Percentages - Astronomy: The Human Quest for Understanding

The use of percentages is very common, not only in the sciences, but in society in general. You routinely see that a sweater is $20\%$ off ($20$ percent off), or that the interest rate is currently at $2.1\%$ ($2.1$ percent). Despite its constant use, percentages can still be confusing to many students.

In order to understand what percentages mean, you need only go to the derivation of the term percent which is actually a concatenated term based on per and cent. The term per means the amount of something being described is in terms of some other amount. That “other amount” is described by cent, meaning $100$ (think of the smallest unit of currency in the United States; $100$ cents = $1$ dollar). Thus, when you measure a quantity as a percentage, you are referring to the number per $100$, rather than per $1$, or per some other number.

When you divide one number by another (fractions in other words) you are essentially computing how many “things” are in the numerator per the number of “things” in the denominator. For example, if you have a pizza pie [standard college student breakfast fare :-)] that originally had $5$ slices,¹ and $4$ remain, the fraction $\displaystyle \frac{4}{5}$ tells you what fraction of the number of slices remains, namely $0.8$ ($4 \div 5$). To think about this in terms of percentages, you are basically asking the question, “how many slices would remain out of $100$ original slices?” That answer simply requires converting the denominator of the fraction to $100$ with the corresponding increase in the numerator, or

$$\frac{4}{5} = \frac{4}{5} \times 1 = \frac{4}{5} \times \frac{20}{20} = \frac{4 \times 20}{5 \times 20} = \frac{80}{100}.$$

¹Weird way to divide a pizza, but it’s for science!

(Remember that you can multiply any number by $1$ without changing the answer. In this case we wrote $1$ as $\displaystyle \frac{20}{20}$ so that the denominator would become $100$.)

The fraction is now saying that there would be $80$ slices of pizza remaining out of an original total of $100$; $80$ per $100$ or $80$ per cent ($80\ \mbox{percent} = 80\%)$. Another way to think about it is to evenly slice the same pizza up into $100$ equal slices, and then remove $20$ of them. You would still have removed the same amount of pizza as when one piece was taken out of an original five, the slices would simply be inconvenient, bite-size pieces. The remaining $80$ tiny slices would be the same amount of pizza as the four larger remaining ones.

Of course, you could have easily obtained the same answer by simply multiplying the decimal representation of $\displaystyle \frac{4}{5} = 0.8$ by $100\%$, giving $80\%$. This works because $100\%$ of something is all of that thing, or $1$ rather than some fraction of $1$. To be technical, what you are really doing when you multiply a fraction by $100\%$ is changing the “units” of the quantity you are describing, which is done by writing $1$ in another cleaver way and multiplying the original fraction by that cleaver version of $1$. In the example, this is just

$$\frac{4}{5} \times 1 = \frac{4}{5} \times \frac{100\%}{1} = 80\%.$$

The cleaver version of $1$ is actually $\displaystyle \frac{100\%}{1}$ since by definition $100\% = 1$. If you reviewed the unit conversion tutorial you should realize that what just happened is identical to converting from one unit of measurement to another unit of measurement.

It is worth pointing out that in the last example, rather than saying that $80\%$ of the pizza pie remains, you can also say that $100\% – 80\% = 20\%$ of the pizza has been eaten.

Not only can you have a fraction of what you started with that is less than $1$ (or less than $100\%$), it is also possible to have something that is more than what you started with (or more than $100\%$). What this is saying is that after starting with a certain amount of something, more was accumulated so that later on the total is more than the original. For example, suppose you weighed yourself one morning and found your weight to be $150$ pounds. A few weeks later your weight was $160$ pounds. Clearly there is “more of you to love” later on than during the earlier measurement, but by what fraction? by what percentage? To answer these questions note that $$\frac{160}{150} = \frac{16}{15} = 1.067,$$

meaning that when you measured yourself the second time you were $1.067$ times heavier than when you measured yourself the first time. But that also means that the second time you weighed yourself you were $106.7\%$ of the weight you were during the first measurement. You can also say this another way: After the second weighing you were $106.7\% – 100\% = 6.7\%$ heavier than after the first weighing.

Consider one more example: When you go shopping at a sale, you see a sweater that you would like to purchase marked as $20\%$ off. If the original price is $\$45$, what is the sale price?

Since the sweater is $20\%$ off, that means that the sale price is $100\% – 20\% = 80\%$ of the original price. Since $80\%$ is $80$ per $100$ or $\displaystyle \frac{80}{100} = 0.8$, to find the sale price you need to multiply the original price by $0.8$ or, putting all of the steps together,

$$\$45 \times (100\% – 20\%) = \$45 \times 80\% = \$45 \times 0.8 = \$36.$$

Of course if your state, county, and city assess a sales tax you will still need to pay that as well. If the combined sales tax in your area is $6.850\%$, how much tax will you need to pay?

With a sale price for the sweater of $$36$, you will need to multiply that amount by $6.850\% = 6.850/100 = 0.0685$. Therefore, the amount of sales tax you owe becomes

$$\$36 \times 6.850\% = \$36 \times \frac{6.850}{100} = \$36 \times 0.06850 = \$2.47.$$

The actual result of the multiplication is $\$2.466$, which gets rounded up to $\$2.47$ because United States of America currency does not recognize a fraction of $1$¢; $1$ cent. Any time the third digit after the decimal point is $5$ or higher, the result is rounded up; if the digit is less than $5$ the result is rounded down.

Finally, the total cost of the sweater to you is

$$\$36 + \$2.47 = \$38.47.$$

Note that you could have arrived at this answer directly by multiplying the sale price by $1.06850$, the sale price of the sweater ($1 = 100\%$) plus the sales tax ($0.06850 = 6.850\%$):

$$\$36 \times (100\% + 6.850\%) = \$36 \times 106.850\% = \$36 \times 1.06850 = \$38.47.$$

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where $10$ raised to an integer (either positive or negative) replaces $\times 10^\text{exponent}$ with ${\rm E(exponent)}$ or ${\rm e(exponent)}$; in other words, $3.14159 \times 10^9$ is represented by $3.14159\text{E}9$. If your calculator has the $\Large{\hat{ }}$ symbol, then you would enter the number as $3.14159 \times 10{\Large{\hat{ }}}9$. If your calculator has a $10^x$ button then you would enter the number as $3.14159\times9\,\,10^x$. If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

When the semester started there were $53$ students in the class. After the third week, the total enrollment had decreased to $45$ students. What percentage of the class dropped the class during the first three weeks of the term?
Suppose that a star started its lifetime with a mass that was twice the mass of the Sun. At the end of its lifetime, it’s mass has decreased to $0.85$ times the mass of the Sun. What percentage of the star’s mass was lost during its lifetime?
At one point in its history, a proto-planet had a mass that was $35\%$ the mass of Earth, but due to the accumulation of mass from the pre-planetary system that it was forming in, the fully-formed planet ended up with a mass that was $2.5$ times the mass of Earth. What percentage of the planet’s final total mass was accumulated since it was in that particular proto-planetary stage? Hint: Consider the proto-planetary mass (the $35\%$ of the mass of Earth), so what additional amount of mass must be added to reach the final amount?
Supermassive black holes are known to exit at the centers of most large galaxies. The black hole at the center of our Milky Way Galaxy has a mass of $4.1 \times 10^6$ times the mass of our Sun. The entire Milky Way Galaxy has a mass that is estimated to be $1.3 \times 10^{12}$ times the mass of our Sun. What percentage of the mass of the Milky Way Galaxy is contained in the supermassive black hole at its center?

Answers

$15\%$
$57.5\%$
$86\%$
$0.000\,32\% = 3.2\times10^{-4}\%$