Comments on: GED Math: Distance, Rate, and Time http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/ My Fast and Smart Road to the GED Tue, 20 Sep 2011 13:49:29 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Curtis http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-1430 Curtis Mon, 02 Mar 2009 21:25:49 +0000 http://www.passged.com/student_blogs/curtis/?p=50#comment-1430 Quick answer: Hey Hammond! 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 x time So, bicyclist 1, let's call him "A" ... "A" = 6 mph x time An' bicyclist 2, let's call him "B" ... "B" = 10 mph x (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 x time = 10 x (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! 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. Quick answer: Hey Hammond!

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 x time

So, bicyclist 1, let’s call him “A” … “A” = 6 mph x time
An’ bicyclist 2, let’s call him “B” … “B” = 10 mph x (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 x time = 10 x (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!

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.

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By: hammond http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-1429 hammond Mon, 02 Mar 2009 19:57:28 +0000 http://www.passged.com/student_blogs/curtis/?p=50#comment-1429 Tow 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 Tow 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

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By: Zaher http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-930 Zaher Fri, 12 Dec 2008 01:42:39 +0000 http://www.passged.com/student_blogs/curtis/?p=50#comment-930 Hey man ! Dat makes sense for me . I never thought it's that easy . As soon as you get the time the dog ran between them , then it's easy for you to calculate the distance . Don't you think we can solve it this way : She started to walk at 4 mph , while there are 2 miles distance between them . Well , dat means she only needs one hour to overtook the man . ( 2 miles distance between them + the 2 miles the man walked ) . Total time for dog is : 1 hour and now we're done with it . d = rt . d = 5mph x 1hour = 5 miles distance the dog ran . Thank you very much for your continued support dear Curtis . I will contact you again for another one if you don't mind .. Zaher Hey man !

Dat makes sense for me . I never thought it’s that easy . As soon as you get the time the dog ran between them , then it’s easy for you to calculate the distance . Don’t you think we can solve it this way :

She started to walk at 4 mph , while there are 2 miles distance between them . Well , dat means she only needs one hour to overtook the man . ( 2 miles distance between them + the 2 miles the man walked ) . Total time for dog is : 1 hour and now we’re done with it . d = rt . d = 5mph x 1hour = 5 miles distance the dog ran .

Thank you very much for your continued support dear Curtis . I will contact you again for another one if you don’t mind ..

Zaher

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