9.1 Use a Problem Solving Strategy

Learning objectives

Learning objectives

• Approach word problems with a positive attitude

• Use a problem solving strategy for word problems

• Solve number problems

Approach Word Problems with a Positive Attitude

The world is full of word problems. How much money do I need to fill the car with gas? How much should I tip the server at a restaurant? How many socks should I pack for vacation? How big a turkey do I need to buy for Thanksgiving dinner, and what time do I need to put it in the oven? If my sister and I buy our mother a present, how much will each of us pay?

Now that we can solve equations, we are ready to apply our new skills to word problems. Do you know anyone who has had negative experiences in the past with word problems? Have you ever had thoughts like the student in Figure 9.2?

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts like the student in Figure 9.3. Read the positive thoughts and say them out loud.

If we take control and believe we can be successful, we will be able to master word problems.

Think of something that you can do now but couldn't do three years ago. Whether it's driving a car, snowboarding, cooking a gourmet meal, or speaking a new language, you have been able to learn and master a new skill. Word problems are no different. Even if you have struggled with word problems in the past, you have acquired many new math skills that will help you succeed now!

Use a Problem-solving Strategy for Word Problems

In earlier chapters, you translated word phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. Since then you've increased your math vocabulary as you learned about more algebraic procedures, and you've had more practice translating from words into algebra.

You have also translated word sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. You had to restate the situation in one sentence, assign a variable, and then write an equation to solve. This method works as long as the situation is familiar to you and the math is not too complicated.

Now we'll develop a strategy you can use to solve any word problem. This strategy will help you become successful with word problems. We'll demonstrate the strategy as we solve the following problem.

Example 9.1

Solution

1. Identify the unknown quantity. Let $p$ be the original price of the shirt. The sale price of $18$ dollars is one-half of $p$, so the relationship is $18=\frac{1}{2}p$.
2. Clear the fraction. Multiply both sides by $2$ to obtain $36=2\left(\frac{1}{2}p\right)$. Dividing out the common factor $2$ gives $36=p$, as the figures illustrate.
3. Check the sale relationship. One-half of $36$ is $18$, so the recovered original price agrees with the stated sale price.
4. State the original price. The original price of the shirt was $36$ dollars.

Takeaway: If the sale price of $18$ dollars is one-half of the original price $p$, then $p=2(18)=36$, so the shirt cost $36$ dollars originally.

Let's use this approach with another example.

Example 9.2

Solution

1. Define the number of bananas. Let $b$ be the number of bananas. The number of apples is three more than twice that amount, so it is $2b+3$.
2. Translate the apple count. Yash brought $11$ apples, so write $2b+3=11$. The first figures connect the wording with the equation $11=2b+3$.
3. Subtract the constant. Subtracting $3$ from both sides gives $11-3=2b+3-3$, and hence $8=2b$, as the intermediate figures display.
4. Divide to solve. Dividing both sides by $2$ gives $\\frac{8}{2}=\\frac{2b}{2}$, so $b=4$. The last figures show the division and the result.
5. State the number of bananas. Yash brought $4$ bananas.

Takeaway: Translate “three more than twice” as $2b+3$, keeping the order of the phrase visible in the algebra.

In Solve Sales Tax, Commission, and Discount Applications, we learned how to translate and solve basic percent equations and used them to solve sales tax and commission applications. In the next example, we will apply our Problem Solving Strategy to more applications of percent.

Example 9.3

Solution

1. Translate the percent relationship. Let $c$ represent the original premium cost. Since the increase of \mathrm{\\$60} is $8\\%$ of the original cost, write $60=0.08c$. The first figures show this translation from the wording to the equation.
2. Divide both sides by the percent factor. Dividing by $0.08$ gives $\\frac{60}{0.08}=\\frac{0.08c}{0.08}$, so $750=c$. The remaining figures display this division and the resulting value of $c$.
3. Check the original cost. Substituting $c=750$ into the percent equation gives $0.08(750)=60$, and the final figure confirms that both sides equal $60$.
4. State the original cost. The original cost of Nga's car insurance premium was \\mathrm{\\$750}.

Takeaway: When a change is a stated percent of an original amount, write the change as a decimal times the original before solving.

Solve Number Problems

Now we will translate and solve number problems. In number problems, you are given some clues about one or more numbers, and you use these clues to build an equation. Number problems don't usually arise on an everyday basis, but they provide a good introduction to practicing the Problem Solving Strategy. Remember to look for clue words such as difference, of, and and.

Example 9.4

Solution

1. Represent the stated difference. Let $n$ be the number. Since the difference of the number and $6$ is $13$, write $n-6=13$. The figures translate the wording directly into this equation.
2. Undo the subtraction. Add $6$ to both sides to obtain $n-6+6=13+6$. The figures highlight the matching $6$ terms on the two sides.
3. Simplify and solve. Combining like terms gives $n=19$, which is the value displayed in the final figure.
4. State the number. The number that differs from $6$ by $13$ is $19$.

Takeaway: Translate “the difference of” as subtraction in the stated order, because reversing the two quantities changes the equation.

Example 9.5

Solution

1. Translate the verbal expression. Let $n$ be the number. Twice the number is $2n$, and adding $7$ gives $2n+7$, so the equation is $2n+7=15$.
2. Subtract the constant. Subtracting $7$ from both sides gives $2n=8$. This balanced step leaves only the term containing the unknown number.
3. Divide to isolate the number. Dividing both sides by $2$ gives $n=4$, which is the value shown in the solution figure.
4. Check the original wording. Substituting $n=4$ gives $2(4)+7=8+7=15$, so the recovered number makes the original statement true.
5. State the number. The number being sought in the original equation is $4$.

Takeaway: Translate “twice a number and” carefully: the coefficient multiplies the number, while the added constant stays separate.

Some number word problems ask you 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. 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 9.6

Solution

1. Express both numbers with one variable. Let $n$ be the smaller number. The larger number is $n+5$, so their sum gives $n+(n+5)=21$. The figures show this translation from the two-number relationship to one equation.
2. Combine like terms. The equation becomes $2n+5=21$, and subtracting $5$ from both sides gives $2n=16$.
3. Solve for the first number. Dividing by $2$ gives $n=8$. Thus the second number is $n+5=13$, as the successive figures illustrate.
4. Check the sum and the relationship. The values $8$ and $13$ add to $21$, and $13$ is $5$ more than $8$; the figures verify both facts.
5. State the numbers. The numbers are $8$ and $13$.

Takeaway: Use one variable for a related pair of quantities, and make sure the final values satisfy both the relationship and the given total.

Example 9.7

Solution

1. Represent the two numbers. Let $n$ be the first number. The second number is $4$ less, so it is $n-4$. Since their sum is $-14$, the equation is $n+(n-4)=-14$, which the figures translate into algebra.
2. Combine like terms. The equation becomes $2n-4=-14$, and adding $4$ to both sides gives $2n=-10$.
3. Solve for the first number. Dividing by $2$ gives $n=-5$. The second number is then $n-4=-9$, as the later figures show.
4. Check both conditions. The sum is $-5+(-9)=-14$, and $-9$ is $4$ less than $-5$; the final figures confirm both equalities.
5. State the numbers. The numbers are $-5$ and $-9$.

Takeaway: Keep negative numbers attached to their operations when translating “less than,” since $-9$ is four less than $-5$.

Example 9.8

Solution

1. Represent both numbers. Let $x$ be the smaller number. The other number is ten more than twice $x$, so it is $2x+10$. Their sum is $1$, giving $x+(2x+10)=1$.
2. Combine like terms. The equation becomes $3x+10=1$, and subtracting $10$ from both sides gives $3x=-9$. The figures display these algebraic changes.
3. Solve for the first number. Dividing by $3$ gives $x=-3$. The second number is $2x+10=2(-3)+10=4$, as the figures show.
4. Check both conditions. The numbers sum to $-3+4=1$, and $4$ is ten more than twice $-3$; the final figures confirm both equalities.
5. State the numbers. The numbers are $-3$ and $4$.

Takeaway: When one number is described as a multiple of another, let the other number be the variable so the relation stays simple.

Consecutive integers are integers that immediately follow each other. Some examples of consecutive integers are:

Notice that each number is one more than the number preceding it. So 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 $n+1+1,$ or $n+2.$

Example 9.9

Solution

1. Write the consecutive integers. Let the first integer be $n$. The second is $n+1$, so their sum gives $n+(n+1)=47$. The figures show the translation from the wording to the equation.
2. Combine like terms. The equation becomes $2n+1=47$, and subtracting $1$ from both sides gives $2n=46$.
3. Solve for the first integer. Dividing by $2$ gives $n=23$. The second integer is $n+1=24$, which the figures display.
4. Check the sum. The consecutive integers satisfy $23+24=47$, and the final figure confirms the equality.
5. State the integers. The two consecutive integers are $23$ and $24$.

Takeaway: Consecutive integers differ by one, so represent them as $n$ and $n+1$ rather than assigning unrelated variables.

Example 9.10

Solution

1. Represent the three integers. Let the first integer be $n$. The next two are $n+1$ and $n+2$, so the sum equation is $n+(n+1)+(n+2)=42$. The figures translate the sum into this expression.
2. Combine like terms. The equation becomes $3n+3=42$, and subtracting $3$ from both sides gives $3n=39$.
3. Solve for the first integer. Dividing by $3$ gives $n=13$. Therefore the other integers are $13+1=14$ and $13+2=15$, as the figures show.
4. Check the sum. The three integers satisfy $13+14+15=42$, and the final figure confirms the equality.
5. State the integers. The three consecutive integers are $13$, $14$, and $15$.

Takeaway: For consecutive integers, add $1$ for each position after the first before writing the equation for their sum.

Section 9.1 Exercises

Section 9.1 Exercises

Use a Problem-solving Strategy for Word Problems

In the following exercises, use the problem-solving strategy for word problems to solve. Answer in complete sentences.

Solve Number Problems

In the following exercises, solve each number word problem.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

1. One number is six more than the other. Their sum is forty-two. Find the numbers.

2. The sum of two numbers is twenty. One number is four less than the other. Find the numbers.

3. A number is one more than twice another number. Their sum is negative five. Find the numbers.

1. Joaquin bought a bookcase on sale for $\text{\textdollar 120},$ which was two-thirds the original price. What was the original price of the bookcase?

2. Pilar's rent increased by $\mathrm{4\%}.$ The increase was $\text{\textdollar 38}.$ What was the original amount of Pilar's rent?

3. Steve saves $\mathrm{12\%}$ of his paycheck each month. If he saved $\text{\textdollar 504}$ last month, how much was his paycheck?

1. One number is six more than another. The sum of the numbers is twenty-four. Find the numbers.

2. The sum of two numbers is fifty-eight. One number is four more than the other. Find the numbers.

3. The sum of two numbers is negative twenty-three. One number is $7$ less than the other. Find the numbers.

1. The sum of two numbers is negative eighteen. One number is $40$ more than the other. Find the numbers.

2. One number is eight more than twice another. Their sum is negative four. Find the numbers.

3. One number is three more than three times another. Their sum is negative five. Find the numbers.

1. The sum of two consecutive integers is $95.$ Find the numbers.

2. The sum of two consecutive integers is $-31.$ Find the numbers.

3. Find three consecutive integers whose sum is $96.$

1. Find three consecutive integers whose sum is $-36.$

2. Two-thirds of the children in the fourth-grade class are girls. If there are $20$ girls, what is the total number of children in the class?

3. Zachary has $25$ country music CDs, which is one-fifth of his CD collection. How many CDs does Zachary have?

1. There are $16$ girls in a school club. The number of girls is $4$ more than twice the number of boys. Find the number of boys in the club.

2. Lee is emptying dishes and glasses from the dishwasher. The number of dishes is $8$ less than the number of glasses. If there are $9$ dishes, what is the number of glasses?

3. After $3$ months on a diet, Lisa had lost $\mathrm{12\%}$ of her original weight. She lost $21$ pounds. What was Lisa's original weight?

1. Tim left a $\text{\textdollar 9}$ tip for a $\text{\textdollar 50}$ restaurant bill. What percent tip did he leave?

2. Yuki bought a dress on sale for $\text{\textdollar 72}.$ The sale price was $\mathrm{60\%}$ of the original price. What was the original price of the dress?

3. Three times the sum of a number and nine is $12.$ Find the number.

1. The sum of two numbers is fourteen. One number is two less than three times the other. Find the numbers.

2. One number is fourteen less than another. If their sum is increased by seven, the result is $85.$ Find the numbers.

3. The sum of two consecutive integers is $77.$ Find the integers.

1. The sum of two consecutive integers is $-23.$ Find the integers.

2. The sum of three consecutive integers is $78.$ Find the integers.

3. Find three consecutive integers whose sum is $-36.$

1. Shopping Patty paid $\text{\textdollar 35}$ for a purse on sale for $\text{\textdollar 10}$ off the original price. What was the original price of the purse?

2. Shopping Minh spent $\text{\textdollar 6.25}$ on $5$ sticker books to give his nephews. Find the cost of each sticker book.

3. Shopping Tom paid $\text{\textdollar 1,166.40}$ for a new refrigerator, including $\text{\textdollar 86.40}$ tax. What was the price of the refrigerator before tax?

1. Write a few sentences about your thoughts and opinions of word problems. Are these thoughts positive, negative, or neutral? If they are negative, how might you change your way of thinking in order to do better?