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While on a straight road, car X and car Y are traveling at

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While on a straight road, car X and car Y are traveling at [#permalink] New post 15 Feb 2007, 22:48
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While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?
(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour
(2) Three minutes ago car X was 1/2 mile ahead of car X.
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Re: DS: Distance and Time [#permalink] New post 16 Feb 2007, 04:03
devilmirror wrote:
While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?
(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour
(2) Three minutes ago car X was 1/2 mile ahead of car X.


Answer is D

Here is my explanation:
Let's say now is at time t = T mins
Cax X's speed = S1
Car Y's speed = S2
Distance travelled by X = d1
Distance travelled by Y = d2

We know, at t = T mins d1 = d2 + 1
We want to find after how many mins d1 = d2 + 2

Condition 1:
Cax X's speed = 50 miles/hr
Car Y's speed = 40 miles/hr

Therefore at t = T mins Car X travels T5/6 miles and Car Y travels T4/6 miles.

we know d1 = d2 + 1 at time t = Tmins
Hence, T5/6 = T4/6 + 1 this give T = 6mins

Now we need to find t = M mins when d1 = d2 + 2
Therefore at t = M mins Car X travels M5/6 miles and Car Y travels M4/6 miles.

we know d1 = d2 + 2 at t = Mmins
Hence M5/6 = M4/6 + 2 this give M = 12 so after 6 min Car X will be 2 miles ahead of Car Y

Condition 1 is sufficient.

Condition 2:
at time t = T mins we know d1 = d2 + 1
Car X will travel d1 = TS1/60 and Car Y will travel d2 = d1 + 1
Therefore, d2 = TS1/60 + 1 ---I

From condition 2 we know,
at time t = T-3 min d1 = d2 + 1/2
at time t= T-3
Car X will travel d1 = (T-3)*S1/60 and Car Y will travel d2 = d1 + 1/2
Therefore, d2 = (T-3)*S1/60 + 1/2 ---II

From I and II we can find out S1 = 10 miles/hr

We need find t = T + x when d1 = d2 + 2
at time t= T+x
Car X will travel d1 = (T+x)*S1/60 and Car Y will travel d2 = d1 + 2
Therefore, d2 = (T+x)*S1/60 + 2 ---III

From II and III we can find x and hence sufficient.

Answer is D
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Last edited by amorpheus on 16 Feb 2007, 05:24, edited 1 time in total.
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 [#permalink] New post 16 Feb 2007, 06:23
Amorpheus, I did the exact same thing when I tried to solve this problem. :-D

However, the explanation of this answer can be extremely short. I want to post the question first and see who can come up with the shortest explanation.

When I read the short explanation, I learned a lot from it. This question is very creative so I want to share it with you guys. 8-)
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 [#permalink] New post 16 Feb 2007, 06:24
devilmirror wrote:
Amorpheus, I did the exact same thing when I tried to solve this problem. :-D

However, the explanation of this answer can be extremely short. I want to post the question first and see who can come up with the shortest explanation.

When I read the short explanation, I learned a lot from it. This question is very creative so I want to share it with you guys. 8-)


Devilmirror: Would you mind sharing the shortest explanation please?
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 [#permalink] New post 16 Feb 2007, 07:04
amorpheus wrote:

Devilmirror: Would you mind sharing the shortest explanation please?


Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.

(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.
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 [#permalink] New post 16 Feb 2007, 07:06
devilmirror wrote:
amorpheus wrote:

Devilmirror: Would you mind sharing the shortest explanation please?


Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.

(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.


You the man.... I like that sometimes I feel stupid, this actually happens to me a lot I always end up solving problems using the longest possible method. What you do to get such short cuts? Any special tricks?
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 [#permalink] New post 16 Feb 2007, 08:00
amorpheus wrote:
devilmirror wrote:
amorpheus wrote:

Devilmirror: Would you mind sharing the shortest explanation please?


Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.

(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.


You the man.... I like that sometimes I feel stupid, this actually happens to me a lot I always end up solving problems using the longest possible method. What you do to get such short cuts? Any special tricks?


No, no, no my friends. I used the longer method at first. But the OE is a lot shorter than I expected.

The explanation is from OE not from me. If I showed you my method, it would be 1 full A4 paper. :lol:
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 [#permalink] New post 16 Feb 2007, 09:33
I look at this problem and I understand that we need to have the information to understand the difference in speed...

The first gives us this 50 mph and 40 mph 10 mph difference suff

second gives us the time it took for the cars to create a 1/2 mile distance suff since I know I can derive the difference in speed from this info.

No math is necessary to solve this problem
  [#permalink] 16 Feb 2007, 09:33
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