Question 328581: Extreme hardship. Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. Her brother Bryan
starts at 7:40 A.M. on his bicycle, traveling 10 mph faster
than Kim. If they get to school at the same time, then how
fast is each one traveling?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. Her brother Bryan
starts at 7:40 A.M. on his bicycle, traveling 10 mph faster
than Kim. If they get to school at the same time, then how
fast is each one traveling?
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Kim DATA:
distance = 3 mi ; rate = x mph ; time = 3/x hrs
==============
Bryan DATA:
distance = 3 mi ; rate = x+10 mph ; tim = 3/(x+10)
-------------------
Equation:
Kim time - Bryan time = 1/6 hr
3/x - 3/(x+10) = 1/6
----
Multiply thru by 6x(x+10) to get:
18(x+10) - 18x = x(x+10)
180 = x^2+10x
------
x^2+10x-180 = 0
-------
Positive solution:
x = [-10+sqrt(100-4*-180]/2
x = 9.32 mph (Kim speed)
---
x+10 = 19.32 mph (Bryan speed)
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Cheers,
Stan H.
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