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15 Dec 2025, 11:08
Kudos
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This is a basic time speed distance problem, you just need to know d=s/t
d= 10
s= 1/3 m/min (3min per mile which means 1/3min per min)
t=0
at 10:48pm, he must travel some distance(lets say x) and be back at home
this means he travel in total d: x+x+10
s= 1/3 m/min (3min per mile which means 1/3min per min)
t = 48min
Hence: (2x+10)/48=1/3 or x=3
d= 10
s= 1/3 m/min (3min per mile which means 1/3min per min)
t=0
at 10:48pm, he must travel some distance(lets say x) and be back at home
this means he travel in total d: x+x+10
s= 1/3 m/min (3min per mile which means 1/3min per min)
t = 48min
Hence: (2x+10)/48=1/3 or x=3
15 Dec 2025, 11:14
Kudos
Bookmarks
The cyclist is 10 miles away from home at 10:00 and he is cycling at 3 miles/min so it will take him 30 mins to reach home. Therefore , reducing 30 mins from 10:48 will leave 18 mins .
The time the cyclist can ride far away from home is 3 miles which will take 9 mins and then he will travel back 9 mins to cover the same distance(that's how 18 mins will be complete) and then 30 mins to reach home at 10:48.
The time the cyclist can ride far away from home is 3 miles which will take 9 mins and then he will travel back 9 mins to cover the same distance(that's how 18 mins will be complete) and then 30 mins to reach home at 10:48.
15 Dec 2025, 11:14
Kudos
Bookmarks
Bunuel
Total Time to be taken from Point A (10 Miles Away from Home) to travel further straight away and Back Home = 48 Minutes
Let the Distance covered from Point A in a straight path until the U-turn is taken be D Miles.
As, Distance = Rate x Time,
The given rate is 3 Minutes per Mile, which can be written as 1/3 Miles per Minute
Hence,
10+2D = 1/3 x 48
2D = 16-10
D = 6/2 = 3
Hence, from Point A starting at 10:00 pm, the cyclist can go 3 Miles away and then take a U-turn to reach home at 10:48 pm.
