SOLUTION: A man walk a certain distance at certain rate. Had he walked 1/2 km an hour faster he would have taken 1 hour less, but he had gone 1 km an hour slower, he would have taken 3 hour
Question 188127: A man walk a certain distance at certain rate. Had he walked 1/2 km an hour faster he would have taken 1 hour less, but he had gone 1 km an hour slower, he would have taken 3 hour longeer. find the distance he travelled and his original rate. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A man walk a certain distance at certain rate. Had he walked 1/2 km an hour
faster he would have taken 1 hour less, but he had gone 1 km an hour slower,
he would have taken 3 hour longer. find the distance he traveled and his original rate.
:
let d = distance traveled
:
let r = original rate of travel
and
d/r = actual time
:
then
(r-1) = the slower rate
and
(r+.5) = the faster rate
:
Write Two time equations; Time =
Eq1; slower time - actual time = 3 hr - = 3
multiply equation by r(r-1) to get rid of the denominators, results:
dr - d(r-1) = 3r(r-1)
dr - dr + d = 3r^2 - 3r
d = 3r^2 - 3r
:
eq2; actual time - fast time = 1 hr - = 1
Multiply equation by r(r+.5) to get rid of the denominator, results
d(r+.5) - dr = r(r+.5)
dr + .5d - dr = r^2 + .5r
.5d = r^2 + .5r
Multiply equation by 2
d = 2r^2 + r
:
d = d, therefore we can say:
3r^2 - 3r = 2r^2 + r
Divide equation by r
3r - 3 = 2r + 1
3r - 2r = 1 + 3
r = 4 km/h actual waking speed
:
Find the distance traveled:
d = 2r^2 + r
Substitute 4 for r
d = 2(4^2) + 4
d = 2(16) + f
d = 36 km
;
;
Check distance in the 2nd equation
d = 3r^2 - 3r
Substitute 4 for r
d = 3(4^2) - 3(4)
d = 48 - 12
d = 36 km
:
:
Confirm times:
Actual time = 36/4 = 9 hrs
Faster time = 36/4.5 = 8 hrs (1 hr faster)
slower time = 36/3 = 12 hrs (3 hrs slower)
:
:
Hope all this made sense to you.
