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Give step-by-step explanation and explain each step ad final answer EV consumes 1 5 k W h 1 0 0 k m , price 1 . 7 M TL . ICE car consumes 7 l n 1 0 0 k m , price 1 . 6 M TL . Electricity cost 3 TL / kWh , Gasoline 4 3 TL / t . What is the minimum distance for EV to be profitable?

Question:

Give step-by-step explanation and explain each step ad final answer EV consumes 1 5 k W h 1 0 0 k m , price 1 . 7 M TL . ICE car consumes 7 l n 1 0 0 k m , price 1 . 6 M TL . Electricity cost 3 TL / kWh , Gasoline 4 3 TL / t . What is the minimum distance for EV to be profitable?

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Asked by: Test User 100192
Created at: 2025-05-17 15:32:25

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Let's break down the problem step by step: ### **Given Data** - **EV (Electric Vehicle):** - Consumption: **15 kWh/100 km** - Price: **1.7 million TL** - **ICE (Internal Combustion Engine) Car:** - Consumption: **7 liters/100 km** - Price: **1.6 million TL** - **Electricity Cost:** **3 TL/kWh** - **Gasoline Cost:** **43 TL/liter** We are to **find the minimum distance (X km)** for which the **total cost of owning and operating the EV becomes less than or equal to the ICE**. --- ## **Step 1: Calculate Energy/Fuel Cost per 100 km** ### **EV running cost per 100 km:** \[ \text{Consumption} = 15~\text{kWh/100 km} \] \[ \text{Electricity Cost} = 3~\text{TL/kWh} \] \[ \text{Cost per 100 km (EV)} = 15 \times 3 = 45~\text{TL} \] ### **ICE running cost per 100 km:** \[ \text{Consumption} = 7~\text{l/100 km} \] \[ \text{Gasoline Cost} = 43~\text{TL/l} \] \[ \text{Cost per 100 km (ICE)} = 7 \times 43 = 301~\text{TL} \] --- ## **Step 2: Setup Total Cost Formulas** Let \( X \) = total distance driven (in km) ### **Total cost for EV:** \[ \text{Total Cost (EV)} = \text{Purchase Price} + \text{Running Cost} \] \[ = 1,700,000~\text{TL} + \left(\frac{X}{100}\right) \times 45 \] ### **Total cost for ICE:** \[ \text{Total Cost (ICE)} = \text{Purchase Price} + \text{Running Cost} \] \[ = 1,600,000~\text{TL} + \left(\frac{X}{100}\right) \times 301 \] --- ## **Step 3: Find the Break-Even Point** Set **Total Cost (EV) ≤ Total Cost (ICE)**: \[ 1,700,000 + \frac{X}{100} \times 45 \leq 1,600,000 + \frac{X}{100} \times 301 \] --- ## **Step 4: Solve for X** First, subtract \(1,600,000\) from both sides: \[ 100,000 + \frac{X}{100} \times 45 \leq \frac{X}{100} \times 301 \] Now bring all terms involving X to one side: \[ 100,000 \leq \frac{X}{100} \times 301 - \frac{X}{100} \times 45 \] \[ 100,000 \leq \frac{X}{100} \times (301 - 45) \] \[ 100,000 \leq \frac{X}{100} \times 256 \] Multiply both sides by 100 to clear the denominator: \[ 10,000,000 \leq X \times 256 \] \[ X \geq \frac{10,000,000}{256} \] \[ X \geq 39,062.5~\text{km} \] --- ## **Step 5: Interpret the Result** - **Minimum distance where the EV is more economical:** **39,062.5 km** --- ### **Final Answer** \[ \boxed{39,063~\text{km}} \] (Rounded to the nearest kilometer) --- ## **Explanation of Steps** 1. **Calculate the running cost per 100 km** for both vehicles using their consumption and energy/fuel prices. 2. **Write the total cost equations** for both vehicles as a function of distance traveled. 3. **Set the total costs equal and solve for distance (X)**: This finds the break-even point where the EV's higher upfront cost is offset by its cheaper running costs. 4. **Solve the equation** to find the minimum X (distance) where the EV is more cost-effective. --- ### **Summary Table** | | Upfront Cost (TL) | Running Cost (TL/100 km) | Running Cost (TL/km) | |--------|-------------------|--------------------------|----------------------| | EV | 1,700,000 | 45 | 0.45 | | ICE | 1,600,000 | 301 | 3.01 | **Break-even distance:** **39,063 km**

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