Looking for a little help with the SAT Math section?

In this video, Magoosh’s SAT expert Chris explains the Distance-Rate Formula!

You don’t have to be scared of these question types anymore, once you’ve got the Distance-Rate Formula in your pocket!

Watch the embedded video below, or scroll down for a full video transcript. 🙂

## What Will I See in the “SAT Math Section: Distance Rate | SAT Practice Questions” Video?

In this 8-minute video, Magoosh’s SAT expert Chris will take you, step-by-step, through three SAT practice questions, explaining how you can identify whether a question calls for the Distance-Rate Formula and how to use the formula to solve even the hardest Distance-Rate problems!

If you liked this video, hit that *Like* button–or better yet, send it to a friend! Let’s all go to college 🎓🙌

## “SAT Math Section: Distance Rate | SAT Practice Questions” Full Transcript

Hi, this is Chris, the SAT expert at Magoosh.

I’ve had over 15 years’ experience helping students ace the SAT, and today, I’m going to talk about the distance rate formula.

Why do you have to know this?Well, there’s a certain question type, a word problem about moving vehicles going in different directions, that freaks a lot of people out.

But this formula can make things a lot easier.

So here’s a distance problem, we have Stephen cycling at a constant rate of 10 miles per hour, and Gertrude cycling at a constant rate of 15 miles per hour.

So we know people are moving at a certain rate for a certain amount of time, and that’s how we know we’re in the distance formula world.

So we’re gonna come back to this question in a second, and we’re gonna go straight to that distance formula.

We have distance = rate x time.

Now, we’re gonna put this to work with us, with that very problem we just saw, but I’m gonna show you an easy way to remember this.

See how D stands for distance, Rate stands for R at the beginning, and Time starts with T, so there it is?

And we’re gonna make that, and simplify that into D = RT, that’s gonna be your magic formula.

You always wanna keep that in mind, but of course, know what those things stand for.

Okay, so let’s take a look back here at Stephen and Gertrude.

And not only am I going to give you that question, but I’m gonna give you some answer choices.

You can try this on your own if you want, but let’s do it together now.

So Stephen cycles at a constant rate of 10 miles an hour for two hours, and then stops, so let’s use our formula.

W know the rate, that’s the speed, so that’s 10 miles an hour for Stephen, let’s get that right there.

And then he goes for how long, for two hours.

So we know that if he’s going at this rate ,10 times 2, we have a distance of 20, and that’s essentially how far Stephen has gone.

Now, if we look at Gertrude, we can do the same math, but she’s going faster, and the question is, how long will it take her?

So we actually have the distance she needs to go, which is 20, and her speed, which is 15.

And now, we suddenly don’t know T, that’s what the question is asking for.

But we can plug this information back into that original equation, and that’s why it’s so handy.

Essentially, we use it twice, once for Stephen, we figured out how far he had gone.

And then we plugged that 20 down into the equation here, with Gertrude, and that gave us 20.

And then that equals 15, being the rate, T is the time.

We solve for T, next step, here, T is 20 over 15.

That is equal to 4 over 3, so what is 4 over 3, it’s the same as 1 and one-third, make that 4 a little bit better there.

1 and one-third, what’s one-third of an hour, 20 minutes, and therefore, 1 hour and 20 minutes it is.

Now, you don’t even have to figure out that one-third being 20 minutes.

If you look at the answer choices, you have 45 minutes ,which is less than an hour.

And you look at your, the number we got, four-thirds, that’s greater than 1 but less than 2.

And the only answer that would work is only answer choice C, so quick way of doing things, as well.

But the focus, of course, here is the distance-rate formula, let’s do some more.

This one’s gonna be a little bit harder, scarier, it’s the two trains, the dreaded scenario.

They are setting off from different cities located 300 miles from each other.

And they’re headed directed towards each other, but don’t worry, on different tracks.

The first train moves at a constant rate of 40 miles an hour, and the second moves at a constant rate of 60 miles an hour.

So what’s the key here, if objects are heading towards each other, we always combine their rate.

Again, if they’re headed directly at each other, you combine their speeds, and so 40 plus 60, that’s gonna be our rate, that’s 100.

And so we have the fact that they are D, 300 miles from each other.

And that they are going at 100 miles an hour, cuz we have to combine the rates, and we have to figure out the time.

And we get 300 = 100T, and then therefore, we can get T = 3.

And so therefore, we can see that in 3 hours, they’ve gone how far?

Well, in three hours, again, you’re combining their speeds together.

So this train over here goes 120 miles, which is 40 times 3, this train over here goes 180 miles, so all in all, they’ve traveled 300 miles.

Now, it would seem like the answer is c, 1 o’clock.

But notice that we’re not waiting for the two trains to get to the other cities, we’re asking when they meet each other.

And so they’re going to meet each other at this halfway point in time.

So in three hours, they’ve covered the total 300 miles, but half of 3 hours, which is 1.5 hours, they’ve actually connected.

That’s the point they intersect, so it’s actually half of 3, which gives us an hour and a half.

If they started at 10, that means that 11:30, they are going to pass each other, and that’s what makes it tricky.

Now, this is definitely a harder question, but the point is that we can still use distance is equal to rate times time.

We can still use that formula when we know that we’re combining rates, things coming together.

But now we’re gonna look at a different problem, the last problem here, which is a tricky one.

Here, we’re going to go back to Stephen and Gertrude, but we’re gonna put them in cars, instead of bicycles, but here, one is chasing the other one.

And so unlike the trains, when you’re combining, because they’re heading towards each other, here they’re in catch-up mode, or at least Gertrude is.

So Stephen, let’s read the problem here, leaves Farm Town, driving at a rate of 40 miles an hour, so where is he in one hour?

Well, he’s 40 hours from Gertrude, who starts driving at that point at 50 miles an hour, so when is she actually gonna reach Stephen?

Well, assuming again that they’re both continuing to drive, after one hour, he is 40 miles an hour, we know that.

But she is, what, 50 miles an hour, which is 10 miles an hour faster, let me get that 0 there, 10 miles an hour faster than Stephen.

So you can think of it this way, for every hour, she catches up.

So in one hour, she catches up 10 miles to him, two hours, she catches up 20 miles to him.

We know that she’s 40 miles behind, so how long will it take her to catch up to him?

Well, she’s gaining 10 miles an hour, that means, and now we can kinda use our distance-rate formula here.

She’s 40 miles away, and she’s catching up at a rate of 10 miles an hour.

We can just say, oh, okay, therefore T is equal to 4 hours, that’s the total number of time it will take her to catch.

So we can still use distance and rate, but we have to think of it differently, in terms of subtracting.

We’re subtracting, not adding together, which is the tricky part, but there’s our answer.

And there you have three different setups where we have distance, rate, and time, and we have need to find an answer, relying on our nifty little formula.

So there we are, you don’t have to be scared anymore of these question types, because you’ve got the distance-rate formula in your back pocket.

If you like this video, then click on the link in the description below.

That will take you to sat.Magoosh.com, where you can join thousands of other students who are prepping for the SAT.

If you want more helpful tips and strategies, then check out the videos on the left, and I will see you next time.

## Want More SAT Study Tips?

Take a look at some of our other useful study tips to help you prepare for the SAT Math section:

Happy studying! 🙂

### More from Magoosh

##### About Molly Kiefer

Molly is one of Magoosh’s Content Creators. She designs Magoosh’s graphic assets, manages our YouTube channels and podcasts, and contributes to the Magoosh High School Blog.

Since 2014, Molly has tutored high school and college students preparing for the SAT, GRE, and LSAT. She began her tutoring journey while in undergrad, helping her fellow students master math, computer programming, Spanish, English, and Philosophy.

Molly graduated from Lewis & Clark College with a B.A. in Philosophy, and she continues to study ethics to this day. An artist at heart, Molly loves blogging, making art, taking long walks and serving as personal agent to her cat, who is more popular on Instagram than she is.

### Leave a Reply

**Magoosh blog comment policy:** To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!