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

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

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

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) \(\frac{4}{3}\)

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

Does 3/2 mean 50% more ? ---> when you multiply 3/2*40

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

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

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.

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.