SSAT Upper Level

Distance, Speed & Time

Solving problems using d = r × t, average speed, and multi-leg journey calculations

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Have you ever raced your best friend to grab the very last slice of pepperoni pizza? 🍕 If you both start from the same spot, the person who runs the fastest (speed) for the shortest amount of time will cover the distance first and win the pizza! Welcome to the world of Distance, Speed, and Time. This is one of the most useful math tricks you will ever learn, and the SSAT test makers absolutely love it. Why? Because we use it every single day!

Whether you are figuring out how long a boring car ride will take, or tracking how fast a cheetah can run, it all comes down to one simple idea. Distance is how far you go. Speed (or rate) is how fast you are moving. Time is how long it takes. To solve these puzzles on the SSAT, you just need to remember the magic formula: $d = r \times t$ (Distance equals Rate times Time). 🚗💨

I like to picture a giant triangle with a 'D' at the top, and 'R' and 'T' at the bottom. If you want to find Distance, you multiply Rate and Time. If you want to find Time, you divide Distance by Rate ( $t = \frac{d}{r}$ ). And if you want to find Rate, you divide Distance by Time ( $r = \frac{d}{t}$ ). Once you master this magic trick, you will be zooming through these SSAT math questions faster than a superhero flying to the rescue! 🦸‍♂️ Let's look at some super fun examples to get you ready.

Key Formula

The magic formula is

d = r \times t

(Distance = Rate × Time). To find Time, use

t = \frac{d}{r}

. To find Rate, use

r = \frac{d}{t}

Practice Questions

3 practice questions for SSAT Upper Level

Q1 Hard

A freight train leaves a station traveling at

40

miles per hour. Two hours later, a passenger train leaves the same station on a parallel track, traveling in the same direction at

60

miles per hour. How many hours will it take for the passenger train to catch up to the freight train?

2

3

4

5

6

Show Solution

First, determine how far the freight train has traveled before the passenger train starts. The freight train has a $2$ -hour head start at $40$ miles per hour, so it has traveled $2 \times 40 = 80$ miles. The passenger train travels at $60$ miles per hour, which is $60 - 40 = 20$ miles per hour faster than the freight train. This means the passenger train closes the gap by $20$ miles every hour. To close the $80$ -mile gap at a relative speed of $20$ miles per hour, it will take the passenger train $\frac{80}{20} = 4$ hours.

Answer: C

Q2 Hard

A family drove

210

miles to a vacation cabin. If the drive took between

3.5

and

4

hours, their average speed, in miles per hour, must have been between which of the following two numbers?

45

and

50

50

and

52.5

52.5

and

60

60

and

65

65

and

70

Show Solution

Average speed is calculated as distance divided by time ( $Speed = \frac{Distance}{Time}$ ). To find the maximum possible average speed, divide the distance by the shortest time: $\frac{210}{3.5} = 60$ miles per hour. To find the minimum possible average speed, divide the distance by the longest time: $\frac{210}{4} = 52.5$ miles per hour. Therefore, the average speed must have been between $52.5$ and $60$ miles per hour.

Answer: C

Q3 Hard

A runner is

24

miles away from the finish line of a trail and is running toward it at a constant rate of

8

miles per hour. At the exact same time, a cyclist starts at the finish line and rides toward the runner at a constant rate of

16

miles per hour. Assuming they stay on the same path, how many miles away from the finish line will they meet?

8

12

16

18

20

Show Solution

Since the runner and cyclist are moving toward each other, their combined rate of travel is $8 + 16 = 24$ miles per hour. The total distance between them is $24$ miles. The time it takes for them to meet is their total distance divided by their combined rate: $\frac{24}{24} = 1$ hour. They will meet exactly $1$ hour after they start. The cyclist started at the finish line and traveled for $1$ hour at $16$ miles per hour. Therefore, the distance they meet from the finish line is $16 \times 1 = 16$ miles.

Answer: C

Tips & Strategies

Draw the D-R-T magic triangle on your scratch paper! Put a 'D' on top, and an 'R' and 'T' on the bottom. Cover up the letter you are looking for, and the triangle tells you exactly what math to do.
Always check your units! If the speed is in miles per hour, but the time given is in minutes, you MUST convert the minutes into hours (like changing 30 minutes into $\frac{1}{2}$ hour) before multiplying.
For Average Speed questions, write down 'Total Distance' and 'Total Time' before doing anything else. Never just add two speeds together and divide by 2!

Common Mistakes

Watch out for mixed-up units! If a question asks how far a train travels in 15 minutes at 60 miles per hour, don't multiply 60 by 15! You must change 15 minutes into $\frac{1}{4}$ hour first.
Don't forget the golden rule of average speed: it is ALWAYS $\frac{Total Distance}{Total Time}$ . The SSAT loves to put the wrong 'simple average' as an answer choice to trick you.

Frequently Asked Questions

Will the SSAT give me the distance formula on the test?

Nope! You have to memorize it. Just remember the word 'Dirt' ( $d = r \times t$ ) to help it stick in your brain!

What if the distance is in kilometers instead of miles?

Don't panic! The math works exactly the same way. Just make sure your speed matches, like kilometers per hour!

Are there really average speed questions on the SSAT?

Yes, especially on the Middle and Upper Level tests. They love to trick students who just try to find the middle of two speeds, so always use your total distance and total time.

Can I use a calculator for the tricky fractions?

Calculators are not allowed on the SSAT. That's why practicing multiplying and dividing fractions, like $\frac{1}{2}$ or $\frac{3}{4}$ , is super important!

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