Car X and Car Y traveled the same 80-mile route. If Car X

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Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3

The OA is the option C.

Experts, what is the equation that helps me to solve this PS question? Can you give me some help? Please.

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by [email protected] » Sat Feb 03, 2018 10:27 am
Hi M7MBA,

We're told that Car X and Car Y traveled the same 80-mile route, Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X. We're asked for the number of hours it took Car Y to travel the route. With a few standard 'math steps', you can answer this question without too much difficulty.

To start, we can calculate Car X's speed using the Distance Formula:
Distance = (Rate)(Time)
80 miles = (Rate)(2 hours)
80/2 = Rate
Rate = 40 miles/hour

Since Car Y's speed was 50% GREATER than Car X's speed, we can now calculate that speed:
(1.5)(40 miles/hour) = 60 miles/hour

Using that speed, we can calculate how long it took Car Y to travel that same distance:
Distance = (Rate)(Time)
80 miles = (60 miles/hour)(Time)
80/60 = Time
Time = 80/60 = 8/6 = 4/3 hours

Final Answer: C

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by EconomistGMATTutor » Sun Feb 04, 2018 11:43 am
Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3

The OA is the option C.
Hi M7MBA,
Let's take a look at your question.

Car X traveled 80 miles in 2 hours.
$$Speed\ =\ \frac{Distance}{Time}$$
$$Speed\ =\ \frac{80}{2}=40 miles/hr\ $$

Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X.
50% of 40 = (50/100)*40 = 20

Therefore, Speed of Car Y = 40 + 20 = 60 miles/hr
Time taken by Car Y to cover 80 miles at a speed of 60 miles/hr can be calculated as:
$$Time\ =\ \frac{Distance}{Speed}$$
$$Time\ =\frac{80}{60}=\frac{4}{3}$$

Therefore, Option C is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Jeff@TargetTestPrep » Tue Feb 06, 2018 5:20 pm
M7MBA wrote:Car X and Car Y traveled the same 80-mile route. If Car X took 2 hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route?

(A) 2/3
(B) 1
(C) 4/3
(D) 8/5
(E) 3
We are given that Car X traveled 80 miles in 2 hours. Thus, the rate of Car X was 80/2 = 40 mph.

We are also given that Car Y traveled 50% faster than Car X. Thus, Car Y traveled at a rate of 1.5 x 40 = 60 mph.

So, it took Car Y 80/60 = 8/6 = 4/3 hours to travel the same route.

Answer:C

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