SOLUTION: Jimmi needs to be home by 6 o�clock. If he travels at an average rate of 15mph, he�ll arrive one hour early, but if he travels at an average rate of 10mph, he�ll arrive one hour la

Question 236349: Jimmi needs to be home by 6 o�clock. If he travels at an average rate of 15mph, he�ll arrive one hour early, but if he travels at an average rate of 10mph, he�ll arrive one hour late, How many miles away is his home?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
rate * time = distance.

r = rate
h = time
d = distance

formula becomes:

r*h = d

if we assume h is the right number of hours, then (h-1) gets him there too early and (h+1) gets him there too late.

we'll want to solve for h.

at 15 miles per hour he gets home 1 hour early.

15*(h-1) = d

at 10 miles per hour he gets home 1 hour late.

10*(h+1) = d

since they both equal the same distance, then they are equal to each other so we get:

15*(h-1) = 10*(h+1)

remove parentheses to get:

15h - 15 = 10h + 10

subtract 10h and add 15 to both sides of the equation to get

15h - 10h = 10 + 15

simplify to get

5h = 25

divide both sides by 5 to get:

h = 5

substitute h = 5 in both original equations to get:

15*(h-1) = d
10*(h+1) = d

become:

15*4 = d
10*6 = d

these equations become:

60 = d
60 = d

looks like the distance is 60 miles.

at 15 miles per hour he gets there in 4 hours which is 1 hour too early.
at 10 miles per hour he gets there in 6 hours which is 1 hour too late.

5 hours is just right which means he should be traveling at 12 miles per hour because 5 * 12 = 60.

you answer is:

the distance is 60 miles.

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