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12 Mar 2026, 05:25
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A motorized boat traveled 30 miles directly against a current of \(c\) miles per hour, and then returned 30 miles in the direction of the current, in a total of 8 hours. If the boat traveled the entire round trip at the same constant speed relative to the water, what is the value of \(c\)?
(1) The boat traveled at a constant speed of \(4c\) miles per hour relative to the water.
(2) The boat traveled at a constant speed of 8 miles per hour relative to the water.
(1) The boat traveled at a constant speed of \(4c\) miles per hour relative to the water.
(2) The boat traveled at a constant speed of 8 miles per hour relative to the water.
12 Mar 2026, 05:25
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Official Solution:
A motorized boat traveled 30 miles directly against a current of \(c\) miles per hour, and then returned 30 miles in the direction of the current, in a total of 8 hours. If the boat traveled the entire round trip at the same constant speed relative to the water, what is the value of \(c\)?
(1) The boat traveled at a constant speed of \(4c\) miles per hour relative to the water.
If the boat’s speed relative to the water is \(4c\) miles per hour, then its actual travel speed for the first leg of the round trip (against a current of \(c\) miles per hour) is \(4c - c = 3c\) miles per hour, and its actual travel speed for the second leg of the round trip (with a current of \(c\) miles per hour) is \(4c + c = 5c\) miles per hour.
The time taken for the first leg of the trip, therefore, is \(=\frac{30\ mi}{3c\ mi/hr}=\frac{10}{c}\) hr, and the time taken for the second leg of the trip is \(=\frac{30\ mi}{5c\ mi/hr}=\frac{6}{c}\) hr.
Since the total duration of the journey was 8 hours, we can write
\(\frac{10}{c} + \frac{6}{c}=8\).
\(\frac{16}{c}=8\).
which can be solved for the unique value \(c = 2\), so statement 1 alone is sufficient.
(2) The boat traveled at a constant speed of 8 miles per hour relative to the water.
If the boat’s speed relative to the water is 8 miles per hour, then its actual travel speed for the first leg of the round trip (against a current of \(c\) miles per hour) is \((8 - c)\) miles per hour, and its actual travel speed for the second leg of the round trip (with a current of \(c\) miles per hour) is \((8 + c)\) miles per hour.
The time taken for the first leg of the trip, therefore, is \(=\frac{30\ mi}{(8 + c)\ mi/hr}=\frac{30}{8 + c}\) hr, and the time taken for the second leg of the trip is \(\frac{30\ mi}{(8 - c)\ mi/hr}=\frac{30}{8 - c}\).
Since the total duration of the journey was 8 hours, we can write
\(\frac{30}{8 + c} + \frac{30}{8 - c}=8\).
Multiply both sides by the common denominator \((8 + c)(8 - c)\) to give
\(30(8 - c) + 30(8 - c)=8(8 + c)(8 - c)\).
\(60=64-c^2\).
which can be solved for the unique positive value \(c = 2\) (\(c\) cannot be negative, as that would contradict the basic meanings of “with” and “against” the current), so statement 2 alone is also sufficient.
Answer: D
A motorized boat traveled 30 miles directly against a current of \(c\) miles per hour, and then returned 30 miles in the direction of the current, in a total of 8 hours. If the boat traveled the entire round trip at the same constant speed relative to the water, what is the value of \(c\)?
(1) The boat traveled at a constant speed of \(4c\) miles per hour relative to the water.
If the boat’s speed relative to the water is \(4c\) miles per hour, then its actual travel speed for the first leg of the round trip (against a current of \(c\) miles per hour) is \(4c - c = 3c\) miles per hour, and its actual travel speed for the second leg of the round trip (with a current of \(c\) miles per hour) is \(4c + c = 5c\) miles per hour.
The time taken for the first leg of the trip, therefore, is \(=\frac{30\ mi}{3c\ mi/hr}=\frac{10}{c}\) hr, and the time taken for the second leg of the trip is \(=\frac{30\ mi}{5c\ mi/hr}=\frac{6}{c}\) hr.
Since the total duration of the journey was 8 hours, we can write
\(\frac{10}{c} + \frac{6}{c}=8\).
\(\frac{16}{c}=8\).
which can be solved for the unique value \(c = 2\), so statement 1 alone is sufficient.
(2) The boat traveled at a constant speed of 8 miles per hour relative to the water.
If the boat’s speed relative to the water is 8 miles per hour, then its actual travel speed for the first leg of the round trip (against a current of \(c\) miles per hour) is \((8 - c)\) miles per hour, and its actual travel speed for the second leg of the round trip (with a current of \(c\) miles per hour) is \((8 + c)\) miles per hour.
The time taken for the first leg of the trip, therefore, is \(=\frac{30\ mi}{(8 + c)\ mi/hr}=\frac{30}{8 + c}\) hr, and the time taken for the second leg of the trip is \(\frac{30\ mi}{(8 - c)\ mi/hr}=\frac{30}{8 - c}\).
Since the total duration of the journey was 8 hours, we can write
\(\frac{30}{8 + c} + \frac{30}{8 - c}=8\).
Multiply both sides by the common denominator \((8 + c)(8 - c)\) to give
\(30(8 - c) + 30(8 - c)=8(8 + c)(8 - c)\).
\(60=64-c^2\).
which can be solved for the unique positive value \(c = 2\) (\(c\) cannot be negative, as that would contradict the basic meanings of “with” and “against” the current), so statement 2 alone is also sufficient.
Answer: D
