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Expert replies
by caramel3536 » Sat May 03, 2008 1:26 pm
A hiker walked for 2 days. On the second day the hiker walked 2 hours longer at an average speed 1m/hr faster than he walked on the first day. If during the 2 days he walked a total of 64 miles and spent a total of 18 hours walking. What was his average speed on the first day? ans=3mph
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Source: — Problem Solving |

by VP_RedSoxFan » Sat May 03, 2008 6:19 pm
For me, the best thing to do is to keep track of what we know about each day's activities as well as the total in a quick chart. For example:

Day 1 Day 2 Total

Rate R1 R1+1

Time T1 T1+2 18 (therefore T1 = 8)

Distance D1 D2 64

Sorry for the crappy formatting!


From here, we can set up D = RT equations. We have 3 unknowns (D1, D2, R1) so we need 3 equations and we can use our chart to make them.

(1) D1 + D2 = 64; or D2 = 64 - D1
(2) D1 = R1 * 8
(3) D2 = (R1+1) * 10

I'll substitute the first equation into the third:

(4) 64 - D1 = (R1+1) * 10

Then equation (2) into equation (4). I choose the methodology so that we can solve for R1, the first day's mph:

(5) 64 - R1 * 8 = (R1+1) * 10
18 * R1 = 54
R1 = 3

There we go. Hope this helps.
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Thanks

by caramel3536 » Sun May 04, 2008 1:52 pm
............
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by ksh » Sun May 04, 2008 11:17 pm
The above post is quite clear. But it can be reduced

T1+(T1+2)=18 => T1=8, T2=10

Now, R1*8+(R1+1)*10=64 =>18R1=54
Hence, R1=3mph
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