QuanticaPrealgebraOpen in the courseWhen the top and bottom of a fraction are the same real amount in different units, that fraction equals \(1\). Such a fraction is a conversion factor. Multiplying by it swaps the unit label without changing the true quantity.
Problem
One hand equals 4 inches. A pony is 36 inches tall. Multiply by \(\frac{1\,\text{hand}}{4\,\text{in}}\) so the inches cancel. How many hands tall is the pony?
Show a hint
- A conversion factor is a fraction equal to \(1\) whose top and bottom are the same height in different units, so both \(\frac{4 \text{ in}}{1 \text{ hand}}\) and \(\frac{1 \text{ hand}}{4 \text{ in}}\) equal \(1\). You start with inches and want hands, so choose the one with inches on the bottom, \(\frac{1 \text{ hand}}{4 \text{ in}}\). That way inches will cancel and hands will be the unit that survives.
- Write the chain as \(36 \text{ in} \times \frac{1 \text{ hand}}{4 \text{ in}}\). Cross out the \(\text{in}\) on top against the \(\text{in}\) on the bottom, then the unit left is hands. Now just do the number part, \(36 \div 4\).
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Put inches on the bottom so they cancel and hands are left. $$36 \text{ in} \times \frac{1 \text{ hand}}{4 \text{ in}} = \frac{36}{4} \text{ hands}$$ The pony is \(\boxed{9}\) hands tall. Flip that factor the other way and the units come out as \(\frac{\text{in}^2}{\text{hand}}\), which is how the units warn you the factor is upside down.
Problem
One fathom equals 6 feet. An anchor rope is 23 fathoms long. Using the factor \(\frac{6\,\text{ft}}{1\,\text{fathom}}\), convert to feet. How many feet long is the rope?
Show a hint
- Write the relationship as a fraction equal to \(1\). You want "fathom" to cancel, so the fathom unit needs to sit on the bottom of your conversion factor while feet sits on top, giving \(\frac{6 \text{ ft}}{1 \text{ fathom}}\).
- Multiply \(23 \text{ fathom} \times \frac{6 \text{ ft}}{1 \text{ fathom}}\). The fathom labels cancel, then multiply \(23 \times 6\). Whatever number you get carries the unit "ft".
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Put fathoms on the bottom so they cancel and feet are left. $$23 \text{ fathom} \times \frac{6 \text{ ft}}{1 \text{ fathom}} = 23 \times 6 \text{ ft} = 138 \text{ ft}$$ The rope is \(\boxed{138}\) feet long. That factor equals \(1\) because 6 feet and 1 fathom are the same distance, so it changes the label without changing the rope.
Problem
270 loose jars need to be packed into crates of 18. Use \(\frac{1\,\text{crate}}{18\,\text{jars}}\) so jars cancel and crates survive. How many full crates do 270 jars make?
Show a hint
- You have two factors to pick from, \(\frac{18 \text{ jars}}{1 \text{ crate}}\) and \(\frac{1 \text{ crate}}{18 \text{ jars}}\), and both equal \(1\). The unit you are starting with is jars, and a unit cancels only when it shows up on the top of one fraction and the bottom of another. Since your \(270\) jars is on top, the factor you choose must carry jars on the bottom so the two jars cancel and crates is left behind. Which of the two factors has jars underneath?
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Jars is the unit you want gone, so put it on the bottom. $$270 \text{ jars} \times \frac{1 \text{ crate}}{18 \text{ jars}} = \frac{270}{18} \text{ crates} = 15 \text{ crates}$$ So 270 jars fill \(\boxed{15}\) full crates. The other orientation would have left units of \(\frac{\text{jars}^2}{\text{crate}}\), which is how the units catch a flipped factor.
Problem
One stone equals 14 pounds. Luggage weighs 91 pounds. Build a conversion factor so the pounds cancel and stone survives. What is the weight in stone? (The answer is not a whole number.)
Show a hint
- You want pounds to cancel and stone to survive. Since 1 stone is 14 pounds, the factor \(\frac{1 \text{ stone}}{14 \text{ pounds}}\) is just a ratio equal to 1. Multiplying by 1 never changes the true weight, it only changes the unit it is written in.
- Write \(91 \text{ pounds} \times \frac{1 \text{ stone}}{14 \text{ pounds}}\) and cross out the matching pound units. That leaves the number \(\frac{91}{14}\) carrying the unit stone. Now reduce \(\frac{91}{14}\) to lowest terms by dividing the top and bottom by their greatest common factor.
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Put pounds on the bottom so they cancel and stone is left. $$91 \text{ pounds} \times \frac{1 \text{ stone}}{14 \text{ pounds}} = \frac{91}{14} \text{ stone}$$ Both numbers share a factor of 7, since \(91 = 7 \times 13\) and \(14 = 7 \times 2\), so the luggage weighs \(\boxed{13/2}\) stone, the same as 6.5 stone. Conversion factors work just as well when the answer lands between two whole numbers.
Problem
A racecourse is 4 furlongs. Use \(1\,\text{furlong} = 220\,\text{yd}\) and \(1\,\text{yd} = 3\,\text{ft}\). Chain two factors through yards. How many feet is 4 furlongs?
Show a hint
- You cannot jump straight from furlongs to feet, so hop through yards in the middle. Build the first conversion factor from \(1 \text{ furlong} = 220 \text{ yards}\), and orient it so the unit you want to get rid of, furlongs, sits on the bottom. That makes it \(\frac{220 \text{ yd}}{1 \text{ furlong}}\), and since top and bottom are the same length, this fraction equals \(1\), so multiplying by it changes the unit without changing the actual distance.
- Now line up the whole chain so each factor cancels the unit the one before it produced. Write \(4 \text{ furlong} \times \frac{220 \text{ yd}}{1 \text{ furlong}} \times \frac{3 \text{ ft}}{1 \text{ yd}}\). The furlongs cancel against the first bottom and the yards cancel against the second bottom, leaving feet. Then just multiply the numbers across, \(4 \times 220 \times 3\).
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No single furlong-to-feet factor exists, so chain two through yards, each turned so the unwanted unit is on the bottom. $$4 \text{ furlong} \times \frac{220 \text{ yd}}{1 \text{ furlong}} \times \frac{3 \text{ ft}}{1 \text{ yd}} = 4 \times 220 \times 3 \text{ ft}$$ Furlong cancels against the first denominator and yard against the second, leaving feet, and \(4 \times 220 \times 3 = 2{,}640\). So 4 furlongs is \(\boxed{2640}\) feet.
Problem
A bolt of cloth is 6 yards long. Using \(1\,\text{yd} = 3\,\text{ft}\) and \(1\,\text{ft} = 12\,\text{in}\), chain two factors. How many inches long is the cloth?
Show a hint
- You have no single yard-to-inch factor, so build a bridge through feet. Start with the factor that kills yards. Since \(1\) yard \(= 3\) feet, write it as \(\frac{3 \text{ ft}}{1 \text{ yd}}\) so that the "yd" on top of \(6 \text{ yd}\) cancels the "yd" on the bottom of the factor, leaving feet.
- Now stack a second factor that kills feet. Use \(\frac{12 \text{ in}}{1 \text{ ft}}\) so "ft" cancels and only "in" survives. The whole chain is \(6 \text{ yd} \times \frac{3 \text{ ft}}{1 \text{ yd}} \times \frac{12 \text{ in}}{1 \text{ ft}}\). Multiply the numbers across, \(6 \times 3 \times 12\).
Show the full solution
There is no yard-to-inch fact, so chain two factors through feet. $$6 \text{ yd} \times \frac{3 \text{ ft}}{1 \text{ yd}} \times \frac{12 \text{ in}}{1 \text{ ft}} = 6 \times 3 \times 12 \text{ in}$$ The yd cancels, then the ft cancels, and only in survives. The cloth is \(\boxed{216}\) inches long.
Problem
A lantern festival lasts 1 fortnight (14 days). Using \(1\,\text{day} = 24\,\text{hr}\) and \(1\,\text{hr} = 60\,\text{min}\), convert through three factors. How many minutes long is 1 fortnight?
Show a hint
- Start with \(1 \text{ fortnight}\) and multiply by a conversion factor that puts \(\text{fortnight}\) on the bottom so it cancels. Since \(1 \text{ fortnight} = 14 \text{ days}\), that first factor is \(\frac{14 \text{ days}}{1 \text{ fortnight}}\), which is just a ratio equal to 1. Now you are holding days, and you need to keep relaying down to minutes.
- Chain all three factors in a row, each one cancelling the unit before it. Write \(1 \text{ fortnight} \times \frac{14 \text{ days}}{1 \text{ fortnight}} \times \frac{24 \text{ hr}}{1 \text{ day}} \times \frac{60 \text{ min}}{1 \text{ hr}}\) and multiply the top numbers, \(14 \times 24 \times 60\). Notice \(14 \times 24 = 336\), then \(336 \times 60\).
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Chain three factors, each turned so the unwanted unit sits on the bottom. $$1 \text{ fortnight} \times \frac{14 \text{ days}}{1 \text{ fortnight}} \times \frac{24 \text{ hr}}{1 \text{ day}} \times \frac{60 \text{ min}}{1 \text{ hr}}$$ Fortnight, day, and hr each cancel in turn, leaving minutes. Then \(14 \times 24 = 336\) and \(336 \times 60 = 20{,}160\), so 1 fortnight is \(\boxed{20160}\) minutes.
Problem
An arcade: 1 dollar buys 4 tokens, and 1 token buys 3 prize tickets. No direct dollar-to-ticket rate is posted. Chain two factors through tokens. How many prize tickets are 9 dollars worth?
Show a hint
- You want to start with \(9 \text{ dollars}\) and end with tickets, so build conversion factors that cancel what you do not want. Each given exchange is a ratio equal to \(1\), and you can flip it either way. Pick the orientation with "dollars" on the bottom for the first factor, \(\frac{4 \text{ tokens}}{1 \text{ dollar}}\), so the dollars divide out.
- Dollars alone cannot reach tickets, so chain a second factor that has tokens on the bottom, \(\frac{3 \text{ tickets}}{1 \text{ token}}\). Now write the whole product, \(9 \text{ dollars} \times \frac{4 \text{ tokens}}{1 \text{ dollar}} \times \frac{3 \text{ tickets}}{1 \text{ token}}\), cross out dollars then tokens, and multiply the numbers \(9 \times 4 \times 3\).
Show the full solution
Chain the two exchanges through tokens, each factor turned so the unwanted unit is on the bottom. $$9 \text{ dollars} \times \frac{4 \text{ tokens}}{1 \text{ dollar}} \times \frac{3 \text{ tickets}}{1 \text{ token}} = 9 \times 4 \times 3 \text{ tickets}$$ Dollars cancel, then tokens cancel, and only tickets survive, so 9 dollars is worth \(\boxed{108}\) prize tickets. A posted exchange rate is a conversion factor like any other.
Problem
A cyclist rides at 45 miles per hour. Use \(1\,\text{mi} = 5{,}280\,\text{ft}\) and \(1\,\text{hr} = 3{,}600\,\text{s}\) to convert. What is the speed in feet per second?
Show a hint
- A conversion factor is a ratio equal to \(1\), so multiplying by it never changes the speed, only its dress. You have miles on top and you want feet on top, so use \(\frac{5{,}280 \text{ ft}}{1 \text{ mi}}\) with the mile on the bottom, ready to cancel the mile in \(\frac{45 \text{ mi}}{1 \text{ hr}}\). Separately, you have hours on the bottom and you want seconds on the bottom, so you will need a factor that puts hours on top to cancel them.
- Chain all three pieces together, \(\frac{45 \text{ mi}}{1 \text{ hr}} \times \frac{5{,}280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3{,}600 \text{ s}}\). The miles cancel and the hours cancel, leaving feet over seconds. Now just the numbers remain, \(\frac{45 \times 5{,}280}{3{,}600}\), which is \(\frac{237{,}600}{3{,}600}\).
Show the full solution
Write the speed as a fraction and attach two factors, one for the top unit and one for the bottom. $$\frac{45 \text{ mi}}{1 \text{ hr}} \times \frac{5{,}280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3{,}600 \text{ s}} = \frac{45 \times 5{,}280}{3{,}600} \text{ ft/s}$$ The mi cancels and the hr cancels, leaving feet over seconds, and \(\frac{237{,}600}{3{,}600} = 66\). The speed is \(\boxed{66}\) ft/s.
Problem
A syrup has density \(1.5\,\text{g/mL}\). A tank holds 4 liters. Convert liters to mL, then apply the density. How many grams of syrup does the tank hold?
Show a hint
- The density \(\frac{1.5 \text{ g}}{1 \text{ mL}}\) is built for milliliters, but your volume is in liters. Before you can use it, turn \(4\) liters into milliliters with a conversion factor that cancels liters, oriented as \(\frac{1000 \text{ mL}}{1 \text{ liter}}\) so "liter" is on the bottom.
- Build the full chain \(4 \text{ liters} \times \frac{1000 \text{ mL}}{1 \text{ liter}} \times \frac{1.5 \text{ g}}{1 \text{ mL}}\). The first factor cancels liters and the second cancels milliliters, leaving grams. Now just multiply the numbers, \(4 \times 1000 \times 1.5\).
Show the full solution
Send liters to milliliters first, then let the density turn milliliters into grams, each factor turned so the unwanted unit is on the bottom. $$4 \text{ liters} \times \frac{1000 \text{ mL}}{1 \text{ liter}} \times \frac{1.5 \text{ g}}{1 \text{ mL}}$$ Liter cancels, then mL cancels, leaving grams. Since \(4 \times 1000 = 4000\) and \(4000 \times 1.5 = 6000\), the tank holds \(\boxed{6000}\) grams. A density is a conversion factor too, one that trades volume for mass.
Problem
A sticker covers 12 square inches. Use \(1\,\text{in} = 2.5\,\text{cm}\). The factor must be applied twice since area is length \(\times\) length. How many square centimeters is the sticker?
Show a hint
- A conversion factor is a ratio equal to \(1\). Here the relationship \(1 \text{ in} = 2.5 \text{ cm}\) gives the factor \(\frac{2.5 \text{ cm}}{1 \text{ in}}\). But the unit you are converting is square inches, which is \(\text{in} \times \text{in}\), so a single \(\frac{\text{cm}}{\text{in}}\) only cancels one of the two inches. What happens to the leftover inch?
- Multiply by the factor \(\frac{2.5 \text{ cm}}{1 \text{ in}}\) twice, once for each side of the little square. Watch both inch units cancel and two centimeter units appear. Then the number part is \(12 \times 2.5 \times 2.5\), so the linear factor \(2.5\) is squared.
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A square inch is \(1 \text{ in} \times 1 \text{ in}\), so the factor \(\frac{2.5 \text{ cm}}{1 \text{ in}}\) gets used twice, once for each side. $$12 \text{ in}^2 \times \frac{2.5 \text{ cm}}{1 \text{ in}} \times \frac{2.5 \text{ cm}}{1 \text{ in}}$$ Both inch units cancel and two centimeter units are left, which make \(\text{cm}^2\). Then \(2.5 \times 2.5 = 6.25\) and \(12 \times 6.25 = 75\), so the sticker is \(\boxed{75}\) square centimeters. Using the linear factor twice is the same as squaring it.
Problem
On a game island: \(1\,\text{brindle} = 5\,\text{floons}\), \(1\,\text{floon} = 4\,\text{ticks}\), \(1\,\text{tick} = 6\,\text{motes}\). A path is 3 brindles long. How many motes?
Show a hint
- You have no idea how big a brindle or a mote really is, and that is fine. The method runs on the relationships alone. Start with \(3 \text{ brindle}\) and pick the first conversion factor so that brindle cancels, which means brindle goes on the bottom, \(\frac{5 \text{ floons}}{1 \text{ brindle}}\).
- Keep going one unit at a time, floon to tick to mote, each new factor oriented so the unit you just made lands on the bottom and cancels. After every cancellation only motes survive, so multiply the top numbers, \(3 \times 5 \times 4 \times 6\).
Show the full solution
Lay the three facts down as factors, each turned so the unwanted unit is on the bottom, and walk brindle to floon to tick to mote. $$3 \text{ brindle} \times \frac{5 \text{ floons}}{1 \text{ brindle}} \times \frac{4 \text{ ticks}}{1 \text{ floon}} \times \frac{6 \text{ motes}}{1 \text{ tick}}$$ Brindle, floon, and tick each cancel, leaving motes, and \(3 \times 5 \times 4 \times 6 = 360\). The path is \(\boxed{360}\) motes long. You never had to picture any of these units, only the relationships between them.
Practice these ideas
Practice
In the highlands, 1 league equals 3 miles. The Ridgeback Trail is 8 leagues long. How many miles long is the trail?
Show the solution
Put leagues on the bottom so they cancel and miles are left. $$8 \text{ leagues} \times \frac{3 \text{ miles}}{1 \text{ league}} = 8 \times 3 \text{ miles} = 24 \text{ miles}$$ The trail is \(\boxed{24}\) miles long.
Practice
One gross equals 144 pencils. A warehouse has 360 loose pencils. How many gross is that?
Show the solution
Put pencils on the bottom so they cancel and gross is left. $$360 \text{ pencils} \times \frac{1 \text{ gross}}{144 \text{ pencils}} = \frac{360}{144} \text{ gross}$$ Both numbers share the factor \(72\), and \(360 \div 72 = 5\) while \(144 \div 72 = 2\), so the warehouse has \(\boxed{5/2}\) gross, which is two and a half gross.
Practice
One ream equals 500 sheets. A dock has 3,500 sheets. Using the factor \(\frac{1\,\text{ream}}{500\,\text{sheets}}\), how many reams is that?
Show the solution
Start with the amount you have and multiply by the conversion factor oriented so sheets cancels. $$3{,}500 \text{ sheets} \times \frac{1 \text{ ream}}{500 \text{ sheets}} = \frac{3{,}500}{500} \text{ reams}$$ The unit sheets appears on top and bottom, so it cancels and reams remains. Now divide. $$\frac{3{,}500}{500} = 7$$ So \(3{,}500\) sheets is \(\boxed{7}\) reams.
Practice
Each brick is 9 inches long. A wall is 162 inches long. Using the factor \(\frac{1\,\text{brick}}{9\,\text{in}}\) so the inches cancel, how many bricks long is the wall?
Show the solution
Start with the wall length and attach the conversion factor with inches on the bottom, so the inches cancel and bricks are left. $$162 \text{ in} \times \frac{1 \text{ brick}}{9 \text{ in}}$$ The inch units cancel, top and bottom. $$162 \cancel{\text{ in}} \times \frac{1 \text{ brick}}{9 \cancel{\text{ in}}} = \frac{162}{9} \text{ bricks}$$ Now divide. $$\frac{162}{9} = 18 \text{ bricks}$$ So the wall is \(\boxed{18}\) bricks long.
Practice
A surveyor has \(1\,\text{mile} = 8\,\text{furlongs}\) and \(1\,\text{furlong} = 40\,\text{rods}\). A boundary is 2 miles. Chain two factors and convert to rods. How many rods long is the boundary?
Show the solution
Chain two factors through furlongs, each turned so the unwanted unit is on the bottom. $$2 \text{ mile} \times \frac{8 \text{ furlong}}{1 \text{ mile}} \times \frac{40 \text{ rod}}{1 \text{ furlong}} = 2 \times 8 \times 40 \text{ rod}$$ Mile cancels, then furlong cancels, leaving rods. The boundary is \(\boxed{640}\) rods long.
Practice
A recipe needs 5 gallons of water, measured in pints. Given \(1\,\text{gal} = 4\,\text{qt}\) and \(1\,\text{qt} = 2\,\text{pt}\), chain two factors. How many pints are in 5 gallons?
Show the solution
Start with \(5\) gallons and chain two conversion factors so the unwanted units cancel. Orient each factor with the unit you want to remove on the bottom. $$5 \text{ gallon} \times \frac{4 \text{ quart}}{1 \text{ gallon}} \times \frac{2 \text{ pint}}{1 \text{ quart}}$$ The gallons cancel against the gallon on the bottom of the first factor, and the quarts cancel against the quart on the bottom of the second factor, leaving only pints. $$5 \times 4 \times 2 = 40 \text{ pint}$$ So \(5\) gallons is \(\boxed{40}\) pints.
Practice
How many seconds are in 1 full day? Chain three factors using \(1\,\text{day} = 24\,\text{hr}\), \(1\,\text{hr} = 60\,\text{min}\), \(1\,\text{min} = 60\,\text{s}\).
Show the solution
Chain three factors, each turned so the unwanted unit is on the bottom. $$1 \text{ day} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 24 \times 60 \times 60 \text{ seconds}$$ Day, hour, and minute each cancel, leaving seconds. One full day holds \(\boxed{86400}\) seconds.
Practice
A collection is 3 stacks. Each stack holds 5 bundles, each bundle holds 4 packs, each pack holds 12 cards. Chain three factors and convert stacks to cards. How many cards is that?
Show the solution
Chain one factor per step, each turned so the old unit cancels. $$3 \text{ stacks} \times \frac{5 \text{ bundles}}{1 \text{ stack}} \times \frac{4 \text{ packs}}{1 \text{ bundle}} \times \frac{12 \text{ cards}}{1 \text{ pack}}$$ Stacks, bundles, and packs all cancel, leaving cards, and \(3 \times 5 \times 4 \times 12 = 720\). The collection holds \(\boxed{720}\) cards.
Practice
A festival booth exchanges 1 euro for 12 ride tokens. A visitor exchanges 15 euros. Using the factor \(\frac{12\,\text{tokens}}{1\,\text{euro}}\), how many tokens does the visitor get?
Show the solution
Put euros on the bottom of the factor so they cancel and tokens are left. $$15 \text{ euros} \times \frac{12 \text{ tokens}}{1 \text{ euro}} = 15 \times 12 \text{ tokens} = 180 \text{ tokens}$$ The visitor gets \(\boxed{180}\) ride tokens.
Practice
A summer fair: 1 dollar buys 20 game chips, and 1 chip earns 3 prize points. A guest exchanges 7 dollars. Chain two factors through chips. How many prize points is that worth?
Show the solution
Chain the two exchanges through chips, each factor turned so the unwanted unit is on the bottom. $$7 \text{ dollars} \times \frac{20 \text{ chips}}{1 \text{ dollar}} \times \frac{3 \text{ points}}{1 \text{ chip}} = 7 \times 20 \times 3 \text{ points}$$ Dollars cancel, then chips cancel, leaving points. So 7 dollars is worth \(\boxed{420}\) prize points.
Practice
A maglev train travels at 72 km/hr. Convert to m/s using \(1\,\text{km} = 1{,}000\,\text{m}\) and \(1\,\text{hr} = 3{,}600\,\text{s}\). What is the speed in meters per second?
Show the solution
Attach one factor for the top unit and one for the bottom, each turned so the unwanted unit cancels. $$\frac{72 \text{ km}}{1 \text{ hr}} \times \frac{1{,}000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3{,}600 \text{ s}} = \frac{72 \times 1{,}000}{3{,}600} \text{ m/s}$$ The km cancels and the hr cancels, leaving meters over seconds, and \(\frac{72{,}000}{3{,}600} = 20\). The train moves at \(\boxed{20}\) meters per second.
Practice
A runner's pace is 15 mph. Convert to feet per minute using \(1\,\text{mi} = 5{,}280\,\text{ft}\) and \(1\,\text{hr} = 60\,\text{min}\). What is the speed in feet per minute?
Show the solution
Use \(\frac{5{,}280 \text{ ft}}{1 \text{ mi}}\) to clear the miles on top and \(\frac{1 \text{ hr}}{60 \text{ min}}\) to clear the hours on the bottom. $$\frac{15 \text{ mi}}{1 \text{ hr}} \times \frac{5{,}280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{60 \text{ min}} = \frac{15 \times 5{,}280}{60} \text{ ft/min}$$ Feet over minutes is what survives, and \(\frac{79{,}200}{60} = 1{,}320\). The runner is moving at \(\boxed{1320}\) feet per minute.
Practice
A jar holds 8 tablespoons of maple syrup. The density is \(\frac{21\,\text{g}}{1\,\text{tablespoon}}\). How many grams of syrup does the jar hold?
Show the solution
Start with the amount you know and attach the density as a conversion factor, oriented so tablespoons cancels and grams stays. $$8 \text{ tablespoons} \times \frac{21 \text{ g}}{1 \text{ tablespoon}}$$ The \(\text{tablespoon}\) on top cancels with the \(\text{tablespoon}\) on the bottom. $$8 \, \cancel{\text{tablespoons}} \times \frac{21 \text{ g}}{1 \, \cancel{\text{tablespoon}}} = (8 \times 21) \text{ g}$$ Now finish the multiplication, \(8 \times 21 = 168\), and the only surviving unit is grams. So the jar holds \(\boxed{168}\) grams of syrup.
Practice
A workshop floor covers 6 square feet. Convert to square inches using \(1\,\text{ft} = 12\,\text{in}\). Remember the linear factor gets applied twice for area. What is the area in square inches?
Show the solution
Use the factor \(\frac{12 \text{ in}}{1 \text{ ft}}\) twice, since an area carries two lengths. $$6 \text{ ft}^2 \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{12 \text{ in}}{1 \text{ ft}} = 6 \times 12 \times 12 \text{ in}^2$$ The \(\text{ft}^2\) cancels against the two \(\text{ft}\) on the bottom, leaving \(\text{in}^2\), and \(6 \times 144 = 864\). The floor is \(\boxed{864}\) square inches.