SOLUTION: A boy walks from his house to school at 2.5km/h and arrives 12 minutes late.
The next day he walks at 4km/h and reaches the school 15 minutes earlier. So determine the distance be
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-> SOLUTION: A boy walks from his house to school at 2.5km/h and arrives 12 minutes late.
The next day he walks at 4km/h and reaches the school 15 minutes earlier. So determine the distance be
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Question 952789: A boy walks from his house to school at 2.5km/h and arrives 12 minutes late.
The next day he walks at 4km/h and reaches the school 15 minutes earlier. So determine the distance between the house and the school. Explain with equation. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A boy walks from his house to school at 2.5km/h and arrives 12 minutes late.
The next day he walks at 4km/h and reaches the school 15 minutes earlier.
So determine the distance between the house and the school.
:
let d = the distance from the house to the school
Convert min to hrs
12 min = 12/60 = .2 hrs late
15 min = 15/60 = .25 hrs early
The time difference between the two trips is .45 hrs
:
2.5 km/h time - 4 km/h time = .45 hrs - = .45
multiply by 20 to clear the denominators
20* - 20* = 20*.45
Cancel the denominators and you have
8d - 5d = 9
3d = 9
d = 9/3
d = 3 km is the distance
