**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 "**K**eep **I**t **S**imple for **S**tudents", 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 __cent__imeters
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 per__cent__ 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 __cent__imeters = 1 meter.

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

Multiply by 236.5:
236.5 __cent__imeters = 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.

An "advanced" ratio problem. At least one Grade 8 textbook would refer to the next problem as an "advanced" ratio problem, because of the additional addition involved.

**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