Question:
A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point?
A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point?
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Let the radius of each circle be \(r\). Find the perimeter of the enclosing rectangle and the length of the figure of eight, and compare the two.
Updated On: Nov 4, 2026
- 3.88%
- 4.22%
- 4.72%
- 4.44%
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Dimensions of the rectangle:
Let each circle have radius \(r\). The two circles touch each other and the rectangle's edges touch both circles, so the rectangle is \(4r\) long (two diameters) and \(2r\) wide (one diameter).
Step 2: Distance run by A:
A runs once around the rectangle: \(2(4r + 2r) = 12r\).
Step 3: Distance run by B:
B runs the figure of eight, which is one full circle for each of the two circles: \(2 \times 2\pi r = 4\pi r\).
Step 4: Speed needed:
Both must take the same time, so the speeds are in the ratio of the distances:
\(\dfrac{v_B}{v_A} = \dfrac{4\pi r}{12 r} = \dfrac{\pi}{3} \approx 1.0472\).
So B must run about \(4.72\%\) faster than A.
Final Answer:
The correct option is (3): 4.72%
Let each circle have radius \(r\). The two circles touch each other and the rectangle's edges touch both circles, so the rectangle is \(4r\) long (two diameters) and \(2r\) wide (one diameter).
Step 2: Distance run by A:
A runs once around the rectangle: \(2(4r + 2r) = 12r\).
Step 3: Distance run by B:
B runs the figure of eight, which is one full circle for each of the two circles: \(2 \times 2\pi r = 4\pi r\).
Step 4: Speed needed:
Both must take the same time, so the speeds are in the ratio of the distances:
\(\dfrac{v_B}{v_A} = \dfrac{4\pi r}{12 r} = \dfrac{\pi}{3} \approx 1.0472\).
So B must run about \(4.72\%\) faster than A.
Final Answer:
The correct option is (3): 4.72%
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