Supposedly Difficult Arithmetic Word Problems

Keep It Simple for Students (KISS)

By Jerome Dancis

Executive Summary, Introduction and Conclusions

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. These better methods of instruction are in the spirit of my version of KISS , that is "Keep It Simple for Students", while emphasizing conceptual-understanding.

This will include several problems, which are normally solved using Algebra; including five of the more difficult problems on the Maryland High School Assessment on Functions, Algebra, Data Analysis and Probability (MD Algebra) sample test, an about-to-be-implemented high school graduation requirement in MD. The arithmetic solutions presented herein, provide more conceptual-understanding of the problems than the expected Algebraic solutions.

We will be solving these types of problems:

(i)   Simple Rate problems, without mentioning the words "ratio" or "proportion". Grade 4.

(ii) Ratio and Proportion problems done arithmetically without having to deal with the full sophistication of proportional reasoning. This will including a "Data Analysis" problem on the MD HSA sample Algebra test, that was solved by few students. Grade 5.

(iii) Catch-up and Overtake problems. Simple arithmetic solutions are presented for four of the more difficult problems on the MD HSA sample Algebra test. Grade 5 or 6.

(iv) Work problems, which were the most difficult word problems in Algebra I courses. Grade 7 or 8.

(v) A sophisticated average problem

These sets of problems epitomize spiral learning, in that they will build a "problem" staircase, in which, doing each set of problems provides useful, if not crucial background for the later sets, including complicated algebraic word problems.

Appendix. Algebraic Word Problems. Translating a word problem into algebraic equations. A problem that stumps college seniors, majoring in engineering: but should be accessible in high school.

Slogan. All elementary school students can learn to solve these Arithmetic Word Problems.

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

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. These better methods of instruction are in the spirit of my version of KISS , that is "Keep It Simple for Students", while emphasizing conceptual understanding.

(i) Simple Rate Problems: The "Unitary Analysis" method will be used to solve many problems that involve proportions, without mentioning the word "proportion". (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 1. 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.

Remark. Of course, this problem should be presented only after the students are fluent with the parts, namely, * If one T-shirt cost 5 dollars, how much do 5 T- shirts cost? And * If three T-shirts cost 15 dollars, how much does one T- shirts cost? Otherwise, the chart will be a mystery.

Problem 2. A child can run 5 blocks in 2 minutes. How long does it take the child to run 8 blocks, at the same speed.

Problem 3. Making 5 apple pies requires 2 pounds of apples. How many pounds of apples are needed to make 8 pies?

Blocks-run/pies                      minutes/pounds-of-apples

5                                              2.00

1                                                 .40

8                                                3.20

Problems 1,2 and 3 provide good background for many problems that follow.

Problem 4. A ball dropped from a tall building, falls 16 feet in the first second. How far does it fall in two seconds? (Ignore friction)

Warning. Multiplying 2 x 16 = 32 whether directly or by setting up a table as in Problems 1,2 and 3, will produce a wrong answer. While the falling distance is a function of time, it is not a linear function; it is not proportional to time. It is crucial check to that the calculations being done are the correct calculations. One needs a reason why multiplying a line, in one of these tables, is a valid operation. This will increase instructional time for these problems. If students go on automatic pilot, while setting up these tables, they will sometimes do it when it is not valid. Correctly, multiplying numbers when multiplication is not justified is wrong.

Darcy Conant 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)":

Problem 5. (Time) A child can run 4 blocks in 2 minutes. How long does it take the child to run 12 blocks?

Problem 6. (Rate) A child can run at a rate of 2 1/2 blocks per minutes. How long does it take the child to run 7 blocks?

Remark. Students should have mastered these in elementary school. Both are the same type as Problems 1,2 and 3. Students will need to be told that "2 1/2 blocks per minutes" means "2 1/2 blocks each minute". It is absurd that they are being taught in high school. It is doubly absurd that they appear in a "regular" HS Geometry class, where they are diversionary.

Decimals to Percents and Centimeters to Meters. "When 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)

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 2.36 x 777.

This instruction to "move the decimal two places" may also be mis-used to teach the conversion of decimals to percents; it will be forgotten or garbled over the summer.

This instruction (above) to "move the decimal two places" is what I call an "Avoid-thinking-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 much, 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 7. 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.

Similarly, 236.5 (236 and a half) cents and 236.5% can be converted to 2.365 dollars and 2.365, resp. Now, lets convert back:

It is absurd that a high school biology teacher had not learned how to change centimeters to meters; -- unless he too, had been taught by the rule for idiots: "move the decimal two places", in which case, it is predictable. The new 1999 California Standards require that students learn this in Grade 4.

Problem 8. Change 2.365 to a percentage.

Start:                         100% = 1

Multiply by 2.365:           236.5% = 2.365

Similarly, 2.365 dollars and 2.365 can be converted to 236.5 cents and 236.5%, resp.

(ii) Ratio and Proportion problems.

"Ratio" and "Proportion" basically, mean that we can set up tables, (as in the previous problems) and then it is valid to multiply or divide a line by a number.

Problem 9. Jack and Jill went up the hill to pick apples and pears. Jack picked 10 apples 15 pears and Jill picked 20 apples and some pears. The ratio of apples to pears picked by both Jack and Jill were the same. Determine how many pears Jill picked.

Solution                Apples        Pears

Jack      10              15

To obtain the 20 apples, Jill picked , we need simply double the 10 apples Jack picked. So we multiply the chart by two.

Apples        Pears

Jack      10              15

Jill       20              30

Thus, Jill picked 30 pears.

Let's redo this problem with less nice numbers:

Problem 10. Jack picked 12 apples 15 pears and Jill picked 16 apples and some pears. The ratio of apples to pears picked by Jack and Jill were the same. Determine how many pears Jill picked.

Solution                Apples        Pears

Jack      12              15

To obtain the 16 apples, Jill picked , we need to find what number multiplied by 12 will yield 16. This is one meaning of division: 16/ 12 = 4/3. So multiply the one-line chart by 4/3:

Apples        Pears

Jack       12              15

Jill       16              20

Thus, Jill picked 20 pears.

Next, we present another solution in the spirit of unitary analysis. First divide the one-line chart by 12, then multiply by 16:

Alternate Solution           Apples        Pears

Jack       12              15

1             15/12

Jill         16            16 x 15/12 = 20

Remark. Of course, these ratio problems should be presented only after the students are fluent with the rate problems in the previous section. Otherwise, multiplying a chart line will be like waving a magic wand, and there will be little understanding as to why the answer that emerges should be correct.

Next is the same type of problem, except that the word "proportion" is used instead of the word "ratio".

Problem 11. Physics tells us that weights of objects on the moon are proportional to their weights on Earth. Suppose an 180 lb man weighs 30 lb on the moon. What will a 60 lb boy weigh on the moon?

Solution:              Earth Weight            Moon Weight

Man               180                          30

To obtain a 60 lb. Earth weight, we divide the 180 lb. Earth weight by 3; so we divide the line on the chart by 3:

Earth Weight            Moon Weight

Man               180                          30
Boy                 60                           10

Remark. Of course, elementary school children should have been taught that the weight of an object is fixed. A bag of apples or a child has the same weight, no matter on which scale [on Earth] the weightings occur. The parenthetical phrase "[on Earth]" is (of course) omitted from instruction. It must be a complete mystery to them as to why a weight on the moon should be any different than the weight on Earth. This is why, Problem 11 should be presented in a science lesson, not in a math lesson. It should be presented only after a long discussion as to why weights on the moon are a fraction of weights on Earth and then why it is always the same fraction. Either is a sophisticated topic for middle school students.

As stated, Problem 11 is a straight-forward and correct proportion problem (albeit an unfortunate one). In contrast, here is a similar, but impossible problem that should not be inflicted on unsuspecting students:

Problem 12. (Impossible) Suppose an 180 lb man weighs 30 lb on the moon. What will a 60 lb boy weigh on the moon?

Remark. There is no way for a student to know about this proportionality of weights of objects on the moon and on Earth Nevertheless, rumor has it that this not an uncommon type of problem. There is the expectation that students will make the completely unwarranted assumption that the weights of objects on the moon are proportional to their weights on Earth. This is training students to make completely unwarranted and sometimes completely incorrect assumptions of proportional in many situations. Just as Problem 4 tempts students. Highly counterproductive!

Problem 13. A sample of 96 light bulbs consisted of 4 defective ones. Assume that today's batch of 6,000 light bulbs has the same proportion of defective bulbs as the sample. Determine the total number of defective bulbs made today. (Ignore the fact that the assumption of exact proportionality is highly unlikely)

Number of Defective Bulbs          Total Bulbs
4                        96

Having the "same proportion" means that it is valid to multiply the chart by a number. There will be 6000 total bulbs. What number do we need to multiply 96 by to obtain 6000? The number 6000 divided by 96 which is 62.5. Thus 62.5 x 96 = 6000. We multiply the chart by 62.5:

Number of Defective Bulbs          Total Bulbs
4                        96
62.5 x 4 = 250          62.5 x 96 = 6000

The answer is 250 Defective Bulbs out of a total of 6000 bulbs.

Problem 14. A very small sample of light bulbs consisted of 4 defective ones and 96 good bulbs. Assume that today's batch of 6,000 light bulbs has the same proportion of defective bulbs as the sample. Determine the total number of defective bulbs made today. (Ignore the fact that the assumption of exact proportionality is highly unlikely)

The difference, in this problem is that 96 is the number of good (Non-Defective) Bulbs in the sample, not the total number. So one adds 4 + 96 =100, to obtain the total number of bulbs in the sample. This yields this table:

Number of Defective Bulbs          Number of Nondefective Bulbs           Total Bulbs
4                                                          96                    100

Again, having the "same proportion" means it is valid to multiply the chart by a number. Here it is useful to multiply by 60, since 60 x 100 is 6,000.

Number of Defective Bulbs          Number of Nondefective Bulbs           Total Bulbs
4                                                          96                    100
60 x 4 = 240                                              60 x 96                60 x 100 = 6000

The answer is 240 defective bulbs out of a total of 6000 bulbs.

Here is how this arithmetic problem appeared on the sample MD Algebra test. Therein, it is listed as Item 15, a Data Analysis item.

MD Algebra Item 15. For quality control, a light bulb company conducted a random sampling of their light bulbs. The results are shown below.

Number of Defective Bulbs 4; Number of Nondefective Bulbs 96

The light bulb company makes 6,000 light bulbs in a day. Based on this sample, how many defective light bulbs can the company expect to make in a day?

A 240, B 250, C 1, 500, D 2, 400

A statistician colleague told me that on 70% of the days, the number of defective light bulbs will range from 225 to 255, but only if it is true that light bulb production is described by a "binomial distribution". Otherwise less is known. Rumors have it that more lemons are, or were, produced on Mondays and Fridays at car factories than during midweek days. (A higher absentee rate being the culprit). The same proportion does not always hold.

As a standarized test-taking tactic, students should think of the "key" word "expect" as if it is basically a synonym to the phrase "when the same ratio/proportion occurs" This converts Item 15 into Problem 14; the same table and calculation yield the same correct answer A 240.

This Item 15 is on the web at http://www.mdk12.org/mspp/high_school/look_like/algebra/v15.html. Clicking on a box on the upper right side, you will read that:

"This item was field-tested in January and May of 2000. [One in eight students] 12.5% omitted this item. The chart shows the percentage of student responses to each choice.

A 35.20% B 23.20% C 29.10% D 12.50%"

Students who solved this Item 15 as if it were Problem 13 would calculate (on their hand calculators) the incorrect answer B 250. It is a mystery to me what the almost 30%, who chose C 1,500 might have been thinking?

As noted, this should be a Grade 5 level proportion-problem. It is absurd that only about one in three students did this problem correctly (in the field-testing of mostly Grade 9 students). It is doubly absurd that the number choosing the absurd answer C 1,500, was not that much less than the number choosing the correct answer. I took the time to write this report in the hope of reducing these absurdities.

Clicking on another box on the upper right side, you will read that: Item 15 Ö

"Indicator 3.2.1: The student will make informed decisions and predictions based upon the results of simulations and data from research."

The "prediction", that the student is expected to make, is that there is the same proportion of defective bulbs in any day's output as there was in the sample. The student is not expected to justify this prediction. In fact, this is the standard prediction that the student should make on many a standardized-test data-analysis item.

In reality, it will be a rare day, when the percent of defective bulbs is exactly the same as in the sample, even if all the sample bulbs were made on the same day. The word "expect" is being mis-used in the wording of the problem. It is used because it is connected to the technical statistical term "expectation". In statistics, the "expectation" (based on a sample) is the best estimate of the average percentage of defective bulbs over many days. This "expectation" will be the same as the percentage of the sample, even though the daily number of defective bulbs will vary considerably.

We highlight the problem with using the word "expect":

Remark. Consider Item 15 when the sample has 97 good bulbs and still 4 defective ones. Redoing the same calculation yields an "expectation" of 237.6, which is fine, since expectation is an average. But what does it mean to "expect" 237.6 defective bulbs?

Remark. Yes, this is quite sophisticated even for Grade 9; this is why I think it is ill-advise to include problems like this Item 15, in a Grade 9 level course. Problem 14 is not sophisticated, it could be taught in Grade 5.

Problem 15. 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?

Remark. First, writing down "verbal equations" is a very good way to understand, explain and justify calculations; they also lead the student to making appropriate calculations. They are also useful later in setting-up Algebraic word problems (As in the appendix, "Algebraic Word Problems ").

Solution.

{The sale price of 1 tube} ? {The cost of making 1 tube} = {The profit for each 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". 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.

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 16, 17, 18 and 19, below.

Problem 16. 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 15 was crucial background for the this problem; useful if it had been 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 16 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 16 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 16. 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 17. As the clock strikes noon, Jogger J is 2500 yards and Walker W is 4000 yards down the road (from here). Jogger J jogs at the constant pace of 10 yard/sec. Walker W walks at the constant pace of 5 yard/sec. How long will it take Jogger J to catch up to Walker B?

Solution: Walker W starts out 1500 yards ahead.

Jogger J is gaining at a rate of 10 ó 5 = 5 yard/sec.

Jogger J will catch up to Walker W in 1500/5 = 300 sec.

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

Problem 18. (Sample MD HSA Algebra test Item #23 "Ace Car Rentals advertises that a rental car costs \$25 per day plus a charge of \$0.10 per mile. For the same car, Better Car Rental advertises a price of \$40 per day plus \$0.05 per mile. Ö

For what number of miles is the cost of renting a car the same at both companies?

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

Ace Car is charging 5 cents per mile more than Better Car.

Ace Car 's price will overtake Better Car's price in 1500/5 = 300 miles.

Thus Ace Car and Better Car charge the same for a 300 mile day.

In contrast, the skipped part, of the text for Item #23 of the sample MD Algebra test, actually presents the bulk of an algebraic solution to Problem 18 using the graphs of two lines. Left to the student is to read the number where the two lines cross. ó ( 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 19. "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. Ö 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 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 sample MD 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 19), 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 16-19 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.

(iv) 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 Problem 20.

Problem 20. (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    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.

Here is the same mathematical problem, but in a different setting:

Problem 21.

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      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

While the work problem was a standard high school Algebra when I attended school; it has been dropped from the curriculum. These arithmetic Problems 20 and 21 are more difficult than any Algebraic problems on the sample MD Algebra Test. A work problem which requires Algebra is included in the appendix.

Now for a slightly different type of work problem, where the same type of ideas are used, but in which multiplying an entire row of the table, by a number is incorrect:

Problem 22. Suppose that it takes Tom and Dick 2 hours to do a certain job. But, today, a friend joins them and works at the same rate as Tom and Dick. How long will it take for the three men, together to do the same job?

Workers           Hours worked            Fraction of job done
2                              2                           1
2                              1                           1/2
1                              1                           1/4
3                              1                           3/4
3                            4/3                          1

Answer: It would it take Tom, Dick and friend 4/3 hours to do the same job together.

(v) A sophisticated average problem Average problems, mainly, require the use of the defining formula for an "average", namely:

Average [of a set of numbers] = Total [or sum of the numbers] / {number of numbers};

Or, in shorthand:                          Average = Total / {number of numbers},

And hence:                                         Total = Average x {number of numbers}

Similarly for gas mileage:           {Average mpg} = { Total mileage} / { Total gallons },

And hence:                                 { Total mileage} = {Average mpg} x { Total gallons }.

Here is a sophisticated average problem, but its solution is basically just repeated use of the defining formula for an "average":

Problem 23. In 1999, suppose that U.S. family small trucks averaged 20 mpg and our family cars averaged 28 mpg. Also suppose that U.S. families drove 100 billion miles in their small trucks and our families drove 84 billion miles in their cars. Find the average mileage of family vehicles, all together.

Solution.

{ Truck mpg [average] } = { Total truck mileage } / { Total truck gas }.

Hence: {Total truck mileage} = { Total truck gas} x { Truck mpg [average] }

Thus: { Total truck gas } = { Total truck mileage } / { Truck mpg [average] }

= 100 billion / 20 = 5 billion gal.

{ Car mpg [average] } = { Total car mileage } / { Total car gas}.

Hence: { Total car mileage } = { Total car gas } x { Car mpg [average] }

Thus: { Total car gas } = { Total car mileage } / { Car mpg [average] }

= 84 billion / 28 = 3 billion gal.

Average mileage all together = {Total mileage} / { Total gas} = 184 billion / 8 billion = 23 mpg.

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.

These sets of problems epitomize spiral learning, in that they will build a "problem" staircase, in which doing each set of problems provides useful, if not crucial background for the later sets.

Having mastery over a repertoire of these 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 including, but not limited to, the ones in the appendix: "Algebraic Word problems".

January, 2003