Question from GRE Quant

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The rate of data transfer, $r$, over a particular network is directly proportional to the bandwidth, $b$, and inversely proportional to the square of the number of networked computers, $n$.

Quantity A = The resulting rate of data transfer if the bandwidth is quadrupled and the number of networked is more than tripled.

Quantity B = $\dfrac{4r}{9}$

Which is greater?

From propotionality I have got that $r=\alpha b$ and $r=\dfrac{\beta}{n^2}$ where $\alpha$ and $\beta$ are constants. Here I am stuck. But in official solution they have written that $r=k\dfrac{b}{n^2}$ where $k$ is also constant. How they got the last relation?

And can anyone explain how they derived it?

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You cannot say $r=\alpha b$ if you know $r$ depends on another variable as well (in this case, $n$). For example, consider that the volume of a cylinder is directly proportional to both the height and the square of the radius. Then it does not follow that $V=\alpha h$ for some constant $\alpha$. Rather, $\alpha$ must itself depend on $r^2$.

For this reason, your $\alpha$ must depend on $n^{-2}$. If we hold $b$ constant, we know $r$ must be directly proportional to $n^{-2}$, so we conclude $\alpha=k\cdot n^{-2}$.

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Quantity B is greater I would think about this in terms of direct and inverse relationships. If there is a direct relationship with bandwidth, then increasing bandwidth, will increase speed. And inverse relationship means that increasing the computers decreases the speed.

You are close to the correct solution, but your equations need a little bit of modification. The rate $r$ depends on both $b$ and $n$. Your first equation would work if you had a fixed number of networked computers. Likewise, your second equation would work if you had a fixed bandwidth.

Since both are variable, it makes sense that our equation should depend on both of them. So, taking the given equation of $r=k\dfrac{b}{n^2}$ This should make sense. $b$ in the numerator is the direct relationship. $n^2$ in the denominator is the inverse relationship with $n^2$.

How does this help with the question? If there is a direct relationship with bandwidth, then quadrupling the bandwidth will increase $r$ by a factor of $4$. We know that the number of networked computers has more than tripled so then. The new rate is $\frac{4}{\gt3^2}r=\frac{4}{\gt9}r$

By common fraction identities, $\frac{4}{\gt9} \lt \frac{4}{9}$