Conversion Factors: Free Response

The starting quantity carries gal on top, so the factor needs gal on the bottom to cancel it, which fixes the orientation as $\dfrac{4 \text{ qt}}{1 \text{ gal}}$, not its flip.

$9 \text{ gal} \times \frac{4 \text{ qt}}{1 \text{ gal}} = 9 \times 4 \text{ qt} = 36 \text{ qt}.$

The gal on top of the quantity and the gal on the bottom of the factor form $\frac{\text{gal}}{\text{gal}} = 1$ and disappear, leaving qt.

This time the starting quantity carries qt on top, so the factor needs qt on the bottom, the OTHER orientation from part A: $\dfrac{1 \text{ gal}}{4 \text{ qt}}$.

$30 \text{ qt} \times \frac{1 \text{ gal}}{4 \text{ qt}} = \frac{30}{4} \text{ gal} = 7.5 \text{ gal}.$

A half gallon is no obstacle: gallons and quarts are measured, not counted, so a fractional amount is ordinary.

Work in letters, not the numbers $4$ and $1$ from parts A and B, since the claim covers every pair of units.

Start from the given equality and divide both sides by $a\,U_1$ (which is not zero):

$a\,U_1 = b\,U_2 \quad\Longrightarrow\quad \frac{a\,U_1}{a\,U_1} = \frac{b\,U_2}{a\,U_1} \quad\Longrightarrow\quad 1 = \frac{b\,U_2}{a\,U_1}.$

Nothing about $a$, $b$, $U_1$, or $U_2$ was special: the same division works for any true equality between two units of the same kind of quantity, so the fraction it produces is always exactly $1$, never merely close to it.

Now take any quantity $c\,U_1$ and multiply by that fraction:

$c\,U_1 \times \frac{b\,U_2}{a\,U_1} = \frac{cb}{a}\left(\frac{U_1}{U_1}\right)U_2 = \frac{cb}{a}\,U_2,$

using $\frac{U_1}{U_1} = 1$, the same cancellation rule that removes a number from a fraction's top and bottom. The multiplicative identity says multiplying by $1$ changes nothing, so the AMOUNT $c\,U_1$ and the amount $\frac{cb}{a}\,U_2$ are the same amount, only carrying different labels; the number in front changed from $c$ to $\frac{cb}{a}$ precisely because the unit changed size, not because anything was added to or taken from the quantity itself.

The starting quantity carries km on top, so the first factor needs km on the bottom: $\dfrac{1000 \text{ m}}{1 \text{ km}}$. That factor now leaves m on top, so the second factor needs m on the bottom: $\dfrac{100 \text{ cm}}{1 \text{ m}}$.

$4.5 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{100 \text{ cm}}{1 \text{ m}}.$

Read the units down the line: km cancels the starting km, and m cancels between the first and second factors, leaving cm.

$4.5 \times 1000 \times 100 = 450{,}000.$

All three factors are numbers now; the units have already done their job of telling you which ones to multiply.

Convert directly with the single fact:

$4.5 \text{ km} \times \frac{100{,}000 \text{ cm}}{1 \text{ km}} = 450{,}000 \text{ cm},$

the same number part B reached through m.

This is not a coincidence to be checked case by case. The two-step chain's combined factor is $1000 \times 100$, and

$1000 \times 100 = 100{,}000,$

which is exactly the single fact's own number. The one-step fact and the two-step chain describe the identical equality between km and cm; the chain merely names m as a stopping point along the way. Since every factor in either route is exactly $1$ (part C of question 1 is the reason why), stringing more of them together, or collapsing them into one, can never change the amount, only how many labels get exchanged before cm is reached.

Square meters is meters times meters, $1 \text{ m}^2 = 1 \text{ m} \times 1 \text{ m}$, so the length factor has to be applied twice, which squares it:

$4.5 \text{ m}^2 \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^2 = 4.5 \text{ m}^2 \times \frac{10{,}000 \text{ cm}^2}{1 \text{ m}^2} = 4.5 \times 10{,}000 = 45{,}000 \text{ cm}^2.$

Cubic meters hides three lengths, $1 \text{ m}^3 = 1 \text{ m} \times 1 \text{ m} \times 1 \text{ m}$, so the factor is applied three times, which cubes it:

$6 \text{ m}^3 \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^3 = 6 \text{ m}^3 \times \frac{1{,}000{,}000 \text{ cm}^3}{1 \text{ m}^3} = 6 \times 1{,}000{,}000 = 6{,}000{,}000 \text{ cm}^3.$

Test the claim against the definition of the unit itself, not against a guessed pattern in the numbers. Since $1 \text{ m}^2 = 1 \text{ m} \times 1 \text{ m}$, substitute $1 \text{ m} = 100 \text{ cm}$ into EACH of the two factors:

$1 \text{ m}^2 = (1 \text{ m}) \times (1 \text{ m}) = (100 \text{ cm}) \times (100 \text{ cm}) = 10{,}000 \text{ cm}^2.$

That is $100^2$, not $2 \times 100$, so doubling the power does not double the number; it SQUARES it. Doubling suggests the two lengths hiding inside the unit add together, but they multiply, because the unit is a product, not a sum.

The same substitution generalizes to any power: $1 \text{ m}^n = (1 \text{ m})^n = (100 \text{ cm})^n = 100^n \text{ cm}^n$. A unit raised to the $n$th power hides $n$ lengths multiplied together, so its numeric conversion factor must be raised to that same $n$th power, one factor of $100$ for every hidden length.

km sits on top of the rate, so its factor needs km on the bottom: $\dfrac{1000 \text{ m}}{1 \text{ km}}$. h sits on the BOTTOM of the rate, so its factor needs h on TOP, the opposite placement: $\dfrac{1 \text{ h}}{3600 \text{ s}}$.

$\frac{108 \text{ km}}{1 \text{ h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}.$

The km on top cancels the km in the second factor's denominator, and the h in the first denominator cancels the h in the third numerator, leaving m over s.

$\frac{108 \times 1000}{3600} = \frac{108{,}000}{3600} = 30.$

A unit cancels only against the same unit sitting on the opposite side of a fraction bar. Kilometer begins on TOP of the rate $\dfrac{108 \text{ km}}{1 \text{ h}}$, so its factor needs kilometer on the BOTTOM to meet it there and form $\dfrac{\text{km}}{\text{km}} = 1$.

Hour begins on the BOTTOM of that same rate, so by the identical logic its factor needs hour on the TOP, to meet the hour already in the denominator and cancel there instead:

$\frac{108 \text{ km}}{1 \text{ h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}: \qquad \frac{\text{km}}{\text{km}} = 1, \qquad \frac{\text{h}}{\text{h}} = 1.$

Nothing about the numbers $1000$ or $3600$ decided this: the position alone did. The rule generalizes to any rate at all: whichever unit starts on top gets its factor's matching unit on the bottom, and whichever unit starts on the bottom gets its factor's matching unit on top, because cancellation only ever happens across the fraction bar, never on the same side of it.

ft$^2$ hides two lengths, so its factor is squared: $\left(\dfrac{12 \text{ in}}{1 \text{ ft}}\right)^2$, oriented with ft on the bottom to cancel the ft$^2$ on top. h sits on the bottom of the rate, so its factor is oriented with h on top, to the FIRST power only: $\dfrac{1 \text{ h}}{60 \text{ min}}$.

$\frac{45 \text{ ft}^2}{1 \text{ h}} \times \left(\frac{12 \text{ in}}{1 \text{ ft}}\right)^{2} \times \frac{1 \text{ h}}{60 \text{ min}}.$

$12^2 = 144, \qquad 45 \times 144 = 6480, \qquad \frac{6480}{60} = 108.$

The power on a conversion factor comes from the EXPONENT its unit carries. Where a unit sits in the product decides that factor's orientation, which is question 4's business; the exponent, and nothing else, decides its power.

A square foot is, by definition, a product of two equal lengths: $1 \text{ ft}^2 = 1 \text{ ft} \times 1 \text{ ft}$. Converting each of those two hidden feet to inches is the same substitution done twice, which is exactly what squaring the factor means:

$1 \text{ ft}^2 = (1 \text{ ft}) \times (1 \text{ ft}) = (12 \text{ in}) \times (12 \text{ in}) = 144 \text{ in}^2.$

In this rate h carries the exponent $1$: the coverage is $\text{ft}^2$ per ONE hour, so there is a single $\text{h}$ to replace with $\text{min}$ and its factor is applied exactly once.

That is a fact about the exponent, not about time. Give a rate a time unit raised to a power and its time factor is raised to that same power. An acceleration of $1 \text{ ft}/\text{s}^2$ converted to $\text{in}/\text{min}^2$ carries $\text{s}^2$, so its time factor is squared exactly as the area factor was here:

$\frac{1 \text{ ft}}{1 \text{ s}^2} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \left(\frac{60 \text{ s}}{1 \text{ min}}\right)^{2} = 12 \times 3600 = 43{,}200 \frac{\text{in}}{\text{min}^2}.$

So the criterion is the one question 3 already reached, and it reads the same for every unit: a unit raised to the $n$th power needs its conversion factor raised to that same $n$th power. Here ft$^2$ carries $n = 2$ and h carries $n = 1$, and that is the whole of the difference. The fact that one of them measures area and the other measures time decides nothing.