Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2 at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

Question

Let \(ΔABC\) be an isosceles triangle such that \(AB\) and \(AC\) are of equal length. \(AD\) is the altitude from \(A\) on \(BC\) and \(BE\) is the altitude from \(B\) on \(AC\) . If \(AD\) and \(BE\) intersect at \(O\) such that \(∠AOB =105\degree\) , then \(\frac{AD}{BE}\) equals

Answer 1

Step 1: Convert speeds from km/h to m/s

Ram's speed = \( \frac{15 \times 1000}{60 \times 60} = \frac{15 \times 5}{18} = \frac{75}{18} \) m/s
Rahim's speed = \( \frac{5 \times 1000}{60 \times 60} = \frac{5 \times 5}{18} = \frac{25}{18} \) m/s

Step 2: Determine the circumference of each circular path

Ram's path circumference: \( C_R = 2\pi \times 100 = 200\pi \) meters
Rahim's path circumference: \( C_H = 2\pi \times 20 = 40\pi \) meters

Step 3: Calculate the time required for one complete round

Ram's time per round = \( \frac{200\pi}{\frac{75}{18}} = \frac{200\pi \times 18}{75} = 48\pi \) seconds
Rahim's time per round = \( \frac{40\pi}{\frac{25}{18}} = \frac{40\pi \times 18}{25} = 28.8\pi \) seconds

Step 4: Find the Least Common Multiple (LCM) of the two round times

The LCM of \( 48\pi \) and \( 28.8\pi \) is calculated as follows: \[ \text{LCM}(48\pi, 28.8\pi) = 144\pi \text{ seconds} \]

Step 5: Determine the number of rounds Ram completes within the LCM time

The number of rounds Ram completes is: \[ \text{Rounds} = \frac{144\pi}{48\pi} = \boxed{3} \]

Answer:

Ram will meet Rahim after completing 3 full rounds.

The Correct Option is B

Solution and Explanation