SOLUTION: on a monday afternoon practice, two cyclists took the same amount of time to ride 39 miles and 30 miles, respectively. if a cyclist rode 3 mph faster than the other, what is the ra

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Question 1043847: on a monday afternoon practice, two cyclists took the same amount of time to ride 39 miles and 30 miles, respectively. if a cyclist rode 3 mph faster than the other, what is the rate of each rider?
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
on a monday afternoon practice, two cyclists took the same amount of time to ride 39 miles and 30 miles, respectively. if a cyclist rode 3 mph faster than the other, what is the rate of each rider?
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Assuming the faster rider went the further distance,
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Faster rider DATA:
rate = x+3 mph ; distance = 39 miles ; time = d/r = 39/(x+3) hrs
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Slower rider DATA:
rate = x mph ; distance = 30 miles ; time = d/r = 30/x hrs
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Equation:
time = time
30/x = 39/(x+3)
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30x + 90 = 39x
9x = 90
x = 10 mph (slower rider rate)
x+3 = 13 mph (faster rider rate)
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Cheers,
Stan H.
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The faster rider rode 39 mi in the same time
Let +s+ = the speed of the slower rider in mi/hr
+s+%2B+3+ = the speed of the faster rider in mi/hr
Let +t+ = time in hrs for both riders
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Equation for slower rider:
(1) +30+=+s%2At+
Equation for faster rider:
(2) +39+=+%28+s%2B3+%29%2At+
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(1) +t+=+30%2Fs+
Substitute (1) into (2)
(2) +39+=+%28+s%2B3+%29%2A%28+30%2Fs+%29+
(2) +39+=+30+%2B+90%2Fs+
(2) +90%2Fs+=+9+
(2) +9s+=+90+
(2) +s+=+10+
and
+s+%2B+3+=+13+
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Their speeds were 10 mi/hr and 13 mi/hr
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check:
(1) +30+=+10%2At+
(1) +t+=+3+ hrs
and
(2) +39+=+%28+10%2B3+%29%2At+
(2) +39+=+13t+
(2) +t+=+3+ hrs
OK