Using Unit Rates to Solve and Compare | Grade 7 Math | LevelKite

A ratio tells how much of one quantity goes with a given amount of another, such as $2$ cups of lemon juice for every $3$ cups of water. A rate is a ratio whose two quantities have different units, such as $12$ posters in $3$ minutes. In this lesson every quantity is positive, and every number keeps its unit.

You already know how to find a unit rate: the amount of the first quantity that goes with exactly $1$ unit of the second quantity. For $12$ posters in $3$ minutes, the unit rate is $12 \div 3 = 4$ posters per minute, which means $4$ posters for each $1$ minute. The quantity you divide by (the divisor) is the per-one quantity. It must never be zero, and here it is always positive.

Now you will put unit rates to work in two ways. You will use a unit rate to solve a word problem, and you will use unit rates to compare two ratios or two rates fairly.

Why can a unit rate solve a problem? A unit rate describes what happens in one unit of the per-one quantity. When a rate is constant, meaning it stays the same the whole time, every unit behaves the same way.

Picture a printer that makes $4$ posters per minute at a constant rate.

You know the minutes and want the posters. In $5$ minutes there are $5$ equal groups of $4$ posters, so you multiply: $4 \times 5 = 20$ posters.
You know the posters and want the minutes. To make $28$ posters, ask how many groups of $4$ posters fit into $28$ posters, so you divide: $28 \div 4 = 7$ minutes.

This reasoning holds when the rate is constant, both quantities are positive, the per-one quantity is not zero, and the units match. A rate in posters per minute needs a time in minutes, not in hours.

Worked example

Scaling up with a unit rate

A pump moves water at a constant rate. It moves $18$ gallons in $4$ minutes. How many gallons does it move in $10$ minutes?

Name the two quantities with their units: gallons of water and minutes of time. You know the minutes and want the gallons, so make minutes the per-one quantity. You may divide by 4 minutes because 4 is positive, so it is not zero.
Divide gallons by minutes to find the unit rate. Because the rate is constant, dividing both quantities by the same positive number, 4, keeps the relationship the same. So 18 gallons in 4 minutes matches 4.5 gallons in 1 minute.
$18 \div 4 = 4.5 \text{ gallons per minute}$
The rate is constant, so each of the 10 minutes moves 4.5 gallons. Multiply the unit rate by 10 minutes. The rate is per minute and the time is in minutes, so the units match and the answer is in gallons.
$4.5 \text{ gallons per minute} \times 10 \text{ minutes} = 45 \text{ gallons}$
Check with a proportion, which is an equation saying that two ratios are equal. Put gallons on top and minutes on the bottom in both ratios, so matching quantities sit in matching positions. Both sides equal 4.5, because 18 divided by 4 is 4.5 and 45 divided by 10 is 4.5.
$\frac{18 \text{ gallons}}{4 \text{ minutes}} = \frac{45 \text{ gallons}}{10 \text{ minutes}}$

Answer

The pump moves $45$ gallons of water in $10$ minutes.

Worked example

Dividing by a unit rate

A candle burns at a constant rate. It burns down $\frac{3}{4}$ inch in $\frac{1}{2}$ hour. How many hours does it take to burn down $6$ inches?

Name the quantities: inches burned and hours of time. Use hours as the per-one quantity, which gives inches per hour. You may divide by 1/2 hour because it is positive, so it is not zero.
Divide the inches by the hours. Dividing by a nonzero fraction gives the same result as multiplying by its reciprocal, the fraction with its top and bottom numbers swapped. So dividing by 1/2 becomes multiplying by 2/1. Then simplify 6/4 to 3/2 by dividing the top and bottom by 2.
$\frac{3}{4} \div \frac{1}{2} = \frac{3}{4} \times \frac{2}{1} = \frac{6}{4} = \frac{3}{2} \text{ inches per hour}$
You know the inches and want the hours. Because the rate is constant, each hour burns 3/2 inches, so ask how many groups of 3/2 inches fit into 6 inches. That means dividing by the unit rate, which is positive, so it is not zero. Inches divided by inches per hour gives hours.
$6 \div \frac{3}{2} = 6 \times \frac{2}{3} = \frac{12}{3} = 4 \text{ hours}$
Check by multiplying the unit rate by your answer. The rate is per hour and the time is in hours, so the units match. The result gives back the 6 inches from the problem.
$\frac{3}{2} \text{ inches per hour} \times 4 \text{ hours} = \frac{12}{2} \text{ inches} = 6 \text{ inches}$

Answer

The candle takes $4$ hours to burn down $6$ inches.

Now for comparing. Suppose you make three lemonade mixes from lemon juice and water, and you want to know which one has the strongest lemon taste. You cannot tell from the cups of lemon juice alone, because the cups of water differ too. Each mix is a relationship between two quantities.

A fair comparison puts every mix on the same per-one basis. Choose one quantity to be the per-one quantity for all the mixes, keep the same units, and find each unit rate. Every amount is positive, so no divisor is zero. Then every unit rate answers the same question: how much lemon juice goes with $1$ cup of water? When answers to the same question have the same units, you can compare them directly.

Key concept

Work on a per-one basis

Let $a$ be the amount of a first quantity and $b$ the amount of a second quantity, each with its unit, where $a > 0$ and $b > 0$ . The unit rate $a \div b$ tells how much of the first quantity goes with each $1$ unit of the second. The second quantity is the per-one quantity: it is the divisor, so it must not be zero.

To solve: when the rate is constant, multiply the unit rate by an amount of the per-one quantity to find the first quantity. Or divide an amount of the first quantity by the unit rate to find the per-one quantity. Keep the units matched.
To compare: give both ratios the same per-one quantity and the same units, with matching quantities in matching positions. Then compare the unit rates.

\text{unit rate} = \frac{a}{b} = a \div b \quad \text{for } a > 0 \text{ and } b > 0

Three lemonade mixes, with water as the per-one quantity

Mix	Lemon juice (cups)	Water (cups)	Cups of lemon juice per 1 cup of water
A	$2$	$3$	$2 \div 3 = \frac{2}{3}$
B	$3$	$5$	$3 \div 5 = \frac{3}{5}$
C	$\frac{3}{4}$	$1\frac{1}{4} = \frac{5}{4}$	$\frac{3}{4} \div \frac{5}{4} = \frac{3}{4} \times \frac{4}{5} = \frac{12}{20} = \frac{3}{5}$

Every unit rate in the table has the same unit: cups of lemon juice per $1$ cup of water. Water is the divisor each time, and it is never zero, so you can compare the numbers.

Mixes B and C both come to $\frac{3}{5}$ . To compare $\frac{2}{3}$ with $\frac{3}{5}$ , rewrite both in fifteenths, because $15$ is a multiple of $3$ and of $5$ . Multiplying the top and bottom of a fraction by the same nonzero number keeps its value.

\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \qquad \text{and} \qquad \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}

Since $\frac{10}{15} > \frac{9}{15}$ , Mix A has the most lemon juice for each cup of water, so Mix A has the strongest lemon taste. A greater unit rate means a stronger taste here because lemon juice is the first quantity. Always read the units to see what a greater unit rate means.

Mixes B and C have the same unit rate, so they taste equally strong. Ratios like these, with equal unit rates in the same order and the same units, are equivalent ratios: they show the same relationship with different amounts.

Worked example

Comparing rates with different time units

Two friends walk at constant rates. Priya walks $\frac{3}{4}$ mile in $\frac{1}{4}$ hour. Omar walks $\frac{1}{2}$ mile in $12$ minutes. Who walks faster?

Decide what to compare. Walking faster means covering more miles in each hour, so find miles per hour for both friends. Time in hours is the per-one quantity in both rates. Both times are positive, so neither divisor is zero.
Match the units before you divide. Omar's time is in minutes, but the per-one unit is the hour. Because 60 minutes make 1 hour, 12 minutes is 12/60 of an hour. Dividing the top and bottom by 12 gives 1/5 of an hour, which is the same amount of time.
$12 \text{ minutes} = \frac{12}{60} \text{ hour} = \frac{1}{5} \text{ hour}$
Find Priya's unit rate by dividing her miles by her hours. Dividing by 1/4 is the same as multiplying by its reciprocal, 4/1.
$\frac{3}{4} \div \frac{1}{4} = \frac{3}{4} \times \frac{4}{1} = \frac{12}{4} = 3 \text{ miles per hour}$
Find Omar's unit rate in the same order, so both rates keep miles in the first position and hours in the second. Dividing by 1/5 is the same as multiplying by 5/1.
$\frac{1}{2} \div \frac{1}{5} = \frac{1}{2} \times \frac{5}{1} = \frac{5}{2} = 2\frac{1}{2} \text{ miles per hour}$
Compare the unit rates. Both are in miles per hour, so the greater number means more miles in each hour. Three is greater than two and a half.
$3 > 2\frac{1}{2}$

Answer

Priya walks faster. She walks $3$ miles per hour, and Omar walks $2\frac{1}{2}$ miles per hour.

Takeaway. A unit rate puts a relationship on a per-one basis. When a rate is constant, you can multiply the unit rate by an amount of the per-one quantity, or divide by the unit rate to go the other way. To compare ratios or rates, write them in the same order with the same per-one quantity and the same units, then compare the unit rates. All of this depends on positive quantities, units that you keep and match, and a per-one quantity that is not zero.

Next, you will move on to error analysis with ratios and unit rates.