Multi-Step Rational Number Problems: Choose the Operation | Grade 7 Math | LevelKite

A rational number is a number you can write as a fraction $\frac{a}{b}$ whose numerator $a$ and denominator $b$ are integers, with $b$ not equal to $0$ . (Integers are the counting numbers, their negatives and $0$ .) For example, $-2\frac{1}{4}$ , $0.75$ and $6$ are rational, because $-2\frac{1}{4} = \frac{-9}{4}$ , $0.75 = \frac{3}{4}$ and $6 = \frac{6}{1}$ . You already know how to add, subtract, multiply and divide these numbers. In this lesson you put those skills to work on real situations.

A multi-step problem needs more than one calculation before you reach the answer. The story usually does not tell you which calculations to do. You read how the quantities are related, choose an operation that matches each relationship, and carry each result into the next step. An operation that matches a relationship is said to model the situation.

Here is a plan you can use for a multi-step problem with rational numbers.

Name the quantities. Write what each number measures, its unit, and what a positive or a negative value means. For example, you might decide that a rise is positive and a fall is negative.
Find the steps. Ask what you need to know before you can answer the question. Each thing you must find first is one step.
Work one step at a time. Write each step as one operation. Keep the sign and the unit with the result, and use that result in the next step.
Check. See whether the sign, the size and the unit of your answer fit the story, or solve the problem a second way.

Each step leans on facts you already know about signs and zero. Here they are with their conditions, for any rational numbers $p$ and $q$ .

Subtracting $q$ gives the same result as adding its additive inverse, the opposite number $-q$ . For example, $2 - 5 = 2 + (-5) = -3$ . A number and its opposite add to $0$ , as in $5 + (-5) = 0$ .
The sum $p + q$ is $\lvert q \rvert$ units from $p$ , where $\lvert q \rvert$ is the absolute value of $q$ , its distance from $0$ . You move in the positive direction when $q$ is positive and in the negative direction when $q$ is negative. So $2 + (-5)$ is $5$ units from $2$ in the negative direction, at $-3$ .
When all the numbers you multiply or divide are nonzero, the result is negative exactly when an odd number of them are negative. So $(-2) \times 3 = -6$ , but $(-2) \times (-3) = 6$ .
If any factor (a number being multiplied) is $0$ , the product is $0$ . Zero divided by a nonzero number is $0$ . Division by $0$ is undefined, so every divisor (the number you divide by) and every denominator must be nonzero.

Worked example

A temperature that falls, then rises

At 6 p.m. the temperature in a town is $3.5$ °C. It then falls $1.25$ °C each hour for $4$ hours. After that it rises $0.75$ °C. What is the temperature at the end?

Name the quantities. Temperature is measured in °C, and a negative value means below zero. A fall makes the temperature smaller, so each hour of falling is a change of −1.25 °C. The rise at the end is a change of +0.75 °C. You must find the total fall before you can find the end temperature.
$\text{start} = 3.5, \quad \text{change each hour} = -1.25, \quad \text{last change} = +0.75$
Find the total change over the 4 hours. The same change happens 4 times, so multiply. Both factors are nonzero and exactly one is negative, an odd count, so the product is negative.
$4 \times (-1.25) = -(4 \times 1.25) = -5$
Apply that change to the start. A starting value together with a change is an addition. Adding −5 moves 5 units from 3.5 in the negative direction: the first 3.5 units reach zero and the other 1.5 units go below it.
$3.5 + (-5) = -1.5$
Apply the last change. Adding 0.75 moves 0.75 units from −1.5 in the positive direction. That is less than the 1.5 units needed to reach zero, so the result is still negative.
$-1.5 + 0.75 = -(1.5 - 0.75) = -0.75$
Check a second way. Add the two changes first, then apply the total change to the start. You get the same temperature. A value a little below zero also fits the story: a fall of 5 °C from 3.5 °C, then a small rise.
$-5 + 0.75 = -4.25, \qquad 3.5 + (-4.25) = -0.75$

Answer

The temperature at the end is $-0.75$ °C, which is $0.75$ °C below zero.

In that problem the operations were easy to spot. Often they are not. To choose an operation, think about what each one does with quantities.

Addition joins amounts, or applies a change to a starting value.
Subtraction compares two values, or finds the change from a start to an end as end minus start.
Multiplication repeats one amount a number of times, or takes a part of an amount.
Division shares an amount equally, or counts how many groups of one size make an amount.

The number of times, the part and the group size do not have to be whole numbers. They can be fractions, mixed numbers such as $2\frac{1}{4}$ , or decimals, and the amounts can be negative. A divisor can be any rational number except $0$ : a fraction, a mixed number or a decimal, positive or negative.

Matching a relationship to an operation

What the situation asks	Operation	Example
Apply a change to a starting value	Addition	Start at $-2.5$ °C, then a change of $+4$ °C: $-2.5 + 4 = 1.5$ °C
Find the change from a start to an end	Subtraction	From $-3.5$ m to $2$ m: $2 - (-3.5) = 2 + 3.5 = 5.5$ m
Repeat the same change several times	Multiplication	$6$ minutes at $-0.5$ m per minute: $6 \times (-0.5) = -3$ m
Take a part of an amount	Multiplication	$\frac{2}{3}$ of a $-4.5$ m change: $\frac{2}{3} \times (-4.5) = -3$ m
Share an amount equally	Division (divisor not $0$ )	A $-7.5$ m change shared over $3$ hours: $-7.5 \div 3 = -2.5$ m per hour
Count how many groups of one size make an amount	Division (divisor not $0$ )	$4\frac{1}{2}$ cups in $\frac{3}{4}$ -cup servings: $4\frac{1}{2} \div \frac{3}{4} = 6$ servings

Key concept

The relationship chooses the operation

Break a multi-step problem into single relationships. For each one, ask how the quantity you want is related to the quantities you know. Is a change applied to a start? Are two values compared? Is one amount repeated, or a part of it taken? Is an amount shared equally or counted out in groups? The answer chooses the operation: addition, subtraction, multiplication, or division by a nonzero number. A keyword alone does not choose it. Keep the sign and the unit with each result and carry it into the next step.

Worked example

One tank, two questions, two operations

A water tank leaks, so its water level changes by $-2\frac{1}{4}$ cm each hour. (a) What is the change in the level after $3\frac{1}{2}$ hours? (b) How many hours does it take for the level to change by $-13\frac{1}{2}$ cm?

Name the quantities. Let r be the rate, which is the change in level for each single hour, in cm per hour. It is negative because the level falls. Let t be the time in hours and let c be the total change in cm. The same change repeats every hour, so the total change is the rate times the time.
$r = -2\frac{1}{4}, \qquad c = r \times t$
Part (a) gives the rate and the time and asks for the total change. One amount is repeated for three and a half hours, so multiply. Write each mixed number as a fraction first.
$c = -2\frac{1}{4} \times 3\frac{1}{2} = -\frac{9}{4} \times \frac{7}{2}$
Multiply the numerators and multiply the denominators. Both factors are nonzero and exactly one is negative, an odd count, so the product is negative.
$-\frac{9}{4} \times \frac{7}{2} = -\frac{9 \times 7}{4 \times 2} = -\frac{63}{8} = -7\frac{7}{8}$
Part (b) gives the total change and the rate and asks for the time. You want to know how many copies of the hourly change make the total change, and that is a division. The divisor is the rate, which is not zero, so you can divide.
$t = c \div r = -13\frac{1}{2} \div \left(-2\frac{1}{4}\right) = -\frac{27}{2} \div \left(-\frac{9}{4}\right)$
Dividing by a nonzero number gives the same result as multiplying by its reciprocal, the number whose product with it is 1. The reciprocal of −9/4 is −4/9. Both numbers are nonzero and both are negative, an even count, so the result is positive.
$-\frac{27}{2} \times \left(-\frac{4}{9}\right) = \frac{27 \times 4}{2 \times 9} = \frac{108}{18} = 6$
Check part (b) by multiplying. Six hours at the rate should give the total change. Both factors are nonzero and one is negative, so the product is negative, and it matches. A positive number of hours also fits the story.
$6 \times \left(-\frac{9}{4}\right) = -\frac{54}{4} = -\frac{27}{2} = -13\frac{1}{2}$

Answer

(a) After $3\frac{1}{2}$ hours the level has changed by $-7\frac{7}{8}$ cm, which is a fall of $7\frac{7}{8}$ cm. (b) The level changes by $-13\frac{1}{2}$ cm in $6$ hours.

A keyword does not choose the operation. The tank problem says each hour. A student who thinks each always means multiply might answer part (b) like this: $-13\frac{1}{2} \times \left(-2\frac{1}{4}\right) = \frac{243}{8} = 30\frac{3}{8}$ hours.

This fails because the rate is multiplied by a time to give a total change. Here $-13\frac{1}{2}$ is not a time. It is the total change, and the question asks how many hours of leaking make it. That is a division by the nonzero rate. A check exposes the mistake: $30\frac{3}{8}$ hours at $-2\frac{1}{4}$ cm per hour gives a change of about $-68$ cm, not $-13\frac{1}{2}$ cm. In the temperature problem each hour led to a multiplication. Here it leads to a division. Read the relationship, not only the word.

Worked example

A climb that uses all four operations

A hiker starts in a valley at an elevation of $-42.5$ m, which means $42.5$ m below sea level. She climbs at a steady rate for $2\frac{1}{2}$ hours and reaches $107.5$ m. She then keeps climbing at the same rate for another $\frac{3}{4}$ hour. What is her final elevation?

Name the quantities and find the steps. Elevation is height in meters measured from sea level, which is 0, so a negative value means below sea level. To find the final elevation you need the change in the last part of the climb. For that you need the rate, and for the rate you need the change in the first part.
Find the change in the first part. A change is the end value minus the start value, so subtract. Subtracting −42.5 gives the same result as adding its opposite, 42.5.
$107.5 - (-42.5) = 107.5 + 42.5 = 150$
Find the rate. It is the change for each single hour, so divide the 150 m by the number of hours. That divisor is not zero, and dividing by it gives the same result as multiplying by its reciprocal. Both numbers are nonzero and neither is negative, so the rate is positive: 60 m per hour.
$150 \div 2\frac{1}{2} = 150 \div \frac{5}{2} = 150 \times \frac{2}{5} = \frac{300}{5} = 60$
Find the change in the last part. The hourly change is repeated for three quarters of an hour, so multiply the rate by the time.
$60 \times \frac{3}{4} = \frac{180}{4} = 45$
Apply that change to the elevation she had reached. A starting value together with a change is an addition, and adding 45 moves 45 units in the positive direction.
$107.5 + 45 = 152.5$
Check a second way. As decimals, the two parts of the climb last 2.5 hours and 0.75 hours. Add them, multiply the total time by the rate, and apply that change to the starting elevation. The result is the same.
$2.5 + 0.75 = 3.25, \quad 3.25 \times 60 = 195, \quad -42.5 + 195 = 152.5$

Answer

The hiker's final elevation is $152.5$ m, which is $152.5$ m above sea level.

Takeaway. A multi-step problem is a chain of single relationships. Name each quantity with its unit and the meaning of its sign. For each link, let the relationship choose the operation: add to apply a change, subtract to find a change, multiply to repeat an amount or take a part of it, and divide by a nonzero number to share equally or count groups. Carry each signed result into the next step, then check the sign, size and unit of your answer against the story.

Next comes Error analysis: rational number computations, where you will find and fix mistakes in worked solutions.