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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?
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10 Apr 2017, 08:37
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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.
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
Ann and Bea leave Townville at the same time and travel [#permalink]
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13 Apr 2017, 15:24
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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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04 Sep 2017, 20:20
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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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04 Sep 2017, 20:42
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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
Re: Ann and Bea leave Townville at the same time and travel [#permalink]
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06 Sep 2017, 03:38
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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
gmatclubot
Re: Ann and Bea leave Townville at the same time and travel
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06 Sep 2017, 03:38