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Updated on: 12 Mar 2026, 05:28
Originally posted by RonPurewal on 18 Feb 2026, 21:16.
Last edited by Bunuel on 12 Mar 2026, 05:28, edited 2 times in total.
Last edited by Bunuel on 12 Mar 2026, 05:28, edited 2 times in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
56% (02:05) correct
44%
(02:20)
wrong
based on 99
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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.
M48-05
23 Feb 2026, 00:15
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RonPurewal
Official Explanation:
Statement 1 alone:
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.
Statement 2 alone:
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.
The answer is D.
General Discussion
19 Feb 2026, 02:14
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Key concept: Distance-Speed-Time with relative velocity
The trap here is thinking Statement (2) is insufficient because knowing the boat's speed doesn't "feel" like enough. But once you have the boat's absolute speed, the round-trip time equation gives you a quadratic in c that has exactly one positive solution. Let's build this carefully.
Setup: Boat speed relative to water = v. Against current: effective speed = v − c. With current: v + c.
Round-trip equation: 30/(v−c) + 30/(v+c) = 8
Multiply through: 30(v+c) + 30(v−c) = 8(v−c)(v+c)
→ 60v = 8(v2 − c2) ... (*)
Step 1 — Statement (1): v = 4c
Substitute into (*): 60(4c) = 8(16c2 − c2) → 240c = 120c2 → 2 = c
c = 2. SUFFICIENT
Step 2 — Statement (2): v = 8
Substitute into (*): 60(8) = 8(64 − c2) → 480 = 512 − 8c2 → 8c2 = 32 → c2 = 4 → c = 2
(c must be positive since it's a current speed)
c = 2. SUFFICIENT
Answer: D
Takeaway: In DS boat/current problems, once the round-trip time is fixed and either the absolute or relative speed is given, you can always solve for the current — don't dismiss a statement just because the setup looks complex.
The trap here is thinking Statement (2) is insufficient because knowing the boat's speed doesn't "feel" like enough. But once you have the boat's absolute speed, the round-trip time equation gives you a quadratic in c that has exactly one positive solution. Let's build this carefully.
Setup: Boat speed relative to water = v. Against current: effective speed = v − c. With current: v + c.
Round-trip equation: 30/(v−c) + 30/(v+c) = 8
Multiply through: 30(v+c) + 30(v−c) = 8(v−c)(v+c)
→ 60v = 8(v2 − c2) ... (*)
Step 1 — Statement (1): v = 4c
Substitute into (*): 60(4c) = 8(16c2 − c2) → 240c = 120c2 → 2 = c
c = 2. SUFFICIENT
Step 2 — Statement (2): v = 8
Substitute into (*): 60(8) = 8(64 − c2) → 480 = 512 − 8c2 → 8c2 = 32 → c2 = 4 → c = 2
(c must be positive since it's a current speed)
c = 2. SUFFICIENT
Answer: D
Takeaway: In DS boat/current problems, once the round-trip time is fixed and either the absolute or relative speed is given, you can always solve for the current — don't dismiss a statement just because the setup looks complex.
