The 4^{th} of July weekend is upon us, and many of us will be taking road trips, or even taking a plane somewhere. To commemorate this collective movement, let’s learn the most fundamental formula when dealing with movement over time. First, let’s meet Bob…

*Bob drives at an average rate of 50 mph from Berkeley to Los Angeles, a distance of 350 miles. How long does it take him to complete the trip?*

* *

*(A) **4 hrs*

*(B) **5 hrs*

*(C) **7 hrs*

*(D) **10 hrs*

*(E) **12 hrs*

When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time.

With the problem above, the distance between Berkeley and Los Angeles is 350 miles. So D = 350. Bob is traveling at 50 mph, so that is his rate. The question is how long will it take him to complete this trip. Therefore, we have to solve for T. Let’s set up the equation, plugging in the values for D and R:

350 = 50T.

Solving for T, we get 7. Because we are dealing with miles per hours, the 7 corresponds to hours. So T = 7. Answer (C).

Now let’s try another problem:

*Charlie takes 2.5 hours to fly from Los Angeles to Mexico City, a distance of 1200 miles. What is the average speed of his plane in miles per hour?*

* *

*(A) **200 mph*

*(B) **240 mph*

*(C) **410 mph*

*(D) **480 mph*

*(E) **533 mph*

Setting up the equation, we get 1200 (Distance) = 2.5 (Time) x R; 1200 = 2.5R. Before solving (assuming this is on the current GRE, in which you do not have a calculator), let’s change 2.5 to 5/2, as it is much easier to do the math with fractions than with decimals. We get 1200 = 5/2R. Solving for R, we multiply both sides by the reciprocal, 2/5. This gives us R on the right-hand side of the equation, and 1200 x 2/5 on the left-hand side. 2400/5 = 480 (D).

Note you could have solved this problem back-solving, in which you put the answer choices back into the question. Or, even better, we could just plug-in 500 mph. Of course that is not one of the answer choices, but using 500 will make the math very easy. If the number is a little too high when we plug-in 500, then the answer must be (D). If it is a little low, then the answer must be (E) 533. Again, you never try to back solve with an ugly number like 533—it will take too long.

Using 500 mph, we get 1200 (Distance) = 500 (Rate) x 2.5 (Time). You can see that multiplying these two numbers gives us 1250. Meaning, we flew too fast and missed Mexico City by 50 miles. Therefore, we have to slow down the plane a little—the answer is 480.

So, this weekend, whether you are traveling abroad or just making a trip to a friend’s BBQ, you can figure out your average distance, rate, or time. Who said GRE doesn’t relate to the real world?

Magoosh students score **12 points better** than average on the GRE. Click here to learn more!

can i have a step y step solution of your given problem please and a formula too

thank you

Hi Ai-ai,

The formula used for these question is Distance= Rate * Time. If you have a more specific question about the step-by-step instructions provided in the blog post, please let us know! 🙂

I just found a sample question in Barron’s (Example 19, pg. 161) that required knowing how to convert pounds into ounces. I am used to the metric system and don’t know by heart the conversions between pounds and ounces and the metric system, inches and feet and miles and km, Fahrenheit to Celsius and so on. I don’t remember the SAT requiring knowing such conversions. Do we have to know them for GRE?

Hi Ana,

Usually the question will give you the conversions. I don’t think they are testing how well you know these conversions. Nor do they want to unfairly bias those who are used to one system and not the other.

Hope that helps :).

Thanks Chris! I’m having trouble with the ones with multiple rates… going in same/opposite directions, and early vs. late..etc can you elucidate how can I understand and train on those? Thanks alot

Hi Mohamed, for same/opposite directions take a look at the link below:

https://magoosh.com/gre/2011/how-a-moving-train-can-be-stationary/

As for early vs. late, do you mean trains (or any other moving things) starting

at different times?

Hi Chris,

I really liked the Magoosh.com and also your maths videos .I am having problem in circular track questions like two persons are running on a circular track i.e. when will they meet and how many times will they meet

i know which formula to apply but the logic behind these questions i am not really care .

So plz help me with that i have also enrolled in Magoosh course Thanks a lot!

Hi Deepak,

The circular track questions are quite tricky :).

The key is thinking in terms of circumference, which often times means dealing with the messy pi (3.14 is not an easy number to calculate).

The other tricky part is if two entities, say race cars, are headed in opposite directions. You will have to combine their rates when figuring out how long it will take them to meet.

So if you have to cars starting at the same point and heading in opposite directions around a circular track, one going 60 km and the other 40km, you will have to combine those rates (60+40 =100). So if the track has diameter of 100 km, then the track is 314 km long, at it will take the cars 3.14 hrs to meet.

Hope that helps!