SOLUTION: A cyclist leaves home at 7:30 am to cycle to school 7km away. He cycles at 10km/h until he gets a flat tire and then walks the rest of the way at 3km/h. He arrives at school at 8

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A cyclist leaves home at 7:30 am to cycle to school 7km away. He cycles at 10km/h until he gets a flat tire and then walks the rest of the way at 3km/h. He arrives at school at 8      Log On


   

Question 1010582: A cyclist leaves home at 7:30 am to cycle to school 7km away. He cycles at 10km/h until he gets a flat tire and then walks the rest of the way at 3km/h. He arrives at school at 8:40 am. How far did he have to push his bicycle?
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
7:30 to 8:40 is 1 hr 10 min
This is +7%2F6+ hrs
Let +t+ = time spent cycling in hrs
+7%2F6+-+t+ = time spent walking
Let +d+ = distance cycled in km
+7+-+d+ = distance walked in km
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Equation for cycling:
(1) +d+=+10t+
Equation for walking:
(2) +7+-+d+=+3%2A%28+7%2F6+-+t+%29+
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Substitute (1) into (2)
(2) +7+-+10t+=+3%2A%28+7%2F6+-+t+%29+
(2) +7+-+10t+=+7%2F2+-+3t+
(2) +14+-+20t+=+7+-+6t+
(2) +14t+=+7+
(2) +t+=+1%2F2+
and
(1) +d+=+10t+
(1) +d+=+10%2A%281%2F2%29+
(1) +d+=+5+
and
+7+-+d+=+7+-+5+
+7+-+d+=+2+
He pushes his bicycle ( walks ) 2 km
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check:
(1) +d+=+10t+
(1) +5+=+10t+
(1) +t+=+1%2F2+
and
(2) +7+-+d+=+3%2A%28+7%2F6+-+t+%29+
(2) +7+-+5+=+3%2A%28+7%2F6+-+t+%29+
(2) +2+=+7%2F2+-+3t+
(2) +4+=+7+-+6t+
(2) +6t+=+3+
(2) +t+=+1%2F2+
OK

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A cyclist leaves home at 7:30 am to cycle to school 7km away. He cycles at 10km/h until he gets a flat tire and then walks the rest of the way at 3km/h. He arrives at school at 8:40 am. How far did he have to push his bicycle?
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1st leg DATA:
distance = x km ; rate = 10 km/hr ; time = x/10 hrs
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2nd leg DATA:
distance = 7-x km ; rate = 3 km/hr ; time = (7-x)/3 hrs
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Equation:
time + time = (7/6)hr
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x/10 + (7-x)/3 = 7/6
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3*6x + 6*10*7 - 6*10x = 10*3*7
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18x + 420 - 60x = 210
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-42x = -210
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x = 5 (cycle rate)
7-x = 2 (walk rate)
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Cheers,
Stan H.
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