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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