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Car X and Car Y traveled the same 80mile route. If Car X to
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Updated on: 01 Mar 2014, 06:50
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Car X and Car Y traveled the same 80mile 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
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Originally posted by seabhi on 01 Mar 2014, 05:02.
Last edited by Bunuel on 01 Mar 2014, 06:50, edited 1 time in total.
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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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01 Mar 2014, 06:58



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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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06 Feb 2017, 10:15
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



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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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05 Jun 2017, 07:05
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



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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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05 Jun 2017, 08:41
seabhi wrote: Car X and Car Y traveled the same 80mile route. If Car X took 2 hours and Car Y travelled 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 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) \(\frac{4}{3}\)
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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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06 Jun 2017, 16:47
seabhi wrote: Car X and Car Y traveled the same 80mile 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 route. Answer: C
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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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28 Jan 2018, 08:48
seabhi wrote: Car X and Car Y traveled the same 80mile 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 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
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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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21 Feb 2018, 08:49
Bunuel wrote: seabhi wrote: Car X and Car Y traveled the same 80mile 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 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. Does 3/2 mean 50% more ? > when you multiply 3/2*40



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Car X and Car Y traveled the same 80mile route. If Car X to
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21 Feb 2018, 08:59
GMATPrepNow wrote: seabhi wrote: Car X and Car Y traveled the same 80mile 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 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 hi Brent yes you GMATPrepNowyou 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



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Re: Car X and Car Y traveled the same 80mile route. If Car X to
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21 Feb 2018, 09:20
dave13 wrote: GMATPrepNow wrote: seabhi wrote: Car X and Car Y traveled the same 80mile 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 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 hi Brent yes you GMATPrepNowyou 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
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Car X and Car Y traveled the same 80mile route. If Car X to
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21 Feb 2018, 11:13
dave13 wrote: Bunuel wrote: seabhi wrote: Car X and Car Y traveled the same 80mile 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 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. Does 3/2 mean 50% more ? :? > when you multiply 3/2*40 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 hereHope that helps.
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