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Ann and Bea leave Townville at the same time and travel

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Ann and Bea leave Townville at the same time and travel [#permalink]

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Question Stats:

62% (02:52) correct 38% (02:35) wrong based on 149 sessions

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Ann and Bea leave Townville at the same time and travel towards Villageton, which is 2K kilometers away. Their individual speeds are constant, but Ann’s speed is four times Bea’s speed. Upon reaching Villageton, Ann immediately turns around and drives toward Townville until she meets Bea. When they meet, how many kilometers has Bea traveled?

A) K/5
B) K/4
C) 2K/3
D) 3K/4
E) 4K/5

* Kudos for all correct solutions
[Reveal] Spoiler: OA

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Re: Ann and Bea leave Townville at the same time and travel [#permalink]

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GMATPrepNow wrote:
Ann and Bea leave Townville at the same time and travel towards Villageton, which is 2K kilometers away. Their individual speeds are constant, but Ann’s speed is four times Bea’s speed. Upon reaching Villageton, Ann immediately turns around and drives toward Townville until she meets Bea. When they meet, how many kilometers has Bea traveled?

A) K/5
B) K/4
C) 2K/3
D) 3K/4
E) 4K/5

* Kudos for all correct solutions


Hi

Distance travelled by Ann on her journey back - c.

Bea's speed - v

Ann's speed - 4v

(2k + c) / 4v = (2k - c) / v

5vc = 6kv

c = 6k/5

Distance travlled by Bea: 2k - 6k/5 = 4k/5

Answer E.

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Re: Ann and Bea leave Townville at the same time and travel [#permalink]

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New post 11 Apr 2017, 07:27
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GMATPrepNow wrote:
Ann and Bea leave Townville at the same time and travel towards Villageton, which is 2K kilometers away. Their individual speeds are constant, but Ann’s speed is four times Bea’s speed. Upon reaching Villageton, Ann immediately turns around and drives toward Townville until she meets Bea. When they meet, how many kilometers has Bea traveled?

A) K/5
B) K/4
C) 2K/3
D) 3K/4
E) 4K/5

* Kudos for all correct solutions


David has nicely demonstrated the input-output approach for this question.
Here's an algebraic solution:

Let B = the distance Bea traveled
Let R = Bea's speed.


NOTE: the total distance from Townville to Villageton and then BACK TO Townville = 4K.

So, 4K - B = the distance Ann traveled
And 4R = Ann's speed (since her speed is 4 times Bea's speed)


From here, let's create a WORD EQUATION that uses distance and speed.
How about: Ann's travel time = Bea's travel time

Time = distance/rate, so we get:
(4K - B)/4R = B/R
Cross multiply to get: (B)(4R) = (R)(4K - B)
Expand: 4BR = 4RK - BR
Add BR to both sides: 5BR = 4RK
Divide both sides by R to get: 5B = 4K
Divide both sides by 5 to get: B = 4K/5

Answer:
[Reveal] Spoiler:
E


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Re: Ann and Bea leave Townville at the same time and travel [#permalink]

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New post 13 Apr 2017, 14:46
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Time is the same, so plug in 1:

Ann: 4R = 2k + x
Bea: R = 2k - x

Add equations: 5R = 4k, R = 4k/5, R x T = D, D = (4k/5) x 1 = 4k/5.

Answer: E

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Ann and Bea leave Townville at the same time and travel [#permalink]

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GMATPrepNow wrote:
Ann and Bea leave Townville at the same time and travel towards Villageton, which is 2K kilometers away. Their individual speeds are constant, but Ann’s speed is four times Bea’s speed. Upon reaching Villageton, Ann immediately turns around and drives toward Townville until she meets Bea. When they meet, how many kilometers has Bea traveled?

A) K/5
B) K/4
C) 2K/3
D) 3K/4
E) 4K/5

* Kudos for all correct solutions


when A reaches V, B has traveled 1/4*2K=K/2 kilometers
when A meets B, B has traveled an additional 1/5*3/4*2K=3K/10 kilometers
K/2+3K/10=4K/5 kilometers
E

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Re: Ann and Bea leave Townville at the same time and travel [#permalink]

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New post 04 Sep 2017, 21:20
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Let K = 10
So total distance between two towns is 2*10 = 20
Let speed of Ann = 4 km/h
Then speed of Bea = 1 km/h (Ann's speed is 4 times of Bea)
Ann will take \(\frac{20}{4}\)= 5 hrs to reach town V
In that time Bea will travel 1*5 = 5 kms
Now both are travelling towards each other so total distance between them is 15 (total 20 minus 5 already traveled by Bea)
While traveling towards each other their relative speed = 5 km/hr (4 km/hr + 1 km/hr)
They will take 3 hrs to meet, \(\frac{15}{5}=3\)
In 3 hrs Bea will further cover 3 more kilometers.
So total distance traveled by Bea = 5 + 3 = 8
Now using K=10 in all option
Option E gives value 8
Thus option E
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New post 04 Sep 2017, 21:42
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let 2k be 100
then Ann did 100 and Bea did 25 (4 times less)
Ann turned around and coming on Bea did 60 and Bea did 15 ( 4 times less
So total Bea did 40
if 2k - 100
then x - 40
80k/100 = 4k/5

easy-peasy

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Re: Ann and Bea leave Townville at the same time and travel [#permalink]

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New post 06 Sep 2017, 04:38
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GMATPrepNow wrote:
Ann and Bea leave Townville at the same time and travel towards Villageton, which is 2K kilometers away. Their individual speeds are constant, but Ann’s speed is four times Bea’s speed. Upon reaching Villageton, Ann immediately turns around and drives toward Townville until she meets Bea. When they meet, how many kilometers has Bea traveled?

A) K/5
B) K/4
C) 2K/3
D) 3K/4
E) 4K/5

* Kudos for all correct solutions

Given the distance is 2k.
Lets assume Beas' speed =v.
Ann's speed =4v

Distance travelled by Ann = 2k+V.
Distance travelled by Bea = 2k-V.
(2k+V)/4v= (2k-V)/v

8k-4V= 2k+V

V= 6k/5.

So Distance by Bea = 4/5 k.

Ans:E

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Re: Ann and Bea leave Townville at the same time and travel   [#permalink] 06 Sep 2017, 04:38
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Ann and Bea leave Townville at the same time and travel

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