Sunday, February 24, 2008

Beyond Mixed Nuts --A More Challenging Ratio Problem

Readers are strongly encouraged to read the extensive discussion of methods in the comments.


In Virtual HS, the ratio of the number of juniors to seniors is 7:5.

The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?


Comments:
(1) Isn't it a shame that the Jn:Sn ratio isn't 5:7!
(2) Algebraic methods seem to be the most reasonable here, but would some students attempt straight numerical methods using common multiples? Can you find such a way?
(3) Does anyone ever use or teach tree models for ratio problems? We know they are useful for probability problems, but isn't there a similarity here?
(4) Would a Singapore bar model approach work here? I'm counting on those who are proficient with this method! This is not my area of expertise...
(5) Is this question too difficult for the SATs? More appropriate for a math contest question?

10 comments:

jd2718 said...

I believe there is nothing easier than finding the denominators, then finding the common denominator, and choosing the smallest whole number that we can run with.

Note, if we choose a "wrong" number, we will notice a fractional person, and the denominator of that 'mistake' will lead us to adjust n up appropriately.

Fair for SAT, maybe middle school math contest?

I don't know what Singapore bars are. I wouldn't consider a tree diagram.

Jonathan

Hypatia said...

I used a common multiple to start and then three simple equations to arrive at the ratio required. I'm not certain whether to call this an algebraic method or not. But certainly given the level of math I used, it would qualify for an SAT question or a middle school math contest question.

That said, given my answer, I would certainly welcome those multiple choices on the SAT to confirm my answer!

Like Jonathan, I wouldn't use a tree diagram, and I'm not familiar enough with using the Singapore bars to even consider using bars as a method.

Dave Marain said...

Jonathan, Hypatia--
I personally don't consider this to be a difficult problem, but, remember, we have more experience with ratio problems and methods of solution than most students!

Having watched many secondary students struggle with questions like these, I would venture to say that this could be a high-end SAT problem or a math contest problem. A strong middle-schooler might find a way to solve it as well, but this discussion is theoretical. Jonathan, try it out with some of your groups and let us know what percent of the group solved it in a reasonable amount of time. I'd love mathmom and others working with middle schoolers to try it as well.

Ok, as far as methods go, I have one favorite method here that i believe is extremely efficient when working with part:whole ratios with different groups, namely, a dimensional analysis set-up (or 'factor-label' if you like that term):
Key: J=Juniors; JM=Junior Males
S=Seniors; SF=Senior Females

Then JM/SF = (JM/J)⋅(J/S)⋅ (S/SF) =
= (3/5)(7/5)(7/3) = 49/25 Voila!

mathmom said...

My middle schoolers couldn't do this; at least not without a lot of hand holding and simpler problems leading up to it.

They could probably (hopefully!) do it if the original ration was 5:7 instead of 7:5

As to how I would teach it, I would have them use fractions:

7/12 of the students are juniors.
3/5 of the juniors are males
so, 3/5 of 7/12 of the students are junior males.
"what does of mean?" "times!"
so 21/60 = 7/20 of the students are junior males

5/12 of the students are seniors
3/7 of the seniors are females
so 15/84 = 5/28 of the students are senior females

Now is when I think I'd start losing them. We want to get a common denominator so that the JM fraction is directly comparable to the SF fraction:

49/140 JMs and 25/140 SFs which gives a ratio of 49:25

I'm not sure, though, how to explain why the need a common denominator in that step.

Dave, your way is more efficient, but I'm not sure I could explain/motivate it for middle schoolers. Do you have a suggestion as to how to do that?

Hypatia said...

Very nicely done, Dave. I wish I had thought of dimensional analysis.

I found a common multiple for
(7)(5)(3)(4) -- 420

Then Jr. to Sr. 7x + 5x =420
x = 12

So Jr/Sr = 245/175

JM to JF 3x +2x = 245
x = 49
So JM/JF = 147/98

SM to SF 4x + 3x = 175
x = 25
So SM/SF = 100/75

JM/SF 147/75 = 49/25

A fairly straight forward method, but yours is ever so much better.

Totally_clueless said...

For the numerical approach, I first said,
the number of juniors has to be a multiple of 5(since 3/5 are males), thus i multiplied the junior:senior ratio by 5 to get 35:25.
Then, we have that the # of seniors has to be a multiple of 7, so we multiply the ratio by 7 to get 245:175.
Now the required ratio reduces to 49*3:25*3 = 49:25.

The algebraic approach was a lot easier, IMHO:
juniors = 7x, seniors =3x
junior males = (3/5)*7x
senior females = (3/7)*5x

Take the ratio to get 49:25.

Dave's approach has the mathemagical tinge to it.

TC

Dave Marain said...

mathmom--
I really like your 'fraction of a fraction' approach for middle schoolers. I'd like to comment on your two points:

(1) There is no mathematical need for the common denominator. In the last step we want the ratio (7/20):(5/28). This is a straight division of fractions calculation which gives the correct result. The common denominator method may help students see this ratio more easily but it is not required.

(2) Forget my reference to dimensional analysis or 'factor-label'. I was making an analogy to a method used in science that, IMO, should be introduced and motivated earlier on in math classes. However, your method can be directly translated to my 'efficient' method. I will demonstrate it with a simpler example:
Suppose a container holds only 2 Red and 3 Blue marbles. If you ask for the ratio of Red to the Total, you should expect virtually everyone to give you 2:5 (at least we hope for 100% accuracy here!).

However, you can show them another way using MULTIPLICATION OF RATIOS, which is equivalent to your way of expressing a 'fraction of a fraction'.

Using your approach:
The Reds are 2/3 of the Blue and the Blues are 3/5 of the Total. Thus, Reds are (2/3) of (3/5) of Total or (2/3) ⋅ (3/5) ⋅ (Total) = (2/5)(Total).

Now we will use ratio notation and symbols to express this in a slightly different form:
(a) What is the ratio of Red to Blue? 2:3
(b) What is the ratio of Blue to Total? 3:5

Now, girls and boys, let's use the letters R, B and T to symbolize these quantities. We can now write:
R:T = 2:5, R:B = 2:3 and B:T = 3:5.

Then 2/5 = (2/3)(3/5) translates to
R:T = (R:B)⋅ (B:T) or in ordinary fraction form:
R/T = (R/B) ⋅ (B/T).
Students should not be confused by the 'cancellation' of the letter B here, since B represents an actual numerical quantity, namely 3. This algebraic multiplication of fractions then develops naturally from numerical considerations.

Of course, the Junior Males:Senior Females problem is a bit more complicated since it involves a longer chain of ratios, so I would start with the simpler problem first!

Also, the multiplication of ratios is precisely what we do if we were to translate these questions into probability form. In that case, we'd be multiplying probabilities!

Maria Miller said...

I think I'll make a blogpost about the bar model approach so I can use pictures; it isn't too easy to make bars in this tiny comment box where only text is allowed.

Dave Marain said...

Thank you, Maria. I need your expertise here. Let me know when you've posted this and I will post a link to your site.

Maria Miller said...

Here's the link:


A bar diagram to solve a ratio problem
Singapore style.