Multi-Step Ratio Problems and Finding Errors | Grade 7 Math | LevelKite

You already know how to solve a word problem with one proportion, an equation that says two ratios are equal. A ratio compares two quantities by division, such as 3 grams of raisins for every 5 grams of nuts. In this lesson, every quantity is a positive amount and every number keeps its unit.

This lesson adds two skills:

You will solve multi-step ratio problems. These need more than one calculation. Often you must first find a quantity that the problem does not give you directly.
You will do error analysis. That means you read a finished solution, find the exact place where it goes wrong, explain why, and fix it.

Both skills grow from one habit: know what every number measures before you calculate with it.

Picture a multi-step ratio problem as a chain. Each link joins two positive quantities that stay in the same ratio, such as liters and minutes. The question asks about a quantity at one end of the chain. The facts you are given may start at the other end.

Some links are rates. A rate is a ratio of two quantities with different units, such as 18 liters in 4 minutes. In these problems, words such as always, for every and at a constant rate tell you that the ratio stays the same in every situation. That is what lets you use a proportion.

Plan the chain before you calculate:

Write what the question asks for, with its unit.
Write each fact as a pair of quantities with their units.
Ask yourself: Do my facts and my question use the same two quantities? If they do not, find the missing quantity first.

Key concept

Match what the numbers measure

In this lesson, each side of a proportion is the ratio for one situation. Put the same quantity, with the same unit, on top on both sides and its partner on the bottom. Then corresponding quantities, which measure the same thing in both situations, sit in matching positions.

Dividing a top by its bottom gives a unit rate: the amount of quantity A for 1 unit of quantity B. Quantity B is the per-one quantity, so you divide by it, and it must not be zero. A correct proportion says that both situations have the same unit rate.

This holds for positive quantities that keep their units and stay in a constant ratio. In a multi-step problem, check the match at every link. If a given number does not measure what a position needs, first find the number that does.

\frac{\text{quantity A, situation 1}}{\text{quantity B, situation 1}}=\frac{\text{quantity A, situation 2}}{\text{quantity B, situation 2}}

Worked example

Example 1: From one part to the whole

A trail mix is made only of raisins and nuts. It always has 3 grams of raisins for every 5 grams of nuts. You want 240 grams of trail mix in all. How many grams of raisins do you need?

Compare the facts with the question. The ratio 3 to 5 compares raisins with nuts, but 240 grams is the mass of the whole mix. These are not the same two quantities, so first find a ratio that uses the whole mix. All the masses are positive and in grams.
The mix is only raisins and nuts, and both are measured in grams, so add the two parts. The recipe never changes, so there are always 3 grams of raisins in every 8 grams of mix.
$3\text{ grams} + 5\text{ grams} = 8\text{ grams of mix}$
Let r be the number of grams of raisins in 240 grams of mix. Put grams of raisins on top and grams of mix on the bottom on both sides. Then corresponding quantities sit in matching positions.
$\frac{3\text{ grams raisins}}{8\text{ grams mix}} = \frac{r\text{ grams raisins}}{240\text{ grams mix}}$
Multiply both sides by 240. Multiplying both sides by the same number that is not zero gives an equation with the same solution, and here it undoes the division by 240 on the right.
$r = \frac{3 \times 240}{8} = \frac{720}{8} = 90$
Check with the original ratio. The nuts are the rest of the mix. Divide both masses by 30, which is not zero, to see that raisins to nuts is still 3 to 5.
$240 - 90 = 150, \quad 90 : 150 = 3 : 5$

Answer

You need $90$ grams of raisins to make $240$ grams of trail mix.

Example 1 as a table: one column for each quantity, one row for each situation

Situation	Raisins (grams)	Nuts (grams)	Whole mix (grams)
Given ratio	$3$	$5$	$3 + 5 = 8$
10 times as much	$30$	$50$	$30 + 50 = 80$
Your batch, 30 times as much	$90$	$150$	$90 + 150 = 240$

Worked example

Example 2: A chain with two links

A rice recipe always uses 2 cups of rice for every 5 servings. One cup of this rice has a mass of 180 grams. How many grams of rice do you need for 35 servings?

Plan the chain. The question asks for grams of rice. One fact links cups to servings and the other links grams to cups. No fact links grams to servings directly, so first find the cups for 35 servings. Then change those cups to grams. All the amounts are positive.
$35\text{ servings} \rightarrow \text{cups} \rightarrow \text{grams}$
Let c be the number of cups of rice for 35 servings. Put cups on top and servings on the bottom on both sides, so corresponding quantities are in matching positions.
$\frac{2\text{ cups}}{5\text{ servings}} = \frac{c\text{ cups}}{35\text{ servings}}$
Multiply both sides by 35, which is not zero, to leave c alone.
$c = \frac{2 \times 35}{5} = \frac{70}{5} = 14$
Let m be the mass, in grams, of 14 cups of rice. One cup has a mass of 180 grams. Put grams on top and cups on the bottom on both sides.
$\frac{180\text{ grams}}{1\text{ cup}} = \frac{m\text{ grams}}{14\text{ cups}}$
Multiply both sides by 14, which is not zero. Split 14 into 10 and 4 to multiply.
$m = 180 \times 14 = 1800 + 720 = 2520$
Check each link with a unit rate. Divide by the per-one quantity: servings in the first link and cups in the second. Neither divisor, the number you divide by, is zero. You get 0.4 cups per serving and 180 grams per cup, the same rates as the given facts.
$\frac{14}{35} = 0.4 = \frac{2}{5}, \quad \frac{2520}{14} = 180$

Answer

You need $2520$ grams of rice for $35$ servings.

Now switch roles. In error analysis, someone has already solved the problem and you are the checker. An error can hide in two places:

The set-up. The set-up is the proportion you write before you solve. With each side written as the ratio for one situation, it is wrong when a quantity is on top on one side and on the bottom on the other, or when a number measures the wrong thing.
The solving. The set-up is right, but a later step does not keep the two sides equal.

Check the set-up first. If it is wrong, correct arithmetic afterward only solves the wrong equation.

When you report an error, say three things: where it is, why it fails, and how to fix it. The three checks in the table below help you find the where.

Three checks for a proportion (one situation per side) with positive quantities and a constant ratio

Check	What you do	What it finds
Position check	Label every number with what it measures and its unit. Each quantity keeps the same position on both sides.	A set-up error
Equal-rate check	Once positions match, put the answer in. Divide each top by its bottom (not zero). The unit rates must be equal.	A solving error
Size check	Compare with the given pair. More of one quantity needs more of its partner; less needs less.	An answer that shrinks when it should grow, or grows when it should shrink

Worked example

Example 3: Find and fix two errors

A pump fills 18 liters in 4 minutes at a constant rate. How many liters does it fill in 14 minutes? Two students let $x$ be that number of liters. Ana writes $\frac{18}{4} = \frac{14}{x}$ and finds that $x$ is about $3.1$ . Ben writes $\frac{18}{4} = \frac{x}{14}$ . Then he says that $4 + 10 = 14$ , so $x = 18 + 10 = 28$ . Find each error and give the correct answer.

Do the position check on Ana's set-up: label each number with its unit. On the left, liters are on top. On the right, liters are on the bottom. Corresponding quantities are not in matching positions, so Ana has a set-up error.
$\text{left: } \frac{18\text{ liters}}{4\text{ minutes}}, \quad \text{right: } \frac{14\text{ minutes}}{x\text{ liters}}$
Do the position check on Ben's set-up. Liters are on top on both sides and minutes are on the bottom on both sides. His set-up is correct, and it is also the fix for Ana's. Look for Ben's error in the solving.
$\text{left: } \frac{18\text{ liters}}{4\text{ minutes}}, \quad \text{right: } \frac{x\text{ liters}}{14\text{ minutes}}$
Ben's 28 liters is more than 18 liters, so the size check does not catch it. Use the equal-rate check. Minutes are the per-one quantity, so divide liters by minutes. Neither divisor is zero. The unit rates, 4.5 and 2 liters per minute, are not equal. Ben has a solving error: adding 10 to both quantities changed the rate.
$\frac{18}{4} = 4.5, \quad \frac{28}{14} = 2$
Solve Ben's correct proportion. Multiply both sides by 14, which is not zero, to leave x alone.
$x = \frac{18 \times 14}{4} = \frac{252}{4} = 63$
Check the new answer. Its unit rate is 4.5 liters per minute, the same as the given rate. It also passes the size check: more minutes at the same rate must give more than 18 liters. Ana's 3.1 liters fails that check.
$\frac{63}{14} = 4.5, \quad 63 > 18$

Answer

Ana's error is in the set-up: liters are on top on one side and on the bottom on the other. Ben's error is in the solving: he added the same amount to both quantities, which changes the rate. The pump fills $63$ liters in $14$ minutes.

Takeaway. A multi-step ratio problem is a chain of links. Plan the chain, find any quantity the problem leaves out, and at every link keep corresponding quantities, with their units, in matching positions. To check a solution, test the set-up first, then test the answer with unit rates, and then ask whether its size makes sense. Use all three checks, because each one alone can miss an error. All of this relies on positive quantities, a constant ratio, and a divisor that is not zero.