SOLUTION: Felipe jobs for 10 miles and then walks another 10 miles. He jobs 2 1/2 miles per hour faster than he walks, and the entire distance of 20 miles takes 6 hours. find the rate at w

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Felipe jobs for 10 miles and then walks another 10 miles. He jobs 2 1/2 miles per hour faster than he walks, and the entire distance of 20 miles takes 6 hours. find the rate at w      Log On


   



Question 187282This question is from textbook
: Felipe jobs for 10 miles and then walks another 10 miles. He jobs 2 1/2 miles per hour faster than he walks, and the entire distance of 20 miles takes 6 hours. find the rate at which he walks and the rate at which he jogs. This question is from textbook

Found 3 solutions by josmiceli, MathTherapy, josgarithmetic:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Jogging:
d%5Bj%5D+=+r%5Bj%5D+%2A+t%5Bj%5D
Walking:
d%5Bw%5D+=+r%5Bw%5D+%2A+t%5Bw%5D
given:
d%5Bj%5D+=+10 mi
d%5Bw%5D+=+10 mi
r%5Bj%5D+=+r%5Bw%5D+%2B+2.5 mi/hr
t%5Bj%5D+%2B+t%5Bw%5D+=+6 hr
t%5Bj%5D+=+6+-+t%5Bw%5D hr
--------------------------
Jogging:
10+=+%28r%5Bw%5D+%2B+2.5%29%2A+%286+-+t%5Bw%5D%29
10+=+6r%5Bw%5D+%2B+15+-+r%5Bw%5D%2At%5Bw%5D+-+2.5t%5Bw%5D
10+=+6r%5Bw%5D+%2B+15+-+10+-+2.5t%5Bw%5D
(1)5+=+6r%5Bw%5D+-+2.5t%5Bw%5D
Walking:
10+=+r%5Bw%5D%2At%5Bw%5D
t%5Bw%5D+=+10%2Fr%5Bw%5D
Substitute this in (1)
(1) 5+=+6r%5Bw%5D+-+2.5%2A%2810%2Fr%5Bw%5D%29
multiply both sides by r%5Bw%5D
(1) 5r%5Bw%5D+=+6%2A%28r%5Bw%5D%29%5E2+-+25
(1) 6%2A%28r%5Bw%5D%29%5E2+-+5r%5Bw%5D+-+25+=+0
Use quadratic formula to solve for r%5Bw%5D
Then find t%5Bw%5D, then find r%5Bj%5D


Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Felipe jobs for 10 miles and then walks another 10 miles.  He jobs 2 1/2 miles per hour faster than he walks, and
the entire distance of 20 miles takes 6 hours.  find the rate at which he walks and the rate at which he jogs.
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The other person's solution is rather lengthy, and in this author's opinion, a few UNNECESSARY variables were introduced. 
Furthermore, he suggests using the quadratic formula to solve his final equation, which can be easily FACTORIZED. 

Let the speed at which he walks, be S
Then, his jogging speed = S+%2B+2%261%2F2, or S + 2.5
Time taken to walk 10 miles: 10%2FS, and time taken to jog 10 miles = 10%2F%28S+%2B+2.5%29
As he takes 6 hours to walk and jog 20 miles, we get the following
TOTAL-TIME equation: 10%2FS+%2B+10%2F%28S+%2B+2.5%29+=+6
                              10(S + 2.5) + 10S = 6S(S + 2.5) ----- Multiplying by LCD, S(S + 2.5)
                                     10S+%2B+25+%2B+10S+=+6S%5E2+%2B+15S
                                              20S+%2B+25+=+6S%5E2+%2B+15S
                           6S%5E2+%2B+15S+-+20S+-+25+=+0
                                      6S%5E2+-+5S+-+25+=+0
                          6S%5E2+-+15S+%2B+10S+-+25+=+0
                        3S%282S+-+5%29+%2B+5%282S+-+5%29+=+0
                               (2S - 5)(3S + 5) = 0
                                                 2S - 5 = 0      or     3S + 5 = 0 ---- Setting FACTORS equal to 0
                                                      2S = 5      or            3S = - 5 (IGNORE)
                  Walking speed, or S = 5%2F2 = 2.5 mph

                 Jogging speed: S + 2.5 = 2.5 + 2.5 = 5 mph

                 You can do the CHECK!!

Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
                     SPEED         TIME           DISTANCE
JOG                  r+2.5         10/(r+2.5)        10
WALK                 r             10/r
TOTAL                                6  

10%2F%28r%2B2%261%2F2%29%2B10%2Fr=6
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