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S97-13

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S97-13 [#permalink]

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New post 16 Sep 2014, 01:51
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Skier Lindsey Vonn completes a straight 300-meter downhill run in \(t\) seconds and at an average speed of \((x + 10)\) meters per second. She then rides a chairlift back up the mountain the same distance at an average speed of \((x - 8)\) meters per second. If the ride up the mountain took 135 seconds longer than her run down the mountain, what was her average speed, in meters per second, during her downhill run?

A. 10
B. 15
C. 20
D. 25
E. 30
[Reveal] Spoiler: OA

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Official Solution:

Skier Lindsey Vonn completes a straight 300-meter downhill run in \(t\) seconds and at an average speed of \((x + 10)\) meters per second. She then rides a chairlift back up the mountain the same distance at an average speed of \((x - 8)\) meters per second. If the ride up the mountain took 135 seconds longer than her run down the mountain, what was her average speed, in meters per second, during her downhill run?

A. 10
B. 15
C. 20
D. 25
E. 30


First, we set up two \(RT = D\) equations, one for the downhill run and one for the ride back up the mountain.

Downhill run: \((x + 10)t = 300\)

Ride back up: \((x - 8)(t + 135) = 300\)

Technically, we just have to do some algebra & arithmetic from here on out. However, these equations are very difficult to solve in their current state. The tipoff for you is that the variable \(x\) does not represent, on its own, either the downhill or the uphill speed. Thus, the equations wind up being thorny (although still solvable).

However, we can reduce the complexity by creating a new variable, say \(r\), that represents the speed on the ride back up. In other words, \(r = x - 8\). We can rewrite this equation as \(r + 8 = x\), and thus the downhill speed, \(x + 10\), can be re-expressed as \(r + 18\). As you’ll see, this simplifies the algebra. In this sort of situation, when a variable such as \(x\) does not represent any real speed in the scenario, our instinct should be to replace \(x\) with another variable that does represent a real speed.

Downhill run: \((r + 18)t = 300\)

Ride back up: \(r(t + 135) = 300\)

Now we can set the expressions on the left side equal to each other, since they both equal 300:
\((r + 18)t = r(t + 135)\)
\(rt + 18t = rt + 135r\)
\(18t = 135r\)
\(2t = 15r\)
\(t = \frac{15}{2}r\)

Finally, we substitute back into either equation (we’ll just pick the first). Since the numbers get large and we can see we’re going to get a quadratic, we might want to leave certain numbers factored as we go.
\((r + 18)(\frac{15}{2})r = 300\)
\((\frac{15}{2})r^2 + 135r = 300\)
\((\frac{15}{2})r^2 + 135r - 300 = 0\)

Now divide by 15 and multiply throughout by 2.
\(r^2 + 18r - 40 = 0\)
\((r + 20)(r - 2) = 0\)

Since \(r\) must be positive (it represents a speed), \(r\) must be 2. Thus, Lindsey’s downhill speed, in meters per second, is \(r + 18 = 20\).


Answer: C
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Collection of Questions:
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Re: S97-13 [#permalink]

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New post 05 Jan 2015, 13:56
I used backsolving for this problem.
Keeping in mind that answer choices represent average speed going downhill, that is (x+10), we must find out if two equations are equal to each other. So, I checked if (x+10)t=(x-8)(t+135)


A. If x+10=10, then x=0.
10*t=300
t=30, so 10*30=(0-8)*(30+135), we get that the equations are not equal. That means the answer is not the correct one.
This particular answer can be crossed out immediately, because of the other reason too. That is, x cannot be equal to 0, because in such a case it would mean that the speed of the chairlift is -8 meters per second.

B. This answer cannot be right as well. We get that x=5 and in such a case the speed of chairlift would be -3 meters per minute.

C.20*t=300
t=15, so 20*15=(10-8)*(15+135)
300=2*150 <----- C is the correct answer.

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Re: S97-13 [#permalink]

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New post 06 Mar 2015, 12:03
Bunuel's way is really good. I got confused initially with "r" being used for the 1st part.

I see that for me if I name the downhill rate r1 and uphill r2 it's better so that I don't get lost :

r1 = x + 10
x = r1 -10

r2 = x -8
x = r2+8

-- r1 = (r2+8) +10 = r2 +18

Hence the equation: (r2+18) t = (r2)(t + 135)

The trick here is to avoid multiplying 2 numbers and get a bigger one, if you try using r1 instead in the equation then you will run into that problem.

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Re: S97-13 [#permalink]

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New post 18 Sep 2017, 08:18
Hi,
The problem states she runs 300 meters. Even if you pick any rate. let's say she runs 10meters per second in downhill. Still it would take her only about 30 seconds. How can it be more than 135 seconds of uphill time. Am i missing something basic here?

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Re: S97-13 [#permalink]

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New post 08 Oct 2017, 19:20
Bunuel wrote:
Skier Lindsey Vonn completes a straight 300-meter downhill run in \(t\) seconds and at an average speed of \((x + 10)\) meters per second. She then rides a chairlift back up the mountain the same distance at an average speed of \((x - 8)\) meters per second. If the ride up the mountain took 135 seconds longer than her run down the mountain, what was her average speed, in meters per second, during her downhill run?

A. 10
B. 15
C. 20
D. 25
E. 30


(300/x-8)-(300/x+10)=135
=>5400=135(x^2-2x-80)
=>40=x^2-2x-80
=>x^2-2x-120=0
=>(x-12)(x+10)=0
=>x=12

Then downhill speed is (x+10)=12+10=22

Bunuel, can you please tell me what am I doing wrong?? :(

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Re: S97-13   [#permalink] 08 Oct 2017, 19:20
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