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Yuriko drove from Ashland to Bayville at an average rate of 40 miles
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29 Jun 2023, 23:28
29 Jun 2023, 23:28
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Yuriko drove from Ashland to Bayville at an average rate of 40 miles per hour and from Bayville to Cabot at an average rate of 50 miles per hour. If Yuriko’s trip from Ashland to Bayville was 90 miles longer and took 3 hours more than her trip from Bayville to Cabot, how many hours did her trip from Ashland to Bayville take?
A. 5
B. 6
C. 7
D. 9
E. 10
A. 5
B. 6
C. 7
D. 9
E. 10
Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles
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02 Jul 2023, 18:14
02 Jul 2023, 18:14
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houston1980 wrote:
Yuriko drove from Ashland to Bayville at an average rate of 40 miles per hour and from Bayville to Cabot at an average rate of 50 miles per hour. If Yuriko’s trip from Ashland to Bayville was 90 miles longer and took 3 hours more than her trip from Bayville to Cabot, how many hours did her trip from Ashland to Bayville take?
A. 5
B. 6
C. 7
D. 9
E. 10
A. 5
B. 6
C. 7
D. 9
E. 10
A faster way to solve is:
distance = rate * time
(1) x = 50h
(2) x + 90 = 40(h+3)
sub in x=50h into (2):
50h + 90 = 40h + 120
10h = 30
h = 3
Total time for location A to location B is h+3
total time = 3 + 3 = 6
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Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles
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30 Jun 2023, 00:09
30 Jun 2023, 00:09
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houston1980 wrote:
Yuriko drove from Ashland to Bayville at an average rate of 40 miles per hour and from Bayville to Cabot at an average rate of 50 miles per hour. If Yuriko’s trip from Ashland to Bayville was 90 miles longer and took 3 hours more than her trip from Bayville to Cabot, how many hours did her trip from Ashland to Bayville take?
A. 5
B. 6
C. 7
D. 9
E. 10
A. 5
B. 6
C. 7
D. 9
E. 10
Let the distance between Bayville and Cabolt = \(x\)
The distance between Ashland and Bayville = \(x + 90\)
Time taken from Ashland to Bayville = \(\frac{x+90 }{ 40}\)
Time taken from Bayville to Cabot = \(\frac{x}{50}\)
\(\frac{x+90}{40} - \frac{x}{50} = 3\)
50x + 4500 - 40x = 6000
10x = 1500
x = 150
Distance from between Ashland and Bayville = \(150 + 90 = 240\)
Speed between Ashland and Bayville = 40
Time taken = \(\frac{240}{40} = 6\) hours
Option B
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