Question 1197382: Every day, Jane and Bill both drive to a parking lot, and then take a subway to get to work. Jane drives 30 km at 80 km/h, then takes a northbound subway that averages 25 km/h for 15 km while Bill drives 60 km at 100 km/h and then takes a southbound subway that averages 35 km/h for 15 km. It takes each of them 12 minutes to change from car to subway, and they arrive at the same destination at 8:30 a.m. At what time, to the nearest minute, does the earliest of them leave home?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
kph = kilometers per hour
Jane drives 30 km at 80 kph.
So,
distance = rate*time
time = distance/rate
time = (30 km)/(80 kph)
time = 3/8 hour
She spent 3/8 of an hour driving to the parking lot.
Then she takes 12 minutes, aka 12/60 = 1/5 of an hour, to go from her car to the subway.
So far the duration is (3/8)+(1/5) = (15/40)+(8/40) = 23/40 of an hour.
Then she takes a subway which travels at a rate of 25 kph and travels a distance of 15 km
time = distance/rate
time = (15 km)/(25 kph)
time = 3/5 hour
Add this to the duration
(23/40)+(3/5) = (23/40)+(24/40) = 47/40
Put another way
3/8 + 1/5 + 3/5 = 47/40
Her total duration is 47/40 of an hour.
1/40 of an hour is (1/40)*60 = 1.5 minutes exactly.
47/40 of an hour is 47*(1/40 hr) = 47*(1.5 min) = 70.5 minutes exactly
Round up to the nearest minute to get 71 minutes.
71 min = 60 min + 11 min
The total duration is 1 hr + 11 min when rounding to the nearest minute.
She arrives at her destination at 8:30 AM
Subtract off 1 hour to get to 7:30 AM
Subtract off a further 11 minutes to get to 30-11 = 19
I.e. 7:30 AM rewinds back to 7:19 AM after subtracting off those additional 11 minutes.
Jane needs to leave her home at 7:19 AM
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Let's move our attention to Bill.
He travels 60 km at 100 kph when driving.
time = distance/rate
time = (60 km)/(100 kph)
time = 3/5 hour
He takes 12 min = 1/5 hour going from his car to the subway
His subway travels 15 km at a speed of 35 kph
time = distance/rate
time = (15 km)/(35 kph)
time = 3/7 hour
The three durations of each sub-interval are
3/5 hour ... home to parking lot
1/5 hour ... car to subway
3/7 hour ... subway to work
Add up those fractions to get the total duration he travels
3/5+1/5+3/7
4/5+3/7
28/35+15/35
43/35
Multiply by 60 to convert this to purely minutes only
43/35 hr = (43/35)*60 = 73.714 min approximately
Let's round this up to the nearest minute to get 74 minutes.
74 min = 60 min + 14 min
74 min = 1 hr + 14 min
Start at 8:30 AM
Subtract off 1 hour to get to 7:30 AM
Subtract off another 14 min to get to 7:16 AM
Or you could note the gap between durations 71 minutes (Jane) and 74 minutes (Bill) is 74-71 = 3 minutes.
Therefore, Bill needs to leave 3 minutes earlier than Jane does.
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Answers:
Jane = 7:19 AM
Bill = 7:16 AM
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