**2.2 Use a Solving Problem Strategy**

Topics covered in this section are:

- Use a problem solving strategy for word problems
- Solve number word problems
- Solve percent applications
- Solve simple interest applications

**2.2.1 Use a Problem Solving Strategy for Word Problems**

Now that we can solve equations, we are ready to apply our new skills to word problems. We will develop a strategy we can use to solve any word problem successfully.

**Example 1**

Normal yearly snowfall at the local ski resort is $12$ inches more than twice the amount it received last season. The normal yearly snowfall is $62$ inches. What was the snowfall last season at the ski resort?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | What was the snowfall last season? |

Step 3. Name what we are looking for andchoose a variable to represent it. | Let $s=$ the snowfall last season. |

Step 4. Translate.Restate the problem in one sentence with all the important information. | |

Translate into an equation. | $62=2s+12$ |

Step 5. Solve the equation. | $62=2s+12$ |

Subtract $12$ from each side. | $62-12=2s+12-12$ |

Simplify. | $50=2s$ |

Divide each side by two. | $\frac{50}{2}=\frac{2s}{2}$ |

Simplify. | $25=s$ |

Step 6. Check: First, is our answer reasonable?Yes, having $25$ inches of snow seems OK. The problem says the normal snowfall is twelve inches more than twice the number of last season. Twice $25$ is $50$ and $12$ more than that is $62$. | |

Step 7. Answer the question. | The snowfall last season was $25$ inches. |

We summarize an effective strategy for problem solving.

**HOW TO: Use a Problem Solving Strategy for word problems.**

**Read**the problem. Make sure all the words and ideas are understood.**Identify**what you are looking for.**Name**what you are looking for. Choose a variable to represent that quantity.**Translate**into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.**Solve**the equation using proper algebra techniques.**Check**the answer in the problem to make sure it makes sense.**Answer**the question with a complete sentence.

**2.2.2 Solve Number Word Problems**

We will now apply the problem solving strategy to “number word problems.” Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

**Example 2**

The sum of seven times a number and eight is thirty-six. Find the number.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | the number |

Step 3. Name what you are looking for andchoose a variable to represent it. | Let $n=$ the number. |

Step 4. Translate:Restate the problem as one sentence. Translate into an equation. | $7n+8=36$ |

Step 5. Solve the equation.Subtract eight from each side and simplify. Divide each side by seven and simplify. | $7n+8=36$ $7n=28$ $n=4$ |

Step 6. Check.Is the sum of seven times four plus eight equal to $36$? $\begin{align*} 7 \cdot 4 +8&=36 \\ 28+8&=36 \\ 36&=36 \end{align*}$ | |

Step 7. Answer the question. | The number is 4. |

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far, we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

**Example 3**

The sum of two numbers is negative fifteen. One number is nine less than the other. Find the numbers.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | two numbers |

Step 3. Name what you are looking for by choosing a variable to represent the first number.“One number is nine less than the other.” | Let $n=$ first number. $n-9=$ second number. |

Step 4. Translate.Write as one sentence. Translate into an equation. | The sum of two numbers is negative fifteen. |

Step 5. Solve the equation.Combine like terms. Add nine to each side and simplify. Simplify. | $n+n-9=-15$ $2n-9=15$ $2n=-6$ $n=-3$ first number $n-9$ second number $-3-9$ $-12$ |

Step 6. Check.Is $-12$ nine less than $-3$? $\begin{align*} -3-9&=-12 \\ -12&=-12 \end{align*}$ Is their sum $-15$? $\begin{align*} -3+(-12)&=-15 \\ -15&=-15 \end{align*}$ | |

Step 7. Answer the question. | The numbers are $-3$ and $-12$. |

Some number problems involve **consecutive integers**. Consecutive integers are integers that immediately follow each other. Examples of **consecutive integers** are:

$1,\ 2, \ 3, \ 4$ | ||

$-10,\ -9, \ -8, \ -7$ | ||

$150,\ 151, \ 152, \ 153$ |

Notice that each number is one more than the number preceding it. Therefore, if we define the first integer as $n$, the next consecutive integer is $n+1$. The one after that is one more than $n+1$, so it is $n+1+1$, which is $n+2$.

$n$ | $1^{st}$ integer |

$n+1$ | $2^{nd}$ consecutive integer |

$n+2$ | $3^{rd}$ consecutive integer etc. |

We will use this notation to represent consecutive integers in the next example.

**Example 4**

Find three consecutive integers whose sum is $-54$.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | three consecutive integers |

Step 3. Name each of the three numbers | Let $n=1^{st}$ integer. $n+1=2^{nd}$ consecutive integer. $n+2=3^{rd}$ consecutive integer. |

Step 4. Translate.Restate as one sentence. Translate into an equation. | The sum of the three integers is $-54$. $n+n+1+n+2=-54$ |

Step 5. Solve the equation.Combine like terms. Subtract three from each side. Divide each side by three. | $\begin{align*} n+n+1+n+2&=-54 \\ 3n+3&=-54 \\ 3n&=-57 \\ n&=-19 \ \ 1^{st} integer \\ n&+1 \ \ \ 2^{nd} integer \\ -19&+1 \\ -&18 \\ n&+2 \ \ \ 3^{rd} integer \\ -19&+2 \\-&17\\ \end{align*}$ |

Step 6. Check.$\begin{align*} -19+(-18)+(-17) & = -54 \\ -54&=-54 \end{align*}$ | |

Step 7. Answer the question. | The three consecutive integers are $-17$, $-18$, and $-19$. |

Now that we have worked with consecutive integers, we will expand our work to include **consecutive even integers** and **consecutive odd integers**. Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

$24$, $26$, $28$

$-12$, $-10$, $-8$

Notice each integer is two more than the number preceding it. If we call the first one $n$, then the next one is $n+2$. The one after that would be $n+2+2$ or $n+4$.

$n$ | $1^{st}$ even integer |

$n+2$ | $2^{nd}$ consecutive even integer |

$n+4$ | $3^{rd}$ consecutive even integer etc. |

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers $63$, $65$, and $67$.

$63$, $65$, $67$

$n$, $n+2$, $n+4$

$n$ | $1^{st}$ odd integer |

$n+2$ | $2^{nd}$ consecutive odd integer |

$n+4$ | $3^{rd}$ consecutive odd integer etc. |

Does it seem strange to have to add two (an even number) to get the next odd number? Do we get an odd number or an even number when we add $2$ to $3$? to $11$? to $47$?

Whether the problem asks for consecutive even numbers or odd numbers, you do not have to do anything different. The pattern is still the same—to get to the next odd or the next even integer, add two.

**Example 5**

Find three consecutive even integers whose sum is $120$.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | three consecutive even numbers |

Step 3. Name. | Let $n=1^{st}$ even integer. $n+2=2^{nd}$ consecutive even integer $n+4=3^{rd}$ consecutive even integer |

Step 4. Translate.Restate as one sentence. Translate into an equation. | The sum of three even integers is $120$. $n+n+2+n+4=120$ |

Step 5. Solve the equation.Combine like terms. Subtract $6$ from each side. Divide each side by $3$. | $\begin{align*} n+n+2+n+4&=120 \\ 3n+6&=120 \\ 3n&=114 \\ n&=38 \ \ \ 1^{st} \ integer \\ \\ n&+2 \ \ 2^{nd} \ integer \\ 38&+2 \\ &40 \\ \\ n&+4 \ \ 3^{rd} \ integer \\ 38&+4 \\ &42 \\ \end{align*}$ |

Step 6. Check.$\begin{align*} 38+40+42&=120 \\ 120&=120 \end{align*}$ | |

Step 7. Answer the question. | The three consecutive even integers are $38$, $40$, and $42$. |

When a number problem is in a real life context, we still use the same strategies that we used for the previous examples.

**Example 6**

A couple together earns $ \$110,000$ a year. The wife earns $ \$ 16,000$ less than twice what her husband earns. What does the husband earn?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what you are looking for. | How much does the husband earn? |

Step 3. Name.Choose a variable to represent the amount the husband earns. The wife earns $ \$16,000$ less than twice that. | Let $h=$ the amount the husband earns. |

Step 4. Translate.Restate the problem in one sentence with all the important information. Translate into an equation. | $2h-16,000 =$ the amount the wife earns. Together the husband and wife earn $ \$110,000$. |

Step 5. Solve the equation.Combine like terms. Add $16,000$ to both sides and simplify. Divide each side by three. | $\begin{align*} h+2h-16,000&=110,000 \\ 3h-16,000&=110,000 \\ 3h&=126,000 \\ h&=42,000 \end{align*}$ |

$ \$ 42,000$ amount husband earns | |

Step 6. Check:If the wife earns $ \$68,000$ and the husband earns $ \$42,000$, is that $ \$110,000$? Yes! | $2h-16,000$ amount wife earns $2(42,000)-16,000$ $84,000-16,000$ $68,000$ |

Step 7. Answer the question. | The husband earns $ \$42,000$ a year. |

**2.2.3 Solve Percent Applications**

There are several methods to solve percent equations. In algebra, it is easiest if we just translate English sentences into algebraic equations and then solve the equations. Be sure to change the given percent to a decimal before you use it in the equation.

**Example 7**

Translate and solve:

- What number is $45\%$ of $84$?
- $8.5\%$ of what amount is $ \$4.76$?
- $168$ is what percent of $112$?

**Solution**

**Part 1.**

Translate into algebra. Let $n=$ the number. | $n = 0.45 \cdot 84$ |

Multiply. | $n=37.8$ |

$37.8$ is $45\%$ of $84$. |

**Part 2.**

Translate. Let $n=$ the amount. | $0.085 \cdot n = \$ 4.76$ |

Multiply. | $0.085n=4.76$ |

Divide both sides by $0.085$ and simplify. | $n=56$ |

$8.5\%$ of $ \$56$ is $ \$4.76$ |

**Part 3.**

We are asked to find percent, so we must have our result in percent form. | |

Translate into algebra. Let $p=$ the percent. | $168 = p \cdot 112$ |

Multiply. | $168=112p$ |

Divide both sides by $112$ and simplify. | $1.5=p$ |

Convert to percent. | $150\%=p$ |

$168$ is $150\%$ of $112$ |

Now that we have a problem solving strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we will solve involve everyday situations, you can rely on your own experience.

**Example 8**

The label on Audrey’s yogurt said that one serving provided $12$ grams of protein, which is $24\%$ of the recommended daily amount. What is the total recommended daily amount of protein?

**Solution**

What are you asked to find? | What total amount of protein is recommended? |

Choose a variable to represent it. | Let $a=$ total amount of protein. |

Write a sentence that gives the information to find it. | |

Translate into an equation. | $12=0.24 \cdot a$ |

Solve. | $50=a$ |

Check: Does this make sense? Yes, $24\%$ is about $\frac{1}{4}$ of the total and $12$ is about $\frac{1}{4}$ of $50$. | |

Write a complete sentence to answer the question. | The amount of protein that is recommended is $50$ g. |

Remember to put the answer in the form requested. In the next example we are looking for the percent.

**Example 9**

Veronica is planning to make muffins from a mix. The package says each muffin will be $240$ calories and $60$ calories will be from fat. What percent of the total calories is from fat?

**Solution**

What are you asked to find? | What percent of the total calories is fat? |

Choose a variable to represent it. | Let $p=$ percent of fat. |

Write a sentence that gives the information to find it. | |

Translate the sentence into an equation. | $p \cdot 240 = 60$ |

Multiply. | $240p=60$ |

Divide both sides by $240$. | $p=0.25$ |

Put in percent form. | $p=25\%$ |

Check: Does this make sense? Yes, $25\%$ is one-fourth; $60$ is one-fourth of $240$. So, $25\%$ makes sense. | |

Write a complete sentence to answer the question. | Of the calories in each muffin, $25\%$ is fat. |

It is often important in many fields—business, sciences, pop culture—to talk about how much an amount has increased or decreased over a certain period of time. This increase or decrease is generally expressed as a percent and called the **percent change**.

To find the percent change, first we find the amount of change, by finding the difference of the new amount and the original amount. Then we find what percent the amount of change is of the original amount.

**HOW TO: Find percent change.**

- Find the amount of change.

change = new amount−original amount - Find what percent the amount of change is of the original amount.

change is what percent of the original amount?

**Example 10**

Recently, the California governor proposed raising community college fees from $ \$36$ a unit to $ \$46$ a unit. Find the percent change. (Round to the nearest tenth of a percent.)

**Solution**

Find the amount of change. | $46-36=10$ |

Find the percent. | Change is what percent of the original amount? |

Let $p=$ the percent. | |

Translate to an equation. | $10=p \cdot 36$ |

Simplify. | $10=36p$ |

Divide both sides by $36$. | $0.278=p$ |

Change to percent form; round to the nearest tenth | $27.8\%=p$ |

Write a complete sentence to answer the question. | The new fees are approximately a $27.8\%$ increase over the old fees. |

Remember to round the division to the nearest thousandth in order to round the percent to the nearest tenth. |

Applications of discount and mark-up are very common in retail settings.

When you buy an item on sale, the original price has been discounted by some dollar amount. The **discount rate**, usually given as a percent, is used to determine the amount of the discount. To determine the **amount of discount**, we multiply the discount rate by the original price.

The price a retailer pays for an item is called the **original cost**. The retailer then adds a **mark-up** to the original cost to get the **list price**, the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

**DISCOUNT**

amount of discount$=$discount rate $\cdot$ original price

sale price$=$original amount$–$discount price

The sale price should always be less than the original price.

**MARK-UP**

amount of mark-up$=$mark-up rate $\cdot$ original price

list price$=$original amount$+$mark-up

The list price should always be more than the original cost.

**Example 11**

Liam’s art gallery bought a painting at an original cost of $ \$750$. Liam marked the price up $40\%$. Find:

- the amount of mark-up
- the list price of the painting

**Solution**

**Part 1.**

Identify what you are asked to find, and choose a variable to represent it. | What is the amount of mark-up? Let $m=$ the amount of mark-up. |

Write a sentence that gives the information to find it. | |

Translate into an equation. | $m=0.40 \times 750$ |

Solve the equation. | $m=300$ |

Write a complete sentence. | The mark-up on the painting was $ \$300$. |

**Part 2.**

Identify what you are asked to find, and choose a variable to represent it. | What is the list price? Let $p=$ the list price. |

Write a sentence that gives the information to find it. | |

Translate into an equation. | $p=750+300$ |

Solve the equation. | $p=1,050$ |

Check. | Is the list price more than the original cost? Is $ \$1,050$ more than $ \$750$? Yes. |

Write a complete sentence. | The mark-up on the painting was $ \$300$. |

**2.2.4 Solve Simple Interest Applications**

Interest is a part of our daily lives. From the interest earned on our savings to the interest we pay on a car loan or credit card debt, we all have some experience with interest in our lives.

The amount of money you initially deposit into a bank is called the **principal**, $P$, and the bank pays you **interest**, $I$. When you take out a loan, you pay interest on the amount you borrow, also called the principal.

In either case, the interest is computed as a certain percent of the principal, called the **rate of interest**, $r$. The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable$t$, (for time) represents the number of years the money is saved or borrowed.

Interest is calculated as simple interest or compound interest. Here we will use simple interest.

**SIMPLE INTEREST**

If an amount of money, $P$, called the principal, is invested or borrowed for a period of *t* years at an annual interest rate $r$, the amount of interest, $I$, earned or paid is

$I=Prt$ | where | $I =$ interest $P=$ principal $r=$ rate $t=$ time |

Interest earned or paid according to this formula is called **simple interest**.

The formula we use to calculate interest is $I=Prt$. To use the formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

**Example 12**

Areli invested a principal of $\$950$ in her bank account that earned simple interest at an interest rate of $3\%$. How much interest did she earn in five years?

**Solution**

$I=$ ?

$P= \$950$

$r= 3\%$

$t=5$ years

Identify what you are asked to find, and choose a variable to represent it. | What is the simple interest? Let $I=$ interest. |

Write the formula. | $I=Prt$ |

Substitute in the given information. | $I=(950)(0.03)(5)$ |

Simplify. | $I=142.5$ |

Check. Is $\$142.50$ a reasonable amount of interest on $\$950$? Yes. | |

Write a complete sentence. | The interest is $\$142.50$. |

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we do not know the rate.

**Example 13**

Hang borrowed $\$7,500$ from her parents to pay her tuition. In five years, she paid them $\$1,500$ interest in addition to the $\$7,500$ she borrowed. What was the rate of simple interest?

**Solution**

$I= \$1500$

$P=\$7500$

$r=$?

$t=5$ years

Identify what you are asked to find, and choose a variable to represent it. | What is the rate of simple interest? Let $r=$ rate of interest. |

Write the formula. Substitute in the given information. Multiply. Divide. Change to percent form. | $\begin{align*} I&=Prt \\ 1,500&=(7,500)r(5) \\ 1,500&=37,500r \\ 0.04&= r \\ 4\% &=r \end{align*}$ |

Check. $\begin{align*} I&=Prt \\ 1,500&=(7,500)(0.04)(5) \\ 1,500&=1,500 \end{align*}$ | |

Write a complete sentence. | The rate of interest was $4\%$. |

In the next example, we are asked to find the principal—the amount borrowed.

**Example 14**

Sean’s new car loan statement said he would pay $\$4,866.25$ in interest from a simple interest rate of $8.5\%$ over five years. How much did he borrow to buy his new car?

**Solution**

$I=4,866.25$

$P=$ ?

$r=8.5\%$

$t=5$ years

Identify what you are asked to find, and choose a variable to represent it. | What is the amount borrowed (the principal)? Let $P=$ principal borrowed. |

Write the formula. Substitute in the given information. Multiply. Divide. | $\begin{align*} I&=Prt \\ 4,866.25&=P(0.085)(5) \\ 4,866.25&=0.425P \\ 11,450&=P \end{align*}$ |

Check. | |

Write a complete sentence. | The principal was $ \$11,450$. |

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*Marecek, L., & Mathis, A. H. (2020). Use a Problem Solving Strategy. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/2-2-use-a-problem-solving-strategy*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/2-introduction*