16π
12π
Bicycle A covers a distance of \(60\pi\) cm per revolution, and bicycle B covers \(80\pi\) cm per revolution. Let B make \(n\) revolutions to cover a certain distance. Then A makes \((n + 5000)\) revolutions to cover the same distance.
Since both bicycles cover the same distance: \[ n \cdot 80\pi = (n + 5000) \cdot 60\pi \]
Divide both sides by \(\pi\): \[ 80n = 60(n + 5000) \] \[ 80n = 60n + 300000 \] \[ 20n = 300000 \Rightarrow n = 15000 \]
Distance travelled by B: \[ 15000 \cdot 80\pi = 1,200,000\pi \text{ cm} \] Convert to kilometers: \[ \frac{1,200,000\pi}{100000} = 12\pi \text{ km} \]
Time taken = 45 minutes = \(\frac{3}{4}\) hour.
Speed of B: \[ \frac{12\pi}{3/4} = 16\pi \text{ km/h} \]
\[ \boxed{16\pi \text{ km/h}} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: