Jimmy walks to his school every morning. If he walks at 40 meters per minute, he will reach his school 5 minutes late. If he walks at 60 meters per minute, he will reach his school 3 minutes before the bell rings How far is the school from Jimmy's house?
Let's call the distance from Jimmy's house to school "d".
Using the formula distance = rate x time, we can set up two equations based on the given information:
d = 40(t + 5) (since Jimmy is 5 minutes late)
d = 60(t - 3) (since Jimmy is 3 minutes early)
We want to solve for "d", so let's eliminate the "t" variable by setting the two equations equal to each other:
40(t + 5) = 60(t - 3)
40t + 200 = 60t - 180
20t = 380
t = 19
Now that we know Jimmy's travel time is 19 minutes, we can plug that back into either equation to find the distance:
d = 40(19 + 5)
d = 960 meters
Therefore, the school is 960 meters away from Jimmy's house.
Let's denote the distance from Jimmy's house to his school as "d" in meters.
Here are the given conditions:
- When Jimmy walks at 40 meters per minute, he reaches his school 5 minutes late.
- When Jimmy walks at 60 meters per minute, he reaches his school 3 minutes before the bell rings.
Let's calculate the time it would take for Jimmy to reach his school at each speed:
Time taken at 40 meters per minute = (d / 40) + 5 ... (1)
Time taken at 60 meters per minute = (d / 60) - 3 ... (2)
We can set up an equation based on the above conditions:
(d / 40) + 5 = (d / 60) - 3
Now, let's solve the equation step by step:
1) Multiply the equation by 120 to remove the fractions:
120 * [(d / 40) + 5] = 120 * [(d / 60) - 3]
3d + 600 = 2d - 360
2) Simplify the equation by combining like terms:
3d - 2d = -360 - 600
d = -960 meters
Since distance cannot be negative, it indicates an error or an inconsistency in the problem. Please double-check the given information to ensure accuracy, as the current problem setup does not yield a valid distance.