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The wheel of bicycles $\text{A}$ and $\text{B}$ have radii $30$ cm and $40$ cm, respectively. While traveling a certain distance, each wheel of $\text{A}$ required $5000$ more revolutions than each wheel of $\text{B}$. If bicycle $\text{B}$ traveled this distance in $45$ minutes, then its speed, in km per hour, was

- $18\pi$
- $12\pi$
- $16\pi$
- $14\pi$

1 vote

Given that,

The wheel of bicycles $A$ and $B$ have radii $r_{A} = 30 \; \text{cm}$ and $r_{B} = 40 \; \text{cm},$ respectively.

- $ \frac{A_{\text{radii}}}{ B_{\text{radii}}} = \frac{30}{40} = \frac{3}{4} $
- $ \frac{A_{\text{rev}}}{ B_{\text{rev}}} = \frac{4}{3} \; ; \boxed{ \text{Radii} \; \propto \frac{1}{\text{Revolution}}}$

While traveling a certain distance, each wheel of *$A$* required $5000$ more revolutions than each wheel of $B$.

Let $k$ be a some constant.

Then $A_{\text{rev}} = 4k, B_{\text{rev}} = 3k$

So, $4k = 3k + 5000$

$ \Rightarrow 4k – 3k = 5000$

$\Rightarrow k = 5000$

Now, revolutions of $A$ and $B$ are:

- $ A_{\text{rev}} = 4k = 4 \times 5000 = 20000$
- $ B_{\text{rev}} = 3k = 3 \times 5000 = 15000$

We know that, circumference of a circle $= \text{1 revolution} = 2 \pi r $

Using the formula $: \boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

Distance traveled by $B = B_{\text{rev}} \times 2 \pi r_{B} = 15000 \times 2 \times \pi \times 40 = 1200000 \pi\;\text{cm}$

Wheel of bicycle $B$ traveled $1200000 \pi\;\text{cm}$ in $45$ minutes, then speed of $B$ are:

$ \text{S}_{B} = \frac{1200000 \pi\;\text{cm}} {45 \; \text{minutes}}$

$ \Rightarrow \text{S}_{B} = \left(\frac{1200000 \pi} { \frac{45}{60} \times 100000}\right) \frac{\text{km}}{\text{hour}}$

$\Rightarrow \text{S}_{B} = \frac{720\pi}{45} \frac{\text{km}}{\text{hour}} = 16\pi \; \frac{\text{km}}{\text{hour}}$

$\therefore$ The speed of the wheel of bicycle $B = 16\pi \; \frac{\text{km}}{\text{hour}}.$

Correct Answer $: \text{C}$