Solve Conversion Problems and Track Units | Grade 6 Math | LevelKite

A measurement has two parts: a number and a unit. The number alone does not tell you the size. A ribbon that is $3$ long could be $3$ inches or $3$ meters.

You can compare, add or subtract the numbers of two measurements only when their units match. Many word problems give you measurements in different units. Before you can answer, you must convert: name the same amount in a different unit without changing its size.

Use this plan for a conversion word problem:

Find what the question asks for and the unit the answer needs.
Write the conversion fact that links the units, such as $1\text{ m} = 100\text{ cm}$ ( $1$ meter is $100$ centimeters).
Convert so that every measurement is in the unit you need.
Finish the problem. Give the answer in a sentence with its unit.

Worked example

Is the ribbon long enough?

Kai needs $2.5$ meters of ribbon for a poster. The store has one piece left, and it is $240$ centimeters long. Is the piece long enough?

Find what is asked. You must compare two lengths. One is in meters and the other is in centimeters, so you cannot compare 2.5 and 240 yet. Choose centimeters for both, because the piece in the store is already measured in centimeters.
Write the conversion fact that links the two units. One meter is the same length as 100 centimeters.
$1\text{ m} = 100\text{ cm}$
Convert the length Kai needs. Each meter is a group of 100 centimeters, and Kai needs 2.5 of those groups, so multiply 2.5 by 100. Split 2.5 into 2 and 0.5 to see the arithmetic.
$2.5 \times 100 = (2 \times 100) + (0.5 \times 100) = 200 + 50 = 250$
Both lengths are now in centimeters, so you can compare the numbers. Kai needs 250 centimeters and the piece is 240 centimeters. Subtract to find how much is missing.
$240 < 250 \qquad 250 - 240 = 10$

Answer

No. Kai needs $250$ centimeters of ribbon and the piece is only $240$ centimeters long, so it is $10$ centimeters too short.

In that example you chose to multiply by thinking about groups. You can also let the units choose the operation for you.

You already use conversion facts such as $1\text{ ft} = 12\text{ in}$ ( $1$ foot is $12$ inches). Both sides of the fact name the same length, and that length is not zero. So when you divide one side by the other, you get exactly $1$ :

$\frac{12\text{ in}}{1\text{ ft}} = 1$ and $\frac{1\text{ ft}}{12\text{ in}} = 1$

A fraction like this is called a conversion factor. Multiplying by $1$ never changes an amount. So when you multiply a measurement by a conversion factor, the amount stays the same. Only the unit that names it changes.

Key concept

Units go through the arithmetic with the numbers

A quantity is a number together with its unit, such as $5$ feet. Treat every unit like a factor, a number that is multiplied. When you multiply or divide quantities, multiply or divide their units too. A unit on the top and the same unit on the bottom divide to $1$ , so they cancel. The unit that is left is the unit of your answer.

To convert, multiply by the conversion factor that makes the unit you want to remove appear once on the top and once on the bottom. For example, $5\text{ ft} \times \frac{12\text{ in}}{1\text{ ft}} = 60\text{ in}$ : feet cancel and inches are left. You may divide only by a quantity that is not zero. The bottom of a conversion factor, such as $12$ inches, is never zero, so dividing by it is always allowed.

\cancel{\text{old unit}} \times \frac{\text{new unit}}{\cancel{\text{old unit}}} = \text{new unit}

Some quantities are rates. A rate compares two quantities that have different units. The word per means for each, and you write it as a fraction bar. So $40$ miles per hour is $\frac{40\text{ mi}}{1\text{ h}}$ .

The table shows what the units do in four calculations. In each division, the quantity you divide by is not zero. Compare the last two rows. The same unit on the top and on the bottom cancels. The same unit multiplied by itself does not cancel. It makes a square unit.

What the units do when you multiply or divide

Calculation	What the units do	Result
$\frac{40\text{ mi}}{1\text{ h}} \times 3\text{ h}$	Hours are on the bottom and on the top, so they cancel. Miles are left.	$120\text{ mi}$
$150\text{ mi} \div 3\text{ h}$	Miles divided by hours make miles per hour.	$\frac{50\text{ mi}}{1\text{ h}}$ , or $50$ miles per hour
$72\text{ in} \div 8\text{ in}$	Inches divided by inches make $1$ , so no unit is left.	$9$ , a count with no unit
$4\text{ ft} \times 3\text{ ft}$	Feet times feet make square feet.	$12\text{ ft}^2$ , or $12$ square feet

Many conversions end in a division, so you need a way to check one. In $7 \div 2$ , the dividend is $7$ , the amount being divided. The divisor is $2$ , the amount you divide by, and it must never be zero. The result is the quotient.

When the divisor is not zero, multiplication and division are inverse operations: each one undoes the other. Dividing $7$ by $2$ asks what number times $2$ makes $7$ . So you check a division by multiplying back. The check depends on the kind of answer, as the table shows. When you round a quotient, use the sign $\approx$ , which means is about equal to.

Give the kind of answer the problem needs. If the amount can be split, like water or time, a fraction or a decimal is a fine answer. If the things must stay whole, like shelves or bottles, give a whole-number quotient and a remainder, the amount left over, written like $3$ R $1$ . If the division comes out even, there is no remainder at all, so say that nothing is left over.

Checking a division by multiplying back (the divisor is never zero)

Kind of answer	The check	Example
Exact quotient	The quotient times the divisor equals the dividend.	$7 \div 2 = 3.5$ and $3.5 \times 2 = 7$
Rounded quotient	The quotient times the divisor is close to the dividend. It may not match exactly.	$7 \div 3 \approx 2.3$ and $2.3 \times 3 = 6.9$
Whole-number quotient with a remainder, for things that must stay whole	The divisor times the whole quotient, plus the remainder, equals the dividend.	$7 \div 2 = 3$ R $1$ and $2 \times 3 + 1 = 7$

Worked example

Cutting shelves from a board

A board is $6$ feet long. You cut it into shelves that are each $16$ inches long. How many whole shelves can you cut, and how much of the board is left over?

The board is in feet and each shelf is in inches, so make the units match. Use the fact that 1 foot is 12 inches. Put feet on the bottom of the conversion factor so that feet cancel and inches are left.
$6\ \cancel{\text{ft}} \times \frac{12\text{ in}}{1\ \cancel{\text{ft}}} = 72\text{ in}$
Divide the board length by the length of one shelf to find how many shelf lengths fit. The divisor, 16 inches, is not zero. Inches on the top and inches on the bottom cancel, so the quotient is a count of shelves, not a length.
$\frac{72\ \cancel{\text{in}}}{16\ \cancel{\text{in}}} = 72 \div 16$
A shelf must stay whole, so find a whole-number quotient and a remainder. Four shelves use 64 inches. Five shelves would need 80 inches, which is more than the board has. Subtract to find the leftover length. Do not report 4.5 shelves, the exact quotient, because half a shelf is not a shelf.
$16 \times 4 = 64 \qquad 72 - 64 = 8 \qquad 72 \div 16 = 4 \text{ R } 8$
Check with the divisor times the whole quotient, plus the remainder. The result must equal the dividend. The remainder is a piece of the board, so it keeps the unit inches.
$16\text{ in} \times 4 + 8\text{ in} = 64\text{ in} + 8\text{ in} = 72\text{ in}$

Answer

You can cut $4$ whole shelves, and $8$ inches of the board is left over.

Worked example

A dripping faucet

A faucet drips $15$ milliliters of water each minute. How many liters of water does it drip in $4$ hours?

Write the rate as a fraction: 15 milliliters (mL) per 1 minute (min). The rate uses minutes, but the time is in hours (h). The rate uses milliliters, but the answer needs liters (L). So you need two conversion facts.
$\frac{15\text{ mL}}{1\text{ min}} \qquad 1\text{ h} = 60\text{ min} \qquad 1\text{ L} = 1{,}000\text{ mL}$
Convert the time to minutes. Put hours on the bottom of the factor so that hours cancel and minutes are left.
$4\ \cancel{\text{h}} \times \frac{60\text{ min}}{1\ \cancel{\text{h}}} = 240\text{ min}$
The faucet drips 15 milliliters in each of the 240 minutes, so multiply the rate by the time. Minutes are on the bottom of the rate and on the top of the time, so they cancel and milliliters are left. For the arithmetic, 15 times 24 is 360, so 15 times 240 is 3,600.
$\frac{15\text{ mL}}{1\ \cancel{\text{min}}} \times 240\ \cancel{\text{min}} = 3{,}600\text{ mL}$
Convert milliliters to liters. Put milliliters on the bottom so that they cancel. The units now tell you to divide by 1,000, which is not zero. Water can be split, so a decimal number of liters is the right kind of answer.
$3{,}600\ \cancel{\text{mL}} \times \frac{1\text{ L}}{1{,}000\ \cancel{\text{mL}}} = \frac{3{,}600}{1{,}000}\text{ L} = 3.6\text{ L}$
Check by multiplying back. The quotient 3.6 is exact, so the quotient times the divisor must equal the dividend exactly.
$3.6 \times 1{,}000 = 3{,}600$

Answer

The faucet drips $3.6$ liters of water in $4$ hours.

Now use the same tools to judge someone else's work. A conversion can go wrong in different places, so ask three questions:

Is the fact true? A conversion factor must be built from two amounts that are equal, such as $1\text{ ft} = 12\text{ in}$ , or about equal when the fact uses $\approx$ . A false fact makes a factor that does not equal $1$ , so it changes the amount.
Do the units work? The unwanted units must cancel. The unit that is left must be the unit the answer needs.
Is the size sensible? For a length, weight, volume or time greater than zero, a smaller unit needs a larger number, and a larger unit needs a smaller number.

Then check any division by multiplying back. Use the check from the table that fits the kind of answer.

Mistake: multiplying by the conversion number every time. To change $48$ inches to feet, a student remembers that $1\text{ ft} = 12\text{ in}$ and writes $48 \times 12 = 576$ feet.

Track the units to see why this fails. In $48\text{ in} \times \frac{12\text{ in}}{1\text{ ft}}$ , inches are on the top twice. Nothing cancels, so the unit that is left is not feet. The size is wrong too. A foot is longer than an inch, so a length greater than zero needs fewer feet than inches.

The factor needs inches on the bottom: $48\text{ in} \times \frac{1\text{ ft}}{12\text{ in}} = 4\text{ ft}$ . The divisor $12$ is not zero and the quotient is exactly $4$ , with no remainder, so $4 \times 12 = 48$ checks it.

Worked example

Find the mistake: a suitcase in kilograms

Omar's suitcase weighs $50$ pounds (lb). The airline limit is $23$ kilograms (kg). A conversion chart says $1\text{ kg} \approx 2.2\text{ lb}$ . Omar writes $50\text{ lb} \times \frac{2.2\text{ kg}}{1\text{ lb}} = 110\text{ kg}$ and decides the suitcase is far too heavy. Find his mistake. Is the suitcase within the limit?

Ask whether the units work. Pounds are on the top and on the bottom, so they cancel, and kilograms are left. The units pass, so the mistake is somewhere else.
$50\ \cancel{\text{lb}} \times \frac{2.2\text{ kg}}{1\ \cancel{\text{lb}}}$
Ask whether the fact is true and the size is sensible. Omar's factor says that 1 pound is 2.2 kilograms. The chart says it the other way around: 1 kilogram is about 2.2 pounds. A kilogram is heavier than a pound, so a weight greater than zero needs fewer kilograms than pounds. An answer of 110 is too large. The mistake is a factor built from a false fact.
$1\text{ kg} \approx 2.2\text{ lb} \quad \text{so} \quad \frac{1\text{ kg}}{2.2\text{ lb}} \approx 1$
Build the factor from the chart's fact, with pounds on the bottom so that pounds cancel. The units now tell you to divide by 2.2, which is not zero. The fact uses the sign for about equal, so this factor is very close to 1 and the result is an estimate.
$50\ \cancel{\text{lb}} \times \frac{1\text{ kg}}{2.2\ \cancel{\text{lb}}} = \frac{50}{2.2}\text{ kg}$
Divide. First multiply the dividend and the divisor by 10 to get whole numbers. Multiplying both by the same nonzero number keeps the quotient the same. Then 22 times 22 is 484, which leaves 16. Dividing on into the decimal places gives the digits 7 and 2 again and again. Round to the nearest tenth.
$50 \div 2.2 = 500 \div 22 = 22.7272\ldots \approx 22.7$
Check by multiplying back. You rounded the quotient, so the result should be close to the dividend, 50. It does not have to match exactly.
$22.7 \times 2.2 = (22.7 \times 2) + (22.7 \times 0.2) = 45.4 + 4.54 = 49.94 \approx 50$
Compare with the limit. Both weights are now in kilograms, so you can compare the numbers. The estimate is below the limit.
$22.7\text{ kg} < 23\text{ kg}$

Answer

Omar built his factor from a false fact: he attached $2.2$ to kilograms instead of pounds. The suitcase weighs about $22.7$ kilograms, so it is within the $23$ -kilogram limit.

Units are part of the arithmetic. To solve a conversion word problem, first put each measurement in the unit you need. Multiply by a conversion factor made from two equal amounts. It equals $1$ , so the amount stays the same and only its unit changes. Cancel a unit that is on the top and on the bottom, and read the unit that is left.

To judge a conversion, ask three questions. Is the fact true? Do the units work? Is the size sensible?

Check each division by multiplying back. The divisor is never zero. An exact quotient times the divisor equals the dividend. A rounded quotient times the divisor is close to the dividend. For whole things with a leftover, the divisor times the whole quotient, plus the remainder, equals the dividend. When a division comes out even, there is no remainder. When the amount can be split, a fraction or a decimal is a fine answer.