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12 Jan 2026, 05:52
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At 10:00 p.m., a cyclist is 10 miles from home on a straight road. From that time on, the cyclist continues riding farther from home at a constant speed of 3 minutes per mile, then turns around and rides back toward home at the same speed along the same road. If the cyclist must arrive back home at 10:48 p.m., how many additional miles can the cyclist travel away from home after 10:00 p.m. before turning around?
A. 2
B. 3
C. 6
D. 9
E. 12
A. 2
B. 3
C. 6
D. 9
E. 12
12 Jan 2026, 05:52
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Official Solution:
At 10:00 p.m., a cyclist is 10 miles from home on a straight road. From that time on, the cyclist continues riding farther from home at a constant speed of 3 minutes per mile, then turns around and rides back toward home at the same speed along the same road. If the cyclist must arrive back home at 10:48 p.m., how many additional miles can the cyclist travel away from home after 10:00 p.m. before turning around?
A. 2
B. 3
C. 6
D. 9
E. 12
At 10:00 p.m., the cyclist is 10 miles from home. On the way back, riding those 10 miles at 3 minutes per mile will take 30 minutes. Since the cyclist must be home at 10:48 p.m., only 18 minutes remain for riding farther away and then returning to the 10-mile point. That gives 9 minutes out and 9 minutes back. At a pace of 3 minutes per mile, 9 minutes corresponds to 3 miles.
Answer: B
At 10:00 p.m., a cyclist is 10 miles from home on a straight road. From that time on, the cyclist continues riding farther from home at a constant speed of 3 minutes per mile, then turns around and rides back toward home at the same speed along the same road. If the cyclist must arrive back home at 10:48 p.m., how many additional miles can the cyclist travel away from home after 10:00 p.m. before turning around?
A. 2
B. 3
C. 6
D. 9
E. 12
At 10:00 p.m., the cyclist is 10 miles from home. On the way back, riding those 10 miles at 3 minutes per mile will take 30 minutes. Since the cyclist must be home at 10:48 p.m., only 18 minutes remain for riding farther away and then returning to the 10-mile point. That gives 9 minutes out and 9 minutes back. At a pace of 3 minutes per mile, 9 minutes corresponds to 3 miles.
Answer: B
06 Jun 2026, 19:52
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I tried solving it this way but did not work, can someone please explain where I'm going wrong here:
At 10 pm, cyclist is 10 miles away from home and starts moving farther away at a speed of 3 miles/minute. so let's say he went on for x minutes. total time he has to take to complete the trip father away and then head for home os 48 minutes. So (3x further + 3x back in + 10 miles)/3 miles/minutes (which is his constant speed) should equate to 48 minutes.
So, 6x + 10 = 144 (48 x 3)
And we need to solve for 3x.
What am I doing wrong here? Should I not be taking total distance over speed?
At 10 pm, cyclist is 10 miles away from home and starts moving farther away at a speed of 3 miles/minute. so let's say he went on for x minutes. total time he has to take to complete the trip father away and then head for home os 48 minutes. So (3x further + 3x back in + 10 miles)/3 miles/minutes (which is his constant speed) should equate to 48 minutes.
So, 6x + 10 = 144 (48 x 3)
And we need to solve for 3x.
What am I doing wrong here? Should I not be taking total distance over speed?
