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# On his trip from Alba to Benton, Julio drove the first x

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Joined: 25 Oct 2010
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On his trip from Alba to Benton, Julio drove the first x  [#permalink]

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13 Nov 2010, 12:15
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Question Stats:

68% (02:11) correct 32% (02:09) wrong based on 449 sessions

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On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first x miles?

(1) On this trip, Julio drove for a total of 10 hours and drove a total of 530 miles.
(2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance.
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13 Nov 2010, 12:29
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On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first X miles?

(1) On this trip, Julio drove for a total of 10 hours and a total of 530 miles --> $$total \ time=10=\frac{x}{50}+\frac{530-x}{60}$$ --> we have the linear equation with one unknown, so we can solve for $$x$$. Sufficient.

(2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance --> $$\frac{x}{50}=\frac{y}{60}+4$$, where $$y$$ is the remaining distance --> we have the linear equation with two unknowns, so we can not solve for $$x$$. Not sufficient.

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23 Nov 2010, 23:24
Bunuel wrote:
On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first X miles?

(1) On this trip, Julio drove for a total of 10 hours and a total of 530 miles --> $$total \ time=10=\frac{x}{50}+\frac{530-x}{60}$$ --> we have the linear equation with one unknown, so we can solve for $$x$$. Sufficient.

(2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance --> $$\frac{x}{50}=\frac{y}{60}+4$$, where $$y$$ is the remaining distance --> we have the linear equation with two unknowns, so we can not solve for $$x$$. Not sufficient.

i've got sufficient with (1)
However with (2), we are told that : "it took Julio 4 more hours to drive the first x miles than to drive the remaining distance " can we have the fomular as folow:
r1 = 50m/h
r2 = 60m/h
1 hour: we have the difference miles between the first x mile and the remaining = 10m
=> with 4 hours: we can find the distance of the remaining distance = x + (10m/h x 4h) = x + 40miles
=> we have the equation with one unknown as we have solved the remaining distance.
Please tell me where i did get the wrong refers?
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24 Nov 2010, 02:20
MICKEYXITIN wrote:
Bunuel wrote:
On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first X miles?

(1) On this trip, Julio drove for a total of 10 hours and a total of 530 miles --> $$total \ time=10=\frac{x}{50}+\frac{530-x}{60}$$ --> we have the linear equation with one unknown, so we can solve for $$x$$. Sufficient.

(2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance --> $$\frac{x}{50}=\frac{y}{60}+4$$, where $$y$$ is the remaining distance --> we have the linear equation with two unknowns, so we can not solve for $$x$$. Not sufficient.

i've got sufficient with (1)
However with (2), we are told that : "it took Julio 4 more hours to drive the first x miles than to drive the remaining distance " can we have the fomular as folow:
r1 = 50m/h
r2 = 60m/h
1 hour: we have the difference miles between the first x mile and the remaining = 10m
=> with 4 hours: we can find the distance of the remaining distance = x + (10m/h x 4h) = x + 40miles
=> we have the equation with one unknown as we have solved the remaining distance.
Please tell me where i did get the wrong refers?

Remaining distance, $$y$$, won't be $$x+40$$. If $$t$$ is the time to cover $$y$$ then the time to cover $$x$$ will be $$t+4$$ then $$x=(t+4)*50=50t+200$$.
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25 Nov 2010, 10:09
Bunuel wrote:
MICKEYXITIN wrote:
i've got sufficient with (1)
However with (2), we are told that : "it took Julio 4 more hours to drive the first x miles than to drive the remaining distance " can we have the fomular as folow:
r1 = 50m/h
r2 = 60m/h
1 hour: we have the difference miles between the first x mile and the remaining = 10m
=> with 4 hours: we can find the distance of the remaining distance = x + (10m/h x 4h) = x + 40miles
=> we have the equation with one unknown as we have solved the remaining distance.
Please tell me where i did get the wrong refers?

Remaining distance, $$y$$, won't be $$x+40$$. If $$t$$ is the time to cover $$y$$ then the time to cover $$x$$ will be $$t+4$$ then $$x=(t+4)*50=50t+200$$.

thanks Bunuel, how foolish i am
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Re: On his trip from Alba to Benton, Julio drove the first x  [#permalink]

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19 Jun 2012, 01:58
1
On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first x miles?
1) On this trip, Julio drove for a total of 10 hours and drove a total of 530 miles.
2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance.

From the statement :

If the total distance is D.

(x/50) + + (D-x)/60 = Total Time

Now from (1), (x/50) + + (D-x)/60 =10
and D = 530
From the above two equation x can be calculated and so x/50. Hence sufficient

From (2), x/50 + 4 = (D-x)/50
One equation and two unknown ,Hence impossible to find out x. ( Not sufficient)

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Re: On his trip from Alba to Benton, Julio drove the first x  [#permalink]

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01 Aug 2013, 15:02
On his trip from Alba to Benton, Julio drove the first x miles at an average rate of 50 miles per hour and the remaining distance at an average rate of 60 miles per hour. How long did it take Julio to drive the first x miles?

average time = t1+t2
time = distance/rate
average time = (y/50) + ([x-y]/60)

(1) On this trip, Julio drove for a total of 10 hours and drove a total of 530 miles.
average time = (y/50) + ([x-y]/60)
10 = (y/50) + ([530-y]/60)
600 = (60y/50) + 530-y
30000 = 60y + 26500 - 50y
3500 = 10y
y = 350

The first portion (t1) of his trip = y/50. Therefore, t1 = (350/50) = 7.
The first portion of his trip took 7 hours.
SUFFICIENT

(2) On this trip, it took Julio 4 more hours to drive the first x miles than to drive the remaining distance.
average time = (y/50) + ([x-y]/60)
We know nothing about how long he spent on the road in total or the number of miles he covered.
The equation would look something like this: (y/50) = [(x-y]/60) + 4 where y/50 represents the first leg of the trip that took four hours more than the second leg. We know nothing about x or y.
INSUFFICIENT

(A)
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Re: On his trip from Alba to Benton, Julio drove the first x  [#permalink]

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13 Sep 2019, 21:29
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Re: On his trip from Alba to Benton, Julio drove the first x   [#permalink] 13 Sep 2019, 21:29
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