The time taken, t in minutes, to download a particular gaming software is related to internet speed, x Mbps. The variables x and t are related by the formula $t = \frac{a}{b + x}$, where a and b are positive constants. The data below shows some measured values of x and t.

x (Mbps)	50	100	200	500	1000
t (minutes)	23	13	7.3	3.2	1.6

1. Read the value of $\frac{1}{t}$ from the graph when $x = 240$

From the straight line graph of $\frac{1}{t}$ against $x$, when $x = 240$, the corresponding value on the vertical axis is approximately $\frac{1}{t} = 0.17$.

2. Calculate the time $t$

$t = \frac{1}{0.17} \approx 5.882 \approx 5.9$ minutes

1. Rearrange the given formula to find an alternative linear form

Given $t = \frac{a}{b + x}$
$t(b + x) = a$
$tb + tx = a$
$tx = -bt + a$
This is in the form $Y = mX + c$ where $Y = tx$, $X = t$, $m = -b$, and $c = a$. Alternatively, $t = -\frac{1}{b}(tx) + \frac{a}{b}$.

1. Convert the given equation into linear form using logarithms

$y = p x^q$
Taking logarithm to base 10 on both sides:
$\lg y = \lg(p x^q)$
$\lg y = q \lg x + \lg p$
This is in the linear form $Y = mX + c$, where $Y = \lg y$, $X = \lg x$, gradient $m = q$, and $Y$-intercept $c = \lg p$.

2. Use the given point and gradient to find the $Y$-intercept

The gradient is given as 3, so $q = 3$.
The line passes through $(2.5, 8.1)$ in the $(\lg x, \lg y)$ plane, so we substitute $X = 2.5$ and $Y = 8.1$ into the linear equation:
$8.1 = 3(2.5) + \lg p$
$8.1 = 7.5 + \lg p$
$\lg p = 0.6$

3. Solve for $p$ and round to the nearest integer

$p = 10^{0.6}$
$p \approx 3.981$
Correct to the nearest integer, $p = 4$.