Roland2rule wrote:Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?
a) 18
b) 25
c) 30
d) 12
e) 40
Here's one approach:
Let's let Car X's original position be the initial starting point.
So, when Car X is at the initial starting point, Car Y has already traveled 40 miles.
My
word equation involves the conditions when Car X
catches up to Car Y.
At that point, we can say:
Car X's TOTAL distance traveled =
Car Y's TOTAL distance traveled
Car Y's total distance
Let V = Car Y's speed (our goal is to find the value of V)
From the time that Car X begins moving, Car Y drives for 2 2/3 hours (2 hours, 40 minutes).
So Car Y's total distance = (time)(speed) = (2 2/3)(V) + 40 miles
Car X's total distance
We know that Car X is going 50% faster than Car Y. If Car Y's rate is V, then Car X's rate must be 1.5V
We also know that Car X travels for 2 2/3 hours.
So Car X's total distance = (time)(speed) = (2 2/3)(1.5V) miles
Simplify: (2 2/3)(1.5V) = (8/3)(3/2) = 4V
We're now ready to write our algebraic equation.
Car X's total distance =
Car Y's total distance
4V =
(2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V = 30
Answer: C
Cheers,
Brent
