Question 1099270: Betty and Carl live in the same apartment building, work at the same office, and set off for work in the morning at the same time. Betty travels by car at 60 km/h, parks at the car park, then walks at 4 km/h the rest of the way to work. Carl takes the bus, which travels at 40 km/h, to the same car park as Betty uses, and then rides his bike the rest of the way at 12 km/h. If they both leave home at 6 a.m., at what time do they arrive at work?
Found 2 solutions by jorel1380, ikleyn:
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! To answer this, we must assume that Betty and Carl both leave home at the same time, and arrive to work at the same time. Let m and n be the distances between apartments to car park, and car park to office, respectively. Then:
m km/60(km/h)+n km/4(km/h)=m km/40(km/h)+n km/12(km/h)
2m km + 30n km=3m km+10n km
m km=20n km
m=20n
Let n=2. Then m=40.
40 km/60 (km/h)=2/3 hr=40 min
2 km/4 (km/h)=1/2 hr=30 min
40+30=70
40 km/40 (km/h)=1 hr=60 min
2 km/12 (km/h)=1/6 hr=10 min
60+10=70 min
If they both leave at 6 a.m., they both get to work at 7:10 a.m.
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Answer by ikleyn(52898)  (Show Source):
You can put this solution on YOUR website! .
The given problem has INFINITELY many solutions, and the condition DOES NOT ALLOW to get a unique solution.
(The condition gives ONLY ONE EQUATION for two unknowns).
In this sense the condition is DEFECTIVE.
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