SOLUTION: Meg traveled 7 miles yesterday walking for one half-hour and biking for one half-hour. Today traveling at the same rates she traveled a total of 6 miles by walking for 40 mintues a
Question 263584: Meg traveled 7 miles yesterday walking for one half-hour and biking for one half-hour. Today traveling at the same rates she traveled a total of 6 miles by walking for 40 mintues and biking for 20 mintues. What was her walking and biking rates? The answer is walking 4mph and biking 10mph. Please help me set up the equations. Found 3 solutions by richwmiller, stanbon, josmiceli:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! rt=d
rt+st=d
r=walking rate
s=biking speed
r*1/2+s*1/2=7
r*4/6+s*1/3=6
You should get 4 and 10. I did.
You can put this solution on YOUR website! Meg traveled 7 miles yesterday walking for one half-hour and biking for one half-hour. Today traveling at the same rates she traveled a total of 6 miles by walking for 40 mintues and biking for 20 mintues. What was her walking and biking rates? The answer is walking 4mph and biking 10mph.
Let walking rate be "w": let biking rate be "b":
Equations:
(1/2)w + (1/2)b = 7
(2/3)w + (1/3)b = 6
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Modify the equations:
w + b = 14
2w + b = 18
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Subtract 1st from 2nd equation to get:
w = 4 mph (walking rate)
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Since w + b = 14, w = 10 mph (walking rate)
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Cheers,
Stan H.
You can put this solution on YOUR website! Let = her rate walking in mi/hr
Let = her rate biking in mi/hr
For yesterday's trip:
(1)
Today's trip:
(2)
Multiply both sides of (1) by
(1)
Multiply both sides of (2) by
(2)
From (1),
Substitute this in (2)
(2)
(2)
And, from (1),
She walks 4 mi/hr and bikes 10 mi/hr
check:
For yesterday's trip:
(1)
Today's trip:
(2)
OK
(