Question

at 11:30 a.m. the bottle is \\(\frac{1}{4}\\) of the way full. at what time will the bottle be \\(\frac{1}{2}\\) full?\
11:35 a.m.\
11:31 a.m.\
11:40 a.m.

Explanation:

Step1: Define exponential growth model

Let $V(t)$ be the volume at time $t$, $V_0$ be the full volume. The model is $V(t) = V_0 \cdot 2^{\frac{t - t_0}{T}}$, where $T$ is the doubling time.

Step2: Substitute initial condition

At $t_0 = 11:30$ a.m., $V(t_0) = \frac{1}{4}V_0$. So $\frac{1}{4}V_0 = V_0 \cdot 2^{\frac{0}{T}}$, which simplifies to $2^{-2} = 2^{0}$ (we need to find when $V(t)=\frac{1}{2}V_0$: $\frac{1}{2}V_0 = V_0 \cdot 2^{\frac{t - 11:30}{T}}$). Cancel $V_0$: $2^{-1} = 2^{\frac{t - 11:30}{T}}$.

Step3: Relate to doubling time

Since $\frac{1}{4}$ to $\frac{1}{2}$ is doubling the current volume. The time to double from $\frac{1}{4}$ to $\frac{1}{2}$ is the doubling time. From $\frac{1}{4}$ to $\frac{1}{2}$ is multiplying by 2, so the exponent goes from -2 to -1, an increase of 1. This means $\frac{t - 11:30}{T} = 1$, and since going from $\frac{1}{4}$ to full would take 2 doubling periods (to $\frac{1}{2}$ then to 1), we know each doubling period is 5 minutes (since 11:30 + 10 minutes would be full, so 5 minutes to double from $\frac{1}{4}$ to $\frac{1}{2}$).

Step4: Calculate target time

Add the doubling time to 11:30 a.m.: $11:30 + 0:05 = 11:35$ a.m.

Answer:

11:35 a.m.