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Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles [#permalink]
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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


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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Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles [#permalink]
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 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
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Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles [#permalink]
Expert Reply
­Bread and butter distance problem:

­
GMAT Club Bot
Re: Yuriko drove from Ashland to Bayville at an average rate of 40 miles [#permalink]
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