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Re: After driving to a riverfront parking lot, Bob plans to run
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22 Oct 2014, 11:33
I'm sorry but I see as 8 beign the speed, and i cant see why we don't use s=d/t...which is 8=d/50...



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Re: After driving to a riverfront parking lot, Bob plans to run
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22 Oct 2014, 11:46



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Re: After driving to a riverfront parking lot, Bob plans to run
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22 Oct 2014, 11:50
1/8. I got it! we always write speed as miles/ hr or minute, and not the other way around. Thanks Bunuel!!



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Re: After driving to a riverfront parking lot, Bob plans to run
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19 Feb 2015, 20:37
Im confused I thought 50 min was added not total time



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Re: After driving to a riverfront parking lot, Bob plans to run
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20 Sep 2015, 11:10
Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(A) 1.5 (B) 2.25 (C) 3.0 (D) 3.25 (E) 4.75 Bob runs at a rate of 8 minutes per mile. So, in 1 hour he will run 60/8 = 7.5 miles So his speed is 7.5 miles/hr Let x be the distance he further runs south before turning back. So the total distance which he must cover in 50 minutes or 50/60 hour is x+x+3.25 or 2x+3.25 Time = Distance / Speed \(\frac{50}{60} = \frac{2x+3.25}{7.5}\) x=1.5 Answer: A



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Re: After driving to a riverfront parking lot, Bob plans to run
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20 Sep 2015, 23:49
Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(A) 1.5 (B) 2.25 (C) 3.0 (D) 3.25 (E) 4.75 My answer : 1.5 Total distance Bob runs = 3.25 miles south + remaining south + remaining north + 3.25 miles north Speed = 8 mins per mile. Miles he can cover in 50 mins = 50/8 = 6.25 miles. => remaining south + remaining north + 3.25 miles north = 6.25 miles Now, remaining distance should be equally divided between north and south = > 2 remaining distance = 6.25 3.25 => remaining distance =3/2 = 1.5 km.



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Re: After driving to a riverfront parking lot, Bob plans to run
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22 Sep 2015, 14:14
rate=1/8 mile per minute time=50 minutes distance=2x+3.25 miles x=1.5 miles



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Re: After driving to a riverfront parking lot, Bob plans to run
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05 Oct 2015, 21:28
Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(A) 1.5 (B) 2.25 (C) 3.0 (D) 3.25 (E) 4.75 I saw that there was a lot of people asking questions about this one and a lot of them were issues with trying to visualize what's actually happening in the scenario. When I was working through through the 2015 OG I completed this question and found it pretty straight forward when I drew the below picture. Maybe someone will find it useful one day.
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Re: After driving to a riverfront parking lot, Bob plans to run
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21 Dec 2015, 19:27
In 50 minutes, he can do the following:
(3.25+x)(8) = 50
26+8x = 50 8x = 24 x = 3 > He can run 3 more miles. Since he has to run up and back, he can run 1.5 more miles South before he has to turn back.



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Re: After driving to a riverfront parking lot, Bob plans to run
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25 Dec 2015, 11:23
good question with lots of traps ok, so he needs to run 50 more minutes all together  that means few more miles south then return all the way back. he runs 1 mile/8 minutes. which means that in 50 minutes he will run for 6.25 miles. so the total distance he wants to cover is 9.50 miles. since the distance to north needs to be the same as distance to south, it means that in one direction, he needs to run 4.75 miles. since he already covered 3.25, he needs to run additional 1.50 miles before returning back.



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Re: After driving to a riverfront parking lot, Bob plans to run
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14 Mar 2016, 10:54
another method he ran for 3.25 miles at rate of 8 min /miles so he ran for 3.25 x 8=26 mins so the next decision in which he will run only 50 mins must have 26 mins spare for 3.25 mile return, so he is left with 24 mins with him to run ahead and come back now at rate of 8 given the distance he can travel in 24 mins is 24/8 =3 miles but it must be split in equal halves so 2d=3 there fore max distance of 1.5 miles



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Re: After driving to a riverfront parking lot, Bob plans to run
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04 May 2016, 10:07
Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(A) 1.5 (B) 2.25 (C) 3.0 (D) 3.25 (E) 4.75 Solution: We are given that Bob plans to run south along the river, turn around, and return to where he started. We can draw this out. We know that his run south (from the parking lot) and his run north (back to the parking lot) are equal in distance. We will use this information later in the solution. We are also given that Bob’s rate is 8 minutes per mile, or, in other words, (since Rate = Distance/Time) his rate is 1 mile per 8 minutes or 1/8. We are told that Bob had already run 3.25 miles south, and he wants to run for 50 minutes more. Thus, we calculate how far Bob will go in the remaining 50 minutes. Distance = Rate x Time Distance = 1/8 x 50 Distance = 50/8 = 25/4 = 6.25 miles Thus, we know that Bob’s total running distance will be 6.25 + 3.25 = 9.5 miles. Because we know the distance is THE SAME both ways, we know that each leg of his trip is 9.5/2 = 4.75 miles. Since Bob has ALREADY RUN 3.25 miles south, he can run 4.75 – 3.25 = 1.5 miles more. At that point he will have to turn around and head back north to the parking lot. Answer A
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Re: After driving to a riverfront parking lot, Bob plans to run
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27 May 2016, 08:19
An easy shortcut: let the distance from after running 3.25 till reaching south point =x. let d1=3.25, let d2=x+x+3.25=2x+3.25 since distance from south to north is the same 3.25+x
then we need to change the rate from 8 min/mile to r =mile /min using proportion: 8/1= 1/r >> 8r=1> r=1/8 mile/min
since rate is constant= 1/8 always and t2=50 min then d2=r2*t2 =1/8*50=6.25. d2= 6.25=2x+3.25 =2x=3 >> x=3/2=1.5 mile



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After driving to a riverfront parking lot, Bob plans to run
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29 May 2017, 04:08
Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(A) 1.5 (B) 2.25 (C) 3.0 (D) 3.25 (E) 4.75 1. Converting speed to miles / minute, speed =1/8 miles/minute 2. Miles run = 3.25 3. To cover 3.25 miles in the return he needs 3.25/(1/8) minutes = 26 min 4. In the remaining 24 minutes, down south is 12 min and up north is 12 minutes 5. So down south further, he can run 12*1/8 = 1.5 miles
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After driving to a riverfront parking lot, Bob plans to run
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15 Jul 2017, 23:02
I found using a table to keep track of information a bit helpful here. Though, I took more time to solve ( = Need Practice ) See the table below. Final Answer = 1.5 Answer choice A
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Re: After driving to a riverfront parking lot, Bob plans to run
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16 Jul 2017, 00:31
Speed = 1 mile per 8 mins = 7.5 mph. Distance that can be run in 50 mins = 50*7.5/60 = 6.25 miles. He can run 6.25 miles in the next 50 minutes, which includes distance that he would cover further in south and running back to the parking lot. Since he has already covered 3.25 miles south, he needs to make time from the 50 mins to cover that, additionally, he can run 1.5 miles. Ans  A.
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Re: After driving to a riverfront parking lot, Bob plans to run
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24 Dec 2017, 18:52
Can you use substitution for this question or the fast way is to solve it?



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Re: After driving to a riverfront parking lot, Bob plans to run
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08 Jan 2018, 18:59
at a rate of 8 mins per mile, it takes him 26 minutes to travel 3.25 miles. He has to travel for 50 more minutes. So his total travel time is 50+26=76 minutes. So half the way takes 38 minutes. He can travel for 12 more minutes one way. at a rate of 8 mins per mile that would be 1.5 miles.



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After driving to a riverfront parking lot, Bob plans to run
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11 Feb 2018, 04:23
Bob has \(50\) more minutes to cover \(3.25\) miles plus some distance, lets call it \(x\), twice (twice  because Bob needs to travel in both directions) At the rate of \(1/8\) miles per minute it would take Bob \(26\) minutes to cover the \(3.25\) miles back, hence subtracting that from \(50\) leaves \(24\) minutes with Bob to cover the distance \(x\) twice. Hence at \(1/8 = 2x/24\)(Speed = Distance/Time) gives \(x = 3/2 = 1.5\) miles. Correct Answer: Option \(A\)
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Re: After driving to a riverfront parking lot, Bob plans to run
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11 Feb 2018, 08:30
khairilthegreat wrote: Bob has only 50 minutes to complete the whole run. With rate of 8 minutes/mile the distance he can cover is 50/8 mile.
We can picture the path as follow
Running south <3.25 miles>< d = ? >
Running north back to origin <3.25 miles>< d >
From this we can conclude if he want to go back to his first point the distance would be d + d + 3.25.
Because he only has 50/8 mile left, we can make equation as follows d+d+3.25 = 50/8; 2d+3.25 = 6.25 > d=1.5




Re: After driving to a riverfront parking lot, Bob plans to run
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