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Let's go through each part of the problem one by one:

### Given Information:
- The fare for the first kilometer (fixed) is Rs 20.
- For each additional kilometer, the fare increases by Rs 15 (arithmetic progression).

### (i) What is the total fare for a 10 km journey?

The first kilometer costs Rs 20, and for each of the remaining 9 kilometers, the fare increases by Rs 15. We can calculate the total fare using this logic:

- For the first km: Rs 20
- For the remaining 9 km: Rs 15 per km

\[
\text{Total fare} = 20 + (9 \times 15) = 20 + 135 =  Rs 155
\]

### (ii) A man paid Rs 395 for his journey. What was the distance covered during his journey?

Let's denote the distance covered as \( x \) kilometers.

- The fare for the first km is Rs 20.
- The fare for the remaining \( x - 1 \) kilometers is Rs 15 per kilometer.
  
The total fare equation would be:

\[
20 + (x - 1) \times 15 = 395
\]

Solving for \( x \):

\[
20 + 15(x - 1) = 395
\]

\[
20 + 15x - 15 = 395
\]

\[
15x + 5 = 395
\]

\[
15x = 395 - 5 = 390
\]

\[
x = \frac{390}{15} = 26
\]

So, the man covered **26 kilometers**.

### (iii) If the taxi company charges Rs 40 for the first km and the fare for a 28 km journey is Rs 256, how much does the company charge for each additional km?

Let's denote the additional fare per kilometer as \( y \). 

- The fare for the first km is Rs 40.
- The fare for the remaining 27 kilometers is \( y \) per km.

The total fare equation would be:

\[
40 + 27y = 256
\]

Solving for \( y \):

\[
27y = 256 - 40 = 216
\]

\[
y = \frac{216}{27} = 8
\]

So, the company charges **Rs 8 per additional kilometer**.

---

Would you like more details or have any questions? Here are 5 related questions to expand on the topic:

1. How would the formula change if the fare increase per kilometer was not constant?
2. What if the fare for the first kilometer was based on distance bands (e.g., 1–5 km)?
3. How would the calculation change if the fare was discounted after a certain distance?
4. How would you model this problem if the increase in fare was exponential instead of arithmetic?
5. Can you derive a formula for the total fare for \( n \) kilometers?

**Tip:** For any arithmetic progression problem, the sum of terms can be calculated quickly by multiplying the number of terms by the average of the first and last term.

A taxi company charges Rs 20 for the first km and Rs 15 for each additional km. (i) What is the total fare for a 10 km journey? (ii) A man paid Rs 395 for his journey. What was the distance covered during his journey? (iii) If the taxi company charges Rs 40 for the first km and the fare for a 28 km journey is Rs 256, how much does the company charge for each additional km?

Learn how to calculate taxi fare based on arithmetic progression. This problem includes calculating fare for a journey of 10 km, determining the distance traveled based on a total fare, and identifying the charge per kilometer. Suitable for Grades 8-10.

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