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

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) (Show Source): You can put this solution on YOUR website!
7:30 to 8:40 is 1 hr 10 min
This is hrs
Let = time spent cycling in hrs
= time spent walking
Let = distance cycled in km
= distance walked in km
------------------------------
Equation for cycling:
(1)
Equation for walking:
(2)
-------------------------
Substitute (1) into (2)
(2)
(2)
(2)
(2)
(2)
and
(1)
(1)
(1)
and

He pushes his bicycle ( walks ) 2 km
-----------------------
check:
(1)
(1)
(1)
and
(2)
(2)
(2)
(2)
(2)
(2)
OK
Answer by stanbon(75887) (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?
------
1st leg DATA:
distance = x km ; rate = 10 km/hr ; time = x/10 hrs
----
2nd leg DATA:
distance = 7-x km ; rate = 3 km/hr ; time = (7-x)/3 hrs
-----
Equation:
time + time = (7/6)hr
-----
x/10 + (7-x)/3 = 7/6
-----
3*6x + 6*10*7 - 6*10x = 10*3*7
-------
18x + 420 - 60x = 210
------
-42x = -210
-----
x = 5 (cycle rate)
7-x = 2 (walk rate)
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Cheers,
Stan H.
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