Time, Speed and Distance Question 5 - CAT 2022

Step-by-Step Explanation

Let the port be the origin point O, the initial position of the slower ship be S, and the faster ship be F. Initially, O, S, and F form an equilateral triangle with sides of 24 km, meaning the angle at the port (∠SOF) is 60°.

Analyze the right-angled state:

The slower ship S travels 8 km toward the port, so its new distance from the port is:

OS′ = 24 − 8 = 16 km

The new triangle OS′F′ is right-angled. Since ∠SOF remains fixed at 60°, the right angle must be at vertex F′.

Using basic right-triangle trigonometry:

cos(60°) = Adjacent / Hypotenuse = OF′/OS′

1/2 = OF′/16 ⟹ OF′ = 8 km

Determine the speed ratio:

During this time, the faster ship traveled from its initial 24 km position to its new 8 km position:

Distance covered by F = 24 − 8 = 16 km

Since the faster ship covered 16 km in the same time the slower ship covered 8 km, their speed ratio is:

Speed Ratio (F : S) = 16 : 8 = 2 : 1

Calculate the final position:

When the faster ship reaches the port (covering its entire 24 km distance), the slower ship will have traveled exactly half that distance due to the 2 : 1 speed ratio:

Distance covered by slower ship = 24/2 = 12 km

Remaining distance to port = 24 − 12 = 12 km

Correct Option: B

💡 FundaMakers Tip:

Since one vertex of the moving triangle remains anchored at the 60° corner of the original equilateral shape, the right triangle configuration must follow a standard 30°-60°-90° layout. The sides of such a triangle always follow the ratio 1 : √3 : 2. Recognizing this allows you to find the intermediate distances instantly without drawing out complex coordinate axes.