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A car traveling uphill will take an hour less to cover the

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SOURH7WK
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A car traveling uphill will take an hour less to cover the [#permalink] New post   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?

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

Source: 4gmat
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B
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   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:

\(tr=(t-1)(r+4)\) --> \(tr=tr+4t-r-4\) --> \(4t-r-4=0\);
\(tr=(t+3)(r-6)\) --> \(tr=tr-6t+3r-18\) --> \(r-2t-6=0\);

Solving gives: \(t=5\) and \(r=16\) --> \(d=tr=80\).

Answer: B.
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   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.
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   17 Aug 2012, 05:46
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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.


We have:
\(tr=(t-1)(r+4)\) --> \(tr=tr+4t-r-4\) --> \(4t-r-4=0\);
\(tr=(t+3)(r-6)\) --> \(tr=tr-6t+3r-18\) --> \(r-2t-6=0\);

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.

Hope it's clear.
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   17 Aug 2012, 22:01
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Hi could you explain why you set the two rate/time combinations equal to 0?

tr=(t-1)(r+4) --> tr=tr+4t-r-4 --> 4t-r-4=0;
tr=(t+3)(r-6) --> tr=tr-6t+3r-18 --> r-2t-6=0
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   17 Aug 2012, 22:09
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arichardson26 wrote:
Hi could you explain why you set the two rate/time combinations equal to 0?

tr=(t-1)(r+4) --> tr=tr+4t-r-4 --> 4t-r-4=0;
tr=(t+3)(r-6) --> tr=tr-6t+3r-18 --> r-2t-6=0

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..
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   19 Aug 2012, 03:48
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arichardson26 wrote:
Hi could you explain why you set the two rate/time combinations equal to 0?

tr=(t-1)(r+4) --> tr=tr+4t-r-4 --> 4t-r-4=0;
tr=(t+3)(r-6) --> tr=tr-6t+3r-18 --> r-2t-6=0


To get 2 simultaneous equation to solve 2 unknown variables.

4t-r=4 Eq 1
r-2t=6 Eq 2
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   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
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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   10 Nov 2014, 21:22
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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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Re: A car traveling uphill will take an hour less to cover the [#permalink] New post   26 Feb 2023, 03:52
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Re: A car traveling uphill will take an hour less to cover the [#permalink]
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