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Two cars A and B start from Boston and New York respectively [#permalink]
28 Apr 2012, 19:50

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95% (hard)

Question Stats:

48% (04:41) correct
52% (03:18) wrong based on 289 sessions

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

Re: speed distance time [#permalink]
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

\(\frac{Sa}{Sb} = \sqrt{\frac{Time taken by B}{Time taken by A}}\)

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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:

Attachment:

Boston - New York.png [ 4.56 KiB | Viewed 6980 times ]

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 cross-multiply: \(t^2=3600\) --> \(t=60\) minutes, hence it took car A 60+40=100 minutes to cover the whole distance.

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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.

Re: Two cars A and B start from Boston and New York respectively [#permalink]
30 Apr 2012, 01:31

1

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kartik222 wrote:

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.

Re: speed distance time [#permalink]
30 Dec 2012, 07:16

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

\(\frac{Sa}{Sb} = \sqrt{\frac{Time taken by B}{Time taken by A}}\)

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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 cross-multiply: \(t^2=3600\) --> \(t=60\) 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.

Re: speed distance time [#permalink]
18 Mar 2013, 01:56

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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, 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. _________________

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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!

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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 vice-verse

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. _________________

--It's one thing to get defeated, but another to accept it.

Re: speed distance time [#permalink]
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'.

Re: speed distance time [#permalink]
17 Mar 2014, 20:19

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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 ... os-in-tsd/

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.

Re: Two cars A and B start from Boston and New York respectively [#permalink]
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.

gmatclubot

Re: Two cars A and B start from Boston and New York respectively
[#permalink]
16 Nov 2014, 23:11

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