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15 Dec 2025, 21:36
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The cyclist has 48 mins to cover some additional distance twice PLUS 10 miles (she is already 10 miles away from home, she will go an additional distance one way and then come back on the same way again).
Consider the additional distance to be x miles.
If you set up the Speed, Time, Distance table, it would look like this.
[1 mile][/3 mins] * 48 mins = 10 + 2x
--> 16 = 10 + 2x
--> x = [16-10][/2]
--> x = 3
Therefore, she can travel an additional 3 miles before turning around.
Consider the additional distance to be x miles.
If you set up the Speed, Time, Distance table, it would look like this.
| Speed | Time | Distance |
| 1 mile / 3mins | 48 mins | 10 + 2x |
[1 mile][/3 mins] * 48 mins = 10 + 2x
--> 16 = 10 + 2x
--> x = [16-10][/2]
--> x = 3
Therefore, she can travel an additional 3 miles before turning around.
Bunuel
15 Dec 2025, 22:04
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Let d be the distance that cyclist travels before turning back.
Then, the total distance to be covered by cyclist onward and back home = d + d +10 = 2d+10
The time taken for cyclist to reach back = 48 minutes.
Velocity of cyclist = 1 mile per 3 min
Appying the distance time velocity formula:
Distance/Time=Velocity
(2d + 10)/48 = 1/3
2d + 10 = 48/3 = 16
2d = 16-10=6
d=3miles
Then, the total distance to be covered by cyclist onward and back home = d + d +10 = 2d+10
The time taken for cyclist to reach back = 48 minutes.
Velocity of cyclist = 1 mile per 3 min
Appying the distance time velocity formula:
Distance/Time=Velocity
(2d + 10)/48 = 1/3
2d + 10 = 48/3 = 16
2d = 16-10=6
d=3miles
Bunuel
15 Dec 2025, 22:22
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Bunuel
Given,
Distance from home - 10 miles
Constant speed of cyclist - 3 minutes per mile -> 20 mph
So,
Time needed to cover the 10 miles to home - 10/20 Hrs -> 1/2 hrs -> 30 mins ( remaining time -> (48mins - 18mins) = 18 mins )
Now, he has 18 minutes remaining as he needs to be back home by 10:48pm -> he can cover 6 miles in 18 minutes (3 miles away from home and 3 miles back towards home)
Therefore, answer B) 3 miles
