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Car X is 40 miles west of Car Y. Both cars are traveling

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Car X is 40 miles west of Car Y. Both cars are traveling

by BTGmoderatorRO » Wed Mar 28, 2018 8:58 pm

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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
OA is c
How can I set up the formula to use here? can any expert help me?
Thanks
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by Vincen » Sat Mar 31, 2018 7:31 am
BTGmoderatorRO 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
OA is c
How can I set up the formula to use here? can any expert help me?
Thanks
Hello.

This is how I would do it:

Since Car X is going 50% faster than Car Y, then if the speed of the Car Y is "y" then the speed of the Car X is "1.5y". The relative speed is $$1.5y\ -\ y\ =\ 0.5y.$$

Now, the distance between Car X and Car Y is 40 miles. The time that took Car X to reach Car Y was 2 hours and 40 minutes; we know must convert it to hours: $$160\min\ =160\ \min\cdot\frac{1\ hour}{60\ \min}=\frac{8}{3}hours.$$

Now, using that d=v*t we get: $$40=0.5y\cdot\frac{8}{3}\ \Rightarrow\ \ 40=\frac{4}{3}y\ \Rightarrow\ \ y=\frac{120}{4}=30\ mph.$$ Therefore, the answer is the option C.

I hope it helps.
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by Scott@TargetTestPrep » Mon May 27, 2019 4:27 am
BTGmoderatorRO 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
OA is c
We can let the rate of Car Y = r and the rate of Car X = 1.5r. Recall that 2 hours and 40 minutes = 2 2/3 hours = 8/3 hours.

We can create the equation:

1.5r(8/3) = r(8/3) + 40

12r/3 = 8r/3 + 40

4r/3 = 40

4r = 120

r = 30

Answer: C

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by Brent@GMATPrepNow » Mon May 27, 2019 5:00 am
BTGmoderatorRO 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
OA is c
How can I set up the formula to use here? can any expert help me?
Thanks
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
Brent Hanneson - Creator of GMATPrepNow.com
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