The charges per hour of internet surfing is increased by `25%` then find the percentage decrease in the time period of surfing of a user (a net savvy)Who can afford only a `10%` increase in expenditure:

Options

A. `22%`
B. `12%`
C. `15%`
D. `9.09%`

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the Initial Charges
Let’s assume the initial charge per hour of internet surfing is `C`.

### Step 2: Calculate the New Charges After Increase
The charges are increased by `25%`. Therefore, the new charge per hour can be calculated as:
\[ \text{New Charge} = C + 0.25C = 1.25C \]

### Step 3: Determine the User's Expenditure Limit
The user can afford only a `10%` increase in expenditure. If the initial expenditure is `E`, then the maximum expenditure after the increase is:
\[ \text{New Expenditure} = E + 0.10E = 1.10E \]

### Step 4: Relate Expenditure to Time and Charges
Let’s denote the initial time of surfing as `T`. The initial expenditure can be expressed as:
\[ E = C \times T \]
After the increase in charges, the new expenditure can be expressed as:
\[ \text{New Expenditure} = (1.25C) \times T' \]
where `T'` is the new time of surfing after the increase in charges.

### Step 5: Set Up the Equation
Since the new expenditure cannot exceed `1.10E`, we can set up the equation:
\[ 1.25C \times T' = 1.10 \times (C \times T) \]

### Step 6: Simplify the Equation
We can cancel `C` from both sides (assuming `C` is not zero):
\[ 1.25T' = 1.10T \]

### Step 7: Solve for the New Time `T'`
Now, we can solve for `T'`:
\[ T' = \frac{1.10T}{1.25} \]
\[ T' = 0.88T \]

### Step 8: Calculate the Percentage Decrease in Time
The percentage decrease in time can be calculated as:
\[ \text{Percentage Decrease} = \frac{T - T'}{T} \times 100 \]
Substituting `T'`:
\[ \text{Percentage Decrease} = \frac{T - 0.88T}{T} \times 100 \]
\[ \text{Percentage Decrease} = \frac{0.12T}{T} \times 100 = 12\% \]

### Final Answer
The percentage decrease in the time period of surfing is **12%**.

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