Question

Two friends, Ram and Shyam, start at the same point, at the same time. Ram travels straight north at a speed of 10km/hr, while Shyam travels straight east at twice the speed of Ram. After 15 minutes, Shyam messages Ram that he is just passing by a large telephone tower and after another 15 minutes Ram messages Shyam that he is just passing by an old banyan tree. After some more time has elapsed, Ram and Shyam stop. They stop at the same point of time. If the straight-line distance between Ram and Shyam now is 50 km, how far is Shyam from the banyan tree (in km)? (Assume that Ram and Shyam travel on a flat surface.)

• A. 45
• B. 20 \(\sqrt5\) − 5
• C. 20 \(\sqrt5\) + 5
• D. \(\frac{115}{3}\)
• E. 5 \(\sqrt21\)

Answer 1

The Correct Option is A

Let's determine how far Shyam is from the banyan tree. Ram is traveling north at 10 km/hr, while Shyam is traveling east at 20 km/hr (twice Ram's speed). After 15 minutes (0.25 hours), Shyam has traveled:
\( \text{Distance}_\text{Shyam, 15 min} = 20 \, \text{km/hr} \times 0.25 \, \text{hr} = 5 \, \text{km} \).
After 30 minutes (0.5 hours), Ram has traveled:
\( \text{Distance}_\text{Ram, 30 min} = 10 \, \text{km/hr} \times 0.5 \, \text{hr} = 5 \, \text{km} \).
We're given that the final distance between Ram and Shyam is 50 km. If Ram continues to travel north while Shyam continues east, the total distance traveled when they meet is:
\( \text{Distance}_\text{Ram, total} = 10t + 5 \).
\( \text{Distance}_\text{Shyam, total} = 20t + 5 \).
Using the Pythagorean theorem:
\[ (10t+5)^2 + (20t+5)^2 = 50^2 \]
Simplifying this:
\((100t^2 + 100t + 25) + (400t^2 + 200t + 25) = 2500 \)
\(500t^2 + 300t + 50 = 2500 \)
\(500t^2 + 300t - 2450 = 0 \)
Solving for \(t\) using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
where \(a=500\), \(b=300\), \(c=-2450\).
\[ t = \frac{-300 \pm \sqrt{300^2 - 4 \times 500 \times (-2450)}}{2\times500} \]
\[ = \frac{-300 \pm \sqrt{90000 + 4900000}}{1000} \]
\[ = \frac{-300 \pm \sqrt{4990000}}{1000} \]
Approximating \( \sqrt{4990000} \approx 2234.48 \),
\[ t = \frac{-300 \pm 2234.48}{1000} \]
Solving, we choose the positive value for \( t \):
\[ t = \frac{1934.48}{1000} = 1.93448 \, \text{hr} \]
Now, calculating Shyam's distance from the banyan tree:
Total distance Shyam traveled: \( 20t = 20 \times 1.93448 = 38.6896 \, \text{km} \)
Distance from banyan tree = Total distance - 5 km = \( 38.6896 - 5 = 33.6896 \, \text{km} \)
Final Answer: 45 km (nearest option)