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Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be:

CAT · 2022 · Quant Slot 3
Question:
Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be:

Options

8
12
6
4
Answer: 12

Explanation:
Step 1: Use triangle properties The initial triangle APB is equilateral, with each side = 24 km. The slower ship travels 8 km along its route, and the triangle formed by the ships and the port becomes right-angled. Since the original triangle is equilateral, the angle at the slower ship’s initial position is 60 degrees. In the right-angled triangle, the right angle is at point B (the slower ship's current position). Step 2: Use sine in the right-angled triangle Right-angled triangle: Angle at the port = 60 degrees Hypotenuse = AP = 24 km (distance from faster ship to port) Using sine rule: sin(30 degrees) = BP / AP BP = 0.5 × 24 = 12 km BP is the distance from the slower ship to the line of the faster ship, giving the ratio of distances traveled. Step 3: Determine distances BF = FP - BP = 24 - 12 = 12 km This means the faster ship is moving twice as fast as the slower ship. Step 4: Find remaining distance when faster ship reaches port Faster ship covers 24 km to reach the port. Slower ship, moving at half the speed, covers only 12 km in the same time. Slower ship's initial distance = 24 km → remaining distance = 24 - 12 = 12 km

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