Proportion Word Problems and Population Estimates | Grade 7 Math | LevelKite

You already know how to solve a proportion that is written out for you. A proportion is an equation that says two ratios are equal, such as $\frac{2}{3} = \frac{8}{12}$ . In this lesson you build the proportion yourself from a situation described in words.

Every ratio here compares two positive quantities, and each quantity keeps its unit, such as dollars or kilograms. A rate is a ratio of two quantities with different units, such as $10$ dollars for $4$ kilograms.

A situation fits a proportion when the two quantities are related in the same way in two cases. Two purchases at the same price per kilogram are one example. Two trips at the same speed are another. Each case gives one ratio. Because the price per kilogram or the speed does not change from one case to the other, the two ratios are equal. That equality is your proportion.

Here is a plan for the word problems in this lesson, where two cases share the same ratio or rate:

Name the two quantities and their units, such as dollars and kilograms.
Define a letter for the unknown. Say what it measures and in which unit.
Set up one ratio for each case. Put one quantity on top in both ratios and the other quantity on the bottom in both. Give each quantity the same unit in both ratios.
Solve the proportion.
Check that both cases give the same ratio or rate, then state the answer with its unit.

Step 3 is where the thinking happens. Quantities that measure the same thing in the two cases, such as the two costs, are called corresponding quantities. They must sit in matching positions: both on top or both on the bottom. If they do not, the equation no longer describes the situation.

Key concept

Matching positions make the proportion true

When two cases share the same ratio or rate, you can write a proportion, as long as all of these hold:

Every quantity is positive and keeps its unit.
Corresponding quantities sit in matching positions: one quantity on top in both ratios, the other on the bottom in both.
Each quantity has the same unit in both ratios, such as minutes in both.

In the formula, $a$ and $c$ measure one quantity, and $b$ and $d$ measure the other. The pair $a$ , $b$ comes from the first case and $c$ , $d$ from the second. Dividing top by bottom gives the amount per one unit of the bottom quantity. The bottom quantity is the divisor (the number you divide by), so it must not be zero.

\frac{a}{b} = \frac{c}{d}, \quad a, b, c, d > 0

Worked example

Same price, more apples

At a market, $4$ kilograms of apples cost $10$ dollars. At the same price per kilogram, how much do $7$ kilograms of apples cost?

Name the two quantities: the cost in dollars and the amount of apples in kilograms. Both are positive. Let $c$ be the cost, in dollars, of $7$ kilograms of apples. The price per kilogram is the same for both purchases, so the two dollars-to-kilograms ratios are equal.
Write one ratio for each purchase, with dollars on top and kilograms on the bottom in both. Now the corresponding quantities sit in matching positions, and each unit matches across the equals sign.
$\frac{10 \text{ dollars}}{4 \text{ kilograms}} = \frac{c \text{ dollars}}{7 \text{ kilograms}}$
Multiply both sides by $7$ . Multiplying both sides of an equation by the same number that is not zero keeps the equation true. Here it leaves $c$ alone on the right.
$\frac{10}{4} \cdot 7 = \frac{c}{7} \cdot 7 \quad\Rightarrow\quad \frac{70}{4} = c$
Work out $70 \div 4$ to write the cost as a decimal number of dollars.
$c = 70 \div 4 = 17.5$
Check the price for $1$ kilogram in each purchase. Divide dollars by kilograms; the kilograms are the divisor and are not zero. Both purchases give $2.5$ , which is $2.50$ dollars per kilogram, so they share the same rate.
$10 \div 4 = 2.5 \qquad 17.5 \div 7 = 2.5$

Answer

Seven kilograms of apples cost $17.50$ dollars.

Worked example

When the unknown is on the bottom

A machine fills $150$ bottles in $4$ minutes. At the same rate, how many minutes does it take to fill $525$ bottles?

Name the two quantities: the number of bottles and the time in minutes. Both are positive. Let $t$ be the time, in minutes, needed to fill $525$ bottles. The filling rate is the same in both cases, so the two bottles-to-minutes ratios are equal.
Write bottles on top and minutes on the bottom in both ratios. The unknown $t$ is a time, so it goes on the bottom, in the same position as the $4$ minutes.
$\frac{150 \text{ bottles}}{4 \text{ minutes}} = \frac{525 \text{ bottles}}{t \text{ minutes}}$
Multiply both sides by $4t$ . The equation stays true because $4t$ is not zero: $t$ is a positive time. On the left the $4$ divides out and leaves $150t$ . On the right the $t$ divides out and leaves $525 \cdot 4$ .
$150t = 525 \cdot 4 = 2100$
Divide both sides by $150$ . Dividing both sides by the same number that is not zero keeps the equation true, and it leaves $t$ alone.
$t = 2100 \div 150 = 14$
Check the number of bottles filled in $1$ minute in each case. Divide bottles by minutes; the minutes are the divisor and are not zero. Both cases give $37.5$ bottles per minute, so they share the same rate.
$150 \div 4 = 37.5 \qquad 525 \div 14 = 37.5$

Answer

The machine takes $14$ minutes to fill $525$ bottles.

A proportion can also help you count a group that is too hard to count one by one, such as the fish in a pond. The whole group you want to count is the population. A sample is a smaller group taken from the population.

Here is the idea. Suppose you catch $20$ fish, put a small marker called a tag on each one, and release them. Later you net a sample of $10$ fish, and $2$ of them have tags. In the sample, $2$ out of $10$ fish are tagged. That is $\frac{1}{5}$ of the sample. If the tagged fish have mixed evenly with the others, then about $\frac{1}{5}$ of all the fish in the pond are tagged too. So the $20$ tagged fish are about $\frac{1}{5}$ of the pond, and the pond holds about $5 \cdot 20 = 100$ fish.

That answer is an estimate: a value you work out from evidence in place of an exact count. The table that follows sorts the same numbers. In it, $N$ stands for the number of fish in the pond.

Each column gives one ratio: tagged fish to all fish

Quantity	In the sample	In the whole pond
Tagged fish	$2$	$20$
All fish	$10$	$N$ (unknown)

Read each column from top to bottom to get a ratio of tagged fish to all fish: $\frac{2}{10}$ for the sample and $\frac{20}{N}$ for the pond. Setting the two ratios equal gives $\frac{2}{10} = \frac{20}{N}$ . The value $N = 100$ makes it true, because $\frac{20}{100} = \frac{2}{10}$ .

The same pattern gives an estimate for other populations, as long as the tagged animals have mixed evenly with the rest and the sample contains at least one tagged animal. Put the tagged animals on top in both ratios and all the animals in that group on the bottom in both:

\frac{\text{tagged animals in sample}}{\text{all animals in sample}} = \frac{\text{tagged animals in population}}{\text{all animals in population}}

Worked example

Estimating turtles in a lake

Rangers catch $60$ turtles at a lake, tag each one, and release them. A month later they catch a sample of $45$ turtles, and $9$ of them have tags. Estimate the number of turtles in the lake.

Let $N$ be the number of turtles in the lake, a positive count. Sort the three given numbers: $60$ tagged turtles in the whole lake, $45$ turtles in the sample, and $9$ tagged turtles in the sample.
Put tagged turtles on top and all turtles on the bottom in both ratios, with the sample on the left and the lake on the right. This treats the sample as a fair picture of the lake, which is reasonable if the tagged turtles have mixed evenly with the others.
$\frac{9 \text{ tagged turtles}}{45 \text{ turtles}} = \frac{60 \text{ tagged turtles}}{N \text{ turtles}}$
Multiply both sides by $45N$ , as you multiplied by $4t$ in the bottle problem. The equation stays true because $45N$ is not zero: $N$ is positive. On the left the $45$ divides out and leaves $9N$ . On the right the $N$ divides out and leaves $60 \cdot 45$ .
$9N = 60 \cdot 45 = 2700$
Divide both sides by $9$ , which is not zero, so the equation stays true and $N$ is alone.
$N = 2700 \div 9 = 300$
Check that the tagged share is the same in both groups. Divide tagged turtles by all turtles; each divisor is not zero. Both shares are $0.2$ , so the proportion is true.
$9 \div 45 = 0.2 \qquad 60 \div 300 = 0.2$

Answer

There are about $300$ turtles in the lake. This is an estimate, not an exact count.

Takeaway. A proportion word problem describes two cases that share the same ratio or rate. Name the positive quantities with their units, define the unknown, and put corresponding quantities in matching positions with matching units. Then solve, and check that both cases give the same ratio or rate.

To estimate a population, use a sample taken after the tagged animals have mixed evenly with the rest. Then the ratio of tagged animals to all animals in the sample is about the same as in the whole population. Set those two ratios equal and solve. The result is an estimate, not an exact count.

Next, you will solve multi-step ratio problems and analyze errors in setting up and solving proportions.