Problem 15

Explanation

Given:

• Initial water volume = $86$ liters.
• Leakage per hour ($x$) is between $0.3$ and $0.7$ liters. This means $0.3 < x < 0.7$.
• The observation time is from $07.00$ to $11.00$, which is $4$ hours.

We calculate the range of water leaked over $4$ hours:

So, the total water leaked is between $1.2$ liters and $2.8$ liters.

Next, we calculate the remaining water volume ($Q$) at $11.00$. The volume $Q$ is the initial volume minus the total leakage:

Since the total leakage is between $1.2$ and $2.8$ liters, the range for $Q$ is:

• Lower bound of $Q$ (maximum leakage): $86 - 2.8 = 83.2$ liters.
• Upper bound of $Q$ (minimum leakage): $86 - 1.2 = 84.8$ liters.

Thus, $83.2 < Q < 84.8$.

Now we compare $P$ and $Q$:

• $P = 83.1$ liters.
• The minimum value of $Q$ is greater than $83.2$ liters.

Since $83.1 < 83.2$, the value of $P$ is always smaller than the minimum possible value of $Q$.

Therefore, $P < Q$.