Hi oquiella,

The answer choices to this question are 'spread out' enough that we can avoid much of the long-winded 'math' and do a couple of calculations (with a little estimation) to get to the solution.

Since we know how long each train takes to travel 100 miles, we can calculate their two speeds:

Train X: 100 miles in 5 hours = 20 miles/hour
Train Y: 100 miles in 3 hours = 33 1/3 miles/hour

Since these trains are approaching one another, they travel a TOTAL of 20 + 33 1/3 = 53 1/3 miles per hour.

The route is 100 miles, so it would take a little less than 2 hours for these two trains to travel that distance (and 'meet up'). We're asked how far Train X would have traveled at that point. Since the travel time is a little less than 2 hours and Train X travels at 20 miles/hour, Train X would have traveled LESS than 40 miles. There's only one answer that 'fits'...

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

oquiella wrote:Two trains X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had train X traveled when it met train Y.

A. 37.5
B. 40
C. 60
D. 62.5
E. 77.5

Another approach:

Train X completed the 100-mile trip in 5 hours
Speed = distance/time
= 100/5
= 20 mph

Train Y completed the 100-mile trip in 3 hours
Speed = distance/time
= 100/3
≈ 33 mph (This approximation is close enough. You'll see why shortly)

How many miles had Train X traveled when it met Train Y?
Let's start with a word equation.

When the two trains meet, each train will have been traveling for the same amount of time
So, we can write: Train X's travel time = Train Y's travel time

time = distance/speed
We know each train's speed, but not the distance traveled (when they meet). So, let's assign some variables.

Let d = the distance train X travels
So, 100-d = the distance train Y travels (since their COMBINED travel distance must add to 100 miles)

We can now turn our word equation into an algebraic equation.
We get: d/20 = (100 - d)/33
Cross multiply to get: (33)(d) = (20)(100 - d)
Expand: 33d = 2000 - 20d
Add 20d to both sides: 53d = 2000
So, d = 2000/53

IMPORTANT: Before you start performing any long division, first notice that 2000/50 = 40
Since the denominator is greater than 50, we can conclude that 2000/53 is LESS THAN 40
Since only one answer choice is less than 40, the correct answer must be A

Cheers,
Brent

oquiella wrote:Two trains X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had train X traveled when it met train Y.

A. 37.5
B. 40
C. 60
D. 62.5
E. 77.5

We are given that train X completed the the 100-mile trip in 5 hours, and that train Y completed the 100-mile trip in 3 hours.

Since rate = distance/time, the rate of train X is 100/5 = 20 mph and the rate of train Y is 100/3 mph.

Since the trains left at the same time, we can let the time of each train = t.

We need to determine the distance traveled by train X when it met train Y. Since the two trains are "converging" we can use the formula:

distance of train X + distance of train Y = total distance

20t + (100/3)t = 100

Multiplying the entire equation by 3, we have:

60t + 100t = 300

160t = 300

t = 300/160 = 30/16 = 15/8.

Thus, train X and Y met each other after 15/8 hours.

Since distance = rate x time, the distance traveled by train X when it met train Y was:

15/8 x 20 = 300/8 = 75/2 = 37.5 miles.

Answer: A