kitchen table math, the sequel: percent problem from Math Notations

Tuesday, August 7, 2007

percent problem from Math Notations

A percent problem at Math Notations:

There are 20% more girls than boys in the senior class. What % of the seniors are girls?

The Comment thread is fantastic, as is the thread on the follow-up post.

And here is Denise, of Let's Play Math, on the subject of percent:

Percents are one of the math monsters, the toughest topics of elementary and junior high school arithmetic.

Now she tells me.

As it happens, I agree. Percent problems are he**. That is my final word on the subject. H-e-double hockey sticks.

I am spending my entire summer trying to teach percent to you-know-who, when I had planned to spend my entire summer practicing up on factoring polynomials and such.

Turns out factoring polynomials is child's play compared to decimals, fractions, and percent.

Back to Denise:

The most important step in solving any percent problem is to figure out what quantity is being treated as the basis, the whole thing that is 100%. The whole is whatever quantity to which the other things in the problem are being compared.

Denise's post, which she calls the search for 100%, puts me in mind of Ron Aharoni's book, Arithmetic for Parents (scroll down). Carolyn and I both loved Aharoni's article for American Educator, What I Learned in Elementary School.

Here's his passage on the fifth operation of arithmetic:

In addition to the four classical operations, there is a fifth one that is even more fundamental and important. That is, forming a unit, taking a part of the world and declaring it to be the “whole.”

I love that!

So, on to our problem. The folks at Math Notations discuss the ambiguities of the English language in percent problems. I found the problem's wording ambiguous myself, and used a bar model to clear it up:

large image here

In the bar model the number of girls is definitely 20% greater than the boys. Plain as day. However confused you may have been by the words, there aren't a lot of different ways to draw the girl bar, assuming you've drawn the boy bar correctly (which you won't have done if you're a middle school student, but never mind).

After you've drawn a 10-square bar model to represent the boys and a 12-square bar model to represent the girls, you see that to get the total class size you're going to have to add 100% to 120%.* There is just no way around it.

Without a bar model (and even with a bar model) this is brutally confusing, because in everyday language "whole" and "all" mean "100%."

It's difficult to break set, as my cognitive psychology professors used to say.

Making boys the "100%" requires big-time set-breaking, because when you're given a problem involving boys and girls in a class at school, you think of the boys as the boy-part of the class and the girls as the girl-part, with the class being everyone, i.e. the whole, i.e. 100%.

The idea of adding 100% to 120% to get a class of 220% is just frankly nuts in terms of everyday language and experience. A class can't be 220%. We know this.


I wonder if this problem would be more solveable if you wrote it in nonsense words?

There are 20% more meps than jeps in the pile. What % of the pile are meps?

None of those words, including "pile," screams out at you: I am always and forever 100%; trifle with me at your peril.


I walked Christopher through this problem.

Just because I could.


update from Susan J:

To me the most natural way of doing the original problem is two equations in two unknowns:
g = 1.2 b
g+b = 1
and then convert the answer to per cent or

100 g = 120 b
g+b = 100

The one encouraging aspect of the two posts at Math Notations was that apparently kids who know algebra pretty well can solve this problem. I assume they use Susan's two equations, but I don't know.

Actually, that was very encouraging.

On a related subject, I'm moving into the Saxon/Dolciani "percent charts," in an effort to bring home the idea that if you have 20% more girls than boys you will have 1.2b -- only I'm starting with a whole of 100% (I wonder if that's a bad idea...)

The charts look like this:

original price 100% $50
discount 10% $5
new price 90% $45

A word problem would be something like this:

Catherine bought a new blouse for $45. It had been reduced by 10%. What was the original price?

original price 100% p
discount 10% $45
new price

The charts help you see that if you're taking 10% off the price, then you're paying 90% of the price (something I didn't realize until two years ago, when I did a percent problem in Saxon).

I'm having a terrible time getting C. to see that we're dealing with proportions here, and that percent problems can always be solved as proportions -- and that the chart is a cool way to show you all the various proportions you can employ to deal with simple problems like these.

Probably time to back up to something simpler.

Tomorrow I'm going to have him do a page of equivalent fractions using percent fractions and non-percent fractions.


75% off $60 means:

25/100 = x/60

I'll try to combine these with the charts and see how that goes. Must re-read Dolciani & Saxon, too.


Fantastic posts and threads:

The search for 100%
wrong answer: girls are 60% of the class
there are 20% more girls than boys
percent word problems revisited

and, at ktm-1 (you may have to hit refresh a couple of times):
Aharoni article, part 1
Aharonic article, part 2
Ron Aharoni on the 5th operation of arithmetic
Ron Aharoni on teaching fractions

order Arithmetic for Parents

Ron Aharoni's home page

* or 10 to 12, as the case may be


le radical galoisien said...

It is an ironic thing that you will probably have my old PSLE textbooks (or at least the content they contained) and I don't. :-P (Naturally I either left them all behind when I moved or I gave them to my younger cousins.)

The PSLE had a tendency to make nasty percentage problems. Though I think you may still just have the P5 books and not the P6 (but the P5 problems could be nasty too, as I found them to be quite challenging when I previously moved back to Singapore). I can't remember exactly what the structure of the problems were, but I want to look at them again, for nostalgia.

When they gave you 2 hours and 15 minutes, you *really* had to use all the time you were given. I hardly got to finish my last problem, with one minute left.

Could you find what you think are some of the hardest problems from the Singapore Maths problems. They were portion-whole problems (or vice versa), only they did them on a whole new scale. I can't remember what they did.

le radical galoisien said...

Okay, I remember one kind (that can be split into 2 kinds -- portion-whole, or whole-portion, as you said). This is the "gift resulting in changed ratios" type problem.

A number of beads were divided among 3 children, Eric, Mary and Jane in the ratio of 5:3:2 respectively. (ah, math exams have the tendency to remind you of the cruelty of property division). When Eric gave 24 beads to Mary, and Mary gave 15 beads to Jane, and Jane gave Eric 7 beads, the new ratio of beads among Eric, Mary and Jane became 3:3:4 respectively. How many beads were there in all?

(Note that I made these numbers up and I didn't check to ensure the numbers added up -- you'll probably get the "whole" being a decimal or something.)

The other type gives you the whole, and then when the ratio changes, you are supposed to find the amount of beads given (usually as one transaction only). This is a bit easier, since I suppose asking children to work out all the transactions could result in multiple correct answers.

Of course, there is the third variation which is the simplest: they give you the whole, the transactions given and you are supposed to work out the new (or the old) ratio.

Or something like that. I haven't done such problems in years. I want to do some for some good old-fashioned nostalgia. (Ah, I didn't appreciate them at the time as I wanted to grow up so badly, but now I look back at my primary school days with fondness.)

le radical galoisien said...

Of course, there are so many ways you could twist this problem too to give students a bit more work (and this is only one genre of ratio type problems), such as giving the new ratio different numbers of units from the old one (where the old ratio was 4:6:7, the new ratio is 3:5:8).

Maybe I'm making the problems harder than they should be. But that is the nature of memory, I suppose! I think when I did some of the problems back when I was a 12 year old, I thought, "Oh the possibilities!"

Catherine Johnson said...

The first time I looked at those changing-ratio problems in Primary Mathematics 6 I felt faint.

Catherine Johnson said...

There may not be a middle school student in America who can do them.

Me said...

To me the most natural way of doing the original problem is two equations in two unknowns:
g = 1.2 b
g+b = 1
and then convert the answer to per cent or

100 g = 120 b
g+b = 100

SteveH said...

"There are 20% more girls than boys in the senior class. What % of the seniors are girls?"

Good old algebra, but the bar model works better in this case. (If you know how to use bar models.)

But first, you have to answer the question I raised in another thread: percent of WHAT number. One person mentioned that virtually no middle schooler would get this correct. I would say that all middle schoolers should get this correct if they were taught properly. All kids should be able to set up Catherine's bar model of 120 girls for every 100 boys. I used 5 boxes for boys and 6 boxes for girls.

What might throw people off is that you don't need to know the total number of boys and girls.

The algebra approach is to start writing equations.

G = 1.2*B

One equation in two unknowns, so I need a second equation, something like:

G + B = ????

Stuck. Then, it might sink in that you can use anything for ????. This problem doesn't seem to happen with a bar model. When I did a bar model, It never occurred to me that I needed to know the total number of kids. All I was looking for was a fraction or a percent.

Back to the algebra. If I can pick any number for G+B, then let's try using the variable 'T' for the total number of kids. It's hard to believe that this will work since I have two equations and three unkowns.

G = 1.2B
G+B = T

But, what am I trying to calculate? I don't care about 'T'. I want the percent of girls in the senior class.

Percent of girls = G/(G+B)*100

I don't need 'T' or the second equation. I just need to substitute the G = 1.2B into the "percent of girls" expression above.

1.2B/(1.2B + B) * 100

= 1.2B/2.2B * 100

The B's cancel out. (Oops, B/B = 1.)

So, the percent of girls in the senior class is about 54 percent. I would have to give the nod in this case to the bar model.

TerriW said...

I apologize for horning in on this comment thread, but speaking of Saxon ....

... I stumbled across this the other day and thought of you.

I had another Think of Catherine Moment when I was unpacking books from the basement and came across a Lucy Calkins book!


Lucy Calkins! Lucy Calkins!

It was Raising Lifelong Learners, purchased at a used bookstore years ago [pre-ktm-reading] because, hey, I want to raise a lifelong learner!

Well, I didn't get any further than the anecdote about a preschool she toured [and rejected] because of the reason why the children there worked with playdough or somesuch. She didn't approve of said reason.

I can't do the anecdote justice here, but it made my teeth scream. If you're interested in the book, drop me a line with your address, or it's going onto the Going to Half Price Pile.

Back to your regularly scheduled comment thread.

Anonymous said...

To me the most natural way of doing the original problem is two equations in two unknowns:

That is the beauty of algebra. All these problems become algebra problems with a standard method of solution. No need to figure out if it is a percent problem, ratio problem, etc. It is an algebra problem whose answer must be in percent. This is the great progression of math.

hard arithmetic problems -> easy algebra problems
Hard algebra problems -> easy calculus problems

Catherine Johnson said...

oh my gosh!!

hi, Terri!

Using John Saxon's Math Books ---- wow!

Catherine Johnson said...


What a great observation!

That's exactly the way I feel about it, but I didn't know if I was hallucinating.

I've been thinking for quite awhile that algebra (beginning algebra) seemed "easier."

Catherine Johnson said...


I can't believe you own a Lucy Calkins!

Catherine Johnson said...


Catherine Johnson said...

If you're still around, you MUST take a look at the "teaching writing & grammar" posts (you can click on the link on the sidebar).

We've found some fantastic stuff.

I'm pretty sure that WRITING WITH A POINT by William Kerrigan is going to be the ultimate buried treasure.

Catherine Johnson said...


I have to have it!

I'll email you my address ----