Understanding the Problem
First, let's break down the billing structure:
1. Fixed Charge: $\(\$ 0.35\)$
- This is a base fee that's always included in the bill, no matter how long the call is.
2. Charge for the First 3 Minutes: $\(\$ 0.35\)$
- This is an additional charge that covers the first 3 minutes of the call.
3. Extra Charge After First 3 Minutes: $\(\$ 0.18\)$ per minute
- For every minute beyond the first 3 minutes, there's an extra charge of $\(\$ 0.18\)$.

Total Bill: Fixed charge + charge for first 3 minutes + extra charge for additional minutes.
We're told that Jane's recent telephone bill exceeds $\$ 2$. We need to find the minimum duration (in seconds) she was on the call to reach this bill.

Defining the Variables
Let's define:
- Let $t$ be the total duration of the call in minutes.

The bill can be calculated as:
- Fixed charge: $\(\$ 0.35\)$
- First 3 minutes: $\(\$ 0.35\)$
- Additional minutes: $t-3$ minutes at $\(\$ 0.18\)$ per minute (if $\(t>3\)$ )

So, the total cost $C$ is:

$$
\(C=0.35(\text { fixed })+0.35(\text { first } 3 \text { minutes })+0.18 \times(t-3)\)
$$


Simplify:

$$
\(\begin{gathered}
C=0.35+0.35+0.18(t-3) \\
\quad C=0.70+0.18(t-3)
\end{gathered}\)
$$


We know that $\(C>2\)$, so:

$$
\(0.70+0.18(t-3)>2\)
$$

Solving the Inequality
Let's solve for $t$ :
1. Subtract 0.70 from both sides:

$$
\(\begin{gathered}
0.18(t-3)>2-0.70 \\
0.18(t-3)>1.30
\end{gathered}\)
$$

2. Divide both sides by 0.18 :

$$
\(t-3>\frac{1.30}{0.18}\)
$$

3. Add 3 to both sides:

$$
\(t>10 . \overline{2}\)
$$


So, $\(t>10 . \overline{2}\)$ minutes. Since $t$ must be greater than approximately 10.222 minutes, the smallest integer minute would be 11 minutes. But since the billing is per minute, and we're looking for the minimum duration where the bill exceeds $\(\$ 2\)$, we need to consider fractional minutes because the bill increases continuously with each additional minute beyond 3 .

However, since the charge is per minute (not per second), the call duration is charged in whole minutes. That is, even if you talk for 10.1 minutes, you're charged for 11 minutes (assuming partial minutes are rounded up). But the problem doesn't specify rounding, so we'll assume that the time can be any real number, and the charge is exactly $\$ 0.18$ per minute beyond the first 3 minutes.

Given that, the smallest $t$ where $C>2$ is $\(t>10 . \overline{2}\)$ minutes. The minimum duration is just over 10.222... minutes.

But the question asks for the duration in seconds, rounded to the nearest integer.


Convertıng Minutes to Seconds
$\(10 . \overline{2}\)$ minutes is:
- 10 minutes $=600$ seconds
- $0.222 \ldots$ minutes $\(=\frac{2}{9}\)$ minutes $\(=\frac{2}{9} \times 60 \approx 13 . \overline{3}\)$ seconds

So, $\(10 . \overline{2}\)$ minutes $\(\approx 600+13.333 \ldots \approx 613.333 \ldots\)$ seconds.
Since the bill must exceed $\$ 2$, the duration must be just over $\(10.222 \ldots\)$ minutes, or just over 613.333... seconds. The smallest integer greater than 613.333... is 614 seconds.

But let's verify the exact point where the bill equals $\$ 2$ to ensure that exceeding $\$ 2$ requires at least 614 seconds.

Calculating Exact Threshold
Set $C=2$ :

$$
\(\begin{gathered}
0.70+0.18(t-3)=2 \\
0.18(t-3)=1.30 \\
t-3=\frac{1.30}{0.18} \\
t-3=\frac{130}{18}=\frac{65}{9} \approx 7.222 \ldots \\
t \approx 10.222 \ldots \text { minutes } \\
t \approx 613 . \overline{3} \text { seconds }
\end{gathered}\)
$$


To exceed \(\$2\), the duration must be more than 613.333... seconds. The smallest integer greater than $613.333 \ldots$ is 614 .