Question:medium

Two ships meet mid-ocean,and then,one ship goes south and the other ship goes west,both travelling at constant speeds.Two hours later,they are 60 km apart.If the speed of one of the ships is 6km per hour more than the other one,then the speed,in km per hour,of the slower ship is

Updated On: Jan 15, 2026

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The Correct Option is B

Solution and Explanation

Step 1: Variable Declaration

  • Slower ship's speed: \( x \) km/h
  • Faster ship's speed: \( x + 6 \) km/h
  • Travel duration for both ships: 2 hours

Step 2: Application of the Pythagorean Theorem

Assuming perpendicular movement of the ships, the separation distance is the hypotenuse: \[ (60)^2 = (2x)^2 + [2(x + 6)]^2 \] Simplified equation: \[ 3600 = 4x^2 + 4(x^2 + 12x + 36) \Rightarrow 3600 = 4x^2 + 4x^2 + 48x + 144 \Rightarrow 3600 = 8x^2 + 48x + 144 \] Rearrangement to standard form: \[ 8x^2 + 48x - 3456 = 0 \Rightarrow x^2 + 6x - 432 = 0 \quad \text{(division by 8)} \]

Step 3: Quadratic Equation Solution

Utilizing the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] With parameters \( a = 1 \), \( b = 6 \), \( c = -432 \): \[ \text{Discriminant} = 6^2 - 4(1)(-432) = 36 + 1728 = 1764 \Rightarrow \sqrt{1764} = 42 \] Solutions for x: \[ x = \frac{-6 \pm 42}{2} \Rightarrow x = \frac{36}{2} = 18 \quad \text{or} \quad x = \frac{-48}{2} = -24 \] The positive solution is retained as speed cannot be negative.

Final Result:

\[ \boxed{18 \text{ km/h}} \quad \text{(Option B)} \]

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