Madelaine88 wrote:
Two trains 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, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5
Train X completed the 100-mile trip in 5 hoursSpeed = distance/time = 100/5 =
20 mphTrain Y completed the 100-mile trip in 3 hoursSpeed = 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 timetime = 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)/33Cross 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 (53) 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
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