Math Problem Statement

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?

Solution

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

Total fare=20+(9×15)=20+135=Rs155\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 xx kilometers.

  • The fare for the first km is Rs 20.
  • The fare for the remaining x1x - 1 kilometers is Rs 15 per kilometer.

The total fare equation would be:

20+(x1)×15=39520 + (x - 1) \times 15 = 395

Solving for xx:

20+15(x1)=39520 + 15(x - 1) = 395

20+15x15=39520 + 15x - 15 = 395

15x+5=39515x + 5 = 395

15x=3955=39015x = 395 - 5 = 390

x=39015=26x = \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 yy.

  • The fare for the first km is Rs 40.
  • The fare for the remaining 27 kilometers is yy per km.

The total fare equation would be:

40+27y=25640 + 27y = 256

Solving for yy:

27y=25640=21627y = 256 - 40 = 216

y=21627=8y = \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 nn 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.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra

Formulas

Fare for first km + Fare for additional km = Total fare
Fare for n kilometers = First km fare + (n-1) × Additional km fare

Theorems

Arithmetic progression for calculating fare over multiple kilometers

Suitable Grade Level

Grades 8-10