In a computer simulation, each player uses an identical pail to pour water repeatedly into his assigned tank, which is initially empty. Two players, Chris and Elliot, participate in this simulation and pour water into their assigned tanks as follows.

• Chris pours 5 litres of water into the tank using the pail for the first time. Subsequently, he pours 0.3 litres less than the preceding volume of water poured.
• Elliot pours 5 litres of water into the tank using the pail for the first time. Subsequently, he pours 85% of the preceding volume of water poured.

The simulation stops when the amount of water that can be poured from the pail by either player is no longer positive.

1. Set up an inequality to find the number of times Chris can pour a positive volume of water.

Let $u_n$ be the volume poured by Chris on the $n$-th time. The volumes form an arithmetic progression with first term $a = 5$ and common difference $d = -0.3$.
For the volume poured to be positive, we need $u_n > 0$.
$5 + (n-1)(-0.3) > 0$

2. Solve the inequality to find the maximum integer value of $n$.

$0.3(n-1) < 5$
$n-1 < \frac{5}{0.3} = 16\frac{2}{3}$
$n < 17\frac{2}{3}$
Since $n$ must be an integer, the maximum number of times Chris can pour water is $17$.

3. Calculate the total volume of water in the tank after 17 pours using the sum of an arithmetic progression formula.

Let $S_{17}$ be the total volume after $17$ pours.
$S_{17} = \frac{17}{2} [2(5) + (17-1)(-0.3)]$
$S_{17} = 8.5 [10 - 16(0.3)]$
$S_{17} = 8.5 [10 - 4.8]$
$S_{17} = 8.5 (5.2) = 44.2$

The volume of water in the tank is $44.2 \text{ litres}$.

1. Calculate the maximum possible total volume of water Elliot can pour using the sum to infinity of a geometric series.

For Elliot, the volumes of water poured form a geometric progression with first term $a = 5$ and common ratio $r = 0.85$.
The maximum volume of water he can pour is the sum to infinity of this series:
$S_\infty = \frac{a}{1 - r}$
$S_\infty = \frac{5}{1 - 0.85}$
$S_\infty = \frac{5}{0.15} = 33\frac{1}{3} \text{ litres}$

2. Compare the maximum volume with the target volume to evaluate Elliot's comment.

Since the maximum possible volume he can pour ($33\frac{1}{3}$ litres) is strictly less than the target volume of $33.5$ litres, Elliot will never reach the target.
Therefore, Elliot's comment is justified.

1. Formulate an inequality for the volume of water Chris has poured after $n$ times.

Let $n$ be the number of times Chris pours water. The volumes poured form an arithmetic progression with first term $a = 5$ and common difference $d = -0.3$.

We require the sum of $n$ terms, $S_n$, to be at least 30:
$S_n = \frac{n}{2}[2(5) + (n-1)(-0.3)] \ge 30$
$\frac{n}{2}(10 - 0.3n + 0.3) \ge 30$
$10.3n - 0.3n^2 \ge 60$
$0.3n^2 - 10.3n + 60 \le 0$
$3n^2 - 103n + 600 \le 0$

2. Solve the inequality to find the minimum number of times Chris needs to pour.

Solving $3n^2 - 103n + 600 = 0$ using the quadratic formula:
$n = \frac{103 \pm \sqrt{(-103)^2 - 4(3)(600)}}{2(3)}$
$n \approx 7.435 \text{ or } n \approx 26.898$

The inequality holds for $7.435 \le n \le 26.898$. Since $n$ must be an integer, the minimum number of pours Chris needs is $n = 8$.

3. Formulate an inequality for the volume of water Elliot has poured after $k$ times.

Let $k$ be the number of times Elliot pours water. The volumes poured form a geometric progression with first term $a = 5$ and common ratio $r = 0.85$.

We require the sum of $k$ terms, $S_k$, to be at least 30:
$S_k = \frac{5(1 - 0.85^k)}{1 - 0.85} \ge 30$
$\frac{5(1 - 0.85^k)}{0.15} \ge 30$

4. Solve the inequality to find the minimum number of times Elliot needs to pour.

$1 - 0.85^k \ge 30 \times \frac{0.15}{5}$
$1 - 0.85^k \ge 0.9$
$0.85^k \le 0.1$
Taking the natural logarithm on both sides:
$k \ln(0.85) \le \ln(0.1)$
Since $\ln(0.85)$ is negative, the inequality sign flips:
$k \ge \frac{\ln(0.1)}{\ln(0.85)}$
$k \ge 14.168$
Since $k$ must be an integer, the minimum number of pours Elliot needs is $k = 15$.

5. Compare the number of pours to determine who reaches the target first.

Chris requires 8 pours, while Elliot requires 15 pours. Because both players pour at the same time each time, Chris will reach or exceed the target volume first.

1. Set up the inequality for the total volume of water poured after 7 times.

Let $R = \frac{r}{100}$. The volume of water poured each time forms a geometric sequence with the first term $a = 5$ and common ratio $R$.

The total volume after 7 pours is the sum of the first 7 terms:
$S_7 = \frac{5(1 - R^7)}{1 - R}$

We require the total volume to reach or exceed 30 litres:
$\frac{5(1 - R^7)}{1 - R} \ge 30$

2. Solve the inequality using a graphing calculator to find the smallest integer value of $r$.

Using a graphing calculator to solve the corresponding equation $\frac{5(1 - R^7)}{1 - R} = 30$:
$1 - R^7 = 6(1 - R)$
$R^7 - 6R + 5 = 0$

From the GC, we find $R \approx 0.9481$ (rejecting $R=1$ as $S_7 = 35 \ne 30$).
Since $S_7$ is an increasing function of $R$, we have:
$R \ge 0.9481$
$\frac{r}{100} \ge 0.9481 \implies r \ge 94.81$

Rounding to the nearest integer, the smallest value of $r$ is $95$.