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A train after traveling at 50kmph meets with an accident and then
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06 Dec 2014, 23:43
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A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train. a) 45 b) 33 c) 48 d) 55 e) 61
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A train after traveling at 50kmph meets with an accident and then
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07 Dec 2014, 01:46
Blackbox wrote: A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.
a) 45 b) 33 c) 48 d) 55 e) 61 here is less mathematical approach. let the two end points between which this train is running be A and B. Also suppose that accident occurs at C which is at a distance of 50km from A. thus we have A50kmCB 1) Time taken to travel the distance after the accident i.e. CB is 35 minutes more than the usual time. now, if the accident were to occur at 24 km farther, then train would have reached only 25 minutes late. i.e. A50 kmC24DB 2) if we look at the situation depicted in 1) and 2) we will see that 10 minute difference in time is occurring because of this 24 Km. which is traveled by its usual speed (=x) in the later case. thus we have \(\frac{24*4}{3x}\)  \(\frac{24}{x}\) = \(\frac{10}{60}\) \(\frac{32}{x}\)  \(\frac{24}{x}\) = \(\frac{1}{6}\) solving this we get x= 48km/h




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Re: A train after traveling at 50kmph meets with an accident and then
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07 Dec 2014, 00:43
Let y be the balance distance to be covered and x be the former speed. A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late so, y/(3x/4)  y/x = 35/60 4y/3x  y/x = 7/12 y/x(4/31)=7/12 y/x*1/3=7/12 y/x=7/4 4y7x=0 ........ 1
Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late so, (y24)/(3x/4)  (y24)/x = 25/60 4(y24)/3x  (y24)/x = 5/12 (y24)/x (4/31) = 5/12 (y24)/x *1/3 = 5/12 (y24)*12 = 3x*5 (y24)*4 = 5x 4y5x = 96 ....... 2
eq2  eq1 2x=96 x=48
Ans = C



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Re: A train after traveling at 50kmph meets with an accident and then
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07 Dec 2014, 01:13
Chinmay Thanks for the explanation. I have a followup question: In Eq 1: You said 'y' is the balance distance . Then in the question  Quote: y/(3x/4)  y/x = 35/60 since you are subtracting the latter time (postaccident) from the former time (preaccident), shouldn't \(\frac{y}{x}\) be \(\frac{50}{x}\)?



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07 Dec 2014, 01:24
Blackbox wrote: Chinmay Thanks for the explanation. I have a followup question: In Eq 1: You said 'y' is the balance distance . Then in the question  Quote: y/(3x/4)  y/x = 35/60 since you are subtracting the latter time (postaccident) from the former time (preaccident), shouldn't \(\frac{y}{x}\) be \(\frac{50}{x}\)? My friend, I am not subtracting the latter time (post accident) from the former time. I am subtracting the time that train would have taken, had there been no accident from the actual time it took. If there had been no accident, the train would have continued at its speed (i.e. x) and would have taken y/x hrs to complete. Since there was an accident, speed was reduced (3x/4) and it actually took y/(3x/4) hrs to cover balance distance. I hope it helps.



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07 Dec 2014, 01:46
Chinmay  Sorry for being thick, but I struggle to understand why you would subtract the times if there were no accident. Is it because they say the train arrived late? What if there was a case when the train arrives early? Would you add times?



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07 Dec 2014, 01:54
Manpreet  Easier solution but I guess I am missing the fundamentals here  "why subtract?"



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Re: A train after traveling at 50kmph meets with an accident and then
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07 Dec 2014, 02:03
Blackbox wrote: Manpreet  Easier solution but I guess I am missing the fundamentals here  "why subtract?" hi, remember the difference of 10 minutes is occurring because the 24km is travelled at two different speed. x is the usual speed and 3/4x is the speed after the accident. and to convert minutes into hour i have divided 10 by 60.



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Re: A train after traveling at 50kmph meets with an accident and then
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07 Dec 2014, 03:22
manpreetsingh86 wrote: Blackbox wrote: Manpreet  Easier solution but I guess I am missing the fundamentals here  "why subtract?" hi, remember the difference of 10 minutes is occurring because the 24km is travelled at two different speed. x is the usual speed and 3/4x is the speed after the accident. and to convert minutes into hour i have divided 10 by 60. I see it now, so you are only making an equation for that 24 km part equating the difference to 10 minutes. The rest of the question is actually irrelevant. Hmm... interesting. Are there any other problems on GMAT club in line with this concept? Any other approaches/discussions are welcome.



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Re: A train after traveling at 50kmph meets with an accident and then
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08 Dec 2014, 04:55
Somebody can explain me why is 4/3x  24/x and not 24/x 4/3x????? 24/x is not bigger?



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Re: A train after traveling at 50kmph meets with an accident and then
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08 Dec 2014, 09:44
marcosgensollen wrote: Somebody can explain me why is 4/3x  24/x and not 24/x 4/3x????? 24/x is not bigger? Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x}  \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept.



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Re: A train after traveling at 50kmph meets with an accident and then
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08 Dec 2014, 10:20
Blackbox wrote: marcosgensollen wrote: Somebody can explain me why is 4/3x  24/x and not 24/x 4/3x????? 24/x is not bigger? Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x}  \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept. I understand. If (3x/4)t=24 and (x)(t)= 24 > (1) t= 4*24/(3x) and (2) t= 24/x. If you analyze each formula, you are going to conclude that 24/x take LESS time than 4*24/3x to complete the 10 minutes. It's a very confusing exercise because you have to realized that youre substracting time (minutes) and no rates.



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A train after traveling at 50kmph meets with an accident and then
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09 Dec 2014, 21:17
Say total distance = d Kms Speed = s Km/hr Refer diagram below: Attachment:
speed.png [ 7.74 KiB  Viewed 81131 times ]
Setting up the time equation for scenario I \(\frac{50}{s} + \frac{d50}{\frac{3s}{4}} = \frac{d}{s} + \frac{35}{60}\) 4d  7s = 200 .................. (1) Setting up the time equation for scenario II \(\frac{74}{s} + \frac{d74}{\frac{3s}{4}} = \frac{d}{s} + \frac{25}{60}\) 4d  5s = 74*4 .................. (2) (2)  (1) 2s = 74*4  200 s = 2(74  50) = 48 Answer = C



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Re: A train after traveling at 50kmph meets with an accident and then
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09 Dec 2014, 21:29
Blackbox wrote: Chinmay  Sorry for being thick, but I struggle to understand why you would subtract the times if there were no accident. Is it because they say the train arrived late? What if there was a case when the train arrives early? Would you add times? That's precisely correct. Please refer my post above. Usual speed was "s" and the usual time taken would had been "d/s" However, due to accident, the speed decreased, so negative sign. Had the question be like "after 50 kms, they added another engine due to which train reached 10 minutes earlier", then it had been positive sign. (Retrospective) Chinmay & I have approached the same way, however the +ve/ve signs are taken at different sides Refer the diagram for more clarity



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Re: A train after traveling at 50kmph meets with an accident and then
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09 Dec 2014, 21:33
marcosgensollen wrote: Somebody can explain me why is 4/3x  24/x and not 24/x 4/3x????? 24/x is not bigger? The equation is setup for time = distance/speed Less the speed, more time would be required as distance remains constant Equation setup accordingly



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Re: A train after traveling at 50kmph meets with an accident and then
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09 Dec 2014, 21:34
marcosgensollen wrote: Blackbox wrote: marcosgensollen wrote: Somebody can explain me why is 4/3x  24/x and not 24/x 4/3x????? 24/x is not bigger? Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x}  \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept. I understand. If (3x/4)t=24 and (x)(t)= 24 > (1) t= 4*24/(3x) and (2) t= 24/x. If you analyze each formula, you are going to conclude that 24/x take LESS time than 4*24/3x to complete the 10 minutes. It's a very confusing exercise because you have to realized that youre substracting time (minutes) and no rates. We cannot add/subtract speeds / rates. However, time can be added/ subtracted. As mentioned in earlier posts, that's why time equation is created to extract values of "assumed variables" to find correct answer



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Re: A train after traveling at 50kmph meets with an accident and then
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09 Dec 2014, 23:32
Blackbox wrote: A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.
a) 45 b) 33 c) 48 d) 55 e) 61 Hi If the answer choices are such then you should directly use 48 and do backward calculation and you can land up to the answer in less than minute. Why I am saying for this particular question is it is given that speed reduces to 3/4th of the orignal Only 48 is the one which can be divided by 4 and have a realisitc reduced speed i.e 36km/hr. In rest you get in fractions. Not full proof, but it works in 90% cases.



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Re: A train after traveling at 50kmph meets with an accident and then
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10 Dec 2014, 07:36
PareshGmat wrote: ...
Setting up the time equation for scenario I \(\frac{50}{s} + \frac{d50}{\frac{3s}{4}} = \frac{d}{s} + \frac{35}{60}\) 4d  7s = 200 .................. (1) Setting up the time equation for scenario II \(\frac{74}{s} + \frac{d74}{\frac{3s}{4}} = \frac{d}{s} + \frac{25}{60}\) ...
This is amazing and exactly how I solved after a lot of digging. Just to add for others, 35/60 is added to d/s (original time taken sans accident) to show the 35 minute delay mentioned. And likewise, for the second equation, 25/60 is added to d/s to show the 25 minute delay. Thanks Paresh!



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10 Dec 2014, 07:47
believer700 wrote: In rest you get in fractions. Not full proof, but it works in 90% cases. You're probably right in this case. The rest of the options were made up by me and I didn't give it enough thought that they would result in a fraction. Back solving isn't a foolproof solution, personally I would do that unless I am in a time crunch.



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27 Dec 2015, 07:22
let d=total distance divide each case into 3 segments: 1. first 50 km 2. next 24 km 3. last d74 km let s=speed of train for segment 1, in each case, time=50/s for segment 3, in each case, time=d74/(3s/4) therefore, segment 2 must account for the 10 minute total time difference between the two cases 24/(3s/4)24/s=1/6 s=48 kmh




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