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Frank and Georgia started traveling from A to B at the same time. Geo

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Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 28 Oct 2014, 10:40
10
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

64% (02:44) correct 36% (02:54) wrong based on 184 sessions

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Frank and Georgia started traveling from A to B at the same time. Georgia’s constant speed was 1.5 times Frank’s constant speed. When Georgia arrived at B, she turned right around and returned by the same route. She cross paths with Frank, who still was coming toward B, when they were 60 miles away from B. How far away are A and B?
(A) 72 mi
(B) 120 mi
(C) 144 mi
(D) 240 mi
(E) 300 mi


For a few more challenging motion questions, as well as the OE for this question, see:
https://magoosh.com/gmat/2014/gmat-prac ... on-motion/

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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 28 Oct 2014, 23:28
Let Sg= Speed of georgia, Sf=Speed of frank.

Since their speeds are constant and in the ratio as, Sg=1.5 * Sf= 3/2 * Sf.

Now Time=Dist/Speed. Let D be distance between A & B.
Since they meet 60miles from end B. So Georgia has travelled distance=D+60 and Frank is short of speed so she traveled distance=D-60.
Equating,

Tg=Tf................. put Sg=1.5 * Sf

[D+60][/1.5 * Sf]=[D-60][/Sf]

Solving we get D=300miles.
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Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 28 Oct 2014, 23:33
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Answer = E = 300

Let the distance between point A & Point B = d

Meeting point M is 60 miles away from B, so distance between M & A = d-60

Distance travelled by Frank = d-60

Lets say speed of Frank = s

Distance travelled by Georgia = d+60

Speed of Georgia = 1.5s

Attachment:
dis.png
dis.png [ 5.28 KiB | Viewed 4170 times ]


Time taken by Georgia to reach point M (After complete trip) = Time taken by Franck to reach point M

Setting up the equation

\(\frac{d-60}{s} = \frac{d+60}{1.5s}\)

d = 300
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 26 Jan 2015, 10:14
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Another way to think of the problem. Georgia travels 1.5 times further than Frank does based on her speed (and the fact they have travelled for the same amount of time).

Given where they meet, Georgia has travelled 120 miles further (60 miles further towards point B, and 60 miles on the way back from point B).

Putting this information together, 120 must be equal to 0.5 * Frank's distance travelled, and so Frank has travelled 240 miles. He has 60 miles further to go in order to reach point B, so the distance from A to B = 300 miles.
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Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 03 Aug 2016, 06:46
Yet another, simple way to think about this problem.

Since Georgia first went \(D\) miles, and then looped around and covered the distance that Frank had remaining, you know that together they traveled \(2D\) miles.

Together, they were traveling at \(f + 1.5 f\)=\(2.5 f\) miles/hour.

So, the total time taken by both of them to travel \(2D\) miles at \(2.5f\) is \(\frac{2D}{2.5f}\) or \(\frac{4D}{5f} hours.\).

Frank, traveling at f mph, goes f * \(\frac{4D}{5f}\) = \(\frac{4D}{5} miles.\). Therefore, he has \(\frac{D}{5}\) miles to go. Since we know that value is 60 miles, \(D\) must equal 300 miles.

Answer: E.
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 03 Aug 2016, 07:10
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mikemcgarry wrote:
Frank and Georgia started traveling from A to B at the same time. Georgia’s constant speed was 1.5 times Frank’s constant speed. When Georgia arrived at B, she turned right around and returned by the same route. She cross paths with Frank, who still was coming toward B, when they were 60 miles away from B. How far away are A and B?
(A) 72 mi
(B) 120 mi
(C) 144 mi
(D) 240 mi
(E) 300 mi


For a few more challenging motion questions, as well as the OE for this question, see:
https://magoosh.com/gmat/2014/gmat-prac ... on-motion/

Mike :-)


Let the total distance=x
Distance covered by G = x+60
Distance covered by F= x-60

Since the time required to cover above distance is same

Tg=Tf
x+60/1.5s = x-60/s
1.5x -90= x+60
.5x= 150
x= 300

E is the answer
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 16 Jan 2017, 23:37
In my vision there is a problem in this question. When we derive the formula D-60=RT and D+60=3/2*RT and substitute the RT by D-60, the answer is 300. But, if we subtract the one equation from the other, the answer will become 240. I checked the answer in another way and now that 300 is correct, but, can you explain why we should substitute the values rather working directly with two equations? Thx
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 17 Jan 2017, 01:07
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aykhan1991 wrote:
In my vision there is a problem in this question. When we derive the formula D-60=RT and D+60=3/2*RT and substitute the RT by D-60, the answer is 300. But, if we subtract the one equation from the other, the answer will become 240. I checked the answer in another way and now that 300 is correct, but, can you explain why we should substitute the values rather working directly with two equations? Thx


Solving D-60=RT and D+60=3/2*RT gives only one value of D, 300 (no matter how you solve). How did you get 240?
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 17 Jan 2017, 01:28
Sorry Sorry, in second solution forgot to add back 60, right, the answer is 300
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 17 Jan 2017, 01:51
aykhan1991 wrote:
Sorry Sorry, in second solution forgot to add back 60, right, the answer is 300


No problem. In future though, if you have a similar doubt it's better to show your work to be able to point out an error right away.
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Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 13 Feb 2018, 07:19
Let \(D_G =\) distance traveled by Georgia and \(D_F =\) distance traveled by Frank

At the point where Frank and Georgia meet, the distance traveled by both of them can be represented by below equation:

\(D_G=D_F+120\) (Georgia has traveled from A to B and is 60 miles away from B when Frank meets, which means Georgia has traveled 60+60 miles in excess to the distance traveled by Frank)

\(S_G*t=S_F*t+120\)
but \(S_G=1.5*S_F\)
\(therefore, 1.5*S_F*t=S_F*t+120\)
\(0.5*S_F*t=120\)
\(S_F*t=\frac{120}{0.5}\)
\(S_F*t=240\)
\(D_F=240\)
\(D_G=240+120=360\)

\(D_{AB}= D_G-60 = 360-60 = 300\)
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Re: Frank and Georgia started traveling from A to B at the same time. Geo  [#permalink]

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New post 15 Feb 2019, 18:29
mikemcgarry wrote:
Frank and Georgia started traveling from A to B at the same time. Georgia’s constant speed was 1.5 times Frank’s constant speed. When Georgia arrived at B, she turned right around and returned by the same route. She cross paths with Frank, who still was coming toward B, when they were 60 miles away from B. How far away are A and B?
(A) 72 mi
(B) 120 mi
(C) 144 mi
(D) 240 mi
(E) 300 mi




We can let r = Frank’s speed, and thus 1.5r = Georgia’s speed. Furthermore, we can let d = the distance between A and B and t = the time Frank has traveled from A to B when he’s still 60 miles from B. Therefore, we can create the following equations:

rt + 60 = d (since Frank has 60 miles more to go to reach B)

and

(1.5r)t - rt = 120 (since when Georgia meets Frank, she has traveled 60 x 2 = 120 miles more than
Frank)

Simplifying the last equation, we have:

0.5rt = 120

rt = 240

Substituting this into the first equation, we have:

240 + 60 = d

d = 300

Answer: E
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Re: Frank and Georgia started traveling from A to B at the same time. Geo   [#permalink] 15 Feb 2019, 18:29
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