Question

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

• A. 12
• B. 18
• C. 20
• D. 24

Answer 1

The Correct Option is B

Step 1: Define Variables

• Let the speed of the slower ship be \( x \) km/h
• Then the speed of the faster ship is \( x + 6 \) km/h
• Time traveled by both ships = 2 hours

Step 2: Use the Pythagorean Theorem

Since the ships move at right angles (assumed from context), the distance between them forms the hypotenuse: \[ (60)^2 = (2x)^2 + [2(x + 6)]^2 \] Simplify: \[ 3600 = 4x^2 + 4(x^2 + 12x + 36) \Rightarrow 3600 = 4x^2 + 4x^2 + 48x + 144 \Rightarrow 3600 = 8x^2 + 48x + 144 \] Bring everything to one side: \[ 8x^2 + 48x - 3456 = 0 \Rightarrow x^2 + 6x - 432 = 0 \quad \text{(divide by 8)} \]

Step 3: Solve the Quadratic Equation

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 6 \), \( c = -432 \) \[ \text{Discriminant} = 6^2 - 4(1)(-432) = 36 + 1728 = 1764 \Rightarrow \sqrt{1764} = 42 \] \[ x = \frac{-6 \pm 42}{2} \Rightarrow x = \frac{36}{2} = 18 \quad \text{or} \quad x = \frac{-48}{2} = -24 \] Only the positive root is valid for speed.

Final Answer:

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

Answer 2

Let the speed of the slower ship be x km/hr.

Then, the speed of the faster ship is x + 6 km/hr.

Time travelled by both ships = 2 hours

Distance travelled by slower ship = 2x km

Distance travelled by faster ship = 2(x + 6) = 2x + 12 km

Using the Pythagorean Theorem:

(2x)² + (2x + 12)² = 60²

4x² + (4x² + 48x + 144) = 3600

8x² + 48x + 144 = 3600

8x² + 48x - 3456 = 0

Divide by 8:

x² + 6x - 432 = 0

Solve using the quadratic formula:

x = [-6 ± √(6² + 4×432)] / 2

x = [-6 ± √1764] / 2 = [-6 ± 42] / 2

x = 18 or x = -24

Since speed cannot be negative, the valid value is 18 km/hr.