At 1:00, Jack starts to bicycle along a 60-mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?
A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00
[spoiler]OA=B[/spoiler]
Source: Veritas Prep
At 1:00, Jack starts to bicycle along a 60-mile road at a constant speed of 15 miles per hour. Thirty minutes earlier,
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Since Scott started 30 minutes earlier than Jack, by 1:00, he would have covered 12/2 = 6 miles. So for Jack to catch Scott, Jack must cover 6 miles at a relative speed of 15 – 12 = 3 mphVincen wrote: ↑Wed Jul 22, 2020 12:31 pmAt 1:00, Jack starts to bicycle along a 60-mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?
A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00
[spoiler]OA=B[/spoiler]
Source: Veritas Prep
Time is taken by Jack to catch Scott = 6/3 = 2 hours; thus, Jack would catch Scott by 3:00.
Correct answer: B
Hope this helps!
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Solution:Vincen wrote: ↑Wed Jul 22, 2020 12:31 pmAt 1:00, Jack starts to bicycle along a 60-mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?
A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00
[spoiler]OA=B[/spoiler]
We can let t = the time, in hours, Jack takes to catch up and meet Scott, and create the equation:
15t = 12(t + 1/2)
15t = 12t + 6
3t = 6
t = 2
Therefore, it takes 2 hours for Jack to catch up and meet Scott. Since he starts at 1:00, he meets Scott at 3:00.
Answer: B
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