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Two cars A and B start from Boston and New York respectively

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cloaked_vessel
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Two cars A and B start from Boston and New York respectively [#permalink] New post   13 Mar 2005, 10:19
Two cars A and B start from Boston and New York respectively simultaneously and travel towards each other at constant speeds. 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: Two cars A and B start from Boston and New York respectively [#permalink] New post   13 Mar 2005, 12:42
With the information given, you can compute the ratio of the rates r1:r2 as 3:2, where r1 is the rate of A and r2 is the rate of B. You also can deduce that it takes 1 hr for them to meet by picking numbers, if the total distance is 50, with the above ratio the 2 will meet after 1 hour and this is true for any distance between Boston and New York if the relative ratios are known. Further, A travels an additional 40 mins, so A travels a total of 100 mins or 1hr 40 mins.
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   Updated on: 13 Mar 2005, 12:58
1h and 40m D but I did a lot of stupid calculations
prep_gmat, how do you find the ratio 3:2?

Originally posted by thearch on 13 Mar 2005, 12:53.
Last edited by thearch on 13 Mar 2005, 12:58, edited 1 time in total.
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   13 Mar 2005, 13:07
prep_gmat, how do you find the ratio 3:2?

The time they both travel at the point they meet are equal:
d=rt
(distance traveled by b at the meeting pt) / time it take A to cover that distance = rate of A or r1

r2.t = dist. traveled by b. r2t/(2/3) = r1 ; t = (2/3)*(r1/r2)
Similarly, t is also equal to (3/2)*(r2/r1), from here, you can calc the ratio of r1:r2 as 3:2.
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   13 Mar 2005, 20:56
"A"

If A and B r their speeds then

Distance(total) = 90B+40A
Time for A = (90B+40A)/A = 90B/A + 40....eq1


If they met after t hrs then

90B = At
40A = Bt

9B/4A = A/B

(B/A)^2 = 4/9 ==> B/A = 2/3....eq2

From eq 1 and 2

90*2/3+40 = 100 mins.....1 hrs 40 mins
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   13 Mar 2005, 22:34
v1t+v2t=90v2+40v1
40v1+v1t=90v2+v2t
1)-2)
v2t-40v1=40v1-v2t
40v1=v2t
same thing v1t=90v2
t=40v1/v2=90v2/v1
9v2^2=4v1^2
3v2=2v1
t=60

Therefore the time A needs is 40+t=100 minute
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   15 Mar 2005, 13:21
Sigh! took me ten minutes but D here too
does neone know a quick way to do this?
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   15 Mar 2005, 13:55
took me less than a minute.

d = distance from boston to NY
x = distance covered by "a" after meeting "b" in 40 minutes.
t = time they travel for before they meet

Va = x/40 min = x/ (2/3) mph = (d-x) /t - (1)
Vb = x/t = (d-x)/ 90 min = (d-x)/ (3/2) mph - (2)

Goal is to find t, so dividing 1 by 2 to eliminate x and d,

6*t^2 = 6 or t= 1 hr

thus total time = 1hr 40 minutes
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Re: Two cars A and B start from Boston and New York respectively [#permalink] New post   25 Mar 2005, 21:22
Could someone explain this a different way? I really don't get this problem.



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Re: Two cars A and B start from Boston and New York respectively [#permalink]
25 Mar 2005, 21:22
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