Proportional Reasoning and Ruth Parker’s Chicken Problem


     Ruth Parker is a regular speaker on the “professional development” circuit for math teachers, an exponent of what she would call the true understanding of mathematical concepts in place of the rigid routines our own childhood.  In particular, there is no longer any need for children to learn the paper-and-pencil arithmetic needed by 19th Century shopkeepers, or even carpenters and accountants for that matter, now that we have calculators and computer programs for convenience and accuracy.  Long division and “invert and multiply” are passé, so that we (children, that is) can now spend valuable school time understanding the nature of real-life problems (“word problems”, as they were called in earlier times), figuring out ways to solve them, talking them over with other children, drawing diagrams, even inventing algorithms of our own as needed.  For daily life, after all, approximations and mental arithmetic will do quite well, and the calculator will do exact computations, but nothing will replace our own minds when it comes to reasoning, seeing the structure of what is being done.


     Ms. Parker has a standardized speech, it appears, for I heard one of her presentations just last month (January 10, 2006, in East Irondequoit, New York) and yet am aware (from a description by Nilanjan Banerjee in the Dartmouth Review of January 13, 1999[1]), that she had been giving much the same speech well over seven years ago.  That alone doesn’t disqualify the message, but since it is my intention to disparage the message on other grounds I find it disheartening to realize that seven years of other people’s efforts in disparagement appear to have been fruitless so far.


     The core of her message is quite easily seen via a problem she has been proposing to audiences all over the country to illustrate her main points, a problem usually known as “the turkey problem”, though it concerned chicken on the night I heard it.  I was sitting in a middle school auditorium where an audience of about a hundred was gathered at 7 pm, an audience mainly made up of teachers from the vicinity. Sitting next to me was Richard Woodward, an acquaintance and a former teacher (among other things).  Ms. Parker was speaking at the invitation of the East Irondequoit school district, in indirect support of a new math program the district was introducing, the TERC Investigations series for primary school mathematics, a program whose philosophical underpinnings are exactly those Ruth Parker had been advocating, and apparently employing, for many years.  In her talk she often referred to “her students”, indicating long experience teaching according to this philosophy herself.


Now to the chicken problem.  It seems Ms. Parker was on a diet, she explained, and was permitted only 1/4 pound of chicken for her lunch one day.  She went to the deli to buy her quota, but the man with the meat slicer gave her three equal slices that added up to 1/3 of a pound whereas she had only wanted 1/4.  She took the package, but since she was a determined dieter she wanted to figure out exactly how many of those slices she needed to suit her diet.  That was the problem, though her wording was not this succinct:  How many slices add up to 1/4 pound if three slices add to 1/3?


     I found it interesting that in posing the problem she told us that “the three slices were 1/3 of a pound”, but didn’t say whether that was 1/3 pound for each slice or 1/3 pound for the lot.  This ambiguity, which I resolved after a minute of cogitation, did cost me some time, a matter of some importance to the Parker presentation it seems to me, for it also fooled Mr. Woodward beside me, to such effect that he didn’t get a reasonable answer at all in the time Ms. Parker gave us to work at it. 


She had first asked the audience to form into small groups to solve the problem she was about to give us.  This is a preferred method of doing such things according to current fashion in school mathematics classes, and encouraged by the National Council of Teachers of Mathematics [NCTM] for many years now as superior to lonely effort at desks facing a blackboard.  Since we were not around tables, as children in a modern math class are supposed to be, and since in our auditorium seats we were all writing on scrap paper – those of us who could find some -- or in notebooks on our laps, “forming a group” was out of the question for most of us.  Woodward has his scrap of paper and I had mine. 


I first tried the problem under the misapprehension that each slice weighed 1/3 pound, but when I realized that this couldn’t have been the intention, I solved the problem mentally, by thinking that if three slices weighed 1/3 pound, then there were 9 slices to the pound, so 1/4 of  a pound required 1/4 of 9 slices, or 9/4 slices, that is 2 and 1/4 slices.  Had the numbers not been so easy as to give me, instantly, an integral number of slices per pound, I would have had to do a bit more arithmetic, of course.  I thought briefly of how I would set up the proportions in a more general case, but first decided to see how my neighbor Mr. Woodward was doing.  He had been writing something, but I didn’t ask until much later, because Ms. Parker was now addressing the audience.


     Bent on showing us the clumsiness of standard methods, Parker had no intention of making them sound easy.  After having given us about three minutes, she asked for volunteer answers from the audience, which she wrote on an overhead projector.  The very first volunteer got it right, which must have disappointed her, but she insisted on more, and indeed got two more.  She took a vote from the audience on which of the three answers displayed on the board was right, but I don’t remember which one won. Most of the audience was quite uneasy at being examined in arithmetic, it seemed to me.


The two losing answers were not sensible on the face of it, but that was not Parker’s point, which was to show that the problem would turn out to be simpler than the audience imagined. First, though, Parker showed us how the problem is solved “by proportional reasoning”, which she says is still the sort of thing inflicted on children in most places these days, and then she was going to show us how modern math instruction, such as the new Investigations program in East Irondequoit, does it. 


A historical note, not part of the Parker talk:  In the old days “proportional reasoning” was called “The Rule of Three”.  In pure arithmetic terms this rule is the method by which a fourth term is found when three terms of a proportion are known.  People unversed in mathematics are still familiar with this idea in the examination questions concerning analogies, e.g. “Book is to Page as Year is to: 1.Calendar; 2.Century; 3.Day; 4.Age.”  Clearly the fourth proportional is “day”; and we might borrow some mathematical symbolism here and write the true proportion as


“book : page = year : day”.


     In arithmetic the idea of comparison is “ratio”, and a “proportion” is the statement that two ratios are equal, e.g. 9:3 = 3:1, pronounced “Nine is to three as three is to one.”  We understand that we have comparison by division in mind here, not subtraction, since 9 – 3 is 6, and not the same as 3 –1, which is 2.  So to make sure the correct comparison (division, not subtraction) is conveyed by our “ratio” symbolism of the colon, we write it “9/3 = 3/1”.  Well, of course:  Both fractions equal the integer 3.  But other ratios don’t reduce to integers; 6/15 = 8/20 is true though not so easily seen so.  In fact, this example makes for a standard “rule of three” problem in arithmetic:  “Find the fourth proportional to 6, 15, and 8.”  That is, “6 is to 15 as 8 is to WHAT?”


     That “20” is the correct answer is easily seen once we have it before us, because we can “reduce” each of the two fractions, 6/15 and 8/20, “to lowest terms.  6/15 = 2/5 is obtained by dividing numerator and denominator by 3, and 8/20 = 2/5 by dividing numerator and denominator by 4.  Both terms of the proportion being equivalent to 2/5, the proportion is true.  But if the problem is given, “If 6/15 = 8/x, find x”, things are not quite so easy.  Without going into the admittedly complicated story of why the rule should be true, the “rule of three” by which x is found out is almost summarized as “the product of the means equals the product of the extremes.”


     If we list the terms in the order we say them, they are {6, 15, 8, x}.  15 and 8 are between the first and fourth terms and are called the “means”, while 6 and x are the “extremes”, so the Rule of Three says in the present case that 15X8 = 6Xx, or in algebraic notation, 120 = 6x, whence x = 20.  It takes only a modicum of algebra to use the rule symbolically for any three terms.  Suppose A, B, and C are any (non-zero) numbers; what is the fourth proportional?  Calling it D, we are looking for D to satisfy “A/B = C/D”.  The Rule of Three says BC=AD, and solving for D by dividing the equation through by A, we get “D =  BC/A”.  


     Nineteenth Century children often merely memorized this last result, without even writing out the proportion itself:  D = BC/A can be pronounced viva voce, “The fourth proportional is the product of the means, divided by the first extreme,” or perhaps, “the product of the second and third terms, divided by the first”. This sort of thing is reprehensible, of course,  when taught as to a parrot, since the algebra of fractions is of importance in other contexts, too, and the solution of the fractional equation which conveys the meaning of the proportionality is itself an important object of study.


     While from an arithmetic point of view the Rule of Three is quite simply used, if not entirely understood, it is still, even when thoroughly understood from an arithmetic point of view, difficult to apply in practical problems.  School children of my time always dreaded “story problems”, that is, problems about situations as they might occur in daily life:  the age of the father six years from now, the amount of interest to be paid in the seventh year and so on, problems involving numbers one had to dig out from a lifelike description if one were to use them to find an “unknown”. 


Since many such problems reduce to finding a fourth proportional, with the other three terms buried somewhere in the tale set up for the purpose, the schools often insist on children’s developing a faculty the education profession likes to call “proportional reasoning”.  By this phrase is meant the ability to recognize, in a situation involving numbers, that there are ratios involved, and that one of the ratios has the “unknown” as one of its terms, and that an equivalent ratio is actually completely given by the data.  In such a case the “unknown is found by “The Rule of Three”.  Thus older schoolbooks included under the rubric “Rule of Three” not only the arithmetic formula for the fourth proportional, but the ability to use it to solve a story problem amenable to its use.


     Though calling all this “proportional reasoning” is recent, that is only because the education system insists on being up to date, and changes the names of things to imply the importance of recent educational research.  A hundred years ago it was common to hear from a workman the confession, referring to his having dropped out of school after the 6th grade, that he “took Arithmetic as far as the Rule of Three”.  Or, “I never got past 8th grade, but I did learn decimals and The Rule of Three.”  Most of today’s college graduates cannot say as much.


To return now to Ms. Parker and her desired 1/4 pound of chicken, let us see how it yields to the Rule of Three.  There are two clear ratios in the problem: firstly, the number of slices (3) and their total weight in pounds (1/3) as she got them at the deli, and secondly, the number of slices (N) and their weight in pounds (1/4) as she intended to use them for lunch.   These numbers are clearly “in proportion”, i.e. the number of slices is to the number of pounds [bought] as the number of slices is to the number of pounds [to be eaten]. This proportionality might be written,  


“slices we have” : ”weight we have” = “slices we need” : “weight we need” , or,


3: (1/3) = N: (1/4), or arithmetically, 3/(1/3) = N/(1/4).


The rest is arithmetic (sometimes called “algebra”).  Rather than remember the formula for “The Rule of Three” we have only to solve for N.  “Cross-multiplying” (i.e. equating products of means and extremes) produces the equation (1/3)N = 3(1/4), thence N/3 = 3/4 and N = 9/4.  Two and one-fourth slices.


     Now Parker, after introducing us to the fact that she was going to show us the (ugh) “traditional” way to solve the chicken problem, rapidly filled the blackboard (overhead display) with equations equivalent to what I have written above, but without the explanatory lecture on proportions.  As she was writing her equations she offered a patter worthy of Gilbert and Sullivan, or Rossini’s Figaro: “Well, let’s see, three pounds divided by a third is the number of slices divided by a fourth so three over a third is nine over one, and four N makes nine over four equals two and a fourth.”  Her accompanying display bristled with equations, more than I managed to write above, and even required the ill-famed “invert and multiply” ritual (to compute the “3/(1/3)” I think, to get 9/1, or 9).  When she reached the final answer (two and a fourth slices) she was virtually sweating with the effort.  Raising her eyes to heaven she heaved an enormous sigh: Well!  (I.e., Even though we math teachers know how to do this if we have to, surely we don’t want to impose it on our children, do we?


Then she explained how clever the children themselves were, when permitted in groups to solve this problem.  They had lots of ways, beautiful ways, real exercises of the imagination.  One of those ways was my way actually, so I suppose I’m a good child:  As I had done it mentally, if the three slices are 1/3 pound, then 9 slices are one full pound, so 1/4 pound is 1/4 of those 9 slices, or 9/4 slices; voila! 


     As Ruth Parker described it, this student solved it her own way, not even needing arithmetic notation:  She pictured the three slices from the deli as three circles in a row, and then adjoined six more slices, in two rows just below them, to form a three-by-three display of nine slices, which made up a pound of sliced chicken.  Then she literally divided the display in four equal parts, thus:















(The picture is a bit asymmetric; sorry).  Then you can actually count the slices in (say) the upper right quadrant:  One full slice, two half-slices and, from the middle one, one 1/4 slice.  Two and a fourth.


Mirabile! What happened to the rule of three?  Where is the equation?  How come we don’t need “divide and invert” – or whatever that awful rule says?  The children’s own algorithms are better than all the crowned heads of Renaissance mathematics!


     Well, yes, as long as the data consisted of thirds and fourths.  I myself, versed in The Rule of Three since 1935, didn’t choose to go through the agony Ms. Parker had done up there with all her equations and unknowns.  I had mentally calculated the result as a matter of course.  Mental arithmetic is something we all should do when it will do the job, and even when it won’t quite do it sometimes suffices for an estimate, that is, if we round the data to simplify the numbers before setting up our mental model.  (But we mustn’t expect always to be as lucky as Ruth Parker in getting exactly three slices adding to 1/3 pound, and a diet needing 1/4 pound!)


     The audience seemed to be awed at the simplicity of this way of looking at things, and perhaps it was convinced that Investigations is thereby proved more powerful than the arithmetic of fractions.  But it isn’t.  Let us change the problem slightly, even without really unwieldy data:  Let us suppose that Ms. Parker’s diet had called for 2/5 of a pound (instead of 1/4), and that the chicken had been cut very thin, with 17 slices totaling .55 pounds in all, what then?  The Rule of Three would give, exactly as before,


“slices we have” : ”weight we have” = “slices we need” : “weight we need”



17  : 55/100  =  x  :  2/5   ; or


17/(55/100)  =  x/(2/5).


     The rest is then arithmetic (or algebra): The given equation tells us that 1700/55 = 5x/2, if we use “invert and multiply” on both sides, so that 3400=275x (“cross multiply”), and so x = [12 and 4/11] slices (“long division”).


The point here is that once you have the proportion set up all numerical examples will behave the same except for the mindless part, the easy part, the mere “algebra”.  The mentions of “Invert and multiply”, “cross multiply”, and “long division” in my calculation above were intended as brief reminders of what mathematicians call theorems, statements or procedures which, once proved, save having to go back to first principles or to carry in one’s head the burden of recalling the meaning of the computation itself, or recall – or have to invent -- the proof that the operations do yield the true results.  Using these devices makes easy what going back to first principles would make a very long road indeed.


For example, when we replace “7X8” by “56” we are using a theorem, one we proved somewhere in our childhood.  Nobody called it a theorem than; most likely it was showed to us as a seven by eight rectangular array of blocks, which we could then count, seeing its truth for ourselves.  Once having memorized our multiplication tables, which is nothing more than 81, or maybe 100, “theorems” of this sort, we no longer have to imagine a 7 by 8 rectangle every time we wish to compute 7X8.  Having done it once is enough, for the rest of our lives.  We don’t scorn the “mindlessness” of the tables of multiplication; we cherish it.  So should it be with “invert and multiply”:


     Theorem:  If a, b, c, and d are any four numbers, with b and c not equal to zero, then


              (a/b)/(c/d) = ad/bc.


     The proof of this theorem is fairly long if one begins from first principles, but the elements of the proof should be learned by all students, not only because they assure us the formula is true, but because the proof centers on the idea of “equivalent fractions”, something of value, indeed something essential, all by itself.  In the sixth grade the development of all this should take weeks, so I won’t give it here.  Besides, Ruth Parker and the National Council of Teachers of Mathematics never need it.  Nor do they need “cross multiplication” and  “long division”, two other procedures that have long been taught in the schools and are today reviled. But the scorn of current educational theorists for such “mindless” procedures and algorithms is misplaced; it should be reserved, rather, for their own principled failure to teach children where these enormously helpful procedures come from.  Mindlessness, habit, make easy our daily life.  Once we learn to tie our shoes we do it without thinking; that is as it should be if we are not to waste our days.  To deprive children of the equivalent knowledge when number is concerned is unconscionable.


     I should mention here that the student of Ms. Parker’s who drew the three circles to represent the slices she took home from the deli would, in the slightly more complicated problem I offered as a variant, have had to draw 17 circles.  Then, if these 17 circles are to represent .55 pounds, as my modified chicken problem demanded, one must ask how many more circles are needed to make a visible representation of one whole pound.  In the Parker problem six adjoined to the original three made a visible pound, and one that could visibly be divided into four parts, one-fourth of a pound being desired.  In the amended problem, even if we knew how many circles to adjoin to the 17 (it would be about 14 more, but only approximately, to picture a pound), how would we picture the 2/5 of the resulting tableau so as to get the number of slices allowed by the Ms. Parker’s diet?  Solving problems of this type by means of such a diagram is plainly impractical, and yet .55 and 2/5 are by no means the only “complicated” fractions that turn up in daily life, where baseball players, if they have any skill, may enjoy batting averages of 291/1000.


     For the chicken problem the use of these theorems, whether for Parker’s simple problem or my variant of it involving less simple fractions, was found only in the arithmetic and algebraic portions of the use of The Rule of Three.  School children, even today, sometimes learn these things very well and still remain unable to solve the chicken problem, even for such simple numbers as presented in the Parker example.  They can solve equations, divide fractions, multiply multidigit numbers; they may even know enough to understand that there are ratios involved (slices per pound) but still not know how to go about setting the information about diets and delis in the arrangement that will point towards an answer.


     Teachers of mathematics in the schools have long fretted about this phenomenon.  “Math” is one thing, “solving problems” seems to be another. How do you go about setting up that sort of thing?


     Ms. Parker’s children and their marvelous “student-invented” solutions are not the way.  They represent a difficult intellectual feat – for those who were successful – with no staying power.  If every day presents another such problem they can easily waste a month grappling with twenty of them and yet not end up any better equipped to solve the next one than they were at the beginning.  The progress of mathematics, at the “children’s level” too, consists in systematizing things in such a way as to lay bare the structures involved, especially when the structures are useful for more than one sort of problem. 


So, let me now organize the structure of the chicken problem in a way that might or might not be familiar to you, to explain a simpler, yet equally systematic, way of using the Rule of Three, and a way that later illuminates problems where proportionality is no longer the case.


In proportional relationships we have a mental image,  “Weight is proportional to number of slices” in the case of the problem at hand.  Everyone who goes to the grocery knows this and feels it.  The more slices you permit the meat slicer at Zabar’s to pile up, the greater the weight.  We can draw a graph, easily, to show the relationship, no matter how tangled the actual numbers are, since a proportionality is always represented graphically as a straight line through the origin.  Take the case I gave above, which was so complicated, where 17 slices came to .55 pounds.  Here is the graph, instantly:


     Clearly, the number of pounds of chicken rises with the number of slices bought, and the graph shows that for 17 slices we have .55 pounds of chicken.  Children learn in “geometry” that this graph has the equation y=mx, with m the “slope” (Slope is the ratio of “rise” to “run”, i.e. .55/17 in this case).  The slope of the pictured line is therefore, as a fraction, 55/1700, simplifying to 11/360, and the equation is therefore:


Pounds = (11/360)(Number of slices).


The only question (see the question mark?) is, how many slices do we need for .2 pounds?  Put .2, or 2/10 on the left side and the solution is N, where 2/10 = (11/360)N, or

N = (2/10)/(11/360) = 720/110 = 72/11 =   [6 and 8/11] slices.


     The exhibition of the relationship between the “number of slices” and the “weight of the result” as a graph with an equation is a relatively modern development in practical algebra, and even the school textbooks of 1930 did not ordinarily have this.  Drawing the graph from knowledge merely that 17 slices weighed .55 pounds was immediate: Plot the point (17.55/100) and you have the whole story.  It was not really until the “New Math” of the 1960s that the idea of “function” (weight here is given “as a function of” the number of slices) became prominent, though a carefully drawn report of the Mathematical Association of America[2], published in 1923, had already recommended that the idea of “functional relationship” be the dominant concept of school algebra for the foreseeable future.


     In the present case, the linear function models all “proportional reasoning” problems, and the data required to get the equation for this function are just the two numbers that, with the origin, define the relationship.  One might call it “the Rule of Two”, so far as getting the function is concerned.  The third term of the proportion comes with the question posed by the particular problem:  What is the other coordinate, x, of a point (x, .20) on that graph? This manner of exhibiting the data and the solution is not only easy, but downright transparent, compared to all the verbiage that usually goes with a “proportional reasoning” problem in 6th grade arithmetic. 


To use this device, however, one must have a bit more mathematics than the 6th grade ordinarily affords.  One needs to know how to plot points on a coordinate system, how to obtain, given two points on it, the equation of a line (and what ‘the equation of’ means in connection with a locus in the plane), and one needs to know enough algebra to perform the calculation that the illustration above required, including “invert and multiply”.  As an intermediate step, of course, children should learn to plot such graphs from known, or given points, and then find other points on it by actual measurement – and to think about the relationships from a bird’s eye perspective well before getting into formulas.


     I see no reason to get into “proportional reasoning” problems before all this becomes possible. To do so, as long experience has shown, is to ask most children to solve such problems blindfold – a tour de force when it can be done, but of no lasting value.  It persists into the 21st Century only because it was a staple of 19th Century arithmetic, a practical necessity in a time when functions and equations were considered arcane.  If long and difficult exercises in The Rule of Three were not demanded at an early stage in the schools there would have been a large number of people who would miss it, since in those days not everyone finished even the 8th grade, especially boys, whose strong hands and backs were needed in the shop or farm, along with what reading and arithmetic they got in school.


              I have, therefore, a suggestion for Ruth Parker.  She should not continue to pretend to have taught 6th grade children how to “create” the mathematics needed to solve the chicken problem, and thereby advertise the TERC Investigations series for the early grades in the schools,  that time-wasting and often frustrating (for children, parents, and teachers, too) series of adventures in “discovering mathematics”.  Discovering a Ruth Parker solution to a chicken (or turkey) problem is not the road to mathematics.  But Ruth Parker is undeniably a skilled lecturer and entertainer who certainly would be of value if she changed her talk to show how children can (and should) learn “long division” and the arithmetic of fractions and decimals.  These are topics that still vex the 6th grade curriculum.  To show how the “chicken problem” and its relatives are solved once this arithmetic is firmly in place, and then with the aid of the concept of function, could be the subject of a second Ruth Parker road show, intended for the teachers of the 8th or 9th grade level, which is where such problems belong.


Ralph A. Raimi

Revised 14 February 2006

[1] still to be found on the web at <>,

[2] MAA, National Committee on Math Requirements, Report: The Reor­ganization of Mathematics in Secondary Education. (Pub­lished by "MAA, Inc." 1923)