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A motorcyclist has to cover a distance of 200 km to reach
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16 Aug 2011, 17:46
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26% (02:24) correct 74% (03:02) wrong based on 23 sessions
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A motorcyclist has to cover a distance of 200 km to reach city B from city A. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, there by he reached B 1 hour late. Had the problem developed 30 KM earlier, he would have reached b 12 mins later. Find the the initial distance traveled without the problem and the speed over that part of the journey. A. 50KM,60KM B. 40KM, 40KM C. 60KM, 30KM D. 50KM, 50 KM E. 60KM 60KM
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Re: time and distance question #8
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16 Aug 2011, 20:43
cleetus wrote: A motorcyclist has to cover a distance of 200 km to reach city B from city A. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, there by he reached B 1 hour late. Had the problem developed 30 KM earlier, he would have reached b 12 mins later. Find the the initial distance traveled without the problem and the speed over that part of the journey.
A) 50KM,60KM B) 40KM, 40KM C)60KM, 30KM D)50KM, 50 KM E) 60KM 60KM This is beyond GMAT because it is more laborious and less logical. Let the initial breakdown occurred at x km Let the initial speed be v km/h \(\frac{x}{v}+\frac{200x}{\frac{3}{4}v}=\frac{200}{v}+1\) 1 \(\frac{x30}{v}+\frac{200(x30)}{\frac{3}{4}v}=\frac{200}{v}+1+\frac{1}{5}\) 2 Solve these equations to get x and v. Ans: "D"
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Re: time and distance question #8
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17 Aug 2011, 05:39
um so happy to read it's beyond GMAT...wish n hope such ques nvr cum



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Re: time and distance question #8
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25 Aug 2012, 11:01
A motorcyclist has to cover a distance of 200 km to reach city B from city A. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, there by he reached B 1 hour late. Had the problem developed 30 KM earlier, he would have reached b 12 mins later. Find the the initial distance traveled without the problem and the speed over that part of the journey.
we can use options to solve this poblem.
D) 50 KM , 50 KM
Ist part says if he travels 3/4 th of original speed he reaches 1 hour late.
3/4 th speed of 50 = 37.5
if 50 KM is already covered as in option, total time taken at reduced speed to cover remaining distance.
150 / 37.5 = 4 total time = 4 + 1 = 5 and if he has travelled at original speed , time taken = 200/ 50 = 4
Thus, the difference = 1 hour. In all the remaining case the difference of time is not 1 hour.
Answer D.



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Re: time and distance question #8
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25 Aug 2012, 11:58
fluke wrote: cleetus wrote: A motorcyclist has to cover a distance of 200 km to reach city B from city A. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, there by he reached B 1 hour late. Had the problem developed 30 KM earlier, he would have reached b 12 mins later. Find the the initial distance traveled without the problem and the speed over that part of the journey.
A) 50KM,60KM B) 40KM, 40KM C)60KM, 30KM D)50KM, 50 KM E) 60KM 60KM This is beyond GMAT because it is more laborious and less logical. Let the initial breakdown occurred at x km Let the initial speed be v km/h \(\frac{x}{v}+\frac{200x}{\frac{3}{4}v}=\frac{200}{v}+1\) 1 \(\frac{x30}{v}+\frac{200(x30)}{\frac{3}{4}v}=\frac{200}{v}+1+\frac{1}{5}\) 2 Solve these equations to get x and v. Ans: "D" This is the advantage of multiple choice questions... ) The first equation gives \(3v+x=200.\) The second equation boils down to \(18v+5x=1150.\) You can try to pick the correct answer based on the first equation and even before you work out the second equation. They can't be so mean to provide more than one solution which fulfill the first equation. Or let's hope... )
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Re: A motorcyclist has to cover a distance of 200 km to reach
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27 Aug 2012, 07:28
Why not solve this problem in different way ? Distance................: A...............................................B Velocity..................: V...............l....3V/4..................... (this takes +1 hr) If 30 KM before......: V.........l..........3V/4............................ (this takes +1 hr + 12 minutes) So, we can deduce that as the cyclist lowered the speed from V to 3V/4 to cover 30 KM, he tool 12 minutes more time. So, forming equation to find the difference of time we get we get \(30*4/3V  30/V = 12\) \(=> 40/V  30/V=12\) \(=> 10/V=12\) \(=> V=10 KM/12 Minutes\) \(=> V=50 KM/Hr\) Only option D has this value. So, it is the right answer
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Re: A motorcyclist has to cover a distance of 200 km to reach
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11 Jan 2019, 21:49
cleetus wrote: A motorcyclist has to cover a distance of 200 km to reach city B from city A. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, there by he reached B 1 hour late. Had the problem developed 30 KM earlier, he would have reached b 12 mins later. Find the the initial distance traveled without the problem and the speed over that part of the journey.
A. 50KM,60KM B. 40KM, 40KM C. 60KM, 30KM D. 50KM, 50 KM E. 60KM 60KM A simpler approach to the questionthe motorcyclist travels at 3/4th of speed and gets late by 1 hoursthe motorcyclist travels 30km more at 3/4th of speed and gets late by 1 hours and 12 mins i.e. the motorcyclist travels 30 km at 3/4th of speed and gets late by 12 minsi.e. to be late by 60 mins i.e. 1 hour (12*5) he has to travel 150 km (30*5) at 3/4th of usual speedhence, he travels the distance of 200150 = 5 0 km at regular speed before the breakdown
Now, the motorcyclist travels at 3/4th of speed and gets late by 1 hoursSince D = S*T therefore, D = (3/4)S* (4/3)T . (to nullify the effect of 3/4 in speed we need to multiply the time by 4/3 so that the distance remains same) i.e. extra time consumed due to reduced speed = (4/3)T  T = (1/3)*T = 1 hour i.e. T = 3 hours to travel the distance 150 km (distance after breakdown) in 3 hours (regular time to travel the distance 150 km), Speed = 150/3 = 50 km/hrAnswer: Option D
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