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Manager  Joined: 10 Sep 2012
Posts: 137
The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 66% (03:20) correct 34% (03:09) wrong based on 285 sessions

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The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical durations, while asteroid Z-15 was observed for 2 seconds longer. During its period of observation, asteroid Y-14 traveled three times the distance X-13 traveled, and therefore Y-14 was found to be faster than X-13 by 2000 kilometers per second. Asteroid Z-15 had an identical speed as that of X-13, but because Z-15 was observed for a longer period, it traveled five times the distance X-13 traveled during X-13's inspection. Asteroid X-13 traveled how many kilometers during its observation?

A. 500
B. 1,600/3
C. 1,000
D. 1,500
E. 2,500

I have the solution and will paste if anyone would like it, but I'm not convinced that that this is doable in under 2 minutes.... if so, what did you guys do?

Originally posted by anon1 on 31 Oct 2012, 17:04.
Last edited by Bunuel on 01 Nov 2012, 06:07, edited 1 time in total.
Edited the question.
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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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2013gmat wrote:
Somebody please explain this to me ?

Without getting into too much maths, lets solve this.
Lets call these asteroids X,Y,Z instead of X-13/14/15 etc for simplicity.

What is given in question:
1. Notice, Z was observed 2 extra seconds - or that mean Z travelled for 2 extra seconds than X and Y.
2. Distance Y traveled was 3 times X , in same duration. This would mean speed of Y is 3 times that of X. and difference is given as 2000.
Thus, Y traveled 3 times faster (say speed =3V) than X (say speed =V) and difference is 2000 => 3V-V =2000
Or V =1000.
3. Distance Z traveled was 5 times what X traveled despite having same speed.
Z traveled 5 times what X traveled despite having same speed because Z traveled 2 exta seconds. That means, extra 2 seconds = extra 4 times distance
=> original duration of travel for X (and for Y) = 0.5 seconds

So question has become, how much X traveled in 0.5 seconds with speed 1000 km/s.
So total distance covered by X = 1000* 0.5 = 500

Ans A it is.

Hope it is clear.
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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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anon1 wrote:
The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical durations, while asteroid Z-15 was observed for 2 seconds longer. During its period of observation, asteroid Y-14 traveled three times the distance X-13 traveled, and therefore Y-14 was found to be faster than X-13 by 2000 kilometers per second. Asteroid Z-15 had an identical speed as that of X-13, but because Z-15 was observed for a longer period, it traveled five times the distance X-13 traveled during X-13's inspection. Astroid X-13 traveled how many kilometers during its observation?

500
1,600/3
1,000
1,500
2,500

I have the solution and will paste if anyone would like it, but I'm not convinced that that this is doable in under 2 minutes.... if so, what did you guys do?

Ans A. Doable in 2 mins.
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Quote:
Ans A. Doable in 2 mins.

How did you do it?

For this problem, for me, it takes roughly a minute for me just to set up the table and go through my head what the heck is going on before I even start writing down one calculation...
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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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anon1 wrote:
Quote:
Ans A. Doable in 2 mins.

How did you do it?

For this problem, for me, it takes roughly a minute for me just to set up the table and go through my head what the heck is going on before I even start writing down one calculation...

Let distances be $$D_x, D_y, D_z$$ and time X was observed be $$t$$

Velocities are $$\frac{D_x}{t}, \frac{D_y}{t}, \frac{D_z}{t+2}$$

$$D_y = 3D_x$$

$$D_z = 5D_x$$

$$\frac{D_y}{t} = \frac{D_x}{t} + 2000$$

$$\frac{D_x}{t} = \frac{D_z}{t+2}$$

Although it took me a good 5 mins typing and formatting all this, I think setting up these equations would not take more than a min and another half for solving. At worst must be doable within 2 mins.

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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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Somebody please explain this to me ?
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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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2
Assume X-13 ==> X, Y -14 ==> Y and Z-15 ==> Z

Formula D =ST

Given

1. Distance traveled by Z = 5 (Distance traveled by X)
2. time of X be T, given that time of Z is T + 2
3. Speed of X and Z are the same say S

S(T+2) = 5(S x T) solving we get T = 0.5 ------ (1)

Also Given that Y traveled 3 time than X in the same time period

Difference in Speed of X and Y is 2000 and speed would be 3S - S = 2000 ==> S = 1000 ------ (2)

Distance X traveled using 1 and 2
D = ST
D = 1000 x 0.5
D = 500

Option A

Hope the thought process is right..
Took some time to get the equations....
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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical durations, while asteroid Z-15 was observed for 2 seconds longer. During its period of observation, asteroid Y-14 traveled three times the distance X-13 traveled, and therefore Y-14 was found to be faster than X-13 by 2000 kilometers per second. Asteroid Z-15 had an identical speed as that of X-13, but because Z-15 was observed for a longer period, it traveled five times the distance X-13 traveled during X-13's inspection. Asteroid X-13 traveled how many kilometers during its observation?

X13: (t, d, s)
Y14: (t, 3d, s+2000mi/hour)
Z15: (t+2 seconds, s, 5d)

d=?

Distance = Speed*Time
x13: d = s*t
x14: 3d = (s+2000)*t ===> 3d = ts+2000t
z15: 5d = s*(t+2t) ===> 5d = st+2st ===> 5d - 2st = st

3d = 5d - 2st + 2000t
-2d = -2st + 2000t
2d = 2st - 2000t
d = st - 1000t

x13: d = s*t
st - 1000t = s*t
s - 1000 = s
-500 = s

I got to this point and couldn't go any further. This seems like a problem where I can set up individual d=r*t formulas and solve but it appears that's not the case. For future reference how would I know not to waste my time setting up this problem in the aforementioned way? Thanks!!!

The distance of Z15 is equal to five times the distance of X13 (we established that x13 is the baseline and thus, it's measurements are d, s, t)

S(T+2) = 5(S*T) What clues would I have to know to set up the equation in this fashion? Is it because I am better off setting two identical distances together?
ST+2S = 5ST
T+2 = 5T
2=4T
t= 1/2

We are looking for distance (d=s*t) so we need to solve for speed now that we have time.
Speed y14 - speed x13
Speed = d/t
3d/t - d/t = 2000 (remember, t is the same because both asteroids were observed for the same amount of time)
2d = 2000
2 = 1000

d=s*t
d=1000*(1/2)
d=500

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Re: The speeds of three asteroids were compared. Asteroids X-13  [#permalink]

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The first time I solved, I used equations. Bad decision. It took 5 mins. But I think that this approach can help us to solve it in 2 or 3 mins.

First: the question asks for the Distance (D) that X traveled. Only look for infornation to solve this.

Second: info.

Asteroids X-13 and Y-14 were observed for identical durations, and:
1) asteroid Y-14 traveled three times the distance X-13 traveled.
2) Y-14 was found to be faster than X-13 by 2000 kilometers per second
Therefore:

r+2000 (rate of speed of Y, per second) = 3r (distance traveled by Y, in one second, remember that both X and Y were observed for indentical durations) --> r= 1000

Third:
Asteroid Z had an identical speed as that of X, but because Z was observed for 2 seconds longer, it traveled five times the distance X traveled during X's inspection.

Therefore, D (distance of X) +2000 (because r=1000, Z in 2 seconds travel 2000) = 5D --> D = 500

Hope it is clear.
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