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

Question

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

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

BrentGMATPrepNow · Accepted Answer

There's a  nice rule  we can use here.

To set up the rule, recognize that if Y travels   2   times as fast as X, then Y's travel time will be   1/2   of X's. 
Similarly, if Y travels   3   times as fast as X, then Y's travel time will be   1/3   of X's.
Or if Y travels   1/4   as fast as X, then Y's travel time will be   4   times X's travel time.

In general,  if Y travels   a/b   times as fast as X, then Y's travel time will be   b/a   of X's travel time.

So, if Y's speed is 50%  more  than X's speed, we can say that Y travels 1.5 times as fast as X. 
In other words, if Y travels   3/2   times as fast as X, which means Y's travel time will be   2/3   that of X's travel time.

Since X's travel time is 2 hours, Y's travel time will be (  2/3  )(2) = 4/3 = 1 1/3

Answer: C

Cheers, 
Brent

Bunuel · Answer

The speed of car X is (distance)/(time) = 80/2 = 40 miles per hour.

The speed of car Y = 3/2*40 = 60 miles per hour --> (time) = (distance)/(speed) = 80/60 = 4/3 hours.

Answer: C.

Or: to cover the same distance at 3/2 as fast rate 2/3 as much time is needed --> (time)*2/3 = 2*2/3 = 4/3 hours.

Answer: C.

habil · Answer

As Y traveled 50 percent faster than X
So, if X needs 1.5 hour Y needs 1 hour
 if X needs 2 hour Y needs 2/1.5 hour = 4/3 hour

Answer : C

nks2611 · Answer

the distance for both the parties is constant so as we can see that speed of y is multiplied by 3/2 then time will also multiplied by reciprocal of 3/2 which will be 2*2/3= 4/3 time taken by y

hence C

Abhishek009 · Answer

For Car X -

Speed = 40 \ miles/hr 
 Time = 2 \ Hours

For Car Y -

Speed = 60 \ miles/hr 
 Time = \frac{80}{60}

So, the time required is  \frac{4}{3}  Hours, answer will be (C)  \frac{4}{3}

JeffTargetTestPrep · Answer

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 route.
 
Answer: C

dave13 · Answer

Does 3/2 mean 50% more ?

---> when you multiply 3/2*40

dave13 · Answer

hi Brent yes you

GMATPrepNow you say " if Y travels 2 times as fast as X, then Y's travel time will be 1/2 of X's. " how can that be possible

if Y travels twice as fast as X that means for example if Y speed is 30 km per hour and X speed is 15 km per hour. no ? but you say i need to multiply \frac{1}{2} * 15 = 7.5

then i get Y =7.5

pls explain

BrentGMATPrepNow · Answer

I think you are mixing up SPEED and TIME

TRUE: If Y travels   2   times as fast as X, then Y's travel time will be   1/2   of X's.

So, if we're talking SPEED, then Y's speed = 30 km per hour and X's speed = 15 km per hour meets the given condition

Now let's compare travel TIMES. 
Let's say both cars are traveling 30 km
Then Y's travel TIME = 1 hour and X's travel TIME = 2 hours

Does that help?

Cheers,
Brent

generis · Answer

Hi dave13 . Yes, 3/2 means 50 percent more. Why? 3/2 = 1.50, and 1.50 (or 1.5) is the multiplier for "50 percent faster." Often fractions are easier than decimals in "percent increase/decrease" problems. Just convert the decimal multiplier to a fraction. Y's rate is a percent increase of X's rate. Here, 50 percent faster = Original speed + 50% of original speed --Original speed (X's speed) = 40 --50 percent of 40 = 20 (Original + 50% of original) = (40+20) = 60 = Y OR Y=1.5X 1.5=1\frac{5}{10}=1\frac{1}{2}=\frac{3}{2} Y = \frac{3}{2}X Here are some common fractions used in percent increase/ decrease problems: 5/4, 6/5, 2/5, 1/4, 1/2, 3/2. Some are increases, some are decreases. Decreases are harder. mikemcgarry explains the issue you asked about here Hope that helps.

ArnauG · Answer

To solve this problem, let's first find the average speed of Car X. We know that Car X took 2 hours to travel 80 miles, so its average speed can be calculated as:

Average speed of Car X = Distance / Time = 80 miles / 2 hours = 40 miles per hour

Now, we need to determine the average speed of Car Y, which is 50 percent faster than the average speed of Car X. To find this, we'll multiply the average speed of Car X by 1.5 (since 50 percent of 40 is 20):

Average speed of Car Y = 1.5 * 40 miles per hour = 60 miles per hour

Next, we can use the average speed of Car Y to calculate the time it takes for Car Y to travel the 80-mile route:

Time taken by Car Y = Distance / Average speed of Car Y = 80 miles / 60 miles per hour

Simplifying this, we get:

Time taken by Car Y = 4/3 hours. Therefore, Car Y takes (C) 4/3 hours to travel the route.

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