Fixed and Running Charges Word Problems | Class 10 Maths

Quick answer: In fixed-plus-running-charge problems, let the fixed charge be ₹$x$ and the charge per unit (per km or per day) be ₹$y$. Each bill gives an equation of the form $x + (\text{number of units}) \times y = \text{total}$. Subtract the two equations to find $y$, then substitute to find $x$. In the NCERT taxi problem, $x = 5$ and $y = 10$, so a $25\text{ km}$ journey costs ₹$255$.

Many bills have two parts: a fixed charge that you pay just for using the service, and a running charge that depends on how much you use it. Taxi fares, library fees and hostel bills often work this way.

These situations lead to a pair of linear equations in two variables, which can be solved quickly by elimination.

In CBSE Class 10 Mathematics, Chapter 3 (Pair of Linear Equations in Two Variables), word problems on taxi fares, library charges and hostel bills are common board exam questions.

What You Will Learn

• How to write equations for fixed and running charges
• Problem 1: the NCERT taxi fare problem
• Problem 2: the library problem, where the fixed charge covers the first three days
• Problem 3: the hostel mess bill problem
• Why the elimination method is the quickest here
• How to check your answers

1. Setting Up the Equations

Let:

• the fixed charge be ₹$x$
• the running charge per unit (per km, per day and so on) be ₹$y$

Then:

$\text{Total bill} = x + (\text{number of units}) \times y$

In both equations, $x$ has coefficient $1$, so subtracting one equation from the other removes $x$ straight away. This makes the elimination method the quickest choice.

2. Problem 1: Taxi Fares (NCERT)

Question

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is ₹$105$, and for a journey of $15\text{ km}$, the charge paid is ₹$155$.

(i) What are the fixed charges and the charge per km?

(ii) How much does a person have to pay for travelling a distance of $25\text{ km}$?

Solution

Step 1: Define the variables. Let the fixed charge be ₹$x$ and the charge per km be ₹$y$.

Step 2: Form the equations.

$x + 10y = 105 \qquad \text{...(1)}$

$x + 15y = 155 \qquad \text{...(2)}$

Step 3: Subtract (1) from (2).

$5y = 50 \implies y = 10$

Step 4: Substitute $y = 10$ in (1).

$x + 100 = 105 \implies x = 5$

Step 5: Fare for $25\text{ km}$.

$x + 25y = 5 + 25 \times 10 = 255$

Answer: (i) The fixed charge is ₹$5$ and the charge per km is ₹$10$. (ii) The fare for $25\text{ km}$ is ₹$255$.

Check: $5 + 15 \times 10 = 155$. ✓

3. Problem 2: Library Charges (NCERT)

Question

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹$27$ for a book kept for seven days, while Susy paid ₹$21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

The Key Step

The fixed charge covers the first three days, so only the days after that are extra:

• Saritha: $7 - 3 = 4$ extra days
• Susy: $5 - 3 = 2$ extra days

Solution

Step 1: Define the variables. Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$.

Step 2: Form the equations.

$x + 4y = 27 \qquad \text{...(1)}$

$x + 2y = 21 \qquad \text{...(2)}$

Step 3: Subtract (2) from (1).

$2y = 6 \implies y = 3$

Step 4: Substitute $y = 3$ in (2).

$x + 6 = 21 \implies x = 15$

Answer: The fixed charge is ₹$15$ and the charge for each extra day is ₹$3$.

Check: $15 + 4 \times 3 = 27$. ✓

4. Problem 3: Hostel Mess Bill (NCERT)

Question

A part of the monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When student A takes food for $20$ days, she has to pay ₹$1000$ as hostel charges, whereas student B, who takes food for $26$ days, pays ₹$1180$. Find the fixed charges and the cost of food per day.

Solution

Step 1: Define the variables. Let the fixed charge be ₹$x$ and the cost of food per day be ₹$y$.

Step 2: Form the equations.

$x + 20y = 1000 \qquad \text{...(1)}$

$x + 26y = 1180 \qquad \text{...(2)}$

Step 3: Subtract (1) from (2).

$6y = 180 \implies y = 30$

Step 4: Substitute $y = 30$ in (1).

$x + 600 = 1000 \implies x = 400$

Answer: The fixed charge is ₹$400$ and the cost of food is ₹$30$ per day.

Check: $400 + 26 \times 30 = 1180$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: Always state what $x$ and $y$ stand for, with units (₹), before writing the equations.

Exam Tip: Check your answer by substituting $x$ and $y$ into the other equation.

Common Mistake: In the library problem, writing $x + 7y = 27$. The fixed charge already covers 3 days, so the correct equation is $x + 4y = 27$.

Common Mistake: Stopping after finding $x$ and $y$. If the question asks for the fare for $25\text{ km}$, you must calculate it.

Frequently Asked Questions

How do you solve fixed and running charge problems?

Let the fixed charge be x and the charge per unit be y, write one equation for each bill, subtract to find y, then substitute to find x.

What are the fixed charge and charge per km in the NCERT taxi problem?

The fixed charge is ₹5 and the charge per km is ₹10, so a 25 km journey costs ₹255.

Why do we subtract 3 days in the library problem?

The fixed charge already covers the first three days, so only the days after that are charged extra.

Which method is best for these word problems?

The elimination method, because x has coefficient 1 in both equations and disappears when you subtract them.

What is the cost of food per day in the hostel problem?

₹30 per day, with a fixed charge of ₹400.