# The Craft of Word Problems

## Word Problems Dealing with Ratio and Proportion

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## Ratio and Proportion Problems with Strategies

The following problems are organized into four distinct types of problems dealing with ratio and proportion calculations. The first type of problems will be standard problems that deal with understanding ratios and proportions as equivalent fractions. The second type of problems will be compound ratios which will involve calculating several ratios in succession. The third type will involve doing addition or subtraction in order to find a number for computing a ratio which will be called computational ratios. Finally, there will be a few problems that are complex ratios, which will involve developing some complex relationship between proportions. One reason for organizing the problems in this way is to illustrate an orderly progression in complexity, while underscoring similarity of structure. Of course, there are more problems in the first category than there are in subsequent categories. These will serve as a basis to help the students become familiar with and later recognize structure and format.

### Standard Ratio Problems

1. “Bobby has a bag full of marbles that he keeps in his room. He has 35 red marbles and 25 green marbles. Find the ratio of red marbles to green marbles, and put it in its simplest form.”

This is a relatively simple ratio word problem that includes the slight wrinkle of including the aspect of fractions that students seem to universally dislike - simplest form. It begins with the relatively harmless ratio of 35/25. The difficulty in this problem is such that there are fraction processes that will be involved like greatest common factor, factoring, and/or the rules of divisibility. Students should have a familiarity with these processes before tackling this type of problem, but even if they do not there is much that they can do with this problem.

Not all students will immediately note that the number five is a common factor in both the numerator and the denominator. In fact most of my students would insist on factoring both the numerator and the denominator to make certain that five is the greatest common factor.

- Factors of 35 = 5 x 7
- Factors of 25 = 5 x 5

However the students choose to get the result does not really matter, what does matter is that students should have various means for finding simplest form, eventually, dividing the denominator and the numerator by five in order to get: simplest form. Here’s another.

2. “To make 20 biscuits, Juanita uses 5 cups of flour to 1 cup of milk. If she uses 3 cups of milk, how many cups of flour will she use?

This is another pretty straightforward problem which sets up a proportion pretty nicely. There is an added bonus that some extraneous information is given which lends itself to the asking of a series of follow-up questions.

- 5/1 = x/3
- 15 = x

Some follow-up questions might include, “If we know that 5 cups of flour make 20 biscuits, can we figure out how many biscuits are made with 15 cups of flour?” Most middle school students can figure pretty quickly that 15 cups are 3 times as much as 5 cups which means that they will have to multiply 20 by 3 to get 60. By playing around with word problems in this way, students begin to realize that once you understand the idea of equivalency, they have won half the battle when it comes to proportions.

3. “When 2,000 pounds of paper are recycled or reused, 17 trees are saved. How many trees are saved if 5,000 pounds of paper is recycled? How many trees are saved if 10,000 pounds of paper is recycled?”

This is another standard ratio question that lends itself to being a series of questions, and in fact, I would have my students strategize as to which of the two questions is easier to answer. Some of them might recognize that the second question is an easier question to answer, and that when that question is answered, it will help to resolve the first question.

- 2000/17 = 10000/x

10,000 is 5 times more than 2,000; therefore x must be 5 times more than 17 or 85.

- 5000/x = 10000/85

5,000 is half of 10,000; therefore x must be half of 85 or 42.5.

4. “If it costs $90 to feed a family of 3 for one week, how much will it cost to feed a family of 5 for one week? How much will it cost to feed a family of six, seven, and eight?

The ratio again is pretty easy to set up. $90/3 = $x/5. However, I think that I would have my students approach this problem a little different. I would want them to see that 90 is divisible by three, and that once I know how much it costs to feed each person in the family ($30), then whenever I’m asked how much would it cost to feed y number of people in a particular family, I would only need to multiply y by $30.

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More Standard Ratio and Proportion Problems
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5. “In a school there are four boy scouts to every three girl scouts. If there are forty-two girl scouts, how many boy scouts are there? If there are 81 girl scouts, how many boy scouts are there?”

6. “To make green paint, a painter mixes yellow paint and blue paint in the ratio of three to two. If he used twelve gallons of yellow paint, how much blue paint did he use?”

7. “A rectangle measures 40 cm at its length and 20 cm at its width. Find the ratio of the length to the width in lowest form.”

8. “When a robin flies, it beats its wings an average of 23 times in ten seconds. How many times will it beat its wings in two minutes?”

9. “Inez is 5’4” tall. At a certain time of day, she measures her shadow, and finds it is 8’ long. She also measures the shadow of a tree which is 40’. How tall is the tree?”

10. “The average human heart beats at 72 beats per minute. How many times does it beat in 15 seconds? How many in an hour? How many in a day? How many in a year? Take your pulse and record the number of beats in 30 seconds. How many times does it beat in an hour? How many times does it beat in a day, a month, a year?”

### Compound Ratios

11. “On a triangle, each side measures 5 cm, 10 cm, and 30 cm, respectively. In lowest terms, find the ratios of the lengths of the sides.”

In this word problem, there are three numbers that must be placed in a ratio. That will look like this:

- 5:10:30

Immediately, it should be noted that the numbers are all divisible by five and that the ratio can better be expressed in its lowest form of 1:2:6.

12. “The church is going on a trip to Niagara Falls via several buses. The ratio of men to women to children is 1:2:3. If there are 120 people going on the trip, how many men are going? How many women are going? How many children are going?”

There are a variety of strategies that can be implemented. One can show that by adding up the initial ratio one can decipher that out of every 6 people, three will be children, two will be women, and one will be a man. Then one can divide 120 by 6 which will show that there are twenty groups of 6 people. Lastly, we multiply by 20 each part of the original ratio. Consequently, 20 men, 40 women, and 60 children went on the trip to Niagara Falls.

13. “Kim mixed seltzer, fruit punch concentrate, and ginger ale in the ratio of 2:2:4 to make a special drink for her friend’s birthday party. To make three gallons, how much of each ingredient should Kim use?”

The complexity of this problem, I believe solely revolves around converting gallons to cups. Generally, I give the students a helpful hint in which I might let them know that they will have to convert gallons to cups and that there are 16 cups in a gallon. Once I have given the students that information, it usually puts them on the right track.

- 1 gallon = 16 cups
- 3 gallons = 48 cups

2:2:4 means that out of every 8 cups, two of them are seltzer, two of them are fruit punch concentrate, and four of them are ginger ale. 48/8 = 6. That means that I now multiply each part of the ratio with 6 to ascertain how many cups of each ingredient Kim will need in order to make three gallons of punch. Kim will need 12 cups of seltzer, 12 cups of fruit punch concentrate, and 24 cups of ginger ale.

Computational Ratios and Proportions

Computational ratio and proportion problems are problems that require more than just dividing, applying rules of divisibility, or seeking out equivalent fractions. They require addition, subtraction, multiplication, and/or division. As a consequence, these will require a little more thought and strategizing on the part of students to correctly identify the methods that they will need to figure out these word problems.

14. “A rectangle measures 20cm at its length and 5 cm at its width. Find the ratio of the length to the width to the perimeter of the rectangle in lowest form.”

The uniqueness of this word problem revolves around the fact that in this ratio there will be three numbers in the comparison as well as figuring out perimeter, and the fraction processes of lowest terms. The formula of perimeter requires addition of all sides.

- P=2 (length) + 2 (width)
- P= 2(20) + 2(5)
- P=50 cm2
- 20:5:50

The number five appears to be the common factor. In order to make the ratio in its lowest terms we need to divide all of the numbers by five. The answer will be: 4:1:10.

15. “David received $50 dollars for working at his father’s store. He spent $20 dollars at the movies, and $10 dollars buying comic books. After buying $5 worth of candy, he saved the rest. What is the ratio of the amount of money David spent buying comic books and candy, to the amount that David saved?”

Again we have a ratio problem that looks pretty simple on the face of it, but one might be surprised to find how many students fall into the subtle snare that this word problem involves. There are several amounts of money communicated and some that are not made explicit. For instance, students are told how much money David earned, but they are not told how much money he saved. The problem therefore requires several steps on the part of students. Most students will try to cut as many steps out of the equation in the attempt to make it easier for themselves and as a consequence they will fall easily into the trap.

The key to this word problem is found in the fact that the ratio that is required does not involve all of the money spent. The final sentence only asks for a ratio that involves only the money that David spent buying comic books and candy, not the money he spent at the movies. Many students in order to make the problem easier for them will make the incorrect assumption that they will have to add up all the money David spent. What is all the more interesting is that they do not need to add up all the numbers for the first part of this ratio, but they will need to do it to figure out how much David saved. The mathematical procedures should look something like this

- Comic books + Candy = 1st Ratio
- Movies + Comic books + Candy = Amount Spent
- Amount Earned - Amount Spent = Amount Saved (2nd Ratio)
- Answer = 15:15

This type of problem will serve to underscore that students will need to really pay attention to all aspects of the problem. On standardized tests word problems tend to have distractors. These multi-step problems will help students develop strategies to weather the multi-step problems found on standardized tests.

16. “78 children attended a trip to Six Flags. 60 of them were boys. Find the ratio of the number of boys to girls, and express the number of boys as a fraction in lowest terms to the number of girls.”

Many of these types of word problems involve a tremendous amount of verbiage that serve to distract and confuse students. As a consequence many students tend to either look for the numbers and then to blindly do some calculation that they hope will net them the answer or they just skip over those problems that have a lot of words. This problem is of the type that most children will want to skip but it is really not a difficult problem that requires much calculation.

78 children on a trip, 60 of them are boys, 18 obviously are girls. The ratio of boys to girls is 60/18. Both numbers are even, therefore a common factor is 2. Divide both numerator and denominator by two. 60/18 = 30/9. I again see that both numbers have a common factor, but this time that common factor is three. 30/9 = 10/3. The answer is that for every ten boys that went on the trip, there are three girls that went on the trip.

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More Computational Ratio and Proportion Problems
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17. “In the United States there is one car for every 1.7 people. How many car tires (on cars) per person are there?”

18. “One out of three students in the school owns a dog. Of these students one out of two owns a cat. What fraction of the students owns both a cat and a dog?”

### Complex Problems Involving Ratios and Proportions

In these problems, there are some complex relationships that need to be discovered. There are few students that I find have the patience to really grasp these problems, however, if the students can get through one or two of these problems, I believe that they will be amply prepared for any standardized test.

19. “There are two sisters named Mary and Sue who need to buy a present for their mother’s birthday. The perfume there mother likes is $50 dollars a bottle. Mary is the older sister and she gets more allowance money than her little sister, so they have decided that Mary will give $3 for every $2 dollars her sister gives. For every $15 Mary (the older sister) gives, how much does the younger sister give?”

In order to work with this problem an initial ratio must be the first thing ascertained. This is clearly elucidated by the statement declaring that Mary gives $3 for every $2 given by Sue. Mathematically it would look something like this:

Mary 3 $15

--- = ------- = ------

Sue 2 $x

My students (as well as most middle school students) tend to be creatures of habit and because this is the way they learned and understood ratio, proportion, and percent, upon seeing the problem formatted this way, their minds instantly recall what they learned about proportions and cross multiplication. They become so focused on finding an answer that they hardly recognize the variable as an algebraic entity. After a very short time my students can figure this out easily.

Because my students are bilingual students, I tend to do quite a few word problems as a way of preparing them for the language of standardized tests. My sense is that because students do not do enough word problems in class, when they are confronted by word problems on standardized exams like the CMT, they get “brain lock”, the equivalent of stage fright to an actor. If, somehow we teachers could find creative ways to include word problems in every unit, as well as the usual strictly numerical problems, students would be better prepared for these exams. The answer to this problem is

x = $10.

An added benefit to these word problems is that these problems as stated above can be further complicated by adding a series of follow-up questions to the first question. For instance, how much of the $50 dollars did Sue pay? How much of the $50 dollars did Mary pay? If they had a little brother named Harry who gave them $5 dollars for their mother’s present, how much would each have to put in? These questions make each word problem unique and can be a subject of profound mathematical discussion.

20. “Guillermo drives between Boston to Pittsburgh, a distance of 600 miles. From Boston to Pittsburgh, he averages 50 miles per hour. On the return trip, he averages 60 miles per hour. How long does the trip take?”

Guillermo must drive each way 600 miles. 600/50 = x hours (the trip to Pittsburgh). 600/60 = y hours (the return trip to Boston). Add the number of hours it took to go in both directions (x + y = ? hours). Once the equation is set up, the rest is rather pedantic. 12 hours one way, 10 hours the other way, 22 hours all told. The same problem can be told a couple of ways.

21. “Guillermo drives between Boston to Pittsburgh, a distance of 600 miles. From Boston to Pittsburgh, he averages 50 miles per hour. He is able to go faster on the return trip, and the total trip takes only 22 hours. What was his average speed on the way back?”

22. “Guillermo drives between Boston to Pittsburgh. From Boston to Pittsburgh, he averages 50 miles per hour. On the return trip, he averages 60 miles per hour. His driving time was 22 hours. How far is it from Boston to Pittsburgh?”