+0  
 
0
780
6
avatar+466 

The first time Henry made the trip he drove at 10mph and got there 5 hours late. The next time he drove at 25 mph and got there 1 hour early. How long was the trip?

indecision

 Jan 20, 2016

Best Answer 

 #4
avatar
+10

x=the distance     t= time

 

x/10 = t + 5

x/25 = t - 1

 

x/10 - x/25 = (t+5) - (t-1)

5x/50 - 2x/50 = 6

3x/50 = 6

3x=300

x=100 miles

 Jan 20, 2016
 #1
avatar+5265 
0

I'm guessing you mean in miles, so let's find out.

 

So, the first time, he drove 10mph and got there 5 hrs late. Then he drove 15mph faster and got there 1hr early. So this corresponds to 5mph=2hrs, or 2.5mph. The trip there, if it lasts an hour, should be 22.5 miles.

 Jan 20, 2016
 #2
avatar+466 
0

That answer doesn't fully cover it, but thanks for trying. The trip cannot be 22.5 miles, because if you were driving 10mph it would never take you 5 hours too long. Thanks for trying though. smiley

 Jan 20, 2016
 #3
avatar+5265 
+5

Yeah, tried my best. 

 Jan 20, 2016
 #4
avatar
+10
Best Answer

x=the distance     t= time

 

x/10 = t + 5

x/25 = t - 1

 

x/10 - x/25 = (t+5) - (t-1)

5x/50 - 2x/50 = 6

3x/50 = 6

3x=300

x=100 miles

Guest Jan 20, 2016
 #5
avatar+466 
+5

Thank you very much, guest! cool

 Jan 20, 2016
 #6
avatar+128598 
+10

Here's another way to do this......let the normal driving time = T

 

And Rate  * Time   = D   .....so....since the distances equate......

 

10(T + 5)  = 25(T - 1)       (1)

 

10T + 50  = 25T - 25

 

75  = 15T      divide both sides by 15

 

T = 5

 

So......using either side of (1), the distance is 10(5 + 5)  = 10 * 10   = 100 miles

 

 

 

cool cool cool

 Jan 21, 2016

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