Question 831299: I can't seem to get this problem to come out right and can't figure out my mistake. Help please?
At 1:00 pm Sue left her home and began walking at 8 km/h towards Sandy's house. Fifteen minutes later, Sandy left her home and walked at 10 km/h wards Sue's house. If Sue lives 10 km from Sandy, at what time did they meet? Who walked further?
Answer by Elomeht(22)   (Show Source):
You can put this solution on YOUR website! First, forget about the numbers and equations, and satisfy yourself that the following statements are true:
1. When they finally meet, the time elapsed since Sue started walking, must be 15 minutes more than the time elapsed since Sandy started walking
2. Wherever they meet, the distance that Sue has walked, plus the distance Sandy has walked, must equal 10
3. 15 minutes is the same as 0.25 hours
Let us now go to the math, keeping the above statements in mind. We will denote the distance Sue walked as d. So the distance Sandy walked must be 10 - d
So it must be true that d/8 = 0.25 + (10 - d)/10
If we multiply the last equation above by 40, we get:
5d = 10 + 4(10 - d), which simplifies to 9d = 50.
d = 50/9 = 5.555555....
This is clearly greater than 10 - d which amounts to 4.444444.....
So Sue walked the longer distance.
The time elapsed is d/8 = 5.555555/8 = 0.694444 hours, or 41.66667 minutes. Another way of seeing this is to calculate 0.25 + 4.44444444/10 which again is 0.6944444
So the time they met was 1:42 pm
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