Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at the uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?
Choices...
A) 10:00
B) 10:15
C) 10:30
D) 11:00
E) 11:30
2007-02-24 03:32:35 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Do not ignore number 62. Brian is going 4 mph more than Cassie, but left one half an hour later, knowing this figure on how soon will they meet.
Add 16
+12=28
Take 6 away from 62, because that is what Cassie traveled in that half of hour when the other biker wasn't biking
62-6=56
Now divide
56/28 which is two hours
Add that to nine when the last biker left.
11:00 would be the time.
2007-02-24 03:45:19 · answer #1 · answered by Carlene W 5 · 0⤊ 1⤋
Cassie @ 9am......
12/60*30 = 6miles away from Escanaba.
The time starts @ 9am but the route is now 62-6 = 56 miles.
Cos both are heading towards each other, we can sum up their speed...
12+16 = 28 miles/hr
how long they took before they met... h = 56/28 = 2hrs.
Therefore, they will meet at 11:00.
Ans: D
2007-02-24 11:42:17 · answer #2 · answered by Anonymous · 0⤊ 0⤋
D - it would be 11 am
2007-02-24 12:39:27 · answer #3 · answered by Malvi 2 · 0⤊ 1⤋
11am.
2007-02-24 11:50:47 · answer #4 · answered by Asmerom 1 · 0⤊ 1⤋