QuanticaAlgebra IOpen in the courseA conversion factor is a ratio of one amount to itself written in two units, so it is worth \(1\). One gallon and four quarts are the same volume, so \(\frac{4\text{ qt}}{1\text{ gal}}\) has equal volumes above and below the bar. Multiplying by it is multiplying by \(1\), which changes the unit and not the amount. The first two problems give the fraction.
Problem
A fuel tank holds 46 gallons, and \(1\text{ gal}=4\text{ qt}\). Multiply \(46\text{ gal}\) by \(\frac{4\text{ qt}}{1\text{ gal}}\), writing the unit inside the fraction so that gal above the bar cancels against gal below it. Enter the tank's capacity in quarts.
Show a hint
- Write \(46\text{ gal}\) as \(\frac{46\text{ gal}}{1}\), so gal sits above the bar in the amount and below the bar in the factor.
- Once gal cancels, what is left is \(46\times 4\) with qt as the only unit remaining.
Show the full solution
Cancelling gal above against gal below leaves \(46\times 4\) quarts, so the capacity is \(\boxed{184}\). A quart is smaller than a gallon, so more of them cover the same volume, and \(184\) is larger than \(46\) as it has to be.
Problem
A second tank on the same lot holds 268 quarts. The same fact \(1\text{ gal}=4\text{ qt}\) also gives the factor \(\frac{1\text{ gal}}{4\text{ qt}}\), which has qt below the bar. Multiply by that factor and enter this tank's capacity in gallons.
Show a hint
- Write the tank as \(\frac{268\text{ qt}}{1}\), so qt is above the bar. It cancels only against a factor with qt below the bar.
- After qt cancels, the arithmetic left is \(268\div 4\) and gal is the only unit remaining.
Show the full solution
Cancelling qt above against qt below leaves \(268\div 4\) gallons, so the capacity is \(\boxed{67}\). The two factors are one fact written two ways, and which one to use depends on which unit has to cancel.
Problem
A pallet of bolts weighs 592 ounces, and \(1\text{ lb}=16\text{ oz}\). The two factors available are \(\frac{16\text{ oz}}{1\text{ lb}}\) and \(\frac{1\text{ lb}}{16\text{ oz}}\), and only one of them cancels the ounces. Enter the weight in pounds.
Show a hint
- Write the weight as \(\frac{592\text{ oz}}{1}\), so oz is above the bar. Only a factor with oz below the bar cancels it.
- A pound is heavier than an ounce, so the number of pounds is smaller than \(592\). Multiplying by \(16\) cannot be right.
Show the full solution
The factor with oz below the bar is \(\frac{1\text{ lb}}{16\text{ oz}}\), and cancelling oz leaves \(592\div 16\), so the weight in pounds is \(\boxed{37}\). The other choice gives \(9472\) and leaves the label \(\frac{\text{oz}^2}{\text{lb}}\), which is not a weight.
Problem
A recording runs 1020 seconds, and \(1\text{ min}=60\text{ s}\). A student multiplies \(1020\text{ s}\) by \(\frac{60\text{ s}}{1\text{ min}}\) and reports that the recording is 61200 minutes long. Enter the correct length in minutes.
Show a hint
- Check the units in the student's product. Multiplying \(1020\text{ s}\) by a factor with s on top gives s times s, so nothing cancels and the label left over is \(\frac{\text{s}^2}{\text{min}}\).
- The s in \(1020\text{ s}\) cancels only against an s on the bottom. The fact \(1\text{ min}=60\text{ s}\) also gives the factor \(\frac{1\text{ min}}{60\text{ s}}\), which has s there.
Show the full solution
Multiplying by \(\frac{1\text{ min}}{60\text{ s}}\) cancels s and leaves \(1020\div 60\), so the length in minutes is \(\boxed{17}\). The student used the reciprocal factor, so the result is \(3600\) times too large, and the leftover label \(\frac{\text{s}^2}{\text{min}}\) is not a time.
Not every pair of units has a fact stated directly between them. Sometimes one fact runs from the starting unit to a middle unit, and a second fact runs from that middle unit to the one wanted. The next two problems take that route, the first through one middle unit and the second through two.
Problem
A gym orders 8 coils of climbing rope, with \(1\text{ coil}=30\text{ m}\) and \(1\text{ m}=100\text{ cm}\). Neither fact is a direct coil-to-centimeter conversion. Multiply the 8 coils by two conversion factors in a row and enter the total length in centimeters.
Show a hint
- The unit meters is common to both given facts, so the chain is coils to meters to centimeters, one factor for each step.
- The chain is \(8\text{ coils}\times\frac{30\text{ m}}{1\text{ coil}}\times\frac{100\text{ cm}}{1\text{ m}}\), with coil below the bar in the first factor and m below the bar in the second.
Show the full solution
Cancelling coils gives \(8\times 30=240\) meters, and cancelling meters gives \(240\times 100\), so the length in centimeters is \(\boxed{24000}\). Both factors equal \(1\), so the two together equal \(1\) as well, and the rope has the same length as before, only measured in a smaller unit.
Problem
A recipe is written in cups and the jug in the kitchen is marked in milliliters. Use \(1\text{ cup}=16\text{ tablespoons}\), \(1\text{ tablespoon}=3\text{ teaspoons}\) and \(1\text{ teaspoon}=5\text{ mL}\). Enter the volume of 6 cups in milliliters.
Show a hint
- Three stated facts give three factors, run in the order cups to tablespoons to teaspoons to milliliters, and each factor carries the unit you already have below the bar so that unit cancels.
- Tablespoons and teaspoons are both intermediate units and both cancel, and what is left is \(6\times 16\times 3\times 5\).
Show the full solution
Cancelling cups, then tablespoons, then teaspoons leaves \(6\times 16\times 3\times 5\), so the volume in milliliters is \(\boxed{1440}\). Multiplying the three factors together first gives \(16\times 3\times 5=240\), so the whole chain is the single factor \(\frac{240\text{ mL}}{1\text{ cup}}\), and any number of cups then takes one step.
A rate is written with one unit above the bar and a different one below, so it carries two units at once. The next problems change both of them. Speeds, flow rates and rates of pay all have this shape, and a pace written as time over distance is the same information as a speed written as distance over time.
Problem
A conveyor belt runs at 48 meters per minute. Use \(1\text{ m}=100\text{ cm}\) and \(1\text{ min}=60\text{ s}\). Both units change, so two factors are needed. Enter the speed in centimeters per second.
Show a hint
- Write the speed as \(\frac{48\text{ m}}{1\text{ min}}\). The m to be removed sits above the bar and the min to be removed sits below it, so the two factors go in the opposite way up from each other.
- The chain is \(\frac{48\text{ m}}{1\text{ min}}\times\frac{100\text{ cm}}{1\text{ m}}\times\frac{1\text{ min}}{60\text{ s}}\).
Show the full solution
Cancelling m above against m below and min below against min above leaves \(\frac{48\times 100}{60}\), so the speed in centimeters per second is \(\boxed{80}\). The minutes factor goes in as \(\frac{1\text{ min}}{60\text{ s}}\) and not the other way up, since the min to be removed starts below the bar and cancels against a copy above it.
Problem
A runner's watch reports a pace of 3 minutes 45 seconds per kilometer, which is \(\frac{225\text{ s}}{1\text{ km}}\). Use \(1\text{ h}=3600\text{ s}\). Enter the speed in kilometers per hour.
Show a hint
- The unit asked for has km above the bar and h below it, so the pace's s has to cancel and an h has to arrive, and the one fact available for both is \(1\text{ h}=3600\text{ s}\).
- Turned over the pace is \(\frac{1\text{ km}}{225\text{ s}}\), and multiplying that by \(\frac{3600\text{ s}}{1\text{ h}}\) cancels s and leaves km above the bar with h below it.
Show the full solution
Turning the pace over gives \(\frac{1\text{ km}}{225\text{ s}}\), and multiplying by \(\frac{3600\text{ s}}{1\text{ h}}\) cancels s and leaves \(3600\div 225\), so the speed in kilometers per hour is \(\boxed{16}\). Turning a rate over states the same information with the two quantities exchanged, and \(16\times 225=3600\) checks it, since sixteen kilometers at \(225\) seconds each take exactly one hour.
An area is a length times a length, so converting an area means converting both lengths, and the same factor is used twice. Write \(1\text{ yd}^2\) as \(1\text{ yd}\times 1\text{ yd}\) and replace each yard on its own. A volume is three lengths multiplied, so the same factor is used three times.
Problem
A patio covers 34 square yards, and \(1\text{ yd}=3\text{ ft}\). One square yard is \(1\text{ yd}\times 1\text{ yd}\), so convert each of the two lengths and see how many square feet one square yard holds. Enter the patio's area in square feet.
Show a hint
- A square yard is a square measuring 3 ft along each side, so count the square feet inside one of them first.
- One square yard is \(3\times 3=9\) square feet, so the factor is \(\frac{9\text{ ft}^2}{1\text{ yd}^2}\).
Show the full solution
One square yard is \(3\text{ ft}\times 3\text{ ft}=9\text{ ft}^2\), so the factor is \(\frac{9\text{ ft}^2}{1\text{ yd}^2}\) and the area is \(34\times 9=\boxed{306}\). Using the length factor once gives \(102\) and leaves the label \(\text{yd}\cdot\text{ft}\), since only one of the two lengths was converted.
Problem
A shipping crate has a volume of 5 cubic feet, and \(1\text{ ft}=12\text{ in}\). Enter the crate's volume in cubic inches.
Show a hint
- One cubic foot is a cube whose edges each measure \(12\) inches, so converting the edge length alone is not enough.
- One cubic foot is \(12\times 12\times 12=1728\) cubic inches.
Show the full solution
One cubic foot is a cube with \(12\)-inch edges, so it is \(12\times 12\times 12=1728\) cubic inches, and the volume is \(5\times 1728=\boxed{8640}\). Using the conversion factor of \(12\) once gives \(60\) and twice gives \(720\), but a volume is three lengths multiplied together, so the factor belongs in the product three times.
A conversion factor can be built from any stated equality between two amounts, and the two amounts need not be the same kind of thing. A price, a rate of pay, and a count per container are all equalities of that kind, so each one has two factors, and the choice between them comes down to the unit that has to cancel.
Problem
A field technician is paid 35 dollars per hour and works 6 hours per shift. Each of those statements gives a conversion factor, one between hours and dollars and one between shifts and hours. Enter the pay in dollars for 8 shifts.
Show a hint
- Start from 8 shifts and pick the factor that cancels shifts, which leaves hours.
- The chain is \(8\text{ shifts}\times\frac{6\text{ hours}}{1\text{ shift}}\times\frac{35\text{ dollars}}{1\text{ hour}}\).
Show the full solution
Cancelling shifts and then hours leaves \(8\times 6\times 35\), so the pay in dollars is \(\boxed{1680}\). Shifts and hours are both units of time, while dollars and hours are not, and nothing about the method changes, since 35 dollars per hour states that one hour of work goes with 35 dollars the way one foot goes with 12 inches.
Problem
A trading post prices goods in four measures, with \(1\text{ varn}=7\text{ sedge}\), \(1\text{ sedge}=3\text{ pell}\) and \(5\text{ pell}=2\text{ quill}\). The last fact has no \(1\) on either side. A merchant holds 15 varn. Enter the value in quill.
Show a hint
- Three facts and three factors. Take varn to sedge, sedge to pell, then pell to quill, each one written with the unit you are getting rid of below the bar.
- \(15\times 7=105\) sedge, then \(105\times 3=315\) pell. Only one orientation of the last factor cancels pell, so use that one.
Show the full solution
Chaining \(15\text{ varn}\times\frac{7\text{ sedge}}{1\text{ varn}}\times\frac{3\text{ pell}}{1\text{ sedge}}\times\frac{2\text{ quill}}{5\text{ pell}}\) cancels varn, sedge and pell in turn and leaves \(\frac{15\times 7\times 3\times 2}{5}\), so the value in quill is \(\boxed{126}\). Using the last factor upside down gives \(315\times\frac{5}{2}=787.5\), with pell uncancelled and no quill in the answer.
Every conversion here was a multiplication by a fraction worth \(1\), chosen so the unwanted unit appeared above the bar and below it, and in the area and volume cases the same factor was used twice or three times. Lesson 6.3, Percents, uses a ratio whose second part is always \(100\). The question there is what an amount is being compared against.
Practice these ideas
Practice
A sack of rice has a mass of 9000 grams, and \(1\text{ kg}=1000\text{ g}\). Enter the mass in kilograms.
Show the solution
Multiplying by \(\frac{1\text{ kg}}{1000\text{ g}}\) cancels the grams and gives \(\frac{9000}{1000}\), so the mass in kilograms is \(\boxed{9}\). The other fraction, \(\frac{1000\text{ g}}{1\text{ kg}}\), also equals \(1\), but it gives \(9000000\) with units of \(\frac{\text{g}^2}{\text{kg}}\), which is not a mass.
Practice
A spool carries 4500 centimeters of thread, and \(1\text{ m}=100\text{ cm}\). Enter the length of the thread in meters.
Show the solution
Cancelling cm leaves \(4500\div 100\), so the length in meters is \(\boxed{45}\). A meter is longer than a centimeter, so the count of meters is the smaller number.
Practice
A bench is 156 inches long, and \(1\text{ ft}=12\text{ in}\). Multiplying by \(\frac{12\text{ in}}{1\text{ ft}}\) gives \(1872\), a number far larger than the length in inches, and leaves the label \(\frac{\text{in}^2}{\text{ft}}\). Pick the factor that cancels inches instead and enter the length in feet.
Show the solution
Multiplying by \(\frac{1\text{ ft}}{12\text{ in}}\) cancels the inches and leaves \(156\div 12\), so the length in feet is \(\boxed{13}\). The student's factor has inches above the bar as well, so nothing cancels and the label on \(1872\) is \(\frac{\text{in}^2}{\text{ft}}\), not feet.
Practice
A bakery counts its rolls in dozens and has 51 dozen, with \(1\text{ dozen}=12\text{ rolls}\). Multiply by a conversion factor that cancels dozen, and enter the number of rolls.
Show the solution
The factor \(\frac{12\text{ rolls}}{1\text{ dozen}}\) cancels dozen, so the number of rolls is \(51\times 12=\boxed{612}\). The flipped factor \(\frac{1\text{ dozen}}{12\text{ rolls}}\) would leave \(51\div 12\), which is why dozen belongs in the denominator.
Practice
A stationery order is 12 boxes, where \(1\text{ box}=20\text{ packs}\) and \(1\text{ pack}=50\text{ envelopes}\). Enter the number of envelopes in the order.
Show the solution
Writing \(12\text{ boxes}\times\frac{20\text{ packs}}{1\text{ box}}\times\frac{50\text{ envelopes}}{1\text{ pack}}\) and cancelling boxes and then packs leaves \(12\times 20\times 50\), so the number of envelopes is \(\boxed{12000}\). Each box holds \(20\times 50=1000\) envelopes, so the order is \(12\times 1000\), a quick check on the size.
Practice
A warehouse holds 8400 bottles, packed at \(1\text{ case}=12\text{ bottles}\) and \(1\text{ pallet}=25\text{ cases}\). Enter the number of full pallets.
Show the solution
Cancelling bottles and then cases leaves \(\frac{8400}{12\times 25}=\frac{8400}{300}\), so the number of full pallets is \(\boxed{28}\). Both factors go in with the old unit underneath, \(\frac{1\text{ case}}{12\text{ bottles}}\) and then \(\frac{1\text{ pallet}}{25\text{ cases}}\), which is the same order the two facts are written in.
Practice
A delivery is 9 crates, with \(1\text{ crate}=6\text{ trays}\), \(1\text{ tray}=8\text{ punnets}\) and \(1\text{ punnet}=15\text{ berries}\). Enter the number of berries.
Show the solution
Cancelling crates, trays and punnets in turn leaves \(9\times 6\times 8\times 15\), so the number of berries is \(\boxed{6480}\). Each of the three factors is worth \(1\), so the delivery is the same size either way and only the thing being counted is different.
Practice
A tap fills a tank at 12 liters per minute, and \(1\text{ h}=60\text{ min}\). Enter the fill rate in liters per hour.
Show the solution
Multiplying by \(\frac{60\text{ min}}{1\text{ h}}\) cancels the min and gives \(12\times 60\), so the rate in liters per hour is \(\boxed{720}\). Dividing by \(60\) instead gives \(0.2\), which is the wrong direction, since more liters come out in an hour than in a minute.
Practice
A snail moves 3 centimeters per second. Use \(1\text{ m}=100\text{ cm}\) and \(1\text{ h}=3600\text{ s}\). Enter the speed in meters per hour.
Show the solution
\(\frac{3\text{ cm}}{1\text{ s}}\times\frac{1\text{ m}}{100\text{ cm}}\times\frac{3600\text{ s}}{1\text{ h}}\) cancels cm and s and leaves \(\frac{3\times 3600}{100}\), so the speed in meters per hour is \(\boxed{108}\). Each conversion factor is a fraction equal to \(1\), so multiplying by it changes the units and not the speed.
Practice
A conveyor carries 900 kilograms per hour. Use \(1\text{ kg}=1000\text{ g}\) and \(1\text{ h}=60\text{ min}\). Enter the rate in grams per minute.
Show the solution
Cancelling kg and h leaves \(\frac{900\times 1000}{60}\), so the rate in grams per minute is \(\boxed{15000}\). The hours factor goes in as \(\frac{1\text{ h}}{60\text{ min}}\) rather than \(\frac{60\text{ min}}{1\text{ h}}\), since the h below the bar cancels only against an h above it.
Practice
A courtyard covers 7 square meters, and \(1\text{ m}=100\text{ cm}\). Enter the area in square centimeters.
Show the solution
One square meter is \(100\text{ cm}\times 100\text{ cm}=10000\text{ cm}^2\), so the area is \(7\times 10000=\boxed{70000}\). Converting only one of the two lengths gives \(700\), which is too small by a factor of \(100\).
Practice
A poster covers 4320 square inches. Use \(1\text{ ft}=12\text{ in}\). Enter the area in square feet.
Show the solution
One square foot is \(144\text{ in}^2\), so the area is \(4320\div 144=\boxed{30}\). Dividing by \(12\) once gives \(360\), which is not an area in square feet.
Practice
A planter holds 2 cubic yards of soil, and \(1\text{ yd}=3\text{ ft}\). Enter the volume of that soil in cubic feet.
Show the solution
One cubic yard is \(3\text{ ft}\times 3\text{ ft}\times 3\text{ ft}=27\text{ ft}^3\), so the volume is \(2\times 27=\boxed{54}\). The factor between cubic yards and cubic feet is \(3^3=27\), not \(3\), since a volume is a product of three lengths.
Practice
Paint costs 38 dollars per can, and one can covers 25 square meters of wall. A job needs 300 square meters of wall painted. Enter the cost of the paint for that job, in dollars.
Show the solution
Cancelling square meters and then cans in \(300\text{ m}^2\times\frac{1\text{ can}}{25\text{ m}^2}\times\frac{38\text{ dollars}}{1\text{ can}}\) leaves \(12\times 38\), so the cost in dollars is \(\boxed{456}\). The unit cans sits above the bar in one factor and below it in the next, so it divides out and dollars is the only unit left.
Practice
A carpenter is paid 26 dollars per hour and works 4 hours per session, and has earned 1560 dollars from those sessions alone. Enter the number of sessions worked.
Show the solution
\(1560\text{ dollars}\times\frac{1\text{ h}}{26\text{ dollars}}\times\frac{1\text{ session}}{4\text{ h}}\) cancels dollars and hours and leaves \(\frac{1560}{26\times 4}\), so the number of sessions is \(\boxed{15}\). Both stated rates go in upside down from how they are written, because the chain runs from dollars back to sessions.
Practice
Each team at a school has 6 students, each bus carries 4 teams, and the school fills 9 buses for a meet. Enter the number of students the school sends.
Show the solution
Chaining the two rates gives \(9\text{ buses}\times\frac{4\text{ teams}}{1\text{ bus}}\times\frac{6\text{ students}}{1\text{ team}}\), and buses and teams cancel to leave \(9\times 4\times 6=\boxed{216}\). Per bus that is \(4\times 6=24\) students, and \(9\times 24=216\).
Practice
On a game map, \(1\text{ span}=9\text{ notches}\), \(4\text{ notches}=3\text{ pips}\) and \(1\text{ pip}=20\text{ steps}\). A road is 8 spans long. Enter its length in steps.
Show the solution
Chaining the three factors leaves \(\frac{8\times 9\times 3\times 20}{4}\), so the length in steps is \(\boxed{1080}\). The middle factor is \(\frac{3\text{ pips}}{4\text{ notches}}\), with notches in the denominator to cancel the \(72\) notches from the first step. The flipped factor \(\frac{4\text{ notches}}{3\text{ pips}}\) gives \(1920\) instead.