SOLUTION: The ratio of Aldo's cycling speed to Jose cycling speed is 6:5. Jose leaves school at 3pm and Aldo leaves at 3:10pm. By 3:30pm Aldo is only 2km behind Jose. How fast is each cycl

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The ratio of Aldo's cycling speed to Jose cycling speed is 6:5. Jose leaves school at 3pm and Aldo leaves at 3:10pm. By 3:30pm Aldo is only 2km behind Jose. How fast is each cycl      Log On


   

Question 260357: The ratio of Aldo's cycling speed to Jose cycling speed is 6:5. Jose leaves school at 3pm and Aldo leaves at 3:10pm. By 3:30pm Aldo is only 2km behind Jose. How fast is each cycling?
Found 2 solutions by richwmiller, ankor@dixie-net.com:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
a/j=6/5
30 minutes is 1/2 hr
20 minutes is 1/3 hr
a*(1/3)+2=j*(1/2)
a=24 j=20
24/20=6/5=a/j
24*1/3+2=20/2
10=10


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The ratio of Aldo's cycling speed to Jose cycling speed is 6:5.
Jose leaves school at 3pm and Aldo leaves at 3:10pm.
By 3:30pm Aldo is only 2km behind Jose. How fast is each cycling?
:
Let x = the multiplier
then
6x = A's speed
and
5x = J's speed
:
At 3:30
30 min = J's travel time, which is 1%2F2 hr
20 min = A's travel time, which is 1%2F3 hr
:
Write a distance equation: Dist = speed * time
:
J's dist - A's dist = 2 km
1%2F2(5x) - 1%2F3(6x) = 2
Multiply equation by 6 to get rid of the denominators, results
3(5x) - 2(6x) = 6(2)
:
15x - 12x = 12
:
3x = 12
x = 12%2F3
x = 4 is the multiplier
:
6*4 = 24 km/hr is A's speed
5*4 = 20 km/hr is J's speed
:
:
Check solution by finding the distance each traveled
A:1%2F3 * 24 = 8 km
J:1%2F2 * 20 = 10 km; 2km difference