Tag Archives: ratios

Party Ratios

I find LOTS of great middle school problems from @Five_Triangles on Twitter.  Their post two days ago was no exception.

The problem requires a little stamina, but can be approached many ways–two excellent criteria for worthy student explorations.  That it has some solid extensions makes it even better.  Following are a few different solution approaches some colleagues and I created.


The most challenging part of this problem is data organization.  My first thoughts were for a 2-circle Venn Diagram–one for gender and one for age.  And these types of Venn Diagrams are often more easily understood, in my experience, in 2×2 Table form with extra spaces for totals.  Here’s what I set up initially.


The ratio of Women:Girls was 11:4, so the 24 girls meant each “unit” in this ratio accounted for 24/4=6 people.  That gave 11*6=66 women and 66+24=90 females.


At this point, my experience working with algebraic problems tempted me to overthink the situation.  I was tempted to let B represent the unknown number of boys and set up some equations to solve.  Knowing that most 6th graders would not think about variables, I held back that instinct in an attempt to discover what a less-experienced mind might try. I present my initial algebra solution below.

The 5:3 Male:Female ratio told me that each “gender unit” represented 90/3=30 people.  That meant there were 5*30=150 men and 240 total people at the party.


Then, the 4:1 Adult:Children ratio showed how to age-divide every group of 5 partygoers.  With 240/5=48 such groups, there were 48 children and 4*48=192 adults.  Subtracting the already known 66 women gave the requested answer:  192-66=126 men.


While this Venn Diagram/Table approach made sense to me, I was concerned that it was a moderately sophisticated and not quite intuitive problem-solving technique for younger middle school students.


A middle school teaching colleague, Becky, offered a different solution I could see students creating.

Completely independently, she solved the problem in exactly the same order I did using ratio tables to manage the scaling at each step instead of my “unit ratios”.  I liked her visual representation of the 4:1 Adults:Children ratio to find the number of adults, which gave the requested number of men.  I suspect many more students would implicitly or explicitly use some chunking strategies like the visual representation to work the ratios.




Math problems involving ratios can usually be opened up to allow multiple, or even an infinite number of solutions.  This leads to some interesting problem extensions if you eliminate the “24 girls” restriction.  Here are a few examples and sample solutions.

What is the least number of partygoers?

For this problem, notice from the table above that all of the values have a common factor of 6.  Dividing the total partygoers by this reveals that 240/6=40 is the least number.  Any multiple of this number is also a legitimate solution.

Interestingly, the 11:4 Women:Girls ratio becomes explicitly obvious when you scale the table down to its least common value.

ratio6My former student and now colleague, Teddy, arrived at this value another way.  Paraphrasing, he noted that the 5:3 Male:Female ratio meant any valid total had to be a multiple of 5+3=8.  Likewise, the 4:1 Adult:Child ratio requires totals to be multiples of 4+1=5.  And the LCM of 8 & 5 is 40, the same value found in the preceding paragraph.

What do all total partygoer numbers have in common?

As explained above, any multiple of 40 is a legitimate number of partygoers.

If the venue could support no more than 500 attendees, what is the maximum number of women attending?

12*40=480 is the greatest multiple of 40 below 500.  Because 480 is double the initial problem’s total, 66*2=132 is the maximum number of women.

Note that this can be rephrased to accommodate any other gender/age/total target.

Under the given conditions, will the number of boys and girls at the party ever be identical?

As with all ratio problems, larger values are always multiples of the least common solution.  That means the number of boys and girls will always be identical or always be different.  From above, you can deduce that the numbers of boys and girls at the party under the given conditions will both be multiples of 4.

What variations can you and/or your students create?


Now to the solution variation I was initially inclined to produce.  After initially determining 66 women from the given 24 girls, let B be the unknown number of boys.  That gives B+24 children.  It was given that adults are 4 times as numerous as children making the number of adults 4(B+24)=4B+96.  Subtracting the known 66 women leaves 4B+30 men.  Compiling all of this gives


The 5:3 Male:Female ratio means \displaystyle \frac{5}{3} = \frac{5B+30}{90} \longrightarrow B=24, the same result as earlier.



Winding through all of that algebra ultimately isn’t that computationally difficult, but it certainly is more than typical 6th graders could handle.

But the problem could be generalized even further, as Teddy shared with me.  If the entire table were written in variables with W=number of women, M=men, G=girls, and B=boys, the given ratios in the problem would lead to a reasonably straightforward 4×4 system of equations.  If you understand enough to write all of those equations, I’m certain you could solve them, so I’d feel confident allowing a CAS to do that for me.  My TI-Nspire gives this.


And that certainly isn’t work you’d expect from any 6th grader.


Given that the 11:4 Women:Girls ratio was the only “internal” ratio, it was apparent in retrospect that all solutions except the 4×4 system approach had to find the female values first.  There are still several ways to resolve the problem, but I found it interesting that while there was no “direct route”, every reasonable solution started with the same steps.

Thanks to colleagues Teddy S & Becky M for sharing their solution proposals.

Graphing Ratios and Proportions

Last week, some colleagues and I were pondering the difficulties many middle school students have solving ratio and proportion problems.  Here are a few thoughts we developed to address this and what we think might be an uncommon graphical extension (for most) as a different way to solve.

For context, consider the equation \displaystyle \frac{x}{6} = \frac{3}{4}.


The default procedure most textbooks and students employ is cross-multiplication.   Using this, a student would get

\displaystyle 4x=18 \longrightarrow x = \frac{18}{4} = \frac{9}{2}

While this delivers a quick solution, we sadly noted that far too many students don’t really seem to know why the procedure works.  From my purist mathematical perspective, the cross-multiplication procedure may be an efficient algorithm, but cross-multiplication isn’t actually a mathematical function.  Cross-multiplication may be the result, but it isn’t what happens.


In every math class I teach at every grade level, my mantra is to memorize as little as possible and to use what you know as broadly as possible.  To avoid learning unnecessary, isolated procedures (like cross-multiplication), I propose “fraction-clearing”–multiplying both sides of an equation by common denominatoras a universal technique in any equation involving fractions.  As students’ mathematical and symbolic sophistication grows, fraction-clearing may occasionally yield to other techniques, but it is a solid, widely-applicable approach for developing algebraic thinking.

From the original equation, multiply both sides by common denominator, handle all of the divisions first, and clean up.  For our example, the common denominator 24 will do the trick.

\displaystyle 24 \cdot \frac{x}{6} = 24 \cdot \frac{3}{4}

4 \cdot x = 6 \cdot 3

\displaystyle x = \frac{9}{2}

Notice that the middle line is precisely the result of cross-multiplication.  Fraction-clearing is the procedure behind cross-multiplication and explains exactly why it works:  You have an equation and apply the same operation (in our case, multiplying by 24) to both sides.

As an aside, I’d help students see that multiplying by any common denominator would do the trick (for our example, 12, 24, 36, 48, … all work), but the least common denominator (12) produces the smallest products in line 2, potentially simplifying any remaining algebra.  Since many approaches work, I believe students should be free to use ANY common denominator they want.   Eventually, they’ll convince themselves that the LCD is just more efficient, but there’s absolutely no need to demand that of students from the outset.


Remember that every equation compares two expressions that have the same measure, size, value, whatever.  But fractions with differing denominators (like our given equation) are difficult to compare.  Rewrite the expressions with the same “units” (denominators) to simplify comparisons.

Fourths and sixths can both be rewritten in twelfths.  Then, since the two different expressions of twelfths are equivalent, their numerators must be equivalent, leading to our results from above.

\displaystyle \frac{2}{2} \cdot \frac{x}{6} = \frac{3}{3} \cdot \frac{3}{4}

\displaystyle \frac{2x}{12} = \frac {9}{12}


\displaystyle x = \frac{9}{2}

I find this approach more appealing as the two fractions never actually interact.  Fewer moving pieces makes this approach feel much cleaner.

UNCOMMON(?) METHOD 4:  Graphing

A fundamental mathematics concept (for me) is the Rule of 4 from the calculus reform movement of the 1990s.  That is, mathematical ideas can be represented numerically, algebraically, graphically, and verbally.  [I’d extend this to a Rule of 5 to include computer/CAS representations, but that’s another post.]  If you have difficulty understanding an idea in one representation, try translating it into a different representation and you might gain additional insights, or even a solution.  At a minimum, the act of translating the idea deepens your understanding.

One problem many students have with ratios is that teachers almost exclusively teach them as an algebraic technique–just as I have done in the first three methods above.  In my conversation this week, I finally recognized this weakness and wondered how I could solve ratios using one of the missing Rules: graphically.  Since equivalent fractions could be seen as different representations of the slope of a line through the origin, I had my answer.

Students learning ratios and proportions may not seen slope yet and may or may not have seen an xy-coordinate grid, so I’d avoid initial use of any formal terminology.  I labeled my vertical axis “Top,” and the horizontal “Bottom”.  More formal names are fine, but unnecessary.  While I suspect most students might think “top” makes more sense for a vertical axis and “bottom” for the horizontal, it really doesn’t matter which axis receives which label.

In the purely numeric fraction in our given problem, \displaystyle \frac{x}{6} = \frac{3}{4}, “3” is on top, and “4” is on the bottom.  Put a point at the place where these two values meet.  Finally draw a line connecting your point and the origin.


The other fraction has a “6” in the denominator.  Locate 6 on the “bottom axis”, trace to the line, and from there over to the “top axis” to find the top value of 4.5.


  Admittedly, the 4.5 solution would have been a rough guess without the earlier solutions, but the graphical method would have given me a spectacular estimate.  If the graph grid was scaled by 0.5s instead of by 1s and the line was drawn very carefully, this graph could have given an exact answer.  In general, solutions with integer-valued unknowns should solve exactly, but very solid approximations would always result.


Even before algebraic representations of lines are introduced, students can leverage the essence of that concept to answer proportion problems.  Serendipitously, the graphical approach also sets the stage for later discussions of the coordinate plane, slope, and linear functions.  I could also see using this approach as the cornerstone of future class conversations and discoveries leading to those generalizations.

I suspect that students who struggle with mathematical notation might find greater understanding with the graphical/visual approach.  Eventually, symbolic manipulation skills will be required, but there is no need for any teacher to expect early algebra learners to be instant masters of abstract notation.

Unexpected Proof of the Pythagorean Theorem

Following is a neat discovery of an alternative proof of the Pythagorean Theorem resulting from the multiple solutions to the Circle and Square problem.  I’m sure someone has seen this before, as there are literally 100s of unique proofs of the Pythagorean Theorem, but this one was new to me.

The intersecting chord relationships in a circle can be proven using only similar triangles.  Proofs of these are at the bottom of this post, if needed.  Using only those, you can prove the Pythagorean Theorem.


The image below–a revision of the diagram from my previous post–shows diameter DE in circle C.  Chord AB is a side of the given square from the Circle and Square problem and is bisected by symmetry into two segments, each of length a.  Let  be the radius of circle C.  Let the portion of DE from point C to chord AB have length b.  Because AB is a chord bisected by diameter DE, two right triangles are created, as shown.


AB and DE are intersecting chords, so a \cdot a = (r-b) \cdot (r+b).  Expanding the right side and moving the b^2 term to the other side gives the Pythagorean Theorem.

Short and sweet once the chord relationships are established.


In the image below, AB and CD are any two chords intersecting at point E.  Vertical angles give \angle DEA \cong \angle BEC .  Because \angle ADE and \angle CBE are inscribed angles sharing arc AC, they are also congruent.


That means \Delta ADE \sim \Delta CBE, which gives \displaystyle \frac{x}{w} = \frac{y}{z}, or x \cdot z = w \cdot y.  QED

Show that if a diameter bisects a chord, the diameter and chord are perpendicular.  Start with the vertical diameter of circle C bisecting chord AB.


It should be straightforward to show \Delta ADC \cong \Delta BDC by SSS.  That means  corresponding angles \angle ADC \cong \angle BDC; as they also from a linear pair, those angles are both right, and the proof is established.