Find practice problems and solutions for these concepts at Fractions, Percents,and Decimals in Word Problems Practice Problems.
The word problems we've looked at so far have involved mostly integers. In this chapter, we'll look at word problems that involve fractions, percents, and decimals. We'll use the same strategies we used to solve integer word problems.
Fractions: Subject Review
To add two like fractions, add the numerators of the fractions and keep the denominator.
To add two unlike fractions, find common denominators and then add the numerators.
To subtract like fractions, subtract the numerator of the second fraction from the numerator of the first and keep the denominator.
To subtract unlike fractions, find common denominators, subtract the numerator of the second fraction from the numerator of the first, and keep the denominator.
To multiply two fractions, like or unlike, multiply the numerators and multiply the denominators.
To divide two fractions, like or unlike, multiply the first fraction by the reciprocal of the second fraction.
To convert an improper fraction into a mixed number, divide the numerator of the fraction by the denominator. The whole number part of that division is the whole number part of the mixed number. The remainder, if any, becomes the numerator of the fraction, and the denominator remains the same.
To reduce or simplify a fraction, find the greatest common factor of the numerator and denominator and divide the numerator and denominator by that number.
Now that we've reviewed how to work with fractions, let's look at some fraction word problems. We can use any of the strategies we've learned to answer these questions.
Example
DeDe pours 1 cup of cereal into a bowl. She adds cup of milk from a container that contains cups of milk. How many cups of milk are left in the container?
Let's use the eightstep process to answer this question.
Read the entire word problem.
We are given the amount of milk in the container and the amount that DeDe pours into her cereal bowl.
Identify the question being asked.
We are looking for how much milk is left in the container.
Underline the keywords.
The keyword left signals subtraction.
Cross out extra information and translate words into numbers.
We do not need to know that there is 1 cup of cereal in the bowl. That information will not help us solve this problem, so cross it out.
List the possible operations.
To find how much milk is left in the container, we have to subtract the original amount in the container from the amount that DeDe poured out.
Write number sentences for each operation.
–
Solve the number sentences and decide which answer is reasonable.
Convert onehalf to fourths and subtract:
– =
Since DeDe used only half a cup of milk, this answer seems reasonable.
Check your work.
We solved this problem using subtraction, so we must use addition to check our work. Add the number of cups of milk DeDe put in her cereal to the new amount of milk in the container. The sum should equal the original volume of milk in the container: + = . Our answer is correct.
Example
Tatiana spends an hour and a half at the gym four days a week. How many hours does she spend at the gym each week?
Let's solve this problem by drawing a picture. Draw a full circle and a half circle for each of the four days Tatiana goes to the gym. The full circle represents a full hour, and the half circle represents half an hour:
There are four full circles and four half circles. Each pair of half circles makes a whole circle, so we have six whole circles. Tatiana spends six hours in the gym each week.
Example
Mark buys pounds of turkey and pounds of bologna. He uses pounds to make a sandwich. How many pounds of meat does he have left?
Read the entire word problem.
We are given the number of pounds of turkey and bologna that Mark buys and the number of pounds of meat he uses to make a sandwich.
Identify the question being asked.
We are looking for how much meat he has left.
Underline the keywords.
The keyword add signals addition, and the keyword left signals subtraction.
Cross out extra information and translate words into numbers.
There is no extra information in this word problem.
List the possible operations.
First, we need to find how much meat Mark bought by adding the weight of the turkey to the weight of the bologna. Then, we will subtract the weight of the meat used to make the sandwich from this total.
Write number sentences for each operation.
Solve the number sentences and decide which answer is reasonable.
Convert eighths and tenths to fortieths and add:
Write number sentences for each operation.
Now that we have the total weight of the meat Mark bought, we can subtract the weight of the meat used to make the sandwich to find how much meat is left:
Solve the number sentences and decide which answer is reasonable.
Convert fourths to fortieths and subtract:
Mark has pounds of meat left.
Check your work.
We solved this problem using subtraction and addition, so we must use addition and subtraction to check our work. Add the number of pounds of meat Mark used for his sandwich to the number of pounds of meat he had left: Subtract from that total the number of pounds of bologna Mark bought, and we should be left with the number of pounds of turkey Mark bought, pounds of meat.
Pace Yourself
Hope buys water by the case. One water company sells a case of 24 bottles, each filled with liters of water. A second company sells a case of 20 bottles, each filled with liters of water. Which case contains more water? How did you figure that out? If you knew the price of each case, what would you do to figure out which case was the better deal?

Decimals: Subject Review
We add and subtract decimals just as we add and subtract whole numbers; we just need to be sure to line up the decimal points of the addends (or minuend and subtrahend). Place trailing and leading zeros to keep your columns straight.
We multiply two decimals just as we multiply two whole numbers, but the number of places to the right of the decimal point of our answer is equal to the sum of the number of places to the right of the decimal point in our factors.
To divide two decimals, shift the decimal point to the right until the divisor is an integer. Shift the decimal point of the dividend the same number of places to the right. Divide and carry the decimal point into the quotient.
To round a decimal to a place, look at the digit to the right of that place. If the digit is 5 or greater, increase the number in the place by one and change every digit to the right to zero. If the digit is 4 or less, leave the number in the place and change every digit to the right to zero.
54.6078 rounded to the thousandths place is 54.6080; 54.6078 rounded to the tenths place is 54.6000.
We've already seen some word problems that involve decimals. Almost every problem that deals with money involves two decimal places.
Example
If a raffle ticket costs $1.75, how many tickets can be bought with a $20 bill?
We could solve this problem using the eightstep process, but let's use a table instead. By multiplying the number of tickets by $1.75, we can see exactly when the total exceeds $20.
Number of Tickets 
Price Per Ticket 
Total 
1 
× $1.75 
$1.75 
2 
× $1.75 
$3.50 
3 
× $1.75 
$5.25 
4 
× $1.75 
$7.00 
5 
× $1.75 
$8.75 
6 
× $1.75 
$10.50 
7 
× $1.75 
$12.25 
8 
× $1.75 
$14.00 
9 
× $1.75 
$15.75 
10 
× $1.75 
$17.50 
11 
× $1.75 
$19.25 
12 
× $1.75 
$21.00 
Eleven tickets can be purchased for less than $20, but 12 tickets would cost more than $20. The most tickets that can be purchased is 11.
Example
What is 4.7 less than 2.9 times the quotient of 5.75 and 0.25?
Break the problem down into single operations, placing parentheses around each. We are looking for 4.7 less than a number: ( ) – 4.7. That number is 2.9 times another number: [2.9 × ( )] – 4.7. The missing number is the quotient of 5.75 and 0.25, which is : [2.9 × ( )] – 4.7. Start by finding the answer to the operation in the innermost parentheses: = 23, so the problem becomes (2.9 × 23) – 4.7. 2.9 × 23 = 66.7, so the problem is now 66.7 – 4.7 = 62.
Example
The atomic mass of one atom of hydrogen is 1.00794 grams, the mass of one atom of sulfur is 32.065 grams, and the mass of one atom of oxygen is 15.994 grams. What is the weight of one molecule of sulfuric acid, which contains two hydrogen atoms, one sulfur atom, and four oxygen atoms?
Read the entire word problem.
We are given the masses of one atom of hydrogen, one atom of sulfur, and one atom of oxygen.
Identify the question being asked.
We are looking for the total weight of two hydrogen atoms, one sulfur atom, and four oxygen atoms.
Underline the keywords.
There are no keywords in this problem.
Cross out extra information and translate words into numbers.
There is no extra information in this problem.
List the possible operations.
We are given the weights of three different elements and the number of atoms we have of each element. To find the total weight, we must multiply the weight of one atom of each element by the number of atoms we have of that element. Since we have only one sulfur atom, we won't need to multiply that weight. We will need two multiplication sentences and one addition sentence.
Write number sentences for each operation.
First, find the weight of the hydrogen atoms. There are two, so multiply the weight of a hydrogen atom by 2:
1.00794 × 2
Solve the number sentences and decide which answer is reasonable.
1.00794 × 2 = 2.01588 grams
Write number sentences for each operation.
Next, find the weight of the oxygen atoms. There are four, so multiply the weight of an oxygen atom by 4:
15.994 × 4
Solve the number sentences and decide which answer is reasonable.
15.994 × 4 = 63.976 grams
Write number sentences for each operation.
Finally, add the weights of the hydrogen, sulfur, and oxygen atoms:
2.01588 + 32.065 + 63.976
Solve the number sentences and decide which answer is reasonable.
2.01588 + 32.065 + 63.976 = 98.05688 grams
Check your work.
We twice used multiplication and addition to solve this problem, so we will twice use subtraction and division to check our work. Subtract the weight of the sulfur atom from the total weight: 98.05688 – 32.065 = 65.99188 grams, the weight of the hydrogen atoms.
Now subtract 2.01588 grams, the combined weight of the two hydrogen atoms: 65.99188 – 2.01588 = 63.976 grams, the weight of the oxygen atoms. Divide the weight of the hydrogen atoms by 2 and divide the weight of the oxygen atoms by 2. We should be left with the atomic mass of each element: = 1.00794 grams and = 15.994 grams.
Inside Track
Decimals can be easier to work with than fractions, especially when you have to add or subtract unlike fractions. When a problem contains fractions that are easy to convert to decimals, such as halves, fourths, or tenths, turn the problem into a decimal problem—it may make your operations a lot easier!

Percents: Subject Review
We can write a decimal as a percent by moving the decimal point two places to the left and adding the percent sign.
0.23 = 23%
We can write a percent as a decimal by moving the decimal point two places to the right and removing the percent sign.
4% = 0.04
We can find a percent of a number by multiplying that number by that percent.
30% of 50 is 15 because 50 × 0.30 = 15.
We can find what percent one number is of a second number by dividing the first number by the second number.
2 is 40% of 5 because = 0.40 = 40%.
The percent increase or decrease from one number to another is the difference between the original number and the new number divided by the original number.
From 10 to 12 is a 20% increase because = 0.20 = 20%.
From 24 to 18 is a 25% decrease because = 0.25 = 25%.
If a number is increased by some percent, we can multiply that number by the percent to find by how much the number was increased.
If 30 is increased by 10%, it is increased by 30 × 0.1 = 3.
If 24 is decreased by 50%, it is decreased by 24 × 0.5 = 12.
Percent word problems almost always contain the word percent. When you see the word percent, use the context of the problem to decide if you must:
 Find a percent of a number.
 Find what percent one number is of a second number.
 Find percent increase, either from one number to another, or a new value after a percent increase.
 Find percent decrease, either from one number to another, or a new value after a percent decrease.
Example
Kevin must trim the grass of the football field, which is 48,000 square feet. If he cuts 35% of the grass before lunch, how much of the grass has he cut?
This is the first type of percent problem: finding the percent of a number. In this example, we must find 35% of 48,000. Convert 35% to a decimal and multiply by 48,000: 35% = 0.35; thus 0.35 × 48,000 = 16,800 square feet.
Percent word problems, like other word problems, can require more than one step.
Example
If Kevin cuts 30% of the grass before lunch and of the grass after lunch, how much grass does he have left to cut?
This is the same type of percent problem, but you must convert both 35% and to decimals, multiply each by 48,000, and subtract both products from 48,000 to find how much grass Kevin has left to cut. 30% = 0.3, 0.3 × 48,000 = 14,400. = 0.4, 0.4 × 48,000 = 19,200. 48,000 – 14,400 – 19,200 = 14,400 square feet.
If that last example seemed a little complicated, don't worry—we can use pictures and the eightstep process to solve these word problems, too. Let's look again at the last example.
The following diagram represents a football field, divided into ten equal sections:
Why did we divide the field into ten sections? Kevin cuts 30%, which is or of the grass, before lunch, and he cuts which is the same as after lunch. Since both 30% and can be written as a fraction over ten, we can divide the field into ten sections, and shade the parts of the field that Kevin has cut. Shade three sections, since Kevin cut of the grass before lunch, and shade another four sections, since Kevin cut after lunch:
That leaves just three sections, or , of the field left to cut. = 0.3, 0.3 × 48,000 = 14,400 square feet. Kevin has 14,400 square feet of grass left to cut.
Now, let's try a similar problem using the eightstep process.
Example
It takes 20 minutes for Frank to get to school every day. Today, it took Frank 15% longer to get to school because of traffic. How many minutes did it take for Frank to get to school?
Read the entire word problem.
We are given how long it usually takes for Frank to get to school, and the percentage increase of that amount of time.
Identify the question being asked.
We are looking for the number of minutes it took Frank to get to school today.
Underline the keywords.
There are no keywords in this problem, but the problem does contain a % symbol, so we will likely have to use a percent formula.
Cross out extra information and translate words into numbers.
There is no extra information in this problem.
List the possible operations.
Since we are told that it will take 15% longer to get to school, the number of minutes it takes Frank to get to school is increasing by 15%. We must use a percent increase formula.
Write number sentences for each operation.
Convert 15% to a decimal and multiply it by the number of minutes it usually takes Frank to get to school:
20 × 0.15
Solve the number sentences and decide which answer is reasonable.
20 × 0.15 = 3
Write number sentences for each operation.
It will take Frank three minutes longer to get to school today. To find out how long it took him to get to school, add three to the usual number of minutes it takes him to get to school:
20 + 3
Solve the number sentences and decide which answer is reasonable.
20 + 3 = 23 minutes
Check your work.
Since we found the increase in the number of minutes by multiplying by 0.15, divide the new number of minutes by 1.15, and we should have the original number of minutes: = 20 minutes.
Inside Track
In the last example, we found that 20 increased by 15% is equal to 23 by multiplying 20 by 0.15 and then adding the result to 20. We can find a 15% increase in 20 in one step if we multiply 20 by 1.15. Multiplying by (1 + 15%) means that our product will be 20 plus 15% of 20, which equals 23.

Pace Yourself
A decade is ten years. What percent of your age is one decade? When you are a decade older than you are now, by what percent will your age have increased? If you compare your age a decade ago to now, by what percent has your age increased?

Summary
In this chapter, you learned how to recognize fraction, decimal, and percent word problems and how to use the strategies we learned in earlier chapters to solve these kinds of problems. We used formulas to solve percent word problems.
Find practice problems and solutions for these concepts at Fractions, Percents,and Decimals in Word Problems Practice Problems.