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Two cars A and B start from Boston and New York respectively [#permalink]
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28 Apr 2012, 19:50
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Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York? A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes
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Re: speed distance time [#permalink]
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29 Apr 2012, 00:24
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If two objects A and B start from opposite points and, after having met en route, reach their respective destinations in a and b mins (or any other measure of time) respectively, then the ratio of their speeds
\(\frac{Sa}{Sb} = \sqrt{\frac{Time taken by B}{Time taken by A}}\)
Hence:
\(\frac{Sa}{Sb} = \sqrt{\frac{90}{40}}\) \(\frac{Sa}{Sb} = \frac{3}{2}\)
Hence, time taken by A to cover the same distance which B covers in 90 minutes is:
Ta = \(\frac{90*2}{3}\) Ta = 60 minutes
Hence, total time of A = 40 minutes + 60 minutes = 1 hour 40 minutes



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Two cars A and B start from Boston and New York respectively [#permalink]
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29 Apr 2012, 23:53
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vdadwal wrote: Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes Look at the diagram below: The rate of car A is \(a\) and the rate of B is \(b\); The distance covered by A before the meeting point is \(x\) and the distance covered by B before the meeting point is \(y\); Time before meeting \(t\). Since car A covered the distance of \(y\) in 40 minutes then the rate of car A = distance/time = \(a=\frac{y}{40}\), but the same distance of \(y\) was covered by car B in \(t\) minutes, so \(y=bt\) > \(a=\frac{bt}{40}\); Since car B covered the distance of \(x\) in 90 minutes then the rate of car B = distance/time = \(b=\frac{x}{90}\), but the same distance of \(x\) was covered by car A in \(t\) minutes, so \(x=at\) > \(b=\frac{at}{90}\); Substitute \(b\) in the first equation: \(a=\frac{at}{90}*\frac{t}{40}\) > reduce by \(a\) and crossmultiply: \(t^2=3600\) > \(t=60\) minutes, hence it took car A 60+40=100 minutes to cover the whole distance. Answer: D. Hope it's clear. Attachment:
Boston  New York.png [ 4.56 KiB  Viewed 11269 times ]
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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30 Apr 2012, 00:28
hi Bunuel, Thanks a lot for the explanation. Is there any general concept I can follow in these speed, rate, and distance questions. I mean looking at the question I had no fixed approach towards the answer because I saw felt information was too little. It would more helpful if you can share how do YOU approach these type questions. p.s: Kudos +1 thanks,



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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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30 Apr 2012, 02:46
kartik222 wrote: hi Paragkan,
how did you get to the formula related to ratio of speed at the first place?
thanks, I read that formula in some book. I have not derived it



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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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Re: speed distance time [#permalink]
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paragkan wrote: If two objects A and B start from opposite points and, after having met en route, reach their respective destinations in a and b mins (or any other measure of time) respectively, then the ratio of their speeds
\(\frac{Sa}{Sb} = \sqrt{\frac{Time taken by B}{Time taken by A}}\)
Hence:
\(\frac{Sa}{Sb} = \sqrt{\frac{90}{40}}\) \(\frac{Sa}{Sb} = \frac{3}{2}\)
Hence, time taken by A to cover the same distance which B covers in 90 minutes is:
Ta = \(\frac{90*2}{3}\) Ta = 60 minutes
Hence, total time of A = 40 minutes + 60 minutes = 1 hour 40 minutes Great formula. I solved the question the conventional way but took a whopping 4.5 minutes to do so. It took 2 minutes to figure out itself. The formula will surely help.
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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14 Mar 2013, 02:48
Bunuel wrote: vdadwal wrote: Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes Look at the diagram below: Attachment: Boston  New York.png The rate of car A is \(a\) and the rate of B is \(b\); The distance covered by A before the meeting point is \(x\) and the istance covered by B before the meeting point is \(y\); Time before meeting \(t\). Since car A covered the distance of \(y\) in 40 minutes then the rate of car A = distance/time = \(a=\frac{y}{40}\), but the same distance of \(y\) was covered by car B in \(t\) minutes, so \(y=bt\) > \(a=\frac{bt}{40}\); Since car B covered the distance of \(x\) in 90 minutes then the rate of car B = distance/time = \(b=\frac{x}{90}\), but the same distance of \(x\) was covered by car A in \(t\) minutes, so \(x=at\) > \(b=\frac{at}{90}\); Substitute \(b\) in the first equation: \(a=\frac{at}{90}*\frac{t}{40}\) > reduce by \(a\) and crossmultiply: \(t^2=3600\) > \(t=60\) minutes, hence it took car A 60+40=100 minutes to cover the whole distance. Answer: D. Hope it's clear. Bunuel, You are a genius Hats off to you. I can die to have brains like you.



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Re: speed distance time [#permalink]
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18 Mar 2013, 01:56
eaakbari wrote: paragkan wrote: If two objects A and B start from opposite points and, after having met en route, reach their respective destinations in a and b mins (or any other measure of time) respectively, then the ratio of their speeds
\(\frac{Sa}{Sb} = \sqrt{\frac{Time taken by B}{Time taken by A}}\)
Hence:
\(\frac{Sa}{Sb} = \sqrt{\frac{90}{40}}\) \(\frac{Sa}{Sb} = \frac{3}{2}\)
Hence, time taken by A to cover the same distance which B covers in 90 minutes is:
Ta = \(\frac{90*2}{3}\) Ta = 60 minutes
Hence, total time of A = 40 minutes + 60 minutes = 1 hour 40 minutes Great formula. I solved the question the conventional way but took a whopping 4.5 minutes to do so. It took 2 minutes to figure out itself. The formula will surely help. There are many formulas that you can learn to cover the various specific question demands but it is not feasible to remember them all. It will be much better if you try to logically figure it out in case you are unable to recall the formula at crunch time. Think of the situation when they meet: (Boston) A>____________________M_________<B (New York) A starts from Boston and B from New York simultaneously. After some time, say t mins of travel, they meet at M. Since A covers the entire distance of Boston to New York in (t + 40) mins and B covers it in (t + 90) mins, A is certainly faster than B and hence the point M is closer to New York. Distance between Boston and M/Distance between M and New York = Time taken to go from Boston to M/Time taken to go from M to New York = t/40 = 90/t t = 60 mins (distance varies directly with time) So A takes 60 mins + 40 mins = 1 hr 40 mins to cover the entire distance.
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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04 Aug 2013, 12:50
Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
Both cars leave at the same time Both cars travel at constant speed
Stealing a useful piece of information from Paragkan:
If two objects A and B start from opposite points and, after having met en route, reach their respective destinations in a and b mins (or any other measure of time) respectively, then the ratio of their speeds
ratio of speed: (a/b) = sq. rt(b/a) sq. rt(b/a) sq. rt(90/40) sq. rt(3/2)
So, for every three units of distance A travels, B travels two. Because we know the ratio of speed and the time it took B to travel the distance A hasn't yet covered, we can find the time it took A to cover the distance B did in 90 minutes.
90*(2/3) where 2/3 represents the lesser amount of time it took A to travel the distance B did in 90 minutes.
= 60 minutes.
Therefore, A took 40 minutes to travel the first portion then 60 minutes to travel the distance B did in 90 minutes. A spent (40+60)=100 minutes on the road.
D. 1 hour 40 minutes
Bunuel, in your explanation, for a you had speed of a=y/40 which I understand, but why then do you solve for the distance of b? Why wouldn't I do this in other similar problems I have solved? thanks!



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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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05 Aug 2013, 10:26
To put it in simpler terms Let x be the distance traveled by A and y be the distance traveled by B and let a and b be the respective rates So as before meeting they travel for the same time. Let that time be t Now, we can say that as distance traveled by A before meeting B is the same as the distance traveled by B after meeting A and viceverse So, 1) a*t=b*90 2) b*t=a*40 Now to solve this, It depends upon what is required: We can either substitute value of b to get t or divide 1)/2) to get ratio of speeds. If we follow the first approach we get a*t=b*90 => a=90b/t (1) from 2nd we get b=a*40/t (2) a=90*a*40/t*t =>t*t=3600 t=60 mins On the other hand if we were to divide it we would get the formula for speed ratios a/b=b/a*(90/40) a^2/b^2=90/40 =>a/b=3/2 and this is how the formula was reached at.
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Re: speed distance time [#permalink]
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17 Mar 2014, 06:42
VeritasPrepKarishma wrote: Distance between Boston and M/Distance between M and New York = Time taken to go from Boston to M/Time taken to go from M to New York = t/40 = 90/t t = 60 mins (distance varies directly with time)
Do you mean to say: \(\frac{Distance Between Boston And M}{Distance Between M And New York} = \frac{Time Taken To Go From Boston To M}{Time Taken To Go From M to New York} = \frac{t}{40} = \frac{90}{t}\) Can you please explain how have you got distance as 't'. Time taken has been assumed as 't'.



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Re: speed distance time [#permalink]
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17 Mar 2014, 20:19
idinuv wrote: VeritasPrepKarishma wrote: Distance between Boston and M/Distance between M and New York = Time taken to go from Boston to M/Time taken to go from M to New York = t/40 = 90/t t = 60 mins (distance varies directly with time)
Do you mean to say: \(\frac{Distance Between Boston And M}{Distance Between M And New York} = \frac{Time Taken To Go From Boston To M}{Time Taken To Go From M to New York} = \frac{t}{40} = \frac{90}{t}\) Can you please explain how have you got distance as 't'. Time taken has been assumed as 't'. I did not take the distance at 't'. 't' has been assumed to be the time taken for them to meet after starting from their respective starting points. When speed of an object stays constant, the ratio of distance traveled is equal to the ratio of time taken. In 1 hr, it travels 20 miles. In 2 hrs, 40 miles and so on... For more on the use of ratios in TSD, check: http://www.veritasprep.com/blog/2011/03 ... osintsd/Now look at only car A. It travels from Boston to M in t mins and from M to New York in 40 mins. \(\frac{Time Taken ByATo Go From Boston To M}{Time TakenByA To Go From M to New York} = \frac{t}{40}\) But, we discussed that ratio of distances will be the same as ratio of Time take so \(\frac{Distance Between Boston And M}{Distance Between M And New York} = \frac{Time Taken ByATo Go From Boston To M}{Time TakenByA To Go From M to New York} = \frac{t}{40}\) Similarly, considering car B, we get \(\frac{Distance Between Boston And M}{Distance Between M And New York} = \frac{Time Taken ByATo Go From Boston To M}{Time TakenByA To Go From M to New York} = \frac{90}{t}\) Note that both t/40 and 90/t are the ratios of \(\frac{Distance Between Boston And M}{Distance Between M And New York}\) so they will be equal. Does it all make sense now?
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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16 Nov 2014, 23:11
Let's say it took A and B each t minutes to reach to the point they meet. It took A 40 minutes to cover the distance it took B in t minutes. Can we derive the distance? Not really, but we can apply the rate formula here: distance covered by A after meeting = distance covered by B before meeting > 90*Sb = t*Sa (whereby Sb is the speed of B and Sa is the speed of A). Similarly distance covered by B after meeting = distance covered by A before meeting > 40*Sa = t*Sb ==> Sb = (t*Sa)/90 ==> substituting Sb into first equation gives 40*Sa = (t^{2}*Sa)/90 ==> 40 = t^2/90 ==> t^2=3600 ==> t=60. Thus, it took a t+40=60+40=100 minutes to travel from Boston to New York.



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Two cars A and B start from Boston and New York respectively [#permalink]
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16 Dec 2016, 07:24
Bluelagoon wrote: Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes since time taken to cover rest distance after they met =A/B=40/90 so thier speed must be in ratio √B/A==3/2 now let the time covered when they met =t then after meeting and reaching the destination will have same distance total distance by A = 3(40+t) total distance by B = 2(90+t) since distance is same 3(40+t)=2(90+t) t=60mints total time by A =40+60 minutes 1hr 40mints Ans D



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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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16 Dec 2016, 11:43
Bluelagoon wrote: Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes The answer choices look pretty easy, so I'm going to plug them in. Start with B. If car A took 1 hr 10 minutes in total, then:  car A spent 30 minutes on the Bostonmeeting leg, and 40 minutes on the meetingNY leg.  car B spent 30 minutes on the NYmeeting leg (since they left at the same time), and 90 minutes on the meetingBoston leg. But that doesn't make sense. Between Boston and the meeting, Car A was 3 times as fast as Car B. Between NY and the meeting, Car A was actually slower than Car B. Let's try another answer choice. Check D. If car A took 1 hr 40 minutes in total, then:  car A spent 60 minutes on the Bostonmeeting leg, and 40 minutes on the meetingNY leg.  car B spent 60 minutes on the NYmeeting leg (since they left at the same time), and 90 minutes on the meetingBoston leg. So between Boston and the meeting, car B took 1.5x as long (90/60) as car A. Between NY and the meeting, car B took 1.5x as long (60/40) as car A. That all checks out, so D is right.
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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17 Dec 2016, 19:14
Solution attached
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Re: Two cars A and B start from Boston and New York respectively [#permalink]
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22 Jan 2018, 21:27
Bunuel wrote: vdadwal wrote: Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds along the same route. After meeting at a point between Boston and New York the two cars A and B proceed to their respective destinations of New York and Boston. Car A reaches New York 40 minutes after the two cars have met and Car B reaches Boston 90 minutes after they have met. How long did Car A take to cover the distance between Boston and New York?
A. 1 hour B. 1 hour 10 minutes C. 2 hours 30 minutes D. 1 hour 40 minutes E. 2 hours 10 minutes Look at the diagram below: Attachment: The attachment Boston  New York.png is no longer available The rate of car A is \(a\) and the rate of B is \(b\); The distance covered by A before the meeting point is \(x\) and the distance covered by B before the meeting point is \(y\); Time before meeting \(t\). Since car A covered the distance of \(y\) in 40 minutes then the rate of car A = distance/time = \(a=\frac{y}{40}\), but the same distance of \(y\) was covered by car B in \(t\) minutes, so \(y=bt\) > \(a=\frac{bt}{40}\); Since car B covered the distance of \(x\) in 90 minutes then the rate of car B = distance/time = \(b=\frac{x}{90}\), but the same distance of \(x\) was covered by car A in \(t\) minutes, so \(x=at\) > \(b=\frac{at}{90}\); Substitute \(b\) in the first equation: \(a=\frac{at}{90}*\frac{t}{40}\) > reduce by \(a\) and crossmultiply: \(t^2=3600\) > \(t=60\) minutes, hence it took car A 60+40=100 minutes to cover the whole distance. Answer: D. Hope it's clear. I have solved using a different approach and am getting E as the answer, can you please let me know what have I done wrong Assume total distance = d Assume total time = t Assume distance covered by A by the time it meets B = kd kd is covered in (t40) min by A and in 90 min by B Similarly the remaining distance covered by A will be (1k)d which is covered in 40 min The same distance (1k)d will be covered by B in (t90) min Now, because the rate is constant for A for the entire distance and similarly for B I came up with below eqns kd/(t40) = (1k)d/40 (1k)d/(t90) = kd/90 On solving these equations, I am getting 2hr 10 min as the answer Can you please help
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Two cars A and B start from Boston and New York respectively [#permalink]
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14 Feb 2018, 23:05
The time to cover the remaining distance varies because the speed of the two trains vary and the remaining distance varies. The remaining distance varies because the speed varies. So t1/t2 = (speed2/speed1)* (distremaining2/distremaining1) As per our reasoning it is: 4/9= (speed2/speed1)*(speed2/speed1) So speed2/speed1=2/3 . So if train B took 90 min to travel its remaining distance, train A would take (2/3)*90=60 min Total time = 60+40=100 min
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