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Tricky - Ann and Bea travel to Villageton

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by Brent@GMATPrepNow » Mon Apr 10, 2017 8:12 am
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

Answer: E

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by DavidG@VeritasPrep » Mon Apr 10, 2017 9:39 am
Brent@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

Answer: E

Source: www.gmatprepnow.com
Difficulty level: 700
Brent original!

Let's pick some nice numbers. Say k = 10, so they're 2k, or 20 miles apart to start.

Let's say that Ann's speed is 20 mph and Bea's speed is 5mph. After 1 hour, Ann has covered the 20 miles to Villageton and Bea has gone 5 miles. So now they're 15 miles apart. If Ann turns around, the two will be moving in opposite directions, and so we can sum their rates. Thus, they're moving towards each other at 20 + 5 = 25mph. So how long will it take to go 15 miles at 25 mph?

R * T = D
25 * T = 15
T = 15/25 = 3/5.

In 3/5 of an hour, Bea will go an additional 5 * (3/5) = 3 miles. She'd already gone 5 miles, so she'll cover a total of 5 + 3 = 8 miles. Now just plug 10 in for k and see what spits out 8. E will do the trick: (4*10/5 = 8.)
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by Brent@GMATPrepNow » Tue Apr 11, 2017 6:25 am
Brent@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

Answer: E

Source: www.gmatprepnow.com
Difficulty level: 700
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: E

Cheers,
Brent
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