Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hur ans start 3 hours after the first cyclist who is traveing at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking

Okay. bicyclists. start 3 hours apart. You want to know when they meet, so you want to know when the distance is the same.

distance = rate × time

So, bicyclist 1, let’s call him “A” … “A” = 6 mph × time
An’ bicyclist 2, let’s call him “B” … “B” = 10 mph × (time – 3)

The minus 3 is cuz he’s travelin’ 3 hours less than the other one. Now, because “A” = “B” (they’ve gone the same distance when they meet), you’ve got an equation your can solve:

6 × time = 10 × (time – 3) …
that’s the same as: 6t = 10(t – 3)

Now, it’s jus’ algebra, right? you multiply the 10 by both the “t” and the 3…

6t = 10t – 30

Now, subtract 10t from both sides to get the “t”s all together… remember, cuz it’s minus 30, your 30’s gonna be negative:

6t – 10t = -30
-4t = -30

Now, divide by -4 to get t all by itself… a negative divided by a negative is a positive, which is good, otherwise they’d be time travelin’ into the past! Keep it real, man!

t = -30/-4 = 7.5 hours

Now, in what I wrote, “t” is the time of the first cyclist. t – 3, or 4.5 hours is the time from when the second cyclist starts to when he catches up. I ain’t too sure, the way the question’s worded, which time it wants. Read the original again an’ see if you can figure it out… is it from when the first guy starts or from when the second guy starts?

Now, the time seems pretty reasonable, but…. let’s check. First cyclist goes for 7.5 hours at 6 mph, that’s 45 miles. Second cyclist goes for 4.5 hours at 10 mph, that’s 45 miles, too. There’s your answer. It’s 7.5 hours from when the first guy started, and 4.5 hours from when the second guy started. There ya go.

For more information about the GED test and GED test preparation, visit the GED Academy at http://www.passGED.com.

Hey Curtis ,

Thanks for your prompt reply man . This is one of the problems I struggled with involving distance :

- A man started walking at 2 mph, while a woman 2 miles behind him began walking at the same time at a rate of 4 mph, and in the same direction. Just then, the man’s dog left him and ran toward the woman. Upon reaching her, it instantly turned around and ran back toward thr man. And so, the dog continued to run back and forth between them, at a constant rate of 5 mph, until the woman finally overtook the man. How far did the dog run?

**** Go enjoy figuring it out and let me know how to do it man !

Zaher

In the GED test, on the page with all the formulas, you got one for distance:

distance = rate x time

So, what’s it mean? Distance is how far you gonna go, rate is how fast you goin’, and time is, well, time. So, how far you go is how fast you go times how long you’re travelin’. In other words, if you get in your car and start drivin’ down the highway, the faster you go, and the longer time you drive, the farther you’ll get. Makes sense, right? If your speed (rate) goes up, the distance you travel goes up. If your time (how long you drive) goes up, the distance you travel goes up.

How does it get applied to the problem? This problem is tough, not because the math is tough, but because it’s tough to figure out how to use the math to solve it. You gotta take it one step at a time and figgure out what’s really goin on here.

Okay, we got a guy walkin’. Same idea as driving, how far = how fast x time. And, we got a woman walking, to. So that’s complicatin’ it up. We got to deal with 2 people walkin’ at once. And a dog, runnin’ back an fo’! Okay, problem with this problem is it’s hard to see. What’s the real question? How far’s the dog go? How do I figure that out?

Hey, this is harder than anythin’ on the GED. So, let’s stretch our minds. To get my mind around it, sketchin’ a picture sometimes helps. Scuse my bad drawin’.

So, the real point is, how far does the dog go? We want to know distance (how far), which is rate times time:

dog distance = rate x time

dog distance = 5 mph x time

Okay, to answer the question, we run into another question… how long does the dog run? He runs from the time they start walkin’ until the two people meet. Now, we gotta answer a second question to answer the first. How long does it take the woman to get to the guy?

We’re usin’ the same formula, an’ tryin’ to find the time it takes for the woman to catch up.

distance = rate x catchup time

The distance they start out is 2 miles:

2 miles = rate x catchup time

What’s the rate? The man is walking at 2 mph, and the woman is walking at 4 mph. She’s catching up to him, but how fast? For every 4 mph she walks, you’ve gotta subtract the 2 mph he’s walked in the same direction during that same time. So, she’s gaining 2 mph on him.

2 miles = (4 mph – 2 mph) x catchup time

2 miles = 2 mph x catchup time

Now, it’s just math. divide each side by 2, and you get:

1 hour = catchup time

How do you know the units are hours? It’s 1 hour, because we’re talkin’ miles and miles per hour. So, one question solved! Now we take the answer and go back to our original question:

dog distance = 5 mph x 1 hour

dog distance = 5 miles

Whew! Dat dog goin’ doggone fast!

For more information about the GED test and GED test preparation, visit The GED Academy at http://www.passGED.com