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houston1980
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

From Bayville to Cabot:

distance = rate × time

distance = 50 × t

From Ashland to Bayville:

distance = 40 × (t + 3)

Since the trip from Ashland to Bayville was 90 miles longer than the trip from Bayville to Cabot, we have:

40(t + 3) = 50t + 90

40t + 120 = 50t + 90

30 = 10t

t = 3 hours

Therefore, the trip from Ashland to Bayville took:

t + 3 = 3 + 3 = 6 hours

Answer: B
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let B to C distance be x and time be t
x+90 = 40 * ( t+3)
x=50*t
solve for t
50t+90 = 90t +120
t = 3
t+3 ; 6 hrs
option B


houston1980
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
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­Bread and butter distance problem:

­
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houston1980
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------40mph------B------50mph------C

A to B at every hour intervals at 40miles/hr
40,80,120,160,200,240
B to C at every hour intervals 50miles /hr
50,100,150

When Yuriko drives A to B for 6 hours, she covers 240
And when she drives B to C for 3 hours, she covers 150.

Thats 90 miles difference and 3 hours difference.
Thats the answer, 6hrs is what she travels for from A to B
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Time ratio = 5:4 for a given distance x (between B to C)
Time taken to cover 90 miles= 90/40= 2.25 hours
Extra time taken to cover A to B compared to B to C= 3 hours
Thus time to cover distance x between A to B= t+3-2.25= t+ 0.75
Thus 5:4= (t+0.75)/t. Solving for t=3 hours
Thus time taken to cover A to B = 3+3=6 hours
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