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A forest ranger travels up a river by boat to collect soil samples. The boat travels at a speed of 
v miles per hour for a distance of 2 miles. After collecting the samples, the ranger travels the same distance back, but the boat travels 5 miles per hour slower than it did on the way up the river. If the ranger spent 3 hours travelling up and back, which of the following equations could be used to determine the speed of the boat on the way up the river?
Choose 1 answer:
(A) 
2v^(2)-13 v+17=0
(B) 
3v^(2)-15 v-4=0
(C) 
3v^(2)-11 v+10=0
(D) 
3v^(2)-19 v+10=0

A forest ranger travels up a river by boat to collect soil samples. The boat travels at a speed of vv miles per hour for a distance of 22 miles. After collecting the samples, the ranger travels the same distance back, but the boat travels 55 miles per hour slower than it did on the way up the river. If the ranger spent 33 hours travelling up and back, which of the following equations could be used to determine the speed of the boat on the way up the river?\newlineChoose 11 answer:\newline(A) 2v213v+17=02v^{2}-13v+17=0\newline(B) 3v215v4=03v^{2}-15v-4=0\newline(C) 3v211v+10=03v^{2}-11v+10=0\newline(D) 3v219v+10=03v^{2}-19v+10=0

Full solution

Q. A forest ranger travels up a river by boat to collect soil samples. The boat travels at a speed of vv miles per hour for a distance of 22 miles. After collecting the samples, the ranger travels the same distance back, but the boat travels 55 miles per hour slower than it did on the way up the river. If the ranger spent 33 hours travelling up and back, which of the following equations could be used to determine the speed of the boat on the way up the river?\newlineChoose 11 answer:\newline(A) 2v213v+17=02v^{2}-13v+17=0\newline(B) 3v215v4=03v^{2}-15v-4=0\newline(C) 3v211v+10=03v^{2}-11v+10=0\newline(D) 3v219v+10=03v^{2}-19v+10=0
  1. Denote Speed and Distance: Let's denote the speed of the boat on the way up the river as vv miles per hour. The distance traveled up the river and back is the same, which is 22 miles each way. The time it takes to travel up the river is the distance divided by the speed, which is 2v\frac{2}{v} hours. On the way back, the boat is traveling 55 miles per hour slower, so the speed is v5v-5 miles per hour. The time it takes to travel back is the distance divided by this slower speed, which is 2v5\frac{2}{v-5} hours. The total time spent traveling is the sum of these two times, which is given as 33 hours. Therefore, we can set up the equation:\newline2v+2v5=3\frac{2}{v} + \frac{2}{v-5} = 3
  2. Set Up Equation: Now we need to solve this equation for vv. To do this, we need to find a common denominator for the two fractions on the left side of the equation. The common denominator is v(v5)v(v-5). We multiply both sides of the equation by this common denominator to eliminate the fractions:\newlinev(v5)(2v)+v(v5)(2v5)=3v(v5)v(v-5)\left(\frac{2}{v}\right) + v(v-5)\left(\frac{2}{v-5}\right) = 3v(v-5)
  3. Solve for vv: Simplifying the equation, we get:\newline2(v5)+2v=3v215v2(v-5) + 2v = 3v^2 - 15v\newlineThis simplifies to:\newline2v10+2v=3v215v2v - 10 + 2v = 3v^2 - 15v
  4. Common Denominator: Combining like terms on the left side gives us:\newline4v10=3v215v4v - 10 = 3v^2 - 15v\newlineNow we want to set the equation to zero by moving all terms to one side:\newline3v215v4v+10=03v^2 - 15v - 4v + 10 = 0
  5. Simplify Equation: Simplifying the equation further, we combine the vv terms: 3v219v+10=03v^2 - 19v + 10 = 0 This is a quadratic equation in standard form.
  6. Combine Like Terms: Looking at the answer choices, we see that the equation we derived, 3v219v+10=03v^2 - 19v + 10 = 0, matches choice (D).

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