Rate Problem

[SCSmith]

Can You Show Me Where I Am Going Wrong.
Juan and Angela are bicycling along the same trail. Juan passes a marker at 9:00AM and Angela passes the same marker at 9:05AM. Juan is traveling at 8 miles per hour while Angela is traveling at 10 miles per hour. What time will Angela catch up to Juan?

D = RxT
Juan's Rate 8mph x Juan's Time t + .08 = Distance 8(t +.08) I reasoned that Juan is always 5 mins faster than Angela which is 5/60mins, then I converted to the same units as mph and came up with 1/12 or .08.
Next:
Angela's Rate 10mph x Angela's Time t = Distance 10t. Since their Distance is equivalent:
8(t + .08) =10t
8t + .64 = 10t
2t =.64
t = .32mins.
Time when Angela catches up 9:32AM
The Correct answer is 9:25AM
Where did I go Wrong?

 

[dagr8est]

Okay I figured out what the problem is. First you rounded off too early. Second you should get 0.333... hours or 20 minutes. Remember that t represents the time from when Angela started which was 9:05AM so 20 minutes from that time is 9:25AM.

 

[Denis]

Can You Show Me Where I Am Going Wrong.
Juan and Angela are bicycling along the same trail. Juan passes a marker at 9:00AM and Angela passes the same marker at 9:05AM. Juan is traveling at 8 miles per hour while Angela is traveling at 10 miles per hour. What time will Angela catch up to Juan?
D = RxT
Juan's Rate 8mph x Juan's Time t + .08 = Distance 8(t +.08) I reasoned that Juan is always 5 mins faster than Angela which is 5/60mins, then I converted to the same units as mph and came up with 1/12 or .08.
Next:
Angela's Rate 10mph x Angela's Time t = Distance 10t. Since their Distance is equivalent:
8(t + .08) =10t
8t + .64 = 10t
2t =.64
t = .32mins.
Time when Angela catches up 9:32AM
The Correct answer is 9:25AM
Where did I go Wrong?

You "reasoned" incorrectly...

Let t = time by Juan; then time by Angela = t - 5 (left 5 minutes after)
Since distance is same:
10(t - 5) = 8t
10t - 50 = 8t
2t = 50
t = 25 ; so 9:25