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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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]

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Vincen wrote:
Wed Jul 22, 2020 12:31 pm
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
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 mph

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!

-Jay
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Vincen wrote:
Wed Jul 22, 2020 12:31 pm
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]

Solution:

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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