Step 1: Understanding the Concept:
We need to find the number of minutes Wanda was charged for. Let's denote this by \(M\). The charging structure involves a rate for the first minute and a different rate for subsequent minutes.
Step 2: Key Formula or Approach:
Let \(C_1\) be the charge for the first minute and \(C_s\) be the charge for each subsequent minute. The total charge \(T\) for \(M\) minutes is given by:
\[ T = C_1 + (M-1)C_s \]
Step 3: Detailed Explanation:
Analyze Statement (1): The total charge for Wanda's telephone call was \$6.50.
This gives us the equation: \(6.50 = C_1 + (M-1)C_s\).
We have one equation with three unknowns (\(M\), \(C_1\), \(C_s\)). We cannot solve for \(M\). For example, if the call was 1 minute (\(M=1\)), the cost for the first minute would be \$6.50. If the call was 2 minutes (\(M=2\)), we have \(6.50 = C_1 + C_s\), which is possible for many different rate combinations. Statement (1) is not sufficient.
Analyze Statement (2): Wanda was charged \$0.50 more for the first minute of the telephone call than for each minute after the first.
This gives us the relationship: \(C_1 = C_s + 0.50\).
This statement provides a relationship between the rates but gives no information about the total cost or the number of minutes. Statement (2) is not sufficient.
Analyze Both Statements Together:
We have two equations:
1) \(6.50 = C_1 + (M-1)C_s\)
2) \(C_1 = C_s + 0.50\)
Substitute the second equation into the first:
\[ 6.50 = (C_s + 0.50) + (M-1)C_s \]
\[ 6.50 = C_s + 0.50 + MC_s - C_s \]
\[ 6.50 = MC_s + 0.50 \]
\[ 6.00 = MC_s \]
We are left with one equation, \(MC_s = 6\), with two unknowns (\(M\) and \(C_s\)). We cannot find a unique value for \(M\). For example:
If \(C_s = \$1.00\), then \(M=6\) minutes.
If \(C_s = \$0.50\), then \(M=12\) minutes.
If \(C_s = \$0.75\), then \(M=8\) minutes.
Since we cannot find a unique value for \(M\), the statements together are not sufficient.
Step 4: Final Answer:
Even with both pieces of information, the number of minutes charged cannot be uniquely determined.