Digital PSAT Math Practice Question 324: Answer and Explanation

Question: 324

John is taking a rowboat both up and down a 16-kilometer length of a river. A constant current of 1 kilometer per hour makes his trip downstream faster than his trip upstream because he is moving with the current downstream and fighting against the current when traveling upstream. If a round-trip journey took him a total of 4 hours, and if he rowed at a constant pace the whole time, what is the rate in kilometers per hour, to the nearest tenth, at which John is rowing independent of the current?

A. 7.3
B. 8.1
C. 8.9
D. 9.7

Correct Answer: B

Explanation:

(B) Use the formula Distance = Rate x Time to make your calculations. The distance is 16 km/hour for the journey in either direction. The rates, however, are different. The rate going upstream is 1 km/hr less than the rate at which John is actually rowing because he is going against the current. The rate going downstream is 1 km/hr more than the rate at which he is actually rowing because he is going with the current. If x is the rate at which John is rowing, the time, u, to go upstream is:

d = r u \to 16 = (x - 1) \to u = \frac{16}{x - 1}

The time going downstream, t, can be calculated in a similar way:

d = r t \to 16 = (x + 1) t \to t = \frac{16}{x + 1}

Since the total time of the journey is 4 hours,combine these two expressions together into one equation:

\frac{16}{x - 1} + \frac{16}{x + 1} = 4

Then solve for x:

\begin{matrix} \frac{16}{x - 1} + \frac{16}{x + 1} = 4 \to \frac{16 (x + 1)}{(x - 1) (x + 1)} + \frac{16 (x - 1)}{(x + 1) (x - 1)} = 4 \to \\ \frac{16 (x + 1) + 16 (x - 1)}{x^{2} - 1} = 4 \to \frac{16 x + 16 + 16 x - 16}{x^{2} - 1} = 4 \to \\ \frac{32 x}{x^{2} - 1} = 4 \to 32 x = 4 x^{2} - 4 \to 4 x^{2} - 32 x - 4 = 0 \to x^{2} - 8 x - 1 = 0 \end{matrix}

Use the quadratic formula to solve:

\begin{array}{l} \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \to \frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1} = \frac{8 \pm \sqrt{68}}{2} = \\ \frac{8 \pm 2 \sqrt{17}}{2} = 4 \pm \sqrt{17} \end{array}

You get two solutions. However, you can use only $4 \pm \sqrt{17}$ , because velocity cannot be negative. The value of $4 \pm \sqrt{17}$ is approximately 8.1.

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