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David and Stacey are riding bicycles on a flat road at a constant rate

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David and Stacey are riding bicycles on a flat road at a constant rate  [#permalink]

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New post 28 Feb 2016, 08:40
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A
B
C
D
E

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  25% (medium)

Question Stats:

78% (01:15) correct 22% (01:23) wrong based on 153 sessions

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David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?

(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.

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Re: David and Stacey are riding bicycles on a flat road at a constant rate  [#permalink]

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New post 02 Mar 2016, 09:55
IMO D

In option A the difference in speed is 2(i.e 12-10= 2 mph). So every hour david is gaining 2 mile.
Since we have to maintain a distance of two miles between David and Stacey(presently three), the 1 mile will be covered by David in half n hour(as david is gaining 2 mile /hr,so to gain 1 mile it will take David 1/2 hr and the dist will be 2mile b/w david and stacey).==> Suff

In option B

45 mins ago the dist was 4.5 mile and now dist is 3 mile. Therefore david covered 1.5(4.5-3) mile in 45 min so speed of david catching stacey is 2mph.
By same logic as for opton A==> Suff

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Re: David and Stacey are riding bicycles on a flat road at a constant rate  [#permalink]

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New post 02 Mar 2016, 22:11
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?

(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.


In the original condition, there are 2 variables(you need to figure out the speed of David and Stacey as you can figure out the time when they are 1 mile apart by knowing their speed). In order to match with the number of equations, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), for 1), they are 2 miles apart at 1 hour. So, to be 1 mile apart, 30 minutes should pass, which is sufficient.
For 2), 45 minutes:1.5 miles=x;1 mile, x=30 minutes is derived, which is unique and sufficient.
Since 1) = 2), the answer is D.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: David and Stacey are riding bicycles on a flat road at a constant rate  [#permalink]

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New post 02 Mar 2016, 22:43
Bunuel wrote:
David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?

(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.


Distance between Stacey and David = 3 miles
Required distance between them = 2 miles
David needs to cover 1 mile. Time required to do this will depend only on the relative speed of David.

(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.

David's speed relative to Stacey in 12 - 10 = 2 mph.
Time taken to cover 1 mile relative distance = 1/2 = 0.5 hrs = 30 mins
Sufficient.

(2) 45 minutes ago Stacey was 4.5 miles ahead of David.
In 45 mins (0.75 hrs), David covered 1.5 miles of relative distance.
So his speed relative to Stacey is 1.5/0/75 = 2 mph (same as given in statement 1)
Sufficient.

Answer (D)
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Re: David and Stacey are riding bicycles on a flat road at a constant rate  [#permalink]

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New post 23 Dec 2018, 02:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: David and Stacey are riding bicycles on a flat road at a constant rate &nbs [#permalink] 23 Dec 2018, 02:18
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