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So here’s the thing with ratios and proportions on the SAT: they’re really easy. No, seriously, where are you going? Come back! They’re easy, I *swear*. All you have to do is keep very close track of your units, and you’ll be good to go. That means when you set up a proportion, actually *write the units* next to each number. Make sure you’ve got the same units corresponding to each other before you solve, and you’re home free. Pass Go, collect your $200, and spend it all on Lik-M-Aid Fun Dip. So uh…let’s try one?

- A certain farm has only cows and chickens as livestock. The ratio of cows to chickens is 2 to 7. If there are 63 livestock animals on the farm, how many cows are there?

(A) 13

(B) 14

(C) 16

(D) 18

(E) 49

The SAT writers would love for you to set up a simple proportion here and solve:

Hooray! *x *= 18! That’s answer choice (D)! **NOT SO FAST, SPANKY.** You just solved for a terrifying hybrid beast, the ** COWNIMALKEN**. Let’s look at that fraction more carefully, with the units included:

So when you casually multiplied by 63 and solved, you solved for a unit that won’t do you any good: the [(cow)(animal)]/(chicken), or * COWNIMALKEN*. That’s terrifying. Nature never intended it to be so. Not only are you unwisely playing God, but you’re getting an easy question wrong. Before we can solve this bad boy, we need to make sure our units line up on both sides of the equal sign. So let’s change the denominator on the left to match the one on the right. Get rid of “7 chickens” and replace it with “9 animals.” Get it? Because cows count as animals, if there are 2 cows for every 7 chickens, that means there are 2 cows for every 9 animals.

*Now*, we can solve: *x* = 14 cows. That’s choice (B). See how the units cancel out nicely when you’ve properly set up a ratio question? That should make the hairs on your neck stand on end.

##### Look out for this tricky crap, too:

But but but! There’s one more thing you need to watch out for. Sometimes they’ll give you units that aren’t quite as easily converted. Like so:

- The ratio of students to teachers at a certain school is 28 to 3. The ratio of teachers to cafeteria workers is 9 to 2. What is the ratio of cafeteria workers to students?

(A) 1 to 42

(B) 2 to 28

(C) 3 to 37

(D) 9 to 56

(E) 3 to 14

Here, we have a few options. It’s not too hard to find a number of teachers that will work with both ratios, so I’ll leave it to you to figure out how to solve it that way if you prefer. Instead let me point out that there’s a pretty elegant solution here that comes from simply multiplying the two ratios together, essentially solving for the expression we’re looking for. Peep the skillz:

What happens to the teachers? * They cancel*! So multiply, and simplify:

Since the question asked for the ratio of cafeteria workers to students, just flip it and you’re done! 1 cafeteria worker to 42 students. That’s choice (A). Ah-mazing.

##### Mind your units, son.

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For number 19, I just used ratios to figure out that it’s D. It isn’t like your example above where you cancel out units, right?

You can actually do it either way. Plug in the number of strike outs to solve it pretty directly, or set up a couple ratios and eliminate unwanted units: [2 Andy HR/7 Andy StrikeOut]*[9 Sean HR/5 Andy HR] = [18 Sean HR/35 Andy StrikeOut]

Since Andy’s players struck out 105 times, you can then solve:

18/35 = x/105

54 = x

So yeah…either way works here. Which do you like better?

Andy: 7 SO/ 2HR

Sean: 9 HR/ 5 Andy Hits

SO: 105?

105/7 = 15 (the ratio)

So, Andy: 105 SO/ 30 HR

and to get a common denominator for Sean (30) i would have to multiply 5 by 6, and i did that for the 9 HR to get 9×6 = 54.

haha, it seems roundabout but i get it better that way…

Could you please do #12 on the math diagnostic drill #1?

Sure. There are 2 ways to think about it. First, you could make both ratios contain the same number of nuts. Since you’ve got 4 nuts in the first ratio and 5 nuts in the second, let’s change both ratios to have 20 nuts. So now you have 15 croutons to 20 nuts, and 12 raisins to 20 nuts. That means you have 15 croutons to 12 raisins, which simplifies to 5 to 4.

You could also flip one of the fractions and just multiply them to eliminate the nuts unit. I’ve attached an image (from my book) of what that would look like. It’s not necessarily more efficient than the above method, but it’s kinda snappy and I like doing these questions that way.

So sorry I didn’t see this comment when it came in! If you’re still looking for a solution, I’ve posted one in response to “Pilot”‘s comment.

For the Math diagnostic test 1, I don’t get how to do #12

Most people, I’ve found, find it most intuitive to just get everything in terms of the same number of nuts. Croutons to nuts is 3:4, so make that 15:20. Raisins to nuts is 3:5, so make that 12:20. Now you’ve got your ratio of croutons to raisins…15:12, which simplifies to 5:4.

You could also do it the way I do it in my book, which I think is snappier but which I won’t force on you if you don’t like it. 🙂

I liked your way better because that’s how I tried doing it before, but wasn’t sure with the canceling out. Thanks 😀

My pleasure.

How do you solve #19?

Does my reply to Echoyjeff222 above help at all?

Yes! I didn’t look at the bottom comments. Thank you for your help, I really appreciate it!

🙂 No problem.

Can you explain no. 10 ?

The ratio of boys to girls is 7 to 5, but we want to know about TOTAL STUDENTS. So we need to convert the boy/girls ratio to boys/total students. So we go to boys/(boys+girls), or 7/(7+5). The ratio of boys to total students is 7/12.

From there, we can set up a proportion and solve: 7/12 = 14/x, x = 24.

PLEASE EXPLAIN NO.14

For this one, you just need to set up the proportion. But there are 2 tricky parts: 1) you don’t need to use the information about the eye of newt, and 2) 1½ is a mixed fraction; you should convert it to 1.5. Generally, mixed fractions appear in recipe questions—that’s about the only place you see them in real life, too.

So your proportion is 1.5/2 = x/7. Solve that and you get x = 5.25 (or, in mixed fraction form, 5¼).

Hi Mike. I’m having a great time with your book, but really curious about your explanation for #15 above involving not cross-multiplication of equivalent ratios (p. 52 in 1st ed), but straight multiplication with cancellation of the units (e.g. teachers). I’ve never seen this used elsewhere and you don’t employ it in your explanations for the following practice examples except #19 which is way difficult. Can you explain further or give another example where this is more obvious and a more helpful/direct solution than cross-multiplication? Many thanks! Karen

A student of mine recently pointed out to me that it’s just like dimensional analysis in chemistry class. I don’t remember it from HS chem, but maybe that’s where I learned it because I do remember doing it a lot in physics when I couldn’t remember a formula but I knew that an answer needed to have certain units, and I probably didn’t come up with it all by myself. 🙂

I think it’s a useful way to go when you have units that aren’t going to matter for your final answer, like the fantasy baseball question (below). It’s certainly not the only way to go, though. Multiple equations with the more conventional cross-multiplication method will land in the same place.