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	<title>Comments on: GED Math: Distance, Rate, and Time</title>
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	<link>http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/</link>
	<description>My Fast and Smart Road to the GED</description>
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		<title>By: Curtis</title>
		<link>http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-1430</link>
		<dc:creator>Curtis</dc:creator>
		<pubDate>Mon, 02 Mar 2009 21:25:49 +0000</pubDate>
		<guid isPermaLink="false">http://www.passged.com/student_blogs/curtis/?p=50#comment-1430</guid>
		<description>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&#039;s call him &quot;A&quot; ... &quot;A&quot; = 6 mph x time
An&#039; bicyclist 2, let&#039;s call him &quot;B&quot; ... &quot;B&quot; = 10 mph x (time - 3)
The minus 3 is cuz he&#039;s travelin&#039; 3 hours less than the other one. Now, because &quot;A&quot; = &quot;B&quot; (they&#039;ve gone the same distance when they meet), you&#039;ve got an equation your can solve:

6 x time = 10 x (time - 3) ... that&#039;s the same as: 6t = 10(t - 3)

Now, it&#039;s jus&#039; algebra, right? you multiply the 10 by both the &quot;t&quot; and the 3...

6t = 10t - 30

Now, subtract 10t from both sides to get the &quot;t&quot;s all together... remember, cuz it&#039;s minus 30, your 30&#039;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&#039;d be time travelin&#039; into the past!

t = -30/-4 = 7.5 hours

Now, in what I wrote, &quot;t&quot; 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&#039;t too sure, the way the question&#039;s worded, which time it wants. Read the original again an&#039; 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&#039;s check. First cyclist goes for 7.5 hours at 6 mph, that&#039;s 45 miles. Second cyclist goes for 4.5 hours at 10 mph, that&#039;s 45 miles, too. There&#039;s your answer. It&#039;s 7.5 hours from when the first guy started, and 4.5 hours from when the second guy started.</description>
		<content:encoded><![CDATA[<p>Quick answer: Hey Hammond!</p>
<p>Okay. bicyclists. start 3 hours apart. you want to know when they meet, so you want to know when the distance is the same.</p>
<p>distance = rate x time</p>
<p>So, bicyclist 1, let&#8217;s call him &#8220;A&#8221; &#8230; &#8220;A&#8221; = 6 mph x time<br />
An&#8217; bicyclist 2, let&#8217;s call him &#8220;B&#8221; &#8230; &#8220;B&#8221; = 10 mph x (time &#8211; 3)<br />
The minus 3 is cuz he&#8217;s travelin&#8217; 3 hours less than the other one. Now, because &#8220;A&#8221; = &#8220;B&#8221; (they&#8217;ve gone the same distance when they meet), you&#8217;ve got an equation your can solve:</p>
<p>6 x time = 10 x (time &#8211; 3) &#8230; that&#8217;s the same as: 6t = 10(t &#8211; 3)</p>
<p>Now, it&#8217;s jus&#8217; algebra, right? you multiply the 10 by both the &#8220;t&#8221; and the 3&#8230;</p>
<p>6t = 10t &#8211; 30</p>
<p>Now, subtract 10t from both sides to get the &#8220;t&#8221;s all together&#8230; remember, cuz it&#8217;s minus 30, your 30&#8217;s gonna be negative:</p>
<p>6t &#8211; 10t = -30<br />
-4t = -30</p>
<p>Now, divide by -4 to get t all by itself&#8230; a negative divided by a negative is a positive, which is good, otherwise they&#8217;d be time travelin&#8217; into the past!</p>
<p>t = -30/-4 = 7.5 hours</p>
<p>Now, in what I wrote, &#8220;t&#8221; is the time of the first cyclist. t &#8211; 3, or 4.5 hours is the time from when the second cyclist starts to when he catches up. I ain&#8217;t too sure, the way the question&#8217;s worded, which time it wants. Read the original again an&#8217; see if you can figure it out&#8230; is it from when the first guy starts or from when the second guy starts?</p>
<p>Now, the time seems pretty reasonable, but&#8230;. let&#8217;s check. First cyclist goes for 7.5 hours at 6 mph, that&#8217;s 45 miles. Second cyclist goes for 4.5 hours at 10 mph, that&#8217;s 45 miles, too. There&#8217;s your answer. It&#8217;s 7.5 hours from when the first guy started, and 4.5 hours from when the second guy started.</p>
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		<title>By: hammond</title>
		<link>http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-1429</link>
		<dc:creator>hammond</dc:creator>
		<pubDate>Mon, 02 Mar 2009 19:57:28 +0000</pubDate>
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		<description>Tow cyclists start biking from a trail&#039;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</description>
		<content:encoded><![CDATA[<p>Tow cyclists start biking from a trail&#8217;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</p>
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		<title>By: Zaher</title>
		<link>http://www.passged.com/student_blogs/curtis/2008/12/03/ged-math-distance-rate-and-time/comment-page-1/#comment-930</link>
		<dc:creator>Zaher</dc:creator>
		<pubDate>Fri, 12 Dec 2008 01:42:39 +0000</pubDate>
		<guid isPermaLink="false">http://www.passged.com/student_blogs/curtis/?p=50#comment-930</guid>
		<description>Hey man !

Dat makes sense for me . I never thought it&#039;s that easy . As soon as you get the time the dog ran between them , then it&#039;s easy for you to calculate the distance . Don&#039;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&#039;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&#039;t mind ..

Zaher</description>
		<content:encoded><![CDATA[<p>Hey man !</p>
<p>Dat makes sense for me . I never thought it&#8217;s that easy . As soon as you get the time the dog ran between them , then it&#8217;s easy for you to calculate the distance . Don&#8217;t you think we can solve it this way :</p>
<p>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&#8217;re done with it . d = rt . d = 5mph x 1hour = 5 miles distance the dog ran . </p>
<p>Thank you very much for your continued support dear Curtis . I will contact you again for another one if you don&#8217;t mind ..</p>
<p>Zaher</p>
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