Question 101782This question is from textbook
: ok, i'm a little stuck finding the equasion. the problem is:
At 1:00 P.M. Sue left her house and began walking at 6 km/h toward Sandy's house. Fifteen minetes later, Sandy left her home and walked at 8 km/h toward Sue's house. If Sue lives 5 km from Sandy, at what tme did they meet? Who walked farther?
thanks.
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! At 1:00 P.M. Sue left her house and began walking at 6 km/h toward Sandy's house. Fifteen minetes later, Sandy left her home and walked at 8 km/h toward Sue's house. If Sue lives 5 km from Sandy, at what tme did they meet? Who walked farther?
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Total distance walked by both is 5 km.
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Sue DATA:
Rate = 6 kph ; distance walked = x km; time = d/r= x/6 hrs
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Sandy DATA:
Rate = 8 kph; distance walked = 5-x km; time = d/r = (5-x)/8 hrs
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EQUATION:
Sue time - Sandy time = (1/4) hr
x/6 - (5-x)/8 = 1/4
Multily thru by 24 to get:
4x - 3(5-x) = 6
4x -15 +3x = 6
7x = 21
x = 3 km (distance Sue walked)
5-x= 2 km (distance Sandy walked)
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Sue time = 5/6 hr = 50 minutes
They meet at 1:50 PM
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Cheers,
Stan H.
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