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Frank and Claire kiss problem

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Frank and Claire kiss problem

by saintforlife » Tue Apr 30, 2013 12:47 am
Frank and Claire are a couple and are meeting at the SFO airport after months of being apart due to Frank's work in Sudan for the House of Cards non-profit organization called the 'Clean Water Initiative'.

Claire is at the waiting area in Terminal 1 when Frank's flight lands. When Frank walks out of baggage claim he immediately spots Claire and they both start running towards each other. 500 feet of airport terminal separate the two. How far does Frank travel before he gets close enough to Claire to be able to hug her?

1. Frank runs 50% faster than Claire
2. Frank runs 4 ft/sec faster than Claire


[spoiler]Answer: A[/spoiler]
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Source: — Data Sufficiency |
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by killerdrummer » Tue Apr 30, 2013 2:55 am
Information :
they both start running towards each other.
500 feet of airport terminal separate the two.
How far does Frank travel before he gets close enough to Claire to be able to hug her?


How far does Frank travel before he gets close enough to Claire to be able to hug her?

Given :
Initial Distance = 500
Distance when the HUG = 0

To find :
distance run by frank
Question type : Value

Assume:
Frank's speed = f and Claire's Speed = c
Running in opposite direction = speed gets added = f+c


1. Frank runs 50% faster than Claire f = (1.5)c


Time =t = Dist/ (relative speed)

t = 500/(f+c)
Plug in the value of f

t = 500/(2.5)c = 200/c

IN THIS TIME frank runs = (frank's speed )*(time) = f*200 /c

as given f = (1.5)c put the value of f

(1.5)c *200/c = (1.5) *200

frank runs 300 m

Sufficient

2. Frank runs 4 ft/sec faster than Claire


As we do not know Claire's Speed, different speed CAN GIVE US MULTIPLE values OF distance traveeled by Frank.

Insufficient [/b]
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by srcc25anu » Tue Apr 30, 2013 3:39 am
St1: Frank runs 50% faster than Claire. Let franks speed be 1.5x and Claire Speed be x
distance covered will be directly proportional to ratio of speeds
Ratio of distances covered by F:C = 1.5:1 or 3:2
Frank must be covering (3 * 500)/5 = 300 ft and Claire must be covering 200 ft
Hence SUFFICIENT

St2: Frank runs 4 ft/sec faster than Claire
If F's speed is 12ft/sec and C's speed 8 ft/sec, F covers 12/20 or 3/5 the distance
BUT If F's speed is 16ft/sec and C's speed 12 ft/sec, F covers 16/28 or 4/7 the distance
Hence INSUFFICIENT

Ans A
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by saintforlife » Tue Apr 30, 2013 9:48 am
srcc25anu wrote:St1: Frank runs 50% faster than Claire. Let franks speed be 1.5x and Claire Speed be x
distance covered will be directly proportional to ratio of speeds
Ratio of distances covered by F:C = 1.5:1 or 3:2
Frank must be covering (3 * 500)/5 = 300 ft and Claire must be covering 200 ft
Hence SUFFICIENT
That is an interesting way to solve the problem. I hadn't thought of that. Thanks!
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