Question 633948: Question: Mary travels from the city to visit her sister in the country. On the way, she travels 55km/hr. She stays with her sister for 20 hours before starting home. On the return trip, she travels 45 km/hr. She returned home a grand total of 41 hours after she left. How far away, in kilometers, is Mary's house from her sister's?
What I've managed to reason:
41 hours total-20 hours visiting=21 hours traveling
From there, I'm clueless. I've no idea how to start building the equation for this. Help, please?
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Mary travels from the city to visit her sister in the country. On the way, she travels 55km/hr. She stays with her sister for 20 hours before starting home. On the return trip, she travels 45 km/hr. She returned home a grand total of 41 hours after she left. How far away, in kilometers, is Mary's house from her sister's?
What I've managed to reason:
41 hours total-20 hours visiting=21 hours traveling
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The avg speed of a round trip = 2*55*45/(55+45) km/hr
= 49.5 km/hr
rt distance = r*t = 49.5*21 km
Each way = 49.5*21/2 = 519.75 km
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The avg speed of a round trip = 2*55*45/(55+45) km/hr
You can trust me on that, or I can prove it.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Mary travels from the city to visit her sister in the country. On the way, she travels 55km/hr. She stays with her sister for 20 hours before starting home. On the return trip, she travels 45 km/hr. She returned home a grand total of 41 hours after she left. How far away, in kilometers, is Mary's house from her sister's?
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City to Country DATA:
rate = 55 kmh ; time = x hrs ; distance = 55x km
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Country to Cith DATA:
rate = 45 kmh ; time = (21-x) hrs ; distance = 45(21-x) km
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Equation:
distance = distance
55x = 45(21-x)
55x = 45*21 - 45x
100x = 45*21
x = 9.45 hrs (time going city to country)
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Distance = 55x = 55*9.45 = 519.75 km
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Cheers,
Stan H.
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