A man walks 12 km at a certain speed and then 8 km at 1 km / hour faster. If the total time is 6 hours,
find his original speed?
Answers
Let r = the rate the man walked
then (r+.5) = 1 hr less rate
and (r-1) = 3 hr more rate
let t = time he actually walked
then (t-1) = time taken at the faster speed
and (t+3) = time at the slower speed
Distance for all three situations is the same (dist = rate * time):
So, rt = (r+.5)(t-1)
rt = rt - r + .5t - .5
rt - rt + r - .5t = -.5
r - .5t = -.5 (1)
Also, for the slower speed case,
rt = (r-1)(t+3)
rt = rt + 3r -t - 3
rt - rt - 3r + t = -3
-3r + t = -3 (2)
multiply the 1st equation by 2 and add to the above equation
2r - t = -1
-3r + t = -3
---------------eliminates t
-r = -4
r = 4 km/h is the actual walking rate
Find t using: r -.5t = -.5
4 - .5t = -.5
-.5t = -.5 - 4
-.5t = - 4.5
t = 9 hrs actual time
Hence, Dist = r*t
4 * 9 = 36 km
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