Using Algebra for a Ratio Problem

(A new question of the week)

Some kinds of problems can be solved at various levels; in particular, when we get a problem about ratios, we often can’t be sure whether the student knows only arithmetic, or can use algebra, which usually makes the problem easier. This is one reason we ask for a category (such as Arithmetic or Algebra), and also for an age range. It’s even more helpful when we see the start of a solution, which is the biggest hint to the appropriate method to use. This one led to an interesting discussion.

A ratio problem

The problem came in last October, from Eric, who put himself in the 10-13 year old category, where students may or may not be doing algebra:

The problem is as follows:

Tom and Jerry paid for a friend’s present in the ratio 7:4. In doing so, Jerry spent 1/4 of his money and Tom had $99 left. If the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present, how much did the present cost?

To be honest, I don’t know where to begin. It throws so much information at you, that is broken and detached from the rest of the problem, and it’s like you need to put it all together and add some pieces. I had a go, then my mom had a go, and we both got nowhere than just writing down the information and staring at the answer key, which didn’t explain how to solve the question. If I could have some intel on how to start my thinking for this and how everything fits together that would be fantastic.

If you want to know the answer it is:

$33

That isn’t a simple problem is it? This will be a lesson on how to gather information and make sense of it.

Answer keys often are not helpful for telling how to solve a problem; but they can help us see whether we are interpreting a problem correctly, and whether the student is confused only because the given answer is wrong. (It happens!) Here, it turns out to be right.

Doctor Rick took the question:

Hi, Eric, thanks for writing to The Math Doctors!

The fact that you put your question in the Algebra category tells me that you expect to be using variables and solving equations. That’s a big help — if I had to tackle the problem solely in terms of ratios, I don’t know what I’d do! Algebra helps us to encode the information in a complex problem like this into variables and equations; once we’ve done that, I can sit back and just solve the equations, and we’ll be done!

Many problems, including ratios, can be solved in creative ways without algebra; what algebra does is to take out the need for creativity and make the main work routine. Well, mostly …

Define some variables

Here’s a suggestion to get you started. I see three main numbers that we don’t know, but that we want to talk about. So we can define three variables to represent these quantities: let’s say

T = the amount of money that Tom had to start with (in dollars)

J = the amount of money that Jerry had to start with (in dollars)

P = the cost of the present (in dollars)

Now we hope that we can write three equations using these three variables. It sometimes turns out that this isn’t necessary, at least when we’re only interested in finding the value of one of our variables (as here, where we want to solve for P). You will find, though, that you can write three equations.

Note how carefully he has defined the variables. It isn’t just “T = Tom”, as many students write; rather, T is an amount of money that Tom has at a particular time in the problem. Doing this makes it easier to keep track of the details. Orderliness is the key to handling a complex problem.

Now I’d like you to give it a try. Can you come up with an equation to represent the information in the first sentence? Can you write two more equations from the other information in the problem? Let me see what you can do, even if it isn’t much or if you aren’t confident that you did it right — because that’s a start, and then we’ll have more to talk about.

We like to get the student involved in carrying out the work, so they’ll have the experience to do it again on the next problem, and the next. If we just demonstrated solutions, we’d be doing no more than a textbook does, without the necessary interaction. (That’s also why you may want to submit a question to us rather than read our pre-solved problems.)

Write some equations

Eric responded with a multi-part reply:

Sorry for the late reply

Thanks for responding so quickly Dr. Rick! I… tried my best with the equations and came up with all sorts of junk like P=T/J, P=7/4, P=T/J-7/4, but then mom was there to save the day, with 7/4=(T-99)/(J/4), which I know isn’t what you asked, but my mom and I were in this together. We spent hours thinking about it. We were invested in this.

The reply came 50 minutes after Doctor Rick wrote, not exactly late. They did some good work in that time!

Anyway, I came up with the next equation (using your variables thank you so much ahhhhhhhhhhhh) that was T/J=5/2. Groundbreaking, I know, but it played a good role in the final result!

The final equation was P=J/4+(T-99), created out of nothingness by my mom… again…

And so we solved it! From T/J=5/2 we got T=5J/2 and plugged that into the first equation and then the last.

Thank you so much for your help with this problem!

We’ll be seeing how to do this without miraculous intervention.

P.S.: I put this under Algebra because I thought it was the difficulty range for this problem… but it seems that all’s well that ends well…

Is Algebra a difficulty range? It does describe a level of understanding, but it’s largely a way of looking at a problem.

P.P.S.: I know that we already solved this together, but I was wondering: how would you go about this problem with ratios? This problem was listed in the ratio section of my math workbook. I know that you said that you wouldn’t know where to begin with using ratios, and I am totally fine with you not knowing how to use ratios for this problem! It’s just the thought of using ratios was nagging at the back of my mind for a while… It’s fine though! We got the answer (which was Jerry spending 12 dollars and Tom spending 21 dollars)!

We’ll see that the comment was misunderstood.

How to write the equations

Doctor Rick answered each piece, starting with the “lateness”:

Hi, Eric. I don’t consider a response sent 50 minutes after I wrote you to be “late” (but then I’m from the old generation that remembers writing letters) … you seem to have worked quite hard in that time. Also, you probably live not far from my time zone since I received your last message at almost 10:00 pm. I live on the east coast of the USA. Now it’s morning and I’m ready to help

And our generation also remembers submitting a computer program on a stack of cards and waiting a day to get the results …

If you know algebra and you are expected to use it to solve problems, then “algebra” is the correct category to use. Subject matter matters more than difficulty level.

Anyway, let’s talk about the problem and your solution. You (and your mom) did good work. What I said was that I wouldn’t know what to do if I had to solve the problem solely in terms of ratios, without using algebra. However, we must certainly use ratios because two pieces of information are stated in terms of ratios.

Before we close, I will try solving the problem with only ratios and arithmetic (and a picture), as is taught at the elementary level. But for our purposes, the problem is, at heart, an algebra problem about ratios.

You sound like you were able to come up with one equation, but your mom got the other two, and you aren’t sure how she got them. Let’s go over the problem in detail to see how you can develop the equations yourself.

I defined three variables:

T = the amount of money that Tom had to start with (in dollars)

J = the amount of money that Jerry had to start with (in dollars)

P = the cost of the present (in dollars)

The first sentence, “Tom and Jerry paid for a friend’s present in the ratio 7:4,” can’t be written yet – I shouldn’t have asked you about that to start with. Before we can write an equation to represent that, we need to find expressions in terms of T, J, and P for these two quantities:

the amount of money Tom paid for the present (in dollars)

the amount of money Jerry paid for the present (in dollars)

When you try to write an equation all at once, it is easy to make mistakes. It is best to take it slowly, phrase by phrase. Here we want to write an expression for the ratio of the two amounts paid; since these are not variables, and no information given yet tells us how much was paid, we could only write an equation yet if we defined two more variables for these quantities. Since (having preread the problem before starting to work on it) we know that more information is coming, we just put this on hold for now, and wait patiently.

So we look further into the problem, and find this: “Jerry spent 1/4 of his money and Tom had $99 left.” If Jerry spent 1/4 of his money on the present, then he spent (1/4)J. If Tom had 99 dollars left, then he spent (T – 99) dollars, since subtracting (T – 99) from T leaves 99. Thus we can now “translate” the two expressions above into algebra:

the amount of money Tom paid for the present (in dollars) = T – 99

the amount of money Jerry paid for the present (in dollars) = J / 4

As soon as we write this, we should be checking that it makes sense. So we look at the expression we wrote, T – 99; that means $99 less than T, which is the amount Ted had; that makes sense. Similarly, the expression J / 4 means 1/4 of what Jerry started with, which is just what the problem says he spent. (We need to read carefully and make sure that’s what he spent, not what he had left, which a different problem might have said.)

And now we can translate the first sentence:

Tom and Jerry paid for a friend’s present in the ratio 7:4.

The ratio of what Tom paid to what Jerry paid is 7:4.

(T – 99) : (J / 4) = 7:4

T – 99 = (7 / 4)(J / 4)

The first version is written in terms of ratios, using the colon; then it has been rewritten in more algebraic form, and the division restated as a multiplication (that is, both sides were multiplied by J / 4). One reason for doing the latter is that a complex fraction (a fraction of fraction) is hard to understand, and easy to make mistakes with. If you want, you can simplify it now.

We now have one equation, which involves two variables.

The last piece of information, “the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present“, translates to:

T : J = 5 : 2

T / J = 5 / 2

T = (5/2)J

which is the equation you found. Hmm … we used all the information in the problem, but we only have two equations! That’s OK, since these two equations only use two of my three variables: T and J. So you can solve for T and J. However, the problem asked you to find P!

Solving the equations

To recap, we now have two equations in two variables: $$T – 99 = \frac{7}{4}\frac{J}{4}$$ $$T = \frac{5}{2}J$$

While we’re looking at them, let’s go ahead and solve for T and J:

Clearing fractions, the first equation becomes $$16T – 1584 = 7J$$

Substituting in this from the second equation, we have $$40J – 1584 = 7J$$

Solving, $$J = \frac{1584}{33} = 48$$

Then $$T = \frac{5}{2}J = \frac{5\cdot 48}{2} = 120$$

The ratio of these, as required, is 120:48 = 5:2.

Where does P come in?

Well, your mom came to the rescue and solved this dilemma, by writing an expression for P. That’s her last equation:

P=J/4 + (T – 99)

Do you see now how she did that? I think you do.

Recall our definition of P: “the cost of the present (in dollars)”. Here we have to think about how things work: If Tom and Jerry split the cost of the present, then its total cost is the sum of what each of them paid. This is a good example of how some equations are implied by knowledge of what is called “the problem domain”. (It can also sometimes be called “common sense”. You didn’t think math had anything to do with that, did you?)

In other words, the Jerry spent \(\displaystyle\frac{J}{4} = \frac{48}{4} = \$12\), and Tom spent \(T – 99 = 120 – 99 = \$21\). These are the answers Eric gave; and if we check, the ratio of these is 21:12 = 7:4. And the total cost of the present is \(12 + 21 = \$33\).

Eric replied:

Ah ok! I think I get it now! Thank you for all your help!

Doing it without algebra

Pondering the problem now, I’ve seen one way it can be solved just by thinking about ratios arithmetically. I started by drawing a picture, which helped largely by making me aware of what we know. Look at the problem again:

Tom and Jerry paid for a friend’s present in the ratio 7:4.

In doing so, Jerry spent 1/4 of his money and Tom had $99 left.

If the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present, how much did the present cost?

Thinking in terms of “parts”, and starting with the last fact we are told, Tom had 5 “parts” to Jerry’s 2 “parts”. If we can find how big a “part” is, we’ve solved the problem.

Jerry spent 1/4 of his money. That’s 1/4 of 2 parts, which means half a part.

The ratio of what they spent was 7:4. That’s 7 “little parts” from Tom and 4 “little parts” from Jerry. So Jerry’s 1/2 “big part” is 4 “little parts”; a “little part” is 1/8 of a “big part.

Now, Tom has $99 left after taking 7 “little parts” from the 5 “big parts” he had. But 5 “big parts” means 5 times 8 “little parts”. Subtracting 7 “little parts” from those 40, the $99 he has left is 33 “little parts”. So one “little part” is $3, and one “big part” is 8 times that, $24. So Tom started with 5 times that, $120, and Jerry started with 2 times $24, or $48. From there, we just find how much each spent and we have the answer.

What I just did reminds me of the way algebra was done before symbols, just talking through it with words. It also reminds me of the way fractions were handled before fractions were invented, by naming units so the fractions would just be whole numbers of smaller units. (If I wanted, I could have found better names!)

Which do you prefer, algebra, or thinking?

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