In this Algebra word problem solution, an non-algebraic approach would be looking at far Bob goes in 1 hour, since he has an hour head start. He's going 60 mi/hr so he went 60 miles in 1 hour. Steve is going 90 mi/hr or 30 mi/hr faster. So how long would it take someone going 30 mi/hr to go 60 miles? Well the 1st hour he went 30 miles, the 2nd hour he goes the other 30 or 2 hours total until he catches up to Bob. If your going 90 miles an hour for 2 hours-- you have gone 90 * 2 or 180 miles.

Algebraic approach:

x = time

rate * time = distance

60 = # of miles per hour Bob is going

90 = # of miles per hour Steve is going

x + 1 = the hour head start Bob had

Bob = 60(x+1)

Steve = 90x

Since Bob & Steve are going the same distance we can set their respective distance * time equal to one another.

60(x+1) = 90x

60x + 60 = 90x

60 = 30x

30x = 60

x =60/30

x = 2 or 2 hours

check

60(2+1) = 90(2)

180 = 180 --- rate times time = distance which is the answer to the 2nd part of the problem they went 180 miles total.....

## Sunday, September 28, 2008

### Time Distance Problem 6 Solution

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thank you for the problem, i like it!!!

ReplyDeletei like it..

ReplyDeletebut can you please post more distance problems..?

thanks..

i understnd it but maybe u shall given more problems.!

ReplyDelete