SOLUTION: A man covers a certain distance on scootr. Had he moved 3kmph faster, he would have taken 40min less,if he had moved 2kmph slower he would have taken 40 min more.The distance in (k

Question 291540: A man covers a certain distance on scootr. Had he moved 3kmph faster, he would have taken 40min less,if he had moved 2kmph slower he would have taken 40 min more.The distance in (km)is
Found 2 solutions by ankor@dixie-net.com, happyhaps:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A man covers a certain distance on scooter.
Had he moved 3 kmph faster, he would have taken 40 min less,
if he had moved 2 kmph slower he would have taken 40 min more.
The distance in (km)is:
:
I don't think there is a unique solution to this, time, speed, and distance
are all unknown, but we can try and find at least one integer solution
:
We can say the faster time is 80 min less than the slower time
80 min = 80/60 = hrs
and
fast speed is 5 mph more than the slow speed
:
s = slow speed
(s+5) = fast speed
:
Let d = the distance
:
Write a time equation: Time = dist/speed
slow time - fast time = 4/3 hr
-
Multiply equation by 3s(s+5), results
3d(s+5) - 3ds = 4s(s+5)
:
3ds + 15d - 3ds = 4s^2 + 20s
:
d(3s+15-3s) = 4s^2 + 20s
:
d(15) = 4s^2 + 20s
d =
One integer solution to d results when s = 10
d =
d =
d =
d = 40 km is the distance when the slower speed = 10 km/hr
:
See if that is true,
his original speed, 2 mi faster than the slowest speed then s= 12 km/hr
:
If he went 3 km/hr faster = 15 km/hr
:
Check the time difference when 3 km/hr faster
- =
3.33 - 2.66 = .67 hrs; which is .67*60 = 40.2 ~ 40 min
and when it's 2 km/hr slower
- =
4 - 3.33 = .67 hrs; which is .67*60 = 40.2 ~ 40 min
;
:
Bear in mind this is only one of many solutions

Answer by happyhaps(2) (Show Source): You can put this solution on YOUR website!
First of all, let us assume normal speed to be s and normal time be t.
Now the distance become st(Distance = speed x time).
According to the first statement, increased speed becomes s+3 and therefore time decreases by (t - 40/60) that is t - 2/3.
therefore the distance by new speed and time becomes (s + 3)(t - 2/3).
Now let us equate the first distance by the second one above since distance is same.
it becomes
st = (s+3)(t - 2/3)
Rearranging this becomes,
2s -9t = -6..........(I)
Now According to the second statement, decreased speed becomes s-2 and therefore increased time becomes t + 2/3
Again the distance by new speed and time becomes (s-2)(t + 2/3).
Now, again equate this distance by original distance st.
st = (s-2)(t + 2/3)
Rearranging this,
-2s + 6t = -4 ...... (II)
Now you have two linear equation (I) and (II)
Solve these to get values of s and t
which is s = 12 and t = 10/3
Now distance is st therefore 12 x 10/3 = 40 km
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