NCERT Class 7 Maths Solutions provide comprehensive step-by-step explanations for all 15 chapters, helping students master fundamental mathematical concepts aligned with the CBSE 2025-26 syllabus.
Mathematics at the Class 7 level marks a crucial transition from basic arithmetic to more abstract concepts like algebra, rational numbers, and coordinate geometry. The NCERT textbook for this class introduces students to mathematical reasoning and logical thinking that forms the foundation for higher secondary mathematics. Understanding these concepts thoroughly is essential not just for school examinations but also for competitive tests like the Mathematics Olympiad.
These NCERT Solutions for Class 7 Maths have been prepared by experienced educators following the latest guidelines from NCERT.nic.in and CBSE.gov.in. Each solution includes detailed working, alternative methods where applicable, and tips to avoid common mistakes. Whether you are revising for unit tests or preparing for annual examinations, these chapterwise solutions serve as your complete study companion.
Why This Matters: Class 7 Mathematics introduces algebraic thinking, negative numbers, and proportional reasoning—skills that directly impact performance in Class 8, 9, and 10 board examinations. Students who build a strong foundation here consistently perform better in higher classes.
NCERT Class 7 Maths Solutions – Complete Chapter Overview
The Class 7 Mathematics curriculum is thoughtfully designed to progress from number systems to geometry, ensuring students develop both computational fluency and spatial reasoning. Below is a comprehensive overview of all chapters with their key concepts and learning objectives.
| Operation | Rule | Example |
|---|---|---|
| Addition | Same signs – add and keep sign; Different signs – subtract and take sign of larger number | \((-3) + (-5) = -8\), \((-6) + (4) = -2\) |
| Subtraction | Subtracting a number is same as adding its additive inverse | \(5 – (-3) = 5 + 3 = 8\) |
| Multiplication | Product of two integers with same sign is positive; with different signs is negative | \((-4) imes (-2) = 8\), \((-3) imes 2 = -6\) |
Each chapter builds upon concepts learned in Class 6 while introducing new mathematical tools. For instance, Chapter 1 extends natural number operations to integers, Chapter 9 further generalises this to rational numbers, and Chapter 12 introduces symbolic representation through algebraic expressions. This spiral approach ensures deep conceptual understanding. If you are looking to strengthen your algebraic foundations further, you may find our Class 11 Maths NCERT Book PDF useful for advanced reference.
Key Insight: The 15 chapters of Class 7 Maths can be grouped into four main strands: Number System (Chapters 1, 2, 9), Algebra (Chapters 4, 12), Geometry (Chapters 5, 6, 10, 14, 15), and Data Handling with Applications (Chapters 3, 7, 8, 11, 13). Understanding this structure helps in organised revision.
Integers – Operations, Properties, and Solved Examples
Chapter 1 on Integers forms the backbone of Class 7 Mathematics. Students learn to perform all four arithmetic operations on positive and negative numbers, understand the properties that govern these operations, and apply integers to real-world situations like temperature changes, profit-loss calculations, and altitude measurements.
| Operation | Rule | Example |
|---|---|---|
| Addition/Subtraction | Find LCM of denominators and then add or subtract numerators | \(\frac{2}{5} + \frac{3}{10} = \frac{7}{10}\) |
| Multiplication | Multiply numerators and denominators directly | \(\frac{2}{3} imes \frac{4}{5} = \frac{8}{15}\) |
| Division | Multiply the first fraction by reciprocal of the second | \(\frac{3}{4} \div \frac{2}{5} = frac{3}{4} imes frac{5}{2} = frac{15}{8}\) |
The most important concept in this chapter is understanding that subtracting a negative number is equivalent to adding its positive counterpart. This rule, expressed as a − (−b) = a + b, eliminates confusion in complex calculations and is fundamental to algebraic manipulation in later chapters.
Important: When multiplying or dividing integers, count the negative signs. An even number of negative signs gives a positive result; an odd number gives a negative result. For example, (−2) × (−3) × (−4) = −24 because there are three negative signs (odd count).
Worked Example: Integer Operations
Problem: Simplify the expression (−15) + (−8) − (−12) × 2 ÷ (−4)
Step 1: Apply BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction). First, resolve multiplication and division from left to right.
Step 2: Calculate (−12) × 2 = −24
Step 3: Calculate (−24) ÷ (−4) = +6 (negative divided by negative equals positive)
Step 4: Now the expression becomes (−15) + (−8) − (+6)
Step 5: Simplify left to right: (−15) + (−8) = −23
Step 6: Finally, (−23) − (+6) = −23 − 6 = −29
Answer: −29
This systematic approach using BODMAS and integer rules ensures accuracy in complex calculations. Students should practise similar problems from Exercise 1.3 and 1.4 of the NCERT textbook to build confidence.
Fractions, Decimals, and Rational Numbers
Chapters 2 and 9 deal with the representation and manipulation of fractions, decimals, and rational numbers. While fractions and decimals were introduced in earlier classes, Class 7 focuses on operations involving multiplication and division, which require deeper conceptual understanding.
| Concept | Explanation | Example |
|---|---|---|
| Monomial | Expression with one term | \(5x\), \(-3y^2\) |
| Binomial | Expression with two terms | \(x + 4\), \(2a – 5b\) |
| Polynomial | Expression with more than two terms | \(3x^2 + 2x + 1\) |
The connection between fractions, decimals, and percentages is particularly important for real-life applications. Students learn that 1/4 = 0.25 = 25%, and this equivalence helps in solving problems related to discounts, interest rates, and data interpretation. Understanding these relationships early prepares students for the more rigorous treatment of NCERT Solutions Class 10 Maths Chapter on Real Numbers.
Rational Numbers Definition: A rational number is any number that can be expressed as p/q where p and q are integers and q ≠ 0. This includes all integers (since 5 = 5/1), all fractions, and terminating or repeating decimals. Examples: −3/4, 0, 7, 2.5 (= 5/2), 0.333… (= 1/3).
Worked Example: Fraction Division
Problem: A rope of length 7½ metres is cut into pieces of length ¾ metre each. How many pieces are obtained?
Step 1: Convert the mixed number to improper fraction: 7½ = 15/2 metres
Step 2: Divide total length by piece length: (15/2) ÷ (3/4)
Step 3: Apply the division rule (multiply by reciprocal): (15/2) × (4/3)
Step 4: Multiply numerators and denominators: (15 × 4)/(2 × 3) = 60/6 = 10
Answer: 10 pieces can be obtained from the rope.
Such word problems test both computational skills and the ability to identify the correct operation. Students often confuse when to multiply and when to divide—remember that “how many pieces” or “how many times” typically indicates division.
Algebraic Expressions and Simple Equations
Chapters 4 and 12 introduce students to the powerful world of algebra. This is where mathematics transitions from dealing with specific numbers to working with variables and generalised patterns. Understanding algebra is crucial as it forms the foundation for all higher mathematics studied in Classes 8 through 12.
An algebraic expression consists of constants, variables, and operations. For example, 3x + 5y − 7 contains two variables (x and y), coefficients (3 and 5), and a constant term (−7). Students learn to identify like terms, combine them, and simplify expressions systematically.
| Construction | Steps | Use |
|---|---|---|
| Perpendicular Bisector | Draw an arc from both ends of a line segment intersecting above and below; join intersection points | Used to divide a segment into equal halves |
| Angle Bisector | Draw an arc cutting both arms; from intersections draw another arc to intersect previous arc; join with vertex | Used in geometry and engineering drawing |
| Triangle Construction | Based on SSS, SAS, or ASA criteria | Used for geometrical proofs and architecture |
Important: When solving equations, whatever operation you perform on one side must be performed on the other side to maintain equality. This is called the balance principle and is the foundation of all equation solving.
Worked Example: Solving Linear Equations
Problem: The sum of three consecutive odd numbers is 51. Find the numbers.
Step 1: Let the smallest odd number be x. Then the three consecutive odd numbers are x, x + 2, and x + 4.
Step 2: Form the equation: x + (x + 2) + (x + 4) = 51
Step 3: Combine like terms: 3x + 6 = 51
Step 4: Subtract 6 from both sides: 3x = 45
Step 5: Divide both sides by 3: x = 15
Step 6: Find all three numbers: 15, 17, and 19
Verification: 15 + 17 + 19 = 51 ✓
Answer: The three consecutive odd numbers are 15, 17, and 19.
This problem-solving approach—defining variables, forming equations, solving systematically, and verifying—is essential for success not just in Class 7 but throughout mathematical education. Students can explore similar techniques in our NCERT Solutions Class 10 Maths Chapter on Polynomials.
Geometry – Practical Constructions and Visualisation
Class 7 includes substantial geometry content across Chapters 5, 6, 10, 14, and 15. Students learn about lines and angles, properties of triangles and quadrilaterals, practical geometric constructions using compass and ruler, and three-dimensional visualisation of solid shapes.
The geometry chapters require both theoretical understanding and practical skills. Students must know angle relationships (complementary, supplementary, vertically opposite) and apply them to solve problems. They also learn that the sum of angles in a triangle is always 180° and the sum of angles in a quadrilateral is always 360°.
Triangle Inequality Property: The sum of any two sides of a triangle must be greater than the third side. This means if a triangle has sides a, b, and c, then a + b > c, b + c > a, and a + c > b must all be true. Use this to check if three given lengths can form a valid triangle.
Worked Example: Angle Relationships
Problem: Two supplementary angles are in the ratio 4:5. Find both angles.
Step 1: Supplementary angles add up to 180°.
Step 2: Let the angles be 4x and 5x.
Step 3: Form equation: 4x + 5x = 180°
Step 4: Simplify: 9x = 180°, therefore x = 20°