Shaurya and Arjit take a straight route to the same terminal point and travel with constant speeds. At the initial moment, the positions of the two and the terminal point form an equilateral triangle. When Arjit covered a distance of 80 km, the triangle becomes right- angled. When Arjit was at a distance of 120 km from the terminal point, the Shaurya arrived at the point. Find the distance between them at the initial moment assuming that there are integral distances throughout the movements described
A) 300 km
B) 240 km
C) 200 km
D) 225 km
Answers
Answer:
distance between them at the initial moment = 240 km
Step-by-step explanation:
Let say initial Distance = D km
Shaurya Covered D km When Arjit covere D - 120 km
their Speed ratio Shaurya : Arjit :: D : D - 120
Let say their speeds DK & (D - 120)K
When Arjit covered 80 km
Arjit Distance remained = D - 80 km
Arjit covered 80 km => Shaurya Covered = 80D/(D - 120)
Shaurya Distance Remained = D - 80D/(D - 120)
= (D² - 120D - 80D)/(D - 120)
= (D² - 200D)/(D - 120)
Cos 60 = (Shaurya remaining distance )/( Arjit remaining distance)
=> 1/2 = {(D² - 200D)/(D - 120)} / (D - 80)
=> (D - 80)(D - 120) = 2(D² - 200D)
=> D² - 200D + 9600 = 2D² - 400D
=> D² - 200D - 9600 = 0
=> D² - 240D + 40D - 9600 = 0
=> D (D - 240) + 40(D - 240) = 0
=> D = 240 Km
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