? To do this we first need to know the relationship between the different units. We know the following conversion ratio between inches and centimeters:

$1 \; inch = 2.54 \; centimeters$ $\color{blue}{Conversion}$ $\color{blue}{ratio}$

This means that there are $2.54$ cm in each inch. We can see then that $70$ inches is equivalent to $$70 × 2.54 = 178 \; cm$$

Let's analyze the reasoning we used in this example more carefully in order to help us to use the process in other situations. If we write down the units and the conversion ratio (in the appropriate form) in the above calculation we see how the inches "cancel out" to allow us to arrive at an equivalent number in centimeters.$$70 \cancel{in} × \left(\frac{2.54 \; cm}{1 \cancel{in}}\right) = 70 × 2.54 \; cm \approx 178 \; cm$$

Notice how the original units are crossed out in the calculation to obtain a number in terms of the new units. In the calculation we put the conversion ratio in the appropriate form so that the original units cancel, leaving the units that we want. The process can be described schematically as follows:

$\fbox{   }$ $\color{blue}{\cancel{Original \; units}}$ $×$ $\left(\frac{\fbox{   } \color{red}{ New \; units      }}{\fbox{   } \color{blue}{\cancel{Original \; units}}}\right)$ $=$ $\fbox{   }$ $\color{red}{New \; units}$
Number we want to convert
Conversion ratio
Number in the new units

Let's try some other examples that involve converting units. In each example we need to know the conversion ratios. You probably already know many of these. For example, $1$ hour is the same as $60$ minutes, $1$ foot is the same as $12$ inches, and so on. But you may have to look up some other conversion ratios. For instance, $1$ mile is about $1.6$ kilometers. We give the appropriate conversion ratios in the examples and exercises that follow, but if you need others they are readily available on the internet.

Converting Units

We begin with simple examples of unit conversion. In these examples we convert a quantity that is expressed in one unit into another unit. Examples that involve compund units will be given in the next subsection.

EXAMPLE 1 | Converting Units
  1. How many liters are in a $5$ gallon gas container?
  2. How many gallons are in a $100$ liter container?

SOLUTION

We first need to know the conversion ratio between gallons and liters:

$1 \; gal = 3.875 \; L$ $\color{blue}{Conversion}$ $\color{blue}{ratio}$
  1. We want to convert $5$ gallons to the equivalent amount in liters. We multiply by the conversion ratio ($3.875 \; L/ \; 1 \;gal$) so that the appropriate units cancel out, leaving the answer in the units that we want. In this case the "gallons" cancel out and the answer is in liters. $$5 \cancel{gal} × \left(\frac{3.875 \; L}{1 \cancel{gal}}\right) = 5 × 3.875 \; L \approx 19.375 \; L$$
    So $5$ gallons is the same amount as $19.375$ liters.
  2. We want to convert $100$ liters to the equivalent amount in gallons. Again, we multiply by the conversion ratio ($1 \; gal \;/ \;3.875 \;L$) so that the appropriate units cancel out, leaving the answer in the units that we want. In this case the "liters" cancel out and the answer is in gallons. $$100 \cancel{L} × \left(\frac{1 \; gal}{3.875 \cancel{L}}\right) = 100 × \frac{1}{3.875} \; gal \approx 25.806 \; gal$$
    So $5$ gallons is the same amount as $19.375$ liters.
EXAMPLE 2 | Converting Units

Convert $10$ days to the equivalent number of minutes.

SOLUTION

We do this conversion by first converting days to hours and then hours to minutes. The conversion ratios are

$1 \; day = 24 \; h$ $\color{blue}{Conversion}$ $\color{blue}{ratios}$
$1 \; h = 60 \; min$

We want to convert $10$ days into the equivalent number of minutes. We proceed as follows: $$10 \cancel{day} × \left(\frac{24 \cancel{h}}{1 \cancel{day}}\right) × \left(\frac{60 \; min}{1 \cancel{h}}\right) = 10 × 24 × 60 \; min = 14,400 \; min$$
So $10$ days is the same as $14,400$ minutes.

Converting Compound Units

Many real world quantities are expressed in compound units, that is, in terms of more than one unit. One familiar quantity that is described by a compound unit is speed. For example, a speed of $60$ miles per hour involves two units—miles and hours. We write this as $60$ mi/h. We may want to convert this speed into feet per minute, kilometers per hour, or meters per second. The procedure for doing such conversions is the same as the procedure we used in the preceding examples.

EXAMPLE 3 | Converting Compound Units
  1. Convert a speed of $60$ mi/h into the equivalent speed in kilometers per hour.
  2. Convert a speed of $60$ mi/h into the equivalent speed in feet per second.

SOLUTION

We first need to know the conversion ratio between miles and kilometers:

$1 \; mi = 1.609 \; km$ $\color{blue}{Conversion}$ $\color{blue}{ratio}$
  1. We want to convert a speed of $60$ mi/h to the equivalent speed in kilometers per hour. We multiply by the conversion ratio ($1.609 \;km/1 \;mi$) so that the appropriate units cancel out, leaving the answer in the units that we want. In this case the "miles" cancel out and the answer is in kilometers per hour. $$60 \; \frac{\cancel{mi}}{h} × \left(\frac{1.609 \; km}{1 \cancel{mi}}\right) = 60 × 1.609 \; \frac{km}{h} \approx 96.54 \; km/h$$
    So $60$ mi/h is the same $96.54$ km/h.
  2. We want to convert a speed of $60$ mi/h to the equivalent speed in feet per second. We multiply by the conversion ratios ($5280 \; ft/1 \;mi$, $60 \;min/1 \;h$, and $60 \;sec/1 \;min$) so that the appropriate units cancel out, leaving the answer in the units that we want. In this case the "miles" and "hours" cancel out and the answer is in feet per second. $$60 \; \frac{\cancel{mi}}{\cancel{h}} × \left(\frac{5280 \; ft}{1 \cancel{mi}}\right) × \left(\frac{1 \cancel{h}}{60 \cancel{min}}\right) × \left(\frac{1 \cancel{min}}{60 \; sec}\right) = \frac{60 × 5280}{60 × 60} \frac{ft}{sec} = 88 \; ft/sec$$
    So $60$ mi/h is the same as $88$ ft/sec.

In the U.S. blood sugar levels are measured in mg/dL (milligrams per deciliter), whereas in the rest of the world blood sugar levels are measured in mmol/L (millimoles per liter). In order to make the conversion from one measure to the other we need to know the following conversion ratios:

$1 \; mmol \; of \; sugar = 180 \; mg$
$1 \;L = 10 \;deciliters$
EXAMPLE 4 | Converting Compound Units
  1. A desirable blood sugar reading is $5.50$ mmol/L. Find the equivalent blood sugar level in the U.S. measurement system.
  2. In a recent lab test, Jason's fasting blood sugar level is found to be $220$ mg/dL, which his doctor thinks is far too high. Jason wishes to tell his friend Susanna in Germany about his problematic reading, so he converts it to mmol/L, units that she is more familiar with. What is his reading in mmol/L?

SOLUTION

  1. We want to convert a reading of $5.50$ mmol/L to the equivalent reading in mg/dL. To do this we multiply by the appropriate conversion factors to obtain an answer in mg/dL. $$5.50 \; \frac{\cancel{mmol}}{\cancel{L}} × \left(\frac{180 \; mg}{1 \cancel{mmol}}\right) × \left(\frac{1 \cancel{L}}{10 \; dL}\right) = 99 \; mg/dL$$
    So the desirable blood sugar reading in the U.S. is $99$ mg/dL.
  2. To find Jason's blood sugar reading in mmol/L we multiply by the appropriate conversion ratios: $$220 \; \frac{\cancel{mg}}{\cancel{dL}} × \left(\frac{1 \; mmol}{180 \cancel{mg}}\right) × \left(\frac{10 \cancel{dL}}{1 \; L}\right) = 12.2 \; mmol/L$$
    So Jason's blood sugar reading is $12.2$ mmol/L.

The fuel efficiency of cars can be measured in one of two possible ways: how far the car can travel using a fixed amount of fuel, or how much fuel the car consumes over a fixed distance. In the U.S. the first method is used; we measure fuel efficiency in miles per gallon (mi/gal, often abbreviated as mpg), the distance the car can travel on one gallon of gasoline. In most of the rest of the world fuel efficiency is measured in liters per $100$ kilometers (L/100 km), the amount of gasoline the car consumes when traveling $100$ km. To convert from one system to the other we need the following conversion ratios (which we have already encountered in the first two examples).

$1 \; gallon = 3.875 \; liters$
$1 \; mile = 1.609 \; kilometers$

Note that the two ways of measuring fuel efficiency are, in a sense, reciprocal concepts: distance driven per amount of gas or amount of gas per distance driven.

EXAMPLE 5 | Converting Compound Units
  1. A car getting $12$ mi/gal would be considered a "gas guzzler" by any standard. Convert $12$ mi/gal to L/100 km.
  2. Minori has purchased a new car and she finds that it uses $6.8$ L/100 km in highway driving. She wishes to tell her friend Chad who lives in the U.S. the amount of gas her car uses. Help her communicate her car's gas usage to Chad by changing $6.8$ L/100 km to mi/gal.
  3. Complete the following table for various equivalent gas usages in mi/gal and L/100 km. What do you notice?
$mi/gal$ $10$ $16$ $32$
$L/100$ $km$ $11$ $8.6$ $5.1$

SOLUTION

  1. We want to convert $12$ mi/gal to the equivalent in L/100 km. Since we want to convert "distance per amount" to "amount per distance", we begin with the reciprocal of $12$ mi/gal, and then multiply by the appropriate conversion factors to obtain the value in L/km.$$\frac{1 \cancel{gal}}{12 \cancel{mi}} × \left(\frac{3.875 \; L}{1 \cancel{gal}}\right) × \left(\frac{1 \cancel{mi}}{1.609 \; km}\right) = 0.201 \; L/km$$
    Since we want to use L/100 km we multiply the number we obtained above by 100 km/100 km.$$0.201\frac{L}{\cancel{km}} × \left(\frac{100 \cancel{km}}{100 \; km}\right) = 20.1 \; L/100 \; km$$
    So $12$ mi/gal is equivalent to about $20$ L/100 km.
  2. We want to convert $6.8$ L/100 km to the equivalent in mi/gal. Since we want to convert "amount per distance" to "distance per amount," we begin with the reciprocal of $6.8$ L/100 km, and then multiply by the appropriate conversion factors to obtain the value in mi/gal: $$\frac{100 \cancel{km}}{6.8 \cancel{L}} × \left(\frac{1 \; mi}{1.609 \cancel{km}}\right) × \left(\frac{3.875 \cancel{L}}{1 \; gal}\right) = 35.4 \; mi/gal$$
    So Minori's car gets about $35$ mi/gal.
  3. Using the procedures we described in parts (a) and (b), we complete the table as follows.
    $mi/gal$ $10$ $16$ $\color{red}{22}$ $\color{red}{28}$ $32$ $\color{red}{47}$
    $L/100 \;km$ $\color{red}{24}$ $\color{red}{15}$ $11$ $8.6$ $\color{red}{7.5}$ $5.1$
    As miles per gallon increase, liters per $100$ kilometers decrease.
Problems
  1. Installing Wood Flooring Camille is installing new wood flooring in her house. She estimates that she needs $170$ linear feet of wood moldings and $1990$ square feet of wood flooring. Her friend Hugh has a cottage on a lake in Canada that has $48$ meters of wood moldings $160$ square meters of wood flooring.
    1. How many meters of wood moldings does Camille need to install?
    2. How many square meters of wood flooring does Camille needs to install?
    3. How many feet of wood moldings does Hugh's cottage have?
    4. How many square feet of wood flooring does Hugh's cottage have?
  2. Measuring Speed Anika rides her bike at a speed of $20$ km/h, Jun drives to work at a speed of $70$ mi/h, and a peregrine falcon flies at a speed of $0.08$ km/s.
    1. Find the peregrine falcon's speed in m/s and mi/h.
    2. Find Anika's bike speed in m/s and mi/h.
    3. Find Jun's driving speed in km/h and m/s.
    4. How much faster is the falcon's flying speed than Jun's driving speed?
  3. Area of Land In the United States the area of land is usually measured in acres, and in Europe it is usually measured in hectares (ha). (Note that $1 \;acre = 43,560 \; ft^2, \;1 \;ha = 10,000 \; m^2, and \;1 \;ft = 0.3048 \;m$.)
    1. Camille owns $4$ hectares of farmland in the south of France. Find the area of Camille's farmland in square meters, square feet, and acres.
    2. A soccer field is $110$ m by $70$ m. Find the area of the soccer field in square meters, hectares, square feet, and acres.
  4. Blood Sugar Measurements According to a recent lab test (called an A1C test), Zoe's average blood sugar level over the last three months was $8.2$ mmol/L. Her doctor warns that this is too high, and reminds Zoe that the target level for her blood sugar is $110$ mg/dL. Find the blood test result in mg/dL. How much higher is her A1C test result than her target of $110$ mg/dL?
  5. Measuring Cholesterol In the United States, India, and some other countries serum cholesterol values are measured in mg/dL. In most countries serum cholestrol is measure in mmol/L. (Note that $1$ mmol of serum cholesterol is equal to $387$ mg.)
    1. A desirable reading for cholesterol is $4.1$ mmol/L. Find the equivalent cholesterol level in mg/dL.
    2. According to her latest blood test, Daniela has a cholesterol level of $186$ mg/dL. She wants to tell her Norwegian friend Leif her cholesterol level in units that he is more familiar with. Find Daniela's result in mmol/L.
  6. Automobile Gas Consumption Armando is shopping for a new car and is choosing between an SUV and a sedan. The SUV uses $6.1$ L/100 km and the sedan gets $36$ mi/gal.
    1. Convert $36$ mi/gal to L/100 km.
    2. Convert $6.1$ L/100 km to mi/gal.
    3. Which car is more fuel efficient?
  7. Automobile Gas Consumption Karl's father loans him his Mercedes to drive on a road trip to Colorado. According to the manual the car uses $7.8$ L/100 km, and the gas tank capacity is $65.9$ L. During the trip Karl notices that the gas tank is only $15$% full and that the closest gas station is $82$ miles away.
    1. Determine the amount of gas left in the tank, and the distance (in km) that Karl can drive before running out of gas.
    2. Will Karl make it to the closest gas station before running out of gas? (Assume that the gas gauge is accurate.) If not, how many miles will he have to walk to reach the gas station after running out of gas?
  8. A Conversion Formula for Gas Consumption Suppose a car's gas mileage is $x$ mi/gal or equivalently $y$ L/100 km.
    1. Find a formula that gives $y$ if we know $x$. [Hint: Since $x$ and $y$ are in a reciprocal relationship, the formula must be of the form $y = k / x$ ; find the appropriate constant $k$.]
    2. Use your formula to check your answers for Exercise 6.
    3. Use your formula to convert, in your head, $10$ L/100 km to mi/g and $40$ mi/g to L/100 km.