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NCERT Solutions for Class 7 Maths Chapter 4 Expressions Usin...

NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers

Author at PW

January 21, 2026

Class 7 Maths Chapter 4 explains algebraic expressions using letters to represent unknown numbers. These helps students learn how to form expressions, identify variables, terms, coefficients, and constants, and evaluate expressions with given values. The NCERT Solutions for Class 7 Chapter 4 Maths questions and answers provide step-by-step explanations for all textbook problems. These solutions make concepts easy to understand, improve problem-solving skills, and help students practise for exams. This chapter builds a strong base for algebra and higher-level maths concepts.

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Class 7 Maths Chapter 4 Expressions Using Letter Numbers Questions Answers

Below are the NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 4. Each question is solved with the right method, formulas and final answers to help in exam preparation and daily practice.

Q. Example 1: Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be? _______
Solution:
Shabnam is 3 years older than Aftab
Difference in Shabnam’s and Aftab’s age = 3 years
Aftab’s present age = 18 years
Shabnam’s present age = 18 + 3 = 21 years.

Page 82

Q. Use this expression to find Aftab’s age if Shabnam’s age is 20. (Expression: Aftab’s age = Shabnam’s age – 3.)
Solution:
If Shabnam’s age = 20 years
Then,
Aftab’s age = Shabnam’s age – 3
= 20 – 3
= 17 years.

Q. Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60. How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Solution:
Cost of 8 coconuts = 8 × 35 = ₹280
Cost of 9 kg jaggery = 9 × 60 = ₹540
Total money paid = ₹280 + ₹540 = ₹820.

Q. Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery. (Expression: c × 35 + j × 60, where ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery)
Solution:
Number of coconuts (c) = 7
Number of kgs of jaggery (j) = 4kg
Total amount paid = c × 35 + j × 60
= 7 × 35 + 4 × 60
= 245 + 240
= ₹485.

Q. What is the perimeter of a square with side length 7 cm? Use the expression to find out. (Expression: 4 × q, where q stands for the side length.)
Solution:
Side length of square (q) = 7 cm
Perimeter = 4 × q
= 4 × 7
= 28 cm.

Page 84

1. Write formulas for the perimeter of:
(a) triangle with all sides equal.
(b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all side lengths and angle measures are equal)
(c) a regular hexagon
Solution:
(a) Triangle (3 equal sides): Perimeter = 3 × side length.

(b) Regular Pentagon (5 equal sides): Perimeter = 5 × side length.

2. Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.
Solution:
Initial length of pipe = 20 m
Length of pipe joined = ‘k’ m
Expression: 20 + k.

3. What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:

Solution:

No. of ₹100 notes	No. of ₹20 notes	No. of ₹5 notes	Expression and total amount
3	5	6	3×100 + 5×20 + 6×5 = 430
6	4	3	6×100 + 4×20 + 3×5 = 695
8	4	z	8×100 + 4×20 + z×5 = 880 + 5z
x	y	z	100x + 20y + 5z

4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?
(a) 10 + 8 + y (b) (10 + 8) × y (c) 10 × 8 × y (d) 10 + 8 × y (e) 10 × y + 8
Solution:
Time to start the machine = 10 seconds
Time to grind 1 kg of grain = 8 seconds
Quantity of grain = y kg
Total time = Time to start the machine + Time to grind y kg of grain
= 10 + 8 × y
Therefore, 10 + 8 × y is the correct answer.

5. Write algebraic expressions using letters of your choice.
(a) 5 more than a number
(b) 4 less than a number
(c) 2 less than 13 times a number
(d) 13 less than 2 times a number
Solution:
(a) x + 5 (Number = x)

(b) y – 4 (Number = y)

(d) 2 × z – 13 = (Number = z)

6. Describe situations corresponding to the following algebraic expressions:
(a) 8 × x + 3 × y
(b) 15 × j – 2 × k
Solution:
(a) Sum of 8 times x and 3 times y.

(b) Subtract 2 times k from 15 times j.

7. In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Solution:

Q. Let us revise these concepts and find the values of the following expressions:
1. 23 – 10 × 2
2. 83 + 28 – 13 + 32
3. 34 – 14 + 20
4. 42 + 15 – (8 – 7)
5. 68 – (18 + 13)
6. 7 × 4 + 9 × 6
7. 20 + 8 × (16 – 6)
Solution:
1. 23 – 10 × 2 = 23 – 20 = 3.

2. 83 + 28 – 13 + 32 = 83 – 13 + 28 + 32 = 70 + 60 = 130.

3. 34 – 14 + 20 = 30 + 20 = 50.

4. 42 + 15 – (8 – 7) = 42 + 15 – 1 = 57 – 1 = 56.

5. 68 – (18 + 13) = 68 – 31 = 37.

6. 7 × 4 + 9 × 6 = 28 + 54 = 82.

7. 20 + 8 × (16 – 6) = 20 + 8 × 10 = 20 + 80 = 100.

Q. 1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.
Solution:
1. Given equation: 10 – a = 6
Substitute a = -4
10 – (-4) = 6
10 + 4 = 6
14 ≠ 6
Since 14 ≠ 6, a = -4 does not satisfy the equation.
Now, 10 – a = 6
10 – 6 = a
a = 4.
Therefore, the solution to the equation is m = 16/3.

2. Given equation: 3d = 36
Substitute d = 6
3 × 6 = 36
18 ≠ 36
Since 18 ≠ 36, d = 6 does not satisfy the equation.
Now, 3d = 36
d = 36/3
d = 12.
Therefore, the solution to the equation is d = 12.

3. Given equation: 3s – 2 = 15
Substitute s = 7
3 × 7 – 2 = 15
21 – 2 = 15
19 ≠ 15
Since 19 ≠ 14, s = 7 does not satisfy the equation.
Now, 3s – 2 = 15
3s = 15 + 2
3s = 17
s = 17/3.
Therefore, the solution to the equation is s = 17/3.

4. Given equation: 2r + 1 = 29
Substitute r = 8
2 × 8 + 1 = 29
16 + 1 = 29
17 ≠ 29
Since 17 ≠ 29, r = 8 does not satisfy the equation.
Now, 2r + 1 = 29
2r = 29 – 1
2r = 28
r = 28/2 = 14.
Therefore, the solution to the equation is r = 14.

5. Given equation: 2j = 10
Substitute j = 5
2 × 5 = 10
10 = 10
Since 10 ≠ 10, j = 5 does satisfy the given equation.

6. Given equation: 3(m + 1)
Substitute m = -6,
3(-6 + 1) = 19
3(-5) = 19
-15 ≠ 19
Since -15 ≠ 19; m = -6 does not satisfy the equation.
3(m + 1) = 19
3m + 3 = 19
3m = 19 – 3
3m = 16
m = 16/3
Therefore, the solution to the equation is m = 16/3.

7. Given equation: 2f – 2g = 2
Substitute f =3, g = 1,
2 × 3 – 2 × 1 = 2
6 – 2 = 2
4 ≠ 2
Since 4 ≠ 2; f =3, g = 1 do not satisfy the equation.
If f = 2 and g = 1, then
2 × 2 – 2 × 1 = 2
4 – 2 = 2
2 = 2
Therefore, the solution to the equation are f = 2 and g = 1.

8. Given equation: 2t + b = 24
Substitute t = 4, b = 3,
2 × 4 + 3 = 24
8 + 3 = 24
11 ≠ 24
Since 11 ≠ 24, t = 4, b = 3 do not satisfy the equation.
Substitute t = 8, b = 8:
2 × 8 + 8 = 24
16 + 8 = 24
24 = 24
Therefore, the solution to the equation are t = 8, b = 8.

9. Given equation: h – (3 – n) = 4
Substitute h = 5, n = 6,
5 – (3 – 6) = 4
5 – (-3) = 4
5 + 3 = 4
8 ≠ 4
Since 11 ≠ 24, h = 5, n = 6 does not satisfy the equation.
Substitute h = 5, n = 2
5 – (3 – 2) = 4
5 – 1 = 4
4 = 4
Therefore, the solution to the equation are h = 5, n = 2.

Page 88

Q. If c = ₹50, find the total amount earned by the scale of pencils.
Solution:
Total amount earned = Day 1 + Day 2 + Day 3
= 5 × c + 3 × c + 10 × c
= 5c + 3c + 10c = 18c.
If c = ₹50, then
18c = 18 × 50 = ₹900.
Therefore, ₹900 is the total amount earned by the scale of pencils.

Q. Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Solution:
Total amount earned = Day 1 + Day 2 + Day 3
= 4 × d + 6 × d + 1 × d
= 4d + 6d + d
= 11d.

Page 90

Q. A shop rents out chairs and tables for a day’s use. To rent them, one has to first pay the following amount per piece.
When the furniture is returned, the shopkeeper pays back some amount as follows.
Write an expression for the total number of rupees paid if x chairs and y tables are rented.
For x chairs and y tables, let us find the total amount paid at the beginning and the amount one gets back after returning the furniture.
Describe the procedure to get these amounts.

Solution:
Chairs rented = x
Tables rented = y
Total amount paid at the beginning = Amount paid for chairs + Amount paid for tables
= 40 × (x) + 75 × y
= 40x + 7y.
Total amount returned = Amount returned for chairs + Amount returned for tables
= 6 × (x) + 10 × y
= 6x + 10y.
Total amount paid = (40x + 75y) – (6x + 10y)

Q. Can we simplify this expression? If yes, how? If not, why not? [Expression: (40x + 75y) – (6x + 10y)]
Solution:
Yes, the expression can be simplified further as follows:
(40x + 75y) – (6x + 10y) = 40x + 75y – 6x – 10y
= (40x – 6x) + (75y – 10y)
= 34x + 65y.

Page 91

Q. Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)?
Solution:
Yes, the initial expression (40x + 75y) – (6x + 10y) could also be written as
(40x + 75y) + (– 6x – 10y).
This is because subtracting a group of terms is the same as adding their negatives.

Q. What do each of the expressions mean? (Expressions: 7p – 3q, 8p – 4q, and 6p – 2q).
Solution:
Each of these expression involve a difference between two algebraic terms.

Page 92

Q. Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.
Solution:
Krishita’s score in three rounds could be (8p – 3q), (6p – 2q), and (9p – 2q).
Therefore, total score after three rounds = (8p – 3q) + (6p – 2q) + (9p – 2q).
= 8p – 3q + 6p – 2q + 9p – 2q
= 23p – 7q.

Q. Can we say who scored more? Can you explain why? (Charu’ score: 21p – 9q & Krishita’s score: 23p – 7q)
Solution:
Penalties for Charu = 9q
Penalities for Krishita = 7q
Krishita scored more because she had fewer penalties than Charu.

Q. Simplify this expression further. [Expression: (23p – 7q)– (21p – 9q)]
Solution:
23q – 7q – (21p – 9q) = 23q – 7q – 21p + 9q
= (23q – 21q) + (9q – 7q)
= 2q + 2q.

Page 97

Q. Given a position number can we find out the design that appears there? Which Design appears at Position 122? (Position at which the design A, B and C appears for nth time are 3n-2, 3n-1 and 3n)
Solution:
Yes, given a position number can help us find out the design that appears there.
Since 122 ÷ 3 gives 40 as the quotient and a remainder of 2.
Therefore, Design B appears at Position 122.

Q. Can the remainder obtained by dividing the position number by 3 be used for this?

Solution:
Yes, the remainders can be used for this. When the position number is divided by 3: if the remainder is 0, the design at that position is Design C; if the remainder is 1, it’s Design A; and if the remainder is 2, it’s Design B.

Q. Use this to find what design appears at positions 99, 122, and 148.

Solution:
Design C appears at position 99.
Design B appears at position 122.
Design A appears at position 148.

Page 99

Q. Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.
Solution:
Considering the given 2 × 2 square:

Top left number (8) = a
Number to the right 8 (9) = a + 1
Number below 8 (15) = a + 7
Number diagonal to 8 (16) = a + 8

First diagonal sum (8 + 16 = 24) = a + (a + 8) = 2a + 8.
Second diagonal sum (9 + 15 = 24) = (a + 1) + (a + 7) = 2a + 8.
Hence, verified that both diagonal sums are equal to 2a + 8.

Q. Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?
Solution:

Sum of all numbers = 8 + 14 + 22 + 16 + 15 = 75.
Central number = 15.
75 is five times 15.

Repeating this for the above set of numbers:
Sum of all number = 29 + 35 + 43 + 37 + 36 = 180.
Central number = 36.
180 is five times 36.
Observation: The sum of a set of numbers forming a (+) shape is five times the central number of the shape.

Page 100

Q. How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?
Solution:
Matchsticks in Step 33 = (2 × 33) + 1 = 66 + 1 = 67.
Matchsticks in Step 84 = (2 × 84) + 1 = 168 + 1 = 169.
Matchsticks in Step 108 = (2 × 108) + 1 = 216 + 1 = 217.

Q. Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same? [Expression: 3 + 2 × (y – 1) and 2y + 1]
Solution:
Yes, both the expressions 3 + 2 × (y – 1) and 2y + 1 give the number of matchsticks at each step correctly. It is because both are exactly the same.
i.e., 3 + 2 × (y – 1) = 3 + 2y – 2 = 2y + 1.

Q. What are these numbers in Step 3 and Step 4?
Solution:
In Step 3, there are 3 matchsticks placed horizontally and 4 matchsticks placed diagonally.
In step 4, there are 4 matchsticks placed horizontally and 5 matchsticks placed diagonally.

Q. How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1?
Solution:
Number of horizontal matchsticks per step = y
Number of diagonally placed matchsticks per step = y + 1

Total matchsticks for step y:
Horizontal = y
Diagonal = y + 1
Total = y + (y + 1) = 2y + 1
Conclusion: The expressions are correct and they do indeed add up to 2y + 1.

Page 102

Figure it out

For the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described. If required, assume some values for the unknowns and try to find the relationship.

1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30y
(d) (30+20) × x + y
(e) 30x – 20y
Solution:
Jowar roti plate = ₹30
Pulao plate = ₹20
Jowar roti ordered in a day = x
Pulao plate ordered in a day = y
Total amount earned in a day = ₹(x × 30) + (y + 20) = 30x + 20y.
∴ (a) 30x + 20y is the required expression.

2. Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?
(a) p + q + r
(b) p + q + 2r
(c) 2 × (p + q + r)
(d) p + q + r + 2
(e) p + q + r + 1
(f) 2 × (p + q)
Solution:
Customers bought champak = p
Customers bought marigold = q
Customers bought both = r
A national flag was given to every customer.
Total national flags distributed = p + q + r.
∴ (a) p + q + r is the required expression.

3. A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(a) Write an expression describing how far away the snail is from its starting position.
(b) What can we say about the snail’s movement if d > u?
Solution:
(a) Snail’s movement in a day and night = (u – v) cm
Snail’s movement in 10 days and 10 nights = 10(u – v) cm
Total distance moved by the snail from its starting position = 10(u – v) cm

(b) if d > u, then the snail would slip more during night than it climbs each day and can never reach the height of the well.

4. Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?
Solution:
Distance cycled in first week = 7 × 5 = 35 km
Distance cycled in second week = 7 × (5 + z) = (35 + 7z) km
Distance cycled in third week = 7 × (5 + z + z) = 7(5 + 2z) = (35 + 14z) km
Total distance cycled after 3 weeks = 35 + (35 + 7z) + (35 + 14z) = 35 + 35 + 7z + 35 + 14z

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Understanding Chapter 4 of NCERT's 7th Grade Math Class

Starting your journey into algebra can feel like learning a new language. In chapter 4 maths class 7th, we move away from just using numbers to using letters like x, y, or z. This shift is a vital part of your mathematical growth. We use these letters to represent values we don't know yet. The chapter 4 maths class 7th ncert curriculum ensures you understand that these "letter-numbers" behave just like regular numbers when it comes to addition or subtraction.

When you dive into ch 4 maths class 7th, you'll notice the focus is on "Expressions using Letter-Numbers." This means you aren't just calculating; you're translating words into math. If someone says "five more than a number," you now have the power to write x + 5. It's a simple change, but it opens up a whole world of possibilities for solving complex puzzles later on.

Variables and Constants in Chapter 4 Maths Class 7

Every expression you meet in chapter 4 maths class 7 has two main ingredients: variables and constants. A constant is a fixed value, like 5 or 10, that never changes its "personality." However, a variable is the letter-number that can take on different values depending on the situation.

We often see these used together in expressions like 3x + 7. Here, 7 is the constant, while x is the variable that could be anything. Learning to distinguish between these is the first hurdle in mastering chapter 4 maths class 7th ncert. We encourage you to practice identifying these in different contexts to build a strong foundation.

Step-by-Step Guide to Chapter 4 Maths Class 7 Exercise 4.1

The first exercise is usually where students feel the most pressure. In chapter 4 maths class 7 exercise 4.1, the focus is often on setting up equations and checking if a given value is a solution.

Read the Statement: Look for keywords like "sum," "difference," or "product."
Assign a Variable: Pick a letter for the unknown quantity.
Form the Equation: Use the mathematical operations mentioned.
Verification: Plug in the suggested number to see if the Left Hand Side (LHS) equals the Right Hand Side (RHS).

It's not enough to just find the appropriate solution quickly when you solve these difficulties. It's about creating a logical flow that will help you solve even harder challenges in the future without becoming lost or angry.

Making Algebraic Expressions with Letter-Numbers

Forming expressions is the heart of ch 4 maths class 7th. When we say "a number multiplied by 4 and then increased by 6," we write it as 4x + 6. You must pay close attention to the order of operations. If you mix them up, the whole expression changes its meaning entirely.

We often use these expressions to describe patterns. For instance, if you're looking at the number of matchsticks needed to form a shape, a letter-number expression can tell you the total for any number of shapes. This practical application makes chapter 4 maths class 7 much more interesting than just memorizing formulas from a textbook.

Solving Simple Equations in NCERT Solutions

Once you've formed an expression, the next step is often turning it into an equation by adding an equals sign. The chapter 4 maths class 7th syllabus teaches you that an equation is like a balanced scale. Whatever you do to one side, you must do to the other to keep it level.

If you add 5 to the left, add 5 to the right.
If you divide the right by 2, divide the left by 2.

This "balancing act" is the core secret to solving for your variable. It might seem tedious at first, but once you get the hang of it, you'll be solving for x in your sleep. It's all about maintaining that perfect equilibrium between the two sides.

How Expressions Are Used in Real Life

Why do we need to study this? A lot of kids ask this question after they read chapter 4 of the 7th grade math book. We say these things every day without even thinking about it. When you add up the cost of several things at a store or figure out how long it will take to get somewhere, you are employing letter-numbers.

Imagine you had a set amount of money to spend on a party. You know how much the cake will cost, but the appetizers will cost more or less depending on how many friends come. You may plan your spending exactly by using a variable for the amount of pals. This is the real-world magic that chapter 4 math class 7 gives you.

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How to Do Well in Chapter 4 Math Class 7

To truly excel in chapter 4 maths class 7th, you need more than just a calculator. You need a strategy. We've seen that students who write out every step clearly tend to make fewer mistakes than those who try to do everything in their heads.

Don't skip steps: Even if a calculation looks easy, write it down to avoid silly errors.
Check your work: Always plug your final answer back into the original equation.
Practice daily: Algebra is a skill that gets better with repetition and patience.

Using resources can give you access to practice papers and detailed guides that make ch 4 maths class 7th feel like a breeze. We're here to support you with materials that simplify complex ideas into bite-sized, understandable pieces.

Benefits of PW Class 7 Study Material

1. Complete Syllabus Coverage
PW Class 7 study material is designed strictly as per the latest NCERT syllabus. All subjects are covered in a clear and organised manner, helping students prepare thoroughly for school exams.

2. Easy and Simple Explanations
The content is written in simple language, making concepts easy to understand. Important topics are explained step by step with examples, which helps students build strong basics.

3. Concept-Based Learning
PW focuses on concept clarity instead of rote learning. This helps students understand the “why” behind each topic, improving logical thinking and problem-solving skills.

4. Practice Questions and Examples
The study material includes topic-wise questions, examples, and exercises. Regular practice helps students gain confidence and improve accuracy in exams.

5. Exam-Oriented Preparation
Important questions, key points, and revision notes are provided to help students prepare effectively for school tests and competitive exams.

6. Expert-Curated Content
The material is prepared by experienced teachers who understand student learning needs, ensuring high-quality and reliable content.

7. Boosts Confidence and Performance
With structured lessons and regular practice, students feel more confident and perform better in exams.

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FAQs about Chapter 4 Maths Class 7th

Q1:What is a variable in chapter 4 of seventh-grade math?

A variable is a letter, such x or y, that stands for a number that isn't known yet or can change.

Q2: How can I accomplish exercise 4.1 in chapter 4 of math class 7?

Concentrate on the "trial and error" method or the balancing method, which means doing the identical thing on both sides of the equals sign.

Q3: Is Ganita Prakash a different book from the old NCERT?

Ganita Prakash is the latest NCERT textbook for Class 7. It is meant to make learning more engaging and in line with contemporary educational standards.