Arithmetic Word Problems

By Jerome Dancis

November 2002

Executive Summary, Introduction and Conclusions

This report is mostly a collection of useful and important Arithmetic Word problems. This wide variety of problems are not represented on the sample MD Functional Mathematics Test, a current high school graduation requirement. Few are represented on the Maryland High School Assessment on Functions, Algebra, Data Analysis and Probability (MD HSA Algebra) sample test, a proposed high school graduation requirement.

I know that many students are not learning how to solve these problems today. It is important that the next generation learn to do them. If they are placed on a state test, more instructional time and more staff development time will be devoted to them. Hopefully, instruction for many of them is already occurring (even if they do not appear on the MD Functional Mathematics Test or on the MD HSA sample Algebra test).

Here is an introduction to the three main topics to be discussed in this report:

A. Two Big Problems; The "Key-Word" Method and the lack of Two-Step Problems

A "Two-Step" problem is one whose solution requires two steps/calculations. A "Multi-Step" problem is one whose solution requires several steps/calculations. Reading a number off a chart does not count as a step. The ability to cope with multi-step problems is the key to using mathematics in real life. It is also crucial for many non-mathematical problems.

Here is a Two-Step problem, which exceeds the specifications of the MD Functional Mathematics Test.

Problem 1. The price of a tub of ice cream is 2 dollars, and the price of a Sunday newspaper is one dollar. Johnny buys just 5 tubs of ice cream and one Sunday newspaper. How much is the total sale?

There are no two-step problems on the sample MD Functional Mathematics Test. Having most of the word problems be two-step problems would be a worthwhile major upgrading of the MD Functional Mathematics Test.

The "Key-Word" method. The "Key-Word" method is the simplest method of instruction for solving word problems. But, the "Key-Word" method often produces wrong answers. The "Key-Word" method is an example of what I call an "Avoid-the-reasoning" method of mis-education.

The specifications of the MD Functional Mathematics Test are consistent with the "Key-Word" method. There should be an important MD state math or arithmetic test, which is completely inconsistent with the "Key-Word" method and other "non-thinking" instructional methods. After all, MD state tests are powerful drivers of instructional practices in the state.

(B) "Keep It Simple for Students" (KISS) (Better methods of instruction)

An important part of math instruction is to demystify mathematics; thereby making it accessible to more students. This report will present simple, conceptual-understanding based arithmetic methods that will allow students to solve a wide variety of problems, specifically:

1. Certain Proportion problems (without having to deal with the full sophistication of proportional reasoning).

(ii) Catch-up and Overtake problems. Simple arithmetic solutions are presented for four of the more difficult problems on the MD HSA sample Algebra test.

(iii) Work problems, which were the most difficult word problems in Algebra I courses.

1. Other important topics for elementary school -- ones which many Grade 5 students are likely to encounter outside of school.

At the end is a note that reading comprehension is crucial for Arithmetic Word Problems.

I believe that students should receive instruction for the bulk of the arithmetic problems in this report by Grade 5. I am in favor of having a MD state Grade 5 math/arithmetic test in which the word problems are of the types in this report.

After some modification, the following items on the MD HSA sample Algebra test should be included in Grade 5 instruction: Items #2,10,13,15,18,21,23,24,29,32,34,37,40,44,47,48. The modifications would be ones of form, not of mathematical substance. For example, replace: "What is the probability that Michelle [a top basketball player] is successful when she attempts to make a basket?" in sample test Item #47 with "What is Michelle's shooting average?"

Slogan. All elementary school students can learn to solve multi-step Arithmetic Word Problems including all the ones in this report.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ End of Executive Summary

Almost all the arithmetic problems in this report exceed the specifications for the MD Functional Mathematics Test, often for several reasons.

Most real life problems are multi-step and involve interpretation. The difficulty is that multi-step problems usually require at least a modicum of thinking; one has to plan out a set of calculations which will lead to a solution. The ability to cope with two-step problems is essential for being able to do the "minimum" mathematics needed to "function" in our society. It is important to provide students with instruction on how to think through and analyze multi-step problems. It is important to start early with multi-step problems with simple numbers and to have students solving many of them in each grade.

(A) Two Big Problems: The "Key-Word" Method and the lack of Two-Step Problems.

Change (money). Here is a Two-Step problem, which exceeds the specifications of the MD Functional Mathematics Test, which state: "A problem in making change will state the [total] amount of the sale … ". (Hence this problem may not appear on the test.)

Problem 2. The price of a tub of ice cream is \$1.88. Johnny buys 5 tubs of ice cream. Johnny gives the clerk a \$20 bill, how much is the change?

The Two-Step Problem 2 is a combination of the following pair of One-Step Items # 27 and 28 on the sample MD Functional Mathematics Test:

MD Functional Mathematics Test Item #27. A person goes to the food store and buys ice cream for \$1.88. If the person gave the clerk a \$20 bill, how much is the change?

MD Functional Mathematics Test Item #28. A person goes to the food store and buys five gallons of ice cream for \$1.88 a gallon. How much would the five gallons cost?

When Two-Step problems, like Questions 1, 2, 5, and 6 herein, are avoided (or are not a main part of arithmetic instruction), students miss out on an opportunity for training in developing their abilities to solve them. This results in word-problem phobia -- a common, but preventable malady. BOO

Percents. A MD Functional test specifications read: "A problem using a percent will involve finding a given percent of a whole number … ". Hence, Two-step problems like each part of the next Problem 3 exceed it.

Problem 3. The price of a Sunday newspaper is \$2.00. Sales tax in MD is 5%.

(a) How much money is the MD sales tax when two Sunday newspaper are bought?

(b) How much money do you need to pay (including the MD sales tax) when buying a single Sunday newspaper?

A permissible item must be a one-step, for example: How much money is the MD sales tax when a single Sunday newspaper is bought?

With some common sense, simple two step problems with "inequalities" may be easily done.

Problem 3.1. Eva bought 5 pairs of socks for less than \$2 each and a \$6 hairbrush. Find a good (the best) bound on the total cost.

Calculations. Suppose that the cost of each pair of socks was exactly \$2. Then the total cost would be \$16.. Now, common sense tells us that if the cost, of each pair of socks, was less than \$2, then the total cost would be less than \$16.

This common sense approach is appropriate for Problem 3.1, a Grade 3 level arithmetic problem. For more sophisticated problems, in higher grades, more formal methods should be taught.

A variation on this Problem 3.1 appears as Item #32 on the sample MD Algebra Test (on the web at http://www.mdk12.org/mspp/high_school/look_like/algebra/v32.html).

Problem 3.2. Eva bought 5 pairs of identical socks and a \$6. 50 hairbrush. The total cost for the items was less than \$29. Which of these inequalities best describes the cost (c) of each pair of socks?

A c < \$4. 50, B c > \$4. 50, C c < \$7. 10, D c > \$7. 10"

Calculations. Suppose that the total cost is exactly \$29. Then the five pairs of socks cost \$29 - \$6.50 = \$22.50. So one pair of socks costs \$22.50 / 5 = \$4.50. Common sense tells us that when the total cost is less than \$29, then the cost of a pair of socks is less than \$4.50. This translates into Answer A: c < \$4. 50.

Finding that a pair of socks is less than \$4.50 should be a Grade 4 level arithmetic problem. Translating the last part: "less than \$4.50" into "c < \$4.50.", raises it up to a Grade 6 level, background-for-algebra problem.

The "Key-Word" Method when applied to the next two problems will produce wrong answers:

Problem 4. Mr. and Mrs. Jones have 4 sons and 3 cats and 2 dogs (and no other pets). How many pets do they have altogether?

The "Key-Word" Method says

(i) Do not read the problem (or at most speed read it).

(ii) Find the key word in the question.

Here the Key word is "altogether". When you see this phrase "altogether", add up all the numbers in sight. In Problem 4, this means add: 4 + 3 + 2. Sorry, wrong answer.

Of course, textbooks, which "use" the "Key-Word" method, carefully avoid those problems for which the "Key-Word" method would produce a wrong answer. Problems for which the "Key-Word" method works, are rarely two-step problems.

It is my experience that many college students regularly look for key-words -- to their detriment. They will point to a paragraph or more, of explanation and then summarize/trivialize it as "If I see the word '…key..', then I do such and such… " Bad habits are hard to break.

Problem 5. The scores, of the 4 students from John's block, on an exam are:

Students on John's Block: Jennifer, Lawrence, Elizebeth, John

Scores on an exam: 90 70 80 65

What is the average score of the boys on John's block?

Students, using the "Key-Word" method, will spot the key-word "average" and quickly add up all 4 scores and then divide by 4. They will get the wrong answer.

Parents, you need not worry that your child will be tricked by Question 5; it will never appear on the MD Math Test; it is too hard. (The test specifications read: "No numerals except those to be averaged will appear in the table.")

This is in sharp contrast to the1999 CA Math Reasoning Standard 1.1, for Grades 3-7, which states: "Analyze problems by … distinguishing relevant from irrelevant information … " (Mathematics Framework for CA Public Schools).

Problem 6. We traveled from Here to There at an average speed of 50 MPH; we returned from There to Here at an average speed of 40 MPH. The distance from Here to There is 200 miles. What was our average speed for this round-trip?

Students, trained in the "Key-Word" method, and others, are likely to see the word "average" and then simply average 50 and 40, which is 45 MPH, an incorrect answer. This problem requires some simple analysis based on the definition:

{Average speed} = {Distance traveled} / {Time of trip}

Calculations. Time to go: 200/5 = 4 hours. Time to return: 200/4 = 5 hours.

Total travel time: 9 hours.

Average speed = {Total mileage} / {Total time} = 400 / 9 = 44.44 MPH.

The level of difficulty of Problem 6 is higher than that of the problems involving averaging on MD HSA Algebra sample test.

When students use the "Key-Word" Method, they do not read the problem and do not think about what the problem means. Students, who have difficulty understanding what they read, are not at any disadvantage. In this way, "Key-Word" Method questions level the playing field and thereby reduce the measured education gap between weak and strong students. Reducing the education gap is a major objective for school systems. On the other hand, by not reading the problems, students miss out on an important opportunity to practice reading for understanding. This is especially detrimental for weak students.

The "Key-Word" Method is an "Avoid-the-thinking" method of mis-education. It is popular with many textbooks because it is the easiest method of instruction. It avoids all thinking. Ideal for a book entitled "Arithmetic for Dummies". It brings students success, albeit only for textbook and MD Math Test problems. No skin off the textbook's cover/profits, if this mucks up the children later.

The specifications of the MD Math Test are aligned/consistent with the "Key-Word" Method. This is counterproductive! State tests should be inconsistent with the "Key-Word" Method and other "non-thinking" instructional methods. After all, MD state tests are powerful drivers of instructional practices in the state.

The "Key-Word" method contributes to students arriving in middle school very poorly trained in word problems.

Math professor Pat Kenschaft of Montclair State University reports how popular and ingrained the "Key-Word" method is.

"The most important [sad story of the week] is about a prospective elementary school teacher in her LAST math class. The [discussion] was essentially [about], "In 1999 U.S. family small trucks averaged 20 mpg and our family cars averaged 28 mpg. All together the mileage of family vehicles was 23 mpg, … " The student thought a moment and raised her hand. "It can't be 23 mpg all together. 'All together' means you must add. The mileage all together must be 48 mpg." Patiently, repeatedly, I explained [the situation]. She was adamant. I realized she wasn't going to understand today because her indoctrination [in the "Key-Word" method] was so deep, I was mad at a system that sends such people (a nice, dedicated person with all the unchangable characateristics I want in a teacher) into elementary school classrooms to teach innocent little children. … She is smart enough, and extremely conscientious. … The other students assured me that they too had been taught that "all together" means "add." One added, "You know, key words. 'Left' means subtract." Mathematicians and math educators have been inveighing against key words for years, but I didn't realize HOW seriously good students were taking them.

Montclair State University reportedly, admits only one in five who apply. Kenschaft reports that this student was in the top half of those who BEGAN the course.

(B) "Keep It Simple for Students" (KISS) (Better methods of instruction)

(i) Several Proportion-type Problems: The "Unitary Analysis" method will be used to solve many problems that involve proportions, without having to deal with the full sophistication of proportional reasoning. (This is the method that I was taught in Grade 4 at my local P.S. 139 in Brooklyn, New York City, a half century ago when Dr. Dickler, was principal.) In fact, such problems can and should serve as background for formal proportional thinking.

Problem 7. Three T-shirts cost 15 dollars. How much do 5 T- shirts cost?

Method of Unitary Analysis,

Number of T-shirts Cost

3 \$15

1 5

5 25

In the same manner, one can solve the next 2 problems, from two other contexts, which the have the same mathematical structure, and have essentially the same solution.

Problem 8. A turtle can crawl 5 blocks in 2 hours. How long does it take the turtle to crawl 8 blocks, at the same speed. Express answer in hours and minutes.

Problem 9. Making 5 apple pies requires 2 pounds of apples. How much apples, in pounds and ounces, is needed to make 8 pies?

blocks-crawled/pies hours/pounds-of-apples

5 2.00

1 .40

8 3.20

Darcy Conant (Academic Coordinator for Graduate Studies, Department of Mathematics, UMCP) wrote: "Many students -- even 'good' high school students have difficulty with rate problems. Also, many "good" high school students have difficulty in dealing with times (hours and minutes). Look at the types of rate problems in a 'regular' (i.e. not watered down) HS Geometry text (10th grade)":

Time problem: A turtle can crawl 4 blocks in 2 hours. How long does it take the turtle to crawl 12 blocks?

Rate problem: A turtle can crawl at a rate of 2 1/2 blocks per hour. How long does it take the turtle to crawl 7 blocks?

Students should master these in elementary school. In a "regular" HS Geometry class, they are diversionary.

Centimeters to Meters. "When Grant Scott, a biology teacher, had to teach (a chemistry class) at Howard High School how to change centimeters to meters, he just told them to move the decimal two places -- rather than illustrating the concept. ... 'Forty-five minutes later, only three of them got it.' ". ("Right Teacher, Wrong Class", Washington Post, February 15, 1999)

The new California Standards require that students learn this in Grade 4. A student in a Georgia high school Algebra class noted: "I know how to change centimeters to meters [I learned it in middle school], just remind me, do I move the decimal left or right?"

It's predictable that students will forget, over the summer, when to move the decimal left or right.

Procedural instructions, about moving the decimal point, skip the conceptual understanding, namely: Since 100 centimeters make a meter, just like 100 cents make a dollar, not surprisingly 236 centimeters make 2.36 meters, just like 236 cents make \$2.36. (Similarly 236 percent of 777 is 2.36x777.)

This instruction (above) to "move the decimal two places" is what I call the "Avoid-reasoning and-thinking-and-analysis-by-excessive-memorization-of-overly-specialized-procedures" method of mis-education. It is popular with traditional textbooks because it is an easy way to teach. It avoids all thinking. It brings short-term success. That students forget most everything, over the summer, is a good excuse for the next grade's book to be largely a copy of this grade's.

For more complicated numbers, we use the Method of Unitary Analysis.

Problem 10: Change 236.5 centimeters to meters.

Start: 100 centimeters = 1 meter.

Divide by 100: 1 centimeter = 1/100 meter.

Multiply by 236.5: 236.5 centimeters = 236.5/100 m. = 2.365 m.

Problems 7, 8 and 9 provide good background for the next one. Therein the unit will be "The profit on a single tube".

Problem 10.5. The ration of nuts to bolts is 11 to 2 in a pile of 260. How many nuts?

Solution:

Nuts Bolts Total

11 2 13

Think, each set has 13 items; how many sets are needed for 260 items? Answer: 260 ÷ 13 = 20 (and hence 20x13 = 260). So multiply the table by 20.

Nuts Bolts Total

11 2 13

220 40 260

(An involved solution, which takes a full page of calculations, appears (as Example 100.1) in John H. Saxon Jr.'s Algebra 1/2 [pre-algebra] book)

Problem 11. It costs 90 cents for The Striped Toothpaste Company to make, package and ship a tube of toothpaste. The company also has "overhead costs" of \$3000 per month. The company sells (at wholesale) cartons of toothpaste at the price of \$2.50 per tube. This month, the company sold 5000 tubes of toothpaste. What is this month's profit?

Calculations with "verbal equations". The "verbal equations" will justify the calculations:

{The profit for each tube} = {The sale price of 1 tube} - {The cost of making 1 tube}

= \$2.50 - \$0.90 = \$1.60.

{The total gross profit on sale of many tubes} = {Number of tubes} x {The profit for 1 tube} = 5000 x \$1.60 = \$8000.

{This month's net profit} =

{This month's gross profit on sale of 5000 tubes} - {Monthly overhead costs} =

\$8000 - \$3000 = \$5000.

(ii) Catch-up and Overtake problems. I am a strong proponent of and practitioner of the "KISS" slogan, that is "Keep It Simple for Students". The arithmetic method presented here for "Catch-up and Overtake" problems is much simpler than using Algebra.

The Unitary Analysis Method trains students to think in terms of "per unit". This is a flexible technique that may be modified to provide the conceptual understanding for solving a wide variety of problems including Problems 12, 13, 14 and 15, below. These arithmetic solutions are simpler and provide far more conceptual understanding than the algebraic solutions expected on the MD HSA sample Algebra test. In fact, the use of Algebra therein gets in the way of conceptual understanding.

Problem 12. It costs 90 cents for The Striped Toothpaste Company to make, package and ship a tube of toothpaste. The company also has "overhead costs" (machinery or rent or whatever) of \$3000. The Striped Toothpaste Company sells (at wholesale) cartons of toothpaste at the price of \$2.50 per tube. How many tubes of toothpaste does the company need to sell to cover/balance-out the fixed costs?

Problem 11 is background for the next problem; it should be presented in the preceding grade.

The fixed costs of \$3000 will be paid for by the total gross profit on sale of many tubes. We need \$3000 = {fixed costs} = {The total gross profit on sale of many tubes}.

{The profit for each tube} = {The sale price of 1 tube} - {The cost of making 1 tube}

= \$2.50 - \$0.90 = \$1.60.

{The total gross profit on sale of many tubes} = {number of tubes} x {The profit for 1 tube}.

Hence,

{The number of tubes} = {The total gross profit on sale of many tubes} / {The profit for 1 tube}

= \$3000/1.60 = 1875.

Thus, the company will need to sell 1875 tubes to cover/balance-out the overhead.

Problem 12 is an arithmetic version of Item #32 on the sample MD Algebra Test (on the web at http://www.mdk12.org/mspp/high_school/look_like/algebra/v32.html). Problem 12 requires the student to provide the entire solution. In contrast, Item #32 requires the student to provide only a small part of the solution to Problem 12. When Item # 32 was field tested, 70% of the students omitted it.

I took the phrase "Catch-up and Overtake" from the next problem:

Problem 13. As the clock strikes noon, Jogger A is 2500 yards and Walker B is 4000 yards down the road (from here). Jogger A jogs at the constant pace of 10 yard/sec. Walker B walks at the constant pace of 5 yard/sec. How long will it take Jogger A to catch up to Walker B?

Solution: Walker B starts out 1500 yards ahead.

Jogger A is gaining at a rate of 5 yard/sec.

Jogger A will catch up to Walker B in 1500/5 = 300 sec.

We will use the same method to solve a car rental problem (with the same numbers).

Problem 14. (Sample MD HSA Algebra test Item #23) Company A charges \$20/ day plus .20/mi. Company B charges \$30/ day plus .10/mi. Whom to rent from?

Solution. To begin with (before driving) , the one-day price for Company B is \$15 = 1500 cents more (ahead) for a single day.

Company A is charging 5 cents per mile more than Company B.

Company A's price will overtake Company B's price in 1500/5 = 300 miles.

Thus Companies A and B charge the same for a 300 mile day.

In contrast, Item #23 ( of the sample MD HSA Algebra test) presents the bulk of an algebraic solution to Problem #14 using the graphs of two lines. Left to the student is to read the number where the two lines cross. – ( See it on the web at http://www.mdk12.org/mspp/high_school/look_like/algebra/v23.html)

(When this item was field tested, about 2 of every 3 students found the correct answer.)

Problem 15. "Two bicycle shops build custom-made bicycles. Bicycle City charges \$160 plus \$80 for each day that it takes to build the bicycle. Bike Town charges \$120 for each day that it takes to build the bicycle. … For what number of days will the charge be the same at each store?"

(This is essentially Item #18 on the sample MD Algebra test. See it on the web at http://www.mdk12.org/mspp/high_school/look_like/algebra/v18.html)) When Item 18 was field tested, almost 60% of the students omitted it. Item #18 is not a "real-world" problem; bicycles are assembled in hours not days.

Calculations. For each day, Bike Town charges \$120-\$80 = \$40 more than Bicycle City. But Bicycle City starts out higher by the \$160 charge. It takes 160/40 = 4 days for the prices to equilize.

MD HSA Algebra sample test Item #44 is another "Catch-up and Overtake problem". It can be done with the same arithmetic method. When Item 44 was field tested, about a half of the students obtained the correct answer.

These four "Catch-up and Overtake" problems Items #18, 23,32,44 are among the harder problems on the MD HSA sample Algebra test.

Items # 18 and 44 can also be done by making a simple chart of the two sets of prices. For Item #18 (Problem 15), the chart is:

Total Days 1 2 3 4

Bike. City 240 320 400 480

Bike Town 120 240 360 480

The charge will be the same at each store for a four-day bike.

The kinds of skills used in these Problems 12-15 are of real-life benefit and should be developed. Therefore, problems of this nature should be on a MD state math exam without the extensive hints given in the MD HSA sample test. The Method of "Unitary Analysis" allows a down-to-earth approach to all of them.

(iii) Work problems. It seems logical to me that middle school students, who have been trained to think in terms of amount per unit, would be able to learn to do the classic (algebraic) work problems, like this one:

Problem 16. (Work) Suppose that it takes Sally 3 hours to mow a lawn, and it takes Tom 4 hours to mow the same lawn; Tom's mower is less powerful than Sally's. Without using algebra (x or other variables) determine how long it would take Sally and Tom to mow the lawn if they worked together (using both lawn mowers)? (Assume that each works at his/her standard speed and they never get in each others way.)

Method of Unitary Analysis.

Person(s) Hours worked Fraction of job done

Sally 3 1

Sally 1 1/3

Tom 4 1

Tom 1 1/4

Sally and Tom 1 1/3 + 1/4 = 7/12

Sally and Tom 12 12(7/12) = 7

Sally and Tom 12/7 7/7 = 1

Answer. It would take Sally and Tom 12/7 hours to mow the lawn if they worked together.

Problem 17 .

It takes Sally 3 hours to walk from her home to Tom's home. It takes Tom 4 hours to walk from his home to Sally's home. They walk the same road. Suppose they both leave their own homes at noon, walking at their standard constant speeds toward each other. At what time do they meet? (The distance between their homes, who lives uphill from the other and their standard constant speeds shall remain unknown.)

Person(s) Hours walked Fraction of road walked

Sally 3 1

Sally 1 1/3

Tom 4 1

Tom 1 1/4

Sally and Tom 1 1/3 + 1/4 = 7/12

Sally and Tom 12 12(7/12) = 7

Sally and Tom 12/7 7/7 = 1

These arithmetic Problems16 and 17 are more difficult than any Algebraic problems on the MD HSA sample Algebra Test.

1999 CA standards Grades 3-7 Math Reasoning Standard # 1.2 or 1.3 states:

"Determine when and how to break a problem into simpler parts."

All of the solutions presented in this report are demonstrating how to do this.

Having mastery over a repertoire of "Unitary analysis" problems can help students with more complicated algebraic word problems for which various types of "rates" are ideal choices for unknown variables ("Rates" are units.) In fact, such problems can and should serve as an entry to algebraic word problems.

(C) Other Important topics for Elementary school.

Topics which many Grade 5 students are likely to encounter outside of school, including estimates, time, area, English units of measurement, percents, rates. Here is a list of problems that students should be taught how to do.

Estimates

Problem 18 In 1960, 245 children per 100,000, died of measles. This means that roughly/approximately one child in how many died of measles in 1960?

Calculations. 245 is roughly 200.

200 per 100,000 is 200/100,000, which is 2 in 1000 or one in 500.

The rough estimate provides meaning to the otherwise cryptic data 245 per 100,000.

Calculating two estimates, one on each side will show how accurate the estimates are. One could "bound" the estimate, on the other side, as follows.

Calculations. The number 245 is between 200 and 250, (and nearer to 250).

250 per 100,000 is 250/100,000, which is 2.5 in 1000 or one in 400.

Thus 245 per 100,000 is between one in 400 and one in 500.

Later, students can use their hand calculators to find that 245 per 100,000 equals one in 408. I suspect that this may be more difficult than doing the rough estimate of Problem 18.

Most people do not have any sense of what 245 children per 100,000 means. Problem 18, asks for the meaning. The data for Problem 18 came from MD Functional Mathematics sample Test Item #23; which does not ask for the meaning. There is no item which requires an estimate on the MD HSA Algebra sample Test.

Passage of Time.

MD Functional Mathematics Sample Test Item #21. The Orioles game began at 7:30 PM and ended at 10:20 PM. How long did the game last?

A better problem is:

Problem 19. The Wizards game began at 9:30 PM and ended at 1:10 AM. How long did the game last?

Now two questions from a history-teacher friend. These questions stumped her students, at an expensive private middle-school in NY.

Problem 20. The Revolutionary War started in 1776 and ended in 1783. How long did the war last?

Problem 21. Something is known to have occurred in Egypt around 2222 BC. How many centuries ago was that? (Round to the nearest century.)

Area.

Students should know/memorize the formula for the area of a rectangle and should know that area is a "Whole = the Sum of the Parts" type of quantity.

This is similar to the1999 CA Math Grade 4 Standard #1.4 for Measurement and Geometry (Mathematics Framework for CA Public Schools, page 47) which reads: "Use those formulas [for perimeters and areas of rectangles] to find areas of more complex figures by dividing the figures into basic shapes."

Here is the sample problem stated in CA math Grade 4 standards #1.4 (CA had adapted it from TIMSS K-5):

Problem 22. "The length of a rectangle is 6 cm, and its perimeter is 16 cm. What is the area of the rectangle?"

The 1999 CA Math Grade 5 standards #1.1 and 1.2 for Measurement and Geometry (Mathematics Framework for CA Public Schools, page 47) explicitly specify the next problem:

Problem 23. (a) Calculate the area of a block-letter L ,

1. Find the total area of the surface of a (rectangular) box (given the lengths of the various sides/edges).

Doing Problem 22 and 12.2 demonstrates conceptual understanding. Each is a multi-step area problem; They are not permitted on the MD Functional Math test, whose specifications read: "An area problem will show a square or a rectangle…" There are no area problems on the MD HSA Algebra sample test (which does contain two problems involving perimeter).

Formulas for the area of a rectangle and for the total area of the surface of a (rectangular) box were listed on a suggested crib sheet for the proposed MD state "Bridge" Goals exam (Sort of a low level Algebra II exam for high school students going on to college). This avoids basic conceptual understanding of area.

English units of measure. In elementary school, students should become familiar/fluent with the units of measure commonly used in MD, namely: English units. (There is no longer a need to know that a rod is 5 1/2 yards.) This includes changing units as in the next problem and in Problems 8 and 9 listed above. Only Item #38 of the MD HSA sample Algebra test requires the changing units and there Item #38 provides the conversion formula.

Problem 24. You drank (or overdrank) 1 3/8 gallons of beer at a London pub. The bartender requests that you mind your p's and q's. What is your response? (It is closing time, and the bartender is requesting the (whole) numbers of pints and quarts that you consumed.)

Percents and percentiles

Problem 25. (Reportedly from the Grade 5 Singapore textbooks.) There are 500 students at our elementary school (which ends at Grade 5). Of those, 20% are 5th graders and 15% are 4th graders. How many students are in grades K-3?

Problem 26. It says in a newspaper that there were 100 of something last year and that it increased by 300% during this year. How many are there today?

There was an increase of 300, which brings today's total to 100 + 300 = 400.

MD HHS sample Algebra test Item #48 The mathematical content is: "A typical household" spends 44% of its total utility costs for heating and cooling. "A typical household spent \$1,400 on utilities last year. If there are no significant changes in their utility usage for this year, how much should they budget for heating and cooling their home this year?"

Here is my real-world version of Item #48.

Problem 27. The Jones' family used 100 gallons of heating oil last year, the prices paid for a gallon of oil averaged one dollar. The Jones' family considers it possible/reasonable for this coming year, that * the (average) prices for heating oil may increase by as much as 20% and that * the amount of oil used for heating may increase by as much as 20%. How much should they budget for heating their home this year? Include the amount needed to meet the two contingencies listed. This is what percent higher than last year's expense. (Answer is \$144, which is 44% higher.)

Unfortunately, many students do not have a basic grasp of the meaning of percent, let alone the facility to use it in a problem like the Grade 5 Singapore one, written herein as Problem 25. According to a professor of Sociology on my campus, many eyes glaze over when she mentions percents to her college students.

Item #48, mentioned earlier, is the only item involving calculations with percents on the MD HSA Algebra sample test. It is simpler and requires less conceptual understanding than Problems 25, 26 and 27 of this report.

Rates and Arithmetic-based science

Arithmetic-based science. Next is a very practical 'real-world" problem, The skills that it draws on can be used in many situations. Problems of this sort should be presented as soon as students can be taught the concept of the word "per". Problems like this should be an important part of arithmetic-based science lessons.

Problem 28. It is a fact that fat has 9 calories per gram and protein has 4 calories per gram. If a piece of meat consists of 100 grams of protein and 10 grams of fat, how many calories does it have altogether? (Answer 490)

College students receive instruction on how to do Problem 28 in an elementary nutrition course at the Univ. of MD. This should not be necessary.

Of special importance is: {Distance traveled} = { Average speed} / {Time of trip}. There is no need to use this formula on the MD HSA Algebra sample test.

Prof. Steve Wilson, (former chairman, Math Dept. Johns Hopkins Univ.) suggests that there be short sequence of problems, where the answer to one problem is important in the solution of a later problem. Such short sequences of problems, will arise naturally in arithmetic-based science lessons including the math modeling of topics from science.

Sport: Here is a real world problem:

Puzzle: How can the federal government collect a 50-cent/gallon gas tax in 2005 without increasing the annual gas bill of Mr. and Mrs. Average Middle. Also lets reduce pollution from their car by 30% without increasing the costs of air pollution control devices. In addition, lets reduce our national dependency on oil from potentially hostile countries. Assume that today cars average 27 miles/gallon. (Hint: Do the following problem:)

Problem 29. Suppose that Mr. and Mrs. Average Middle are driving 10,000 miles each year. Suppose that in 2002, they are buying gas at \$1.70 per gallon and driving a car which uses 27 miles/gallon. Suppose that by 2005, they will have bought a new car which averages 40 miles/gallon, (as a result of a federal requirement that the average gas mileage for new cars be 40 miles/gallon) Also suppose that the price of gas is \$2.20/gallon, (as a result of a new additional 50-cent/gallon gas tax). What are their total gas bills in 2002 and 2005? Also what is the reduction in their gas consumption for 2005? Assume that they drive the same number of miles in 2005 and continue to be a one-car family.

Answer. \$630 in 2002 and \$550 in 2005, and they use 32% less gas in 2005 than in 2002.

Also since, they will be using 32% less gas in 2005, there will be a 32% reduction in pollution from their car even without any improvement in air pollution control devices.

Answer to the puzzle. The U.S. government needs to both raise the required average gas mileage of new cars to 40 miles/gallon and to collect a 50-cent/gallon gas tax.

I do not expect students to solve the puzzle; but they should be able to do Problem 29 .

The difficult part of the next problem is following directions.

Problem 30. (The Geese Puzzle) A flock of geese on a pond were being observed continuously.

At1:00 P.M., 1/5 of the geese flew away.

At 2:00 P.M., 1/8 of the geese that remained flew away.

At 3:00 P.M., 3 times as many geese as had flown away at 1:00 P.M. flew away,

leaving 28 geese on the pond.

At no other time did any geese arrive or fly away.

How many geese were in the original flock?

Solution.

At 1:00 P.M., 1/5 = 20% of the geese flew away; leaving 80%.

At 2:00 P.M., 1/8 of 80% = 10% flew away.

At 3:00 P.M., 3 times 20% = 60% flew away; leaving 28 geese.

A total of 20% + 10% + 60% = 90% have flown away; leaving 10%.

Thus the remaining 28 geese are 10%.

The original flock was 100% = 10 times 10% = 10 x 28 = 280.

This was question #25 of Section 4 of the May 2000 SAT Math test. On a scale of 1 to 5, the SAT rates this at the # 5 level of difficulty. Only 8 out of the 60 May 2000 SAT Math test questions are in level 5; the questions used to separate the SAT scores of 700 from those of 800.

Students are permitted to use hand calculators to do all the calculations on the Math SAT tests. The Geese problem can also easily be done using fractions or decimals.

Reading for comprehension is necessary for doing Arithmetic as well as other math problems and as well for understanding science and social studies textbooks. Comprehending a word problem correctly is a crucial part of doing math. One small example is that pupils need to understand the passive voice of verbs; this is crucial for understanding how "3 divided by 6" differs grammatically, and hence mathematically, from the common colloquial expression: "3 divides 6". (Note that: 3 pies may be divided by 6 girls; but 3 girls will not be divided by 6 pies.)

Math word problems require slow, careful, and precise reading and rereading of the problems.

Even students, reading on grade level, may have difficulty comprehending math word problems. (More) instruction in precise reading and in analyzing the reading, is needed at all grade levels, at least in math and in reading/English.

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Jerome Dancis is an associate professor of mathematics at the University of Maryland, College Park, MD, 20742-4015. E-mail: jdancis@math.umd.edu