Asteroids X-13,Y-14,Z-15

[amsm25]

The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical duration, 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?

1)500
2)1600/3
3)1000
4)1500
5)2500

OA-1)500

[niketdoshi123]

Let distance traveled by asteroid
X-13 (X) = a km
Y-14 (Y) = 3a km
Z-15 (Z) = 5a km

Let the period of observation for
X = t sec
Y = t sec
Z = t+2 sec

Speed
X = a/t km/sec
Y = 3a/t = a/t+2000
=>a/t = 1000 km/sec = speed of X
Z = 5a/(t+2)

But Speed of Z is same as that of X so,
5a/(t+2)=a/t
=>5at=at+2a
=>4at=2a
=>t=1/2 sec

So distance traveled by X
a/t*t = 1000*1/2 = 500

Hence the answer is A

[GmatKiss]

What is the source of the problem?

[hey_thr67]

Try to use spoiler ..... It otherwise spoils the fun of solving the problem.

[GMATGuruNY]

The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical duration, 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?

1)500
2)1600/3
3)1000
4)1500
5)2500

OA-1)500

We can plug in the answers, which represent the distance traveled by X13.

Answer choice C: distance for X13 = 1000k
Since Z15 travels 5 times the distance, the distance traveled by Z15 = 5*1000 = 5000k.
Since Z15 travels an extra 4000k in 2 seconds, the rate for both Z15 and X13 = d/t = 4000/2 = 2000k per second.
Thus, the time for X13 (and also for Y14, which travels for the same amount of time) = d/r = 1000/2000 = 1/2 of a second.

Since Y14 travels 3 times the distance traveled by X13, the distance traveled by Y14 = 3*1000 = 3000k.
Thus, the rate for Y14 = d/t = 3000/(1/2) = 6000k per second.
Rate for Y14 - Rate for X13 = 6000-2000 = 4000k per second.

Doesn't work: the difference between Y14's rate and X13's rate (4000k per second) is twice the required difference (2000k per second).
Thus, the correct answer must be A (1/2 of answer choice C).

The correct answer is A.

Answer choice A: distance for X13 = 500k
Since Z15 travels 5 times the distance, the distance traveled by Z15 = 5*500 = 2500k.
Since Z15 travels an extra 2000k in 2 seconds, the rate for both Z15 and X13 = d/t = 2000/2 = 1000k per second.
Thus, the time for X13 (and also for Y14, which travels for the same amount of time) = d/r = 500/1000 = 1/2 of a second.

Since Y14 travels 3 times the distance traveled by X13, the distance traveled by Y14 = 3*500 = 1500k.
Since Y14 travels 1500k in 1/2 of a second, the rate for Y14 = d/t = 1500/(1/2) = 3000k per second.
Rate for Y14 - Rate for X13 = 3000-1000 = 2000k per second.
Success!

Note that we had to plug in only ONE answer choice: a very efficient way to solve a complex problem.