Percent: Free Response

Start from the definition and nothing else: the percent sign says 'divide by $100$', so drop the sign and write the number over $100$.

$68\% = \frac{68}{100}.$

That is already correct, and it is not yet in lowest terms. Factorize both numbers to find what they share:

$68 = 2 \times 2 \times 17, \qquad 100 = 2 \times 2 \times 5 \times 5.$

They share two factors of $2$, so the greatest common factor is $4$, and dividing top and bottom by it keeps the ratio unchanged:

$\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}.$

Since $17$ is prime and $25 = 5 \times 5$, nothing is left to cancel.

For the decimal, divide by $100$, which lowers each digit two place columns:

$68\% = 68 \div 100 = 0.68.$

The $6$ that was worth $60$ now sits in the tenths place and the $8$ that was worth $8$ now sits in the hundredths place. That is why the two forms agree: a two place decimal is a count of hundredths, and so is a percent.

Take the fraction first. Scaling to a denominator of $100$ needs a whole number multiplier, and dividing $100$ by the denominator says whether one exists:

$100 \div 16 = 6 \text{ remainder } 4.$

The division does not come out, so no whole number takes $16$ to $100$ and the scaling shortcut has nothing to offer. Fall back on what a percent is: the numerator that sits over $100$ in an equal ratio. Write that as a proportion with the percent unknown,

$\frac{7}{16} = \frac{p}{100},$

and finish it the way any proportion is finished. Multiply the diagonal you know completely, then divide by the number diagonally opposite the blank:

$p = \frac{7 \times 100}{16} = \frac{700}{16} = 43.75.$

The division is worth seeing in full: $16 \times 43 = 688$, which leaves $12$, and $12 \div 16 = 0.75$, so the quotient is $43.75$. Hence $\frac{7}{16} = 43.75\%$. Dividing agrees: $7 \div 16 = 0.4375$, and moving that point two places right gives the same percent.

The decimal is quicker. Going from a decimal to a percent multiplies by $100$, which moves the point two places right:

$0.035 = (0.035 \times 100)\% = 3.5\%.$

Notice that both answers carry a decimal point. A percent is a count out of $100$, and nothing in that definition promises the count is a whole number.

The classmate has mistaken a convenient route for the definition. A percent is the numerator of an equal ratio whose second term is $100$, so asking for a fraction's percent is asking a question the proportion always answers:

$\frac{7}{16} = \frac{p}{100} \quad \Longrightarrow \quad p = \frac{7 \times 100}{16}.$

Cross-multiplying leaves a division, and dividing by a nonzero whole number always has a value. That value may not be a whole number, and it need not even stop, since a denominator of $3$ gives a percent whose digits repeat, but a value it certainly has:

$p = \frac{700}{16} = 43.75.$

Scaling is the special case of that same division coming out even. When the denominator divides $100$ exactly, as $20$ and $25$ do, the multiplier is a whole number and the percent can be read off with no dividing at all. When it does not, as here, the only thing that has failed is the shortcut.

Check that the answer really is the same number: $43.75\% = 43.75 \div 100 = 0.4375$, and dividing the fraction out gives $7 \div 16 = 0.4375$ as well. The two agree, so $\frac{7}{16}$ has been written over $100$ after all.

What the proportion supplies, then, is the general method behind the shortcut, and it does not care whether the arithmetic is tidy. What the classmate has really discovered is that this denominator forces a percent with a decimal point in it, and a percent is as entitled to a decimal point as any other count.

Convert the percent first, because the percent sign is a division by $100$ and a decimal is what multiplication can use:

$46\% = 46 \div 100 = 0.46.$

Now multiply, since 'of' with a percent means multiply. Splitting the $550$ makes the arithmetic easy to see:

$0.46 \times 550 = 0.46 \times 500 + 0.46 \times 50 = 230 + 23 = 253.$

So the riverside garden has $253$ vegetable plots.

Why multiplying is the right move at all: the percent proportion says $\frac{\text{part}}{550} = \frac{46}{100}$, and solving that for the part multiplies $550$ by $\frac{46}{100} = 0.46$. The decimal method is that proportion with the arithmetic already done.

Check the size with friendly pieces. Ten percent of $550$ is $55$, so $40\%$ is $220$; one percent is $5.5$, so $5\%$ is $27.5$; and

$220 + 27.5 + 5.5 = 253,$

which matches. The answer is also believable at a glance, since $46\%$ is a little under half and $253$ is a little under half of $550$.

The percent is the same, so the decimal is the same, and only the whole is different:

$0.46 \times 150 = 0.46 \times 100 + 0.46 \times 50 = 46 + 23 = 69.$

The hillside garden has $69$ vegetable plots.

The percent proportion gives the same figure, which is worth running once to see the two methods meet:

$\frac{\text{part}}{150} = \frac{46}{100} \quad \Longrightarrow \quad \text{part} = \frac{150 \times 46}{100} = \frac{6900}{100} = 69.$

Check with friendly pieces as before: $10\%$ of $150$ is $15$, so $40\%$ is $60$; $1\%$ is $1.5$, so $5\%$ is $7.5$; and $60 + 7.5 + 1.5 = 69$. Once again the count is a little under half the plots.

A percent is a ratio whose second term is fixed at $100$, so what $46\%$ settles is a rate of planting: $46$ vegetable plots for every $100$ plots the garden has. It says nothing whatever about how many hundreds of plots there are, and that is the missing half of the volunteer's claim.

Read the two gardens that way and the arithmetic almost disappears. The riverside garden has five and a half hundreds of plots, since $550 = 5.5 \times 100$, so it gets

$5.5 \times 46 = 253$

vegetable plots. The hillside garden has one and a half hundreds, since $150 = 1.5 \times 100$, so it gets

$1.5 \times 46 = 69.$

One percent, two amounts, and they differ by well over three times. So a percent on its own is an instruction waiting for a whole: give it a whole and it names an amount, withhold the whole and it names nothing.

This is also why two percents may only be added when they belong to the same whole. Adding $46\%$ of the riverside garden to $46\%$ of the hillside garden gives $322$ plots, which is not $46\%$ of anything in the question unless you first say that the two gardens together hold $700$ plots.

Every percent question is the same relationship with a different number missing:

$\frac{\text{part}}{\text{whole}} = \frac{\text{percent}}{100}.$

So the setup is a matter of sorting the given numbers into their places. The graphic novels are a piece of the collection, so $138$ is the part. The collection is what the graphic novels are a percent of, and the number after 'of' is the whole, so the whole is the size of the collection, which the note does not give. That leaves $23$ as the percent.

Name the unknown and write it where it belongs. With $b$ for the number of books in the collection,

$\frac{138}{b} = \frac{23}{100}.$

The unknown sits in a denominator this time, which is the only thing that makes this task look different from the others. It changes nothing about the method, because cross-multiplication does not care where the blank is.

Cross-multiply the diagonal you know completely, $138$ and $100$, then divide by the number sitting diagonally opposite the blank, which is $23$:

$b = \frac{138 \times 100}{23} = \frac{13800}{23} = 600.$

The division is quick to confirm the other way round, since $23 \times 6 = 138$ and therefore $23 \times 600 = 13800$. The collection holds $600$ books.

Now run it forward, which is the check this task deserves because the unknown was underneath and a slip there is easy to miss:

$23\% \text{ of } 600 = 0.23 \times 600 = 138,$

the very count the catalogue note gives. A size check agrees too: $23\%$ is a little under a quarter, so the collection should be a little more than four times the number of graphic novels, and $4 \times 138 = 552$ is indeed a little less than $600$.

First see what the computation is:

$138 \times 0.23 = 31.74.$

That is a perfectly good piece of arithmetic, and it answers the first of the three percent tasks: it finds $23\%$ of $138$. To do so it must treat $138$ as the whole, since the whole is the number a percent is taken of.

But the note says the opposite. The graphic novels are $23\%$ of the collection, so the collection is the whole and $138$ is the part. The student has swapped the two, which turns a question about recovering a whole into a question about sharing a part out.

The size check that kills it costs nothing and needs no arithmetic. A part that is only $23\%$ of its whole is under a quarter of it, so the whole must be more than four times as large as the part:

$4 \times 138 = 552,$

and a correct answer has to be larger even than that. The reported $31.74$ is smaller than the part itself, which is impossible for a whole that contains it. There is a general tell behind that: multiplying a positive number by a decimal between $0$ and $1$ makes it smaller, while recovering a whole from a part has to make it larger, so the operations point in opposite directions. When the unknown is the whole, the arithmetic that finishes the job is a division, not a multiplication.

One more oddity worth naming: a count of books cannot be $31.74$ in the first place. An answer that is not a whole number of books is a signal to reread the setup before trusting the arithmetic.

The compact spaces are the part and the garage's spaces are the whole, so the ratio to rescale is $\frac{51}{300}$. A percent is that same ratio written with $100$ underneath, which makes the percent an unknown numerator:

$\frac{51}{300} = \frac{p}{100}.$

Multiply the diagonal you know, then divide by the number diagonally opposite the blank:

$p = \frac{51 \times 100}{300} = \frac{5100}{300} = 17.$

So $17\%$ of the spaces are compact. Dividing gives the same thing, since $51 \div 300 = 0.17$ and moving that point two places right is $17\%$.

There is a reading of this that needs no arithmetic at all. A percent is a count for every hundred, and $300$ spaces are exactly three hundreds, so spreading $51$ compact spaces evenly over three hundreds puts $51 \div 3 = 17$ in each hundred.

Same setup, different numbers. The part is $33$ and the whole is $200$:

$\frac{33}{200} = \frac{p}{100} \quad \Longrightarrow \quad p = \frac{33 \times 100}{200} = \frac{3300}{200} = 16.5.$

So $16.5\%$ of this garage's spaces are compact. Division agrees: $33 \div 200 = 0.165$, and two places to the right is $16.5\%$.

The counting reading works here too and explains the half. This garage is two hundreds of spaces, so the $33$ compact spaces spread over two hundreds give $33 \div 2 = 16.5$ in each hundred. That is sixteen and a half compact spaces for every hundred spaces, which is not a claim that half a space exists; it is a rate, in the same way that a ratio may be $3 : 2$ without anything being cut in half.

Rounding would be a mistake here rather than a tidy-up, because the two garages' shares are close, and rounding $16.5$ to $17$ would destroy exactly the difference the comparison rests on.

Here $51 : 300$ and $33 : 200$ differ in both terms at once, so a bigger first term proves nothing on its own: it might mean a bigger piece, or it might mean a bigger garage, and the counts cannot tell you which. Rewriting both over the same second term settles that, because it removes one of the two differences and leaves exactly one number to compare:

$\frac{51}{300} = \frac{17}{100}, \qquad \frac{33}{200} = \frac{16.5}{100}.$

Now the comparison is between $17$ and $16.5$ compact spaces per hundred, and the first garage has the larger share, narrowly. That is the whole purpose of fixing every ratio's second term at $100$.

The raw counts $51$ and $33$ are not useless; they answer a different question, namely which garage offers more compact spaces in total. That happens to be the first garage as well, but the agreement is a fact about these particular numbers rather than a reason. Suppose instead that the second garage marked $37$ of its $200$ spaces for compact cars, and the two questions part company:

$\frac{37}{200} = \frac{18.5}{100},$

so the first garage would still have more compact spaces, $51$ against $37$, while giving them the smaller share, $17\%$ against $18.5\%$. A driver hunting for a compact space near this address wants the counts; a planner asking how each garage allots its space wants the percents.

Finally, the condition. Whenever the two garages have the same number of spaces, the counts alone do settle the share, because both ratios already have a common second term and rescaling them to $100$ multiplies both first terms by the same factor, which cannot change their order. Unequal totals may still happen to rank the same way, as they do with the figures the question gives, but that is luck rather than a reason to trust the counts.

The occupied boxes are a piece of the reserve's boxes, so $273$ is the part and $420$, the number the occupied ones are counted out of, is the whole. Put them in the percent proportion with the percent unknown:

$\frac{273}{420} = \frac{p}{100}.$

Multiply the diagonal you know completely, then divide by the number diagonally opposite the blank:

$p = \frac{273 \times 100}{420} = \frac{27300}{420} = 65.$

The division is easy to confirm backwards, since $420 \times 65 = 420 \times 60 + 420 \times 5 = 25200 + 2100 = 27300$. So $65\%$ of the boxes were occupied, and dividing agrees, since $273 \div 420 = 0.65$.

Check the size against a landmark. Two thirds of $420$ is $280$, and $273$ is a little short of that, so the percent should be a little under the roughly $66.7\%$ that two thirds represents. It is.

Start from what $100\%$ means. Since $\frac{100}{100} = 1$, $100\%$ is the whole itself, so a percent above $100\%$ says the part is larger than the whole it is measured against. Occupied boxes are boxes, and the reserve has only $420$ of those, so the occupancy cannot pass $100\%$ however the survey turned out. The figure is therefore not the occupancy, and no arithmetic was needed to know it.

The number is not nonsense, though. It is the answer to the question with the two counts exchanged, which is a legitimate question about a different pair of roles:

$\frac{420}{273} = \frac{p}{100} \quad \Longrightarrow \quad p = \frac{42000}{273} \approx 153.8.$

Read that as: the reserve's total number of boxes is about $153.8\%$ of the number occupied, which is another way of saying the total is a little over one and a half times the occupied count. Here a percent above $100\%$ is entirely proper, because the number playing the part, $420$, really is larger than the number playing the whole, $273$.

So the classmate's arithmetic is sound and their setup is not: they made $273$ the whole. The wording settles which number that should be, since 'what percent of the nesting boxes' names the boxes as the whole, and the whole is the number that goes underneath.

Reading the wording. A percent compares a part with the whole it is a part of, and English marks the whole with the word 'of': in 'what percent of the nesting boxes were occupied', the boxes are what the occupied ones are a percent of. The whole is the second term of the ratio, so it goes underneath. That single reading is what decides the setup, and it is the only decision in the problem that arithmetic cannot repair.

One relationship, three tasks. Write the relationship with all three numbers in place,

$\frac{273}{420} = \frac{65}{100},$

and then cover one of them. Each thing you can cover is one of the three percent tasks.

Cover the part, and the question becomes 'how many of the $420$ boxes are occupied if $65\%$ are', the first task:

$\frac{\text{part}}{420} = \frac{65}{100} \quad \Longrightarrow \quad \text{part} = 0.65 \times 420 = 273.$

Cover the whole, and the question becomes 'if $273$ occupied boxes are $65\%$ of a reserve's boxes, how many boxes has it', the third task:

$\frac{273}{w} = \frac{65}{100} \quad \Longrightarrow \quad w = \frac{273 \times 100}{65} = \frac{27300}{65} = 420.$

Cover the percent and you are back to part A. Three questions that sound quite different are one equation with a different blank, which is why one method serves them all: cross-multiply the diagonal you know completely and divide by the number opposite the blank, wherever the blank happens to sit.

That also explains why the tasks are so easy to confuse and so easy to check. Confuse them, and you have solved the right equation for the wrong letter; check them, and you can always run the relationship forward on your answer and see the given numbers come back, as $0.65 \times 420 = 273$ does here.