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# While on a straight road, car X and car Y are traveling at

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Director
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While on a straight road, car X and car Y are traveling at [#permalink]  26 Apr 2009, 21:46
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While on a straight road, car X and car Y are traveling at different constant rates.. If car X is now 1mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?

1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.

2) Three minutes ago car X was 1/2 mile ahead of car Y.

did any one express this algebraically?
Manager
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Re: Rate Problem..OG #79 [#permalink]  27 Apr 2009, 00:43
You dont need to get the solution. You just need to know whether you can calculate if you had to. And for this question, you can if you have the relative speed of the two vehicles.

(1): D=1, Relative Speed=10-----> Time..1/10 hour
(2): D=1/2 mile T=3 mins, ie, 1/20 hour--->Relative Speed =10. Now use this as you used this in statement (1)

Algebra would waste a lot of time.

Ans: D
Director
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Re: Rate Problem..OG #79 [#permalink]  27 Apr 2009, 09:11
shkusira wrote:
You dont need to get the solution. You just need to know whether you can calculate if you had to. And for this question, you can if you have the relative speed of the two vehicles.

(1): D=1, Relative Speed=10-----> Time..1/10 hour
(2): D=1/2 mile T=3 mins, ie, 1/20 hour--->Relative Speed =10. Now use this as you used this in statement (1)

Algebra would waste a lot of time.

Ans: D

I know we dont need to solve the problem all the way through, but I like to do so in practice, just in case I get a similar PS problem. I try to stick with the same model when it comes to rate problems, using R*T=D. Can you set up an algebraic equation for this problem that would give a solution?

Thanks!
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Re: Rate Problem..OG #79 [#permalink]  28 Apr 2009, 01:20
bigtreezl wrote:

I know we dont need to solve the problem all the way through, but I like to do so in practice, just in case I get a similar PS problem. I try to stick with the same model when it comes to rate problems, using R*T=D. Can you set up an algebraic equation for this problem that would give a solution?

Thanks!

Algebraically,

Say, after h hours, the distance between the two cars is 2 miles.

Thus, 50h - 40h + 1 (initially, the two cars were 1 mile apart) = 2
or, 10h = 1
or, h = 1/10.....sufficient.

For stmt2:
in 3 minutes, car X travelled 3X miles and car Y travelled Y miles. Also, the two cars were 0.5 miles away.

Hence, 3X - 3Y + 0.5 = 1
or, 3X - 3Y = 0.5 -----(1)

Say, the two cars will be 2 miles apart in h hours from now.
Thus, Xh - Yh + 1 = 2
or, h(X-Y) = 1 ------(2)

Using equations (1) and (2), One can get h.....hence, sufficient.

Hence, D should be the answer.

I hope, this is what you were looking for.
Director
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Re: Rate Problem..OG #79 [#permalink]  28 Apr 2009, 08:35
Senior Manager
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Re: Rate Problem..OG #79 [#permalink]  28 Apr 2009, 17:10
I concur on D.

Got B at first, because I missed the 'is now' in the question stem.
I'd pay someone dearly, if that person taught me to be more attentive and concentrated.
Re: Rate Problem..OG #79   [#permalink] 28 Apr 2009, 17:10
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