Abigail lives in an off-campus apartment. Some days she rides her bike to campus, other days she

walks. When she rides her bike, she gets to her first classroom building 36 minutes faster than when she
walks. If her average walking speed is 3 mph and her average biking speed is 12 mph, how far is it from
her apartment to the classroom building?

since time = distance/speed, If she takes m minutes to ride, then if the distance is d miles,

d/3 = d/12 + 36/60
d = 12/5

check:
2.4/3 = 0.8 hours
2.4/12 = 0.2 hours
the difference is 0.6 hours = 36 minutes

Let d be the distance between Abigail's apartment and her campus. Then:

d/3=d/12 +36/60
20d=5d+36
15d=36
d=36/15, or 12/5 miles between Abigail's apartment and the campus
☺☺☺☺

To find the distance from Abigail's apartment to the classroom building, we can use the formula: Distance = Speed × Time.

Let's say the distance is 'd' miles.

When Abigail walks, her average speed is 3 mph. Let's calculate the time it takes for her to walk from her apartment to the classroom building:
Time taken to walk = Distance / Speed = d / 3

When she rides her bike, her average speed is 12 mph. The problem states that she gets to the classroom building 36 minutes faster than when she walks. Since we know that time = distance / speed, we can set up the following equation for the biking journey:

Time taken to bike = Time taken to walk - 36 minutes

Converting 36 minutes to hours: 36 minutes / 60 minutes = 0.6 hours.

So, the equation becomes:
d / 12 = d / 3 - 0.6

To solve this equation, we can simplify it by multiplying both sides by the least common denominator, which is 12:
12 * (d / 12) = 12 * (d / 3 - 0.6)

This simplifies to:
d = 4d - 7.2

Next, let's isolate the variable 'd' on one side of the equation. Subtract 4d from both sides:
d - 4d = -7.2

This simplifies to:
-3d = -7.2

To solve for 'd', divide both sides by -3:
d = (-7.2) / (-3) = 2.4

Therefore, the distance from Abigail's apartment to the classroom building is 2.4 miles.