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A car traveling uphill will take an hour less to cover the [#permalink]

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08 Aug 2012, 06:40

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A car traveling uphill will take an hour less to cover the distance if it moved 4km/hr faster. On the other hand it will take 3 more hours to reach the destination if it moved 6km/hr slower. What is the distance that the car travels?

Re: A car traveling uphill will take an hour less to cover the [#permalink]

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08 Aug 2012, 07:14

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SOURH7WK wrote:

A car traveling uphill will take an hour less to cover the distance if it moved 4km/hr faster. On the other hand it will take 3 more hours to reach the destination if it moved 6km/hr slower. What is the distance that the car travels?

A. 50 B. 80 C. 105 D. 75 E. 65

Source: 4gmat

ALGEBRAIC WAY:

Say it takes the car \(t\) hours to cover the distance at \(r\) kilometers per hour, so \(d=tr\). We are told that:

Re: A car traveling uphill will take an hour less to cover the [#permalink]

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13 Aug 2012, 04:03

Hi bunuel, Is there a simpler way to solve this problem. I am stuck. There are 3 unknowns and only 2 equations can be formed. Unable to comprehend the above method. _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: A car traveling uphill will take an hour less to cover the [#permalink]

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17 Aug 2012, 05:46

Expert's post

rajathpanta wrote:

Hi bunuel, Is there a simpler way to solve this problem. I am stuck. There are 3 unknowns and only 2 equations can be formed. Unable to comprehend the above method.

So, we get two distinct linear equations with two unknowns (\(4t-r-4=0\) and \(r-2t-6=0\)) --> we can solve for \(t\) and \(r\). Now, since \(distance=tr\), then we can get the distance too.

tr=(t-1)(r+4) --> tr=tr+4t-r-4--> Take the tr on the left hand side to the right hand side, the sigh of the moving tr changes and it gets cancelled,whereas the left hand side is left with nothing but zero....Same way for the second part also.. _________________

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.

Re: A car traveling uphill will take an hour less to cover the [#permalink]

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17 Sep 2012, 06:55

My approach d/r - d/(r+4) = 1 ( i) (time lag) d/(r-6) + d/r = 3 (ii) (time lag) From (i) d = r(r+4)/4 (iii) From (ii) d = r(r-6)/2 (iv) (iii) = (iv) => r = 16 and r = 16 in (iii) => d = 80 Brother Karamazov

Re: A car traveling uphill will take an hour less to cover the [#permalink]

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10 Nov 2014, 06:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A car traveling uphill will take an hour less to cover the [#permalink]

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10 Nov 2014, 21:22

Bunuel - in cases such as this when time and rate are unknown but distance is the same, are these always solvable? I'm thinking more along the lines of data sufficiency

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30 May 2016, 22:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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