BTGModeratorVI wrote: ↑Sat Mar 21, 2020 9:52 am
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) 40.0
(C) 60.0
(D) 62.5
(E) 77.5
Answer:
A
Source: Official Guide
One 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
