Skip to main content
Admissions Open 2026-27Ravindra Higher Secondary School (Est. 1988) | Waidhan, Singrauli (MP)
+91 9826986106• Student Portal• Study Notes
Ravindra Higher Secondary School Logo
Ravindra Higher Secondary SchoolWaidhan, Singrauli (M.P.)
Home
Contact
Home
Study Portal
Class 8 Mathematics
Cubes and Cube Roots - Properties of cube numbers, finding cube roots using the prime factorisation method, and estimation of cube roots
Back to All Study GuidesOpen in Interactive App
MathematicsClass 8Cubes and Cube Roots

Cubes and Cube Roots - Properties of cube numbers, finding cube roots using the prime factorisation method, and estimation of cube roots

2026-10-2915 min readRHS Academic Faculty
Overview & Key Summary:Cubes and Cube Roots: Properties, Prime Factorisation, and Estimation In geometry, a cube is a threedimensional solid shape bounded by six square faces of equal side length. If a...

Cubes and Cube Roots: Properties, Prime Factorisation, and Estimation

In geometry, a cube is a three-dimensional solid shape bounded by six square faces of equal side length. If a solid cube has a side length of aaa units, its volume is calculated as the product of its three dimensions: Volume=a×a×a=a3\text{Volume} = a \times a \times a = a^3Volume=a×a×a=a3.

In arithmetic, the process of multiplying a number by itself twice is known as cubing the number. Conversely, if we are given the volume of a cubic region and need to find the edge length, we perform the inverse operation, known as finding the cube root. Understanding cubes and cube roots is a fundamental building block in algebra, spatial geometry, vector calculus, and real-world structural measurement.

1. In-Depth Conceptual Breakdown

What is a Cube Number (Perfect Cube)?

A natural number nnn is called a cube number or a perfect cube if there exists another natural number mmm such that:

n=m3=m×m×mn = m^3 = m \times m \times mn=m3=m×m×m

For example, 272727 is a perfect cube because 27=3×3×3=3327 = 3 \times 3 \times 3 = 3^327=3×3×3=33. On the other hand, 202020 is not a perfect cube because there is no natural number mmm such that m3=20m^3 = 20m3=20.

Table of Cubes (1 to 15)

Memorising the cubes of the first 15 natural numbers is crucial for rapid calculation and estimation:

Natural Number (nnn)Expression (n×n×nn \times n \times nn×n×n)Perfect Cube (n3n^3n3)
1111×1×11 \times 1 \times 11×1×1111
2222×2×22 \times 2 \times 22×2×2888
3333×3×33 \times 3 \times 33×3×3272727
4444×4×44 \times 4 \times 44×4×4646464
5555×5×55 \times 5 \times 55×5×5125125125
6666×6×66 \times 6 \times 66×6×6216216216
7777×7×77 \times 7 \times 77×7×7343343343
8888×8×88 \times 8 \times 88×8×8512512512
9999×9×99 \times 9 \times 99×9×9729729729
10101010×10×1010 \times 10 \times 1010×10×10100010001000
11111111×11×1111 \times 11 \times 1111×11×11133113311331
12121212×12×1212 \times 12 \times 1212×12×12172817281728
13131313×13×1313 \times 13 \times 1313×13×13219721972197
14141414×14×1414 \times 14 \times 1414×14×14274427442744
15151515×15×1515 \times 15 \times 1515×15×15337533753375

Key Properties of Cube Numbers

1. Parity of Cubes

  • Even Numbers: The cube of an even natural number is always even. Example: 43=64(Even),83=512(Even)\text{Example: } 4^3 = 64 \quad \text{(Even)}, \quad 8^3 = 512 \quad \text{(Even)}Example: 43=64(Even),83=512(Even)
  • Odd Numbers: The cube of an odd natural number is always odd. Example: 33=27(Odd),73=343(Odd)\text{Example: } 3^3 = 27 \quad \text{(Odd)}, \quad 7^3 = 343 \quad \text{(Odd)}Example: 33=27(Odd),73=343(Odd)

2. Cubes of Negative Numbers

Unlike square numbers, where the square of any real negative number is positive (e.g., (−3)2=9(-3)^2 = 9(−3)2=9), the cube of a negative number is always negative:

(−n)3=(−n)×(−n)×(−n)=−n3(-n)^3 = (-n) \times (-n) \times (-n) = -n^3(−n)3=(−n)×(−n)×(−n)=−n3

Example: (−5)3=(−5)×(−5)×(−5)=−125\text{Example: } (-5)^3 = (-5) \times (-5) \times (-5) = -125Example: (−5)3=(−5)×(−5)×(−5)=−125

3. Units Digit Pattern of Cubes

Observing the units digit of a number and its cube reveals a unique structural rule:

  • Numbers ending in 0,1,4,5,6,90, 1, 4, 5, 6, 90,1,4,5,6,9 have cubes that end in the same digit:
    • 103=100010^3 = 100\mathbf{0}103=1000
    • 113=133111^3 = 133\mathbf{1}113=1331
    • 143=274414^3 = 274\mathbf{4}143=2744
    • 153=337515^3 = 337\mathbf{5}153=3375
    • 163=409616^3 = 409\mathbf{6}163=4096
    • 193=685919^3 = 685\mathbf{9}193=6859
  • Numbers ending in 222 have cubes ending in 888, and vice versa:
    • 23=8  ⟺  83=5122^3 = \mathbf{8} \iff 8^3 = 51\mathbf{2}23=8⟺83=512
  • Numbers ending in 333 have cubes ending in 777, and vice versa:
    • 33=27  ⟺  73=3433^3 = 2\mathbf{7} \iff 7^3 = 34\mathbf{3}33=27⟺73=343

This reciprocal relationship (2↔82 \leftrightarrow 82↔8 and 3↔73 \leftrightarrow 73↔7) is critical for the estimation method of finding cube roots.

4. Adding Consecutive Odd Numbers

An interesting visual and numerical pattern exists where perfect cubes can be represented as sums of consecutive odd numbers:

1^3 &= 1 &&= 1 \\ 2^3 &= 3 + 5 &&= 8 \\ 3^3 &= 7 + 9 + 11 &&= 27 \\ 4^3 &= 13 + 15 + 17 + 19 &&= 64 \\ 5^3 &= 21 + 23 + 25 + 27 + 29 &&= 125 \end{aligned}$$ #### 5. Prime Factorisation Property If we express a perfect cube as a product of its prime factors, **each prime factor appears in triplets (groups of three)**. Consider $216$: $$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 3^3$$ If any prime factor remains unpaired or forms only a pair instead of a triplet, the given number is **not** a perfect cube. --- ### Finding Cube Roots using the Prime Factorisation Method The **cube root** of a number $n$ is the value $m$ which, when multiplied by itself three times, gives $n$. It is denoted by the radical symbol with an index of 3: $\sqrt[3]{n}$. $$\sqrt[3]{n} = m \iff m^3 = n$$ #### Step-by-Step Algorithm: 1. **Factorise:** Write down the prime factorisation of the given number. 2. **Group:** Arrange the prime factors into groups of three identical factors (triplets). 3. **Extract:** Pick one factor from each triplet. 4. **Multiply:** Multiply these selected factors together to obtain the cube root. $$\text{If } n = (a \times a \times a) \times (b \times b \times b), \quad \text{then } \sqrt[3]{n} = a \times b$$ #### Making Numbers Perfect Cubes (Multiplication / Division Problems) When a number is not a perfect cube, its prime factorisation will leave incomplete triplets. * **To find the smallest number to multiply:** Identify the missing factors needed to complete each triplet. * **To find the smallest number to divide:** Identify the extra factors that prevent the expression from forming complete triplets and divide by their product. --- ### Estimation of Cube Roots For large perfect cubes (up to 6 digits), the prime factorisation method can be time-consuming. The **estimation method** provides a fast, mental alternative using digit grouping and unit patterns. #### Step-by-Step Estimation Algorithm: Given a perfect cube $N$: 1. **Step 1: Grouping** Start from the rightmost digit (units place) and move left to divide the digits into groups of three. * *Group 1 (Rightmost):* Determines the **units digit** of the cube root. * *Group 2 (Leftmost):* Determines the **tens digit** of the cube root. 2. **Step 2: Find Units Digit** Look at the last digit of Group 1 and match it using the unit digit rules: * Ends in $0 \rightarrow 0$ * Ends in $1 \rightarrow 1$ * Ends in $4 \rightarrow 4$ * Ends in $5 \rightarrow 5$ * Ends in $6 \rightarrow 6$ * Ends in $9 \rightarrow 9$ * Ends in $2 \rightarrow 8$ (and $8 \rightarrow 2$) * Ends in $3 \rightarrow 7$ (and $7 \rightarrow 3$) 3. **Step 3: Find Tens Digit** Look at the value of Group 2. Find the largest single-digit perfect cube ($a^3$) that is less than or equal to Group 2: $$a^3 \le \text{Group 2} < (a+1)^3$$ The value $a$ is the **tens digit** of the cube root. --- ## 2. Real-World Applications ### Application 1: Spatial Packaging & Logistics Shipping containers and cardboard packing boxes are produced in standardized 3D volumes. If an industrial manufacturer needs to design a cube-shaped container to store a total fluid volume of $13.824\text{ m}^3$, calculating the side length requires finding the cube root: $$\text{Side Length} = \sqrt[3]{13.824} = 2.4\text{ metres}$$ This ensures exact material budgeting for outer steel panels. ``` +-----------------+ / /| / / | +-----------------+ | | | | Volume = 13.824 m³ | | | Side = ∛(13.824) = 2.4 m | | + | | / | |/ +-----------------+ 2.4 m ``` ### Application 2: Material Density and Metallurgy In metallurgy, the density ($\rho$) of a material relates its total mass ($m$) to its volume ($V$) by $\rho = \frac{m}{V}$. For a cube of solid metal: $$\text{Volume } V = \frac{\text{Mass}}{\text{Density}}$$ Since $V = \text{Side}^3$, a quality control engineer can measure the total mass of a cubic iron casting, divide by iron's density, and take the cube root to check if the side length matches structural tolerances without cutting the sample open. --- ## 3. Step-by-Step Solved Textbook Examples ### Example 1: Finding the Smallest Multiplier **Problem:** Find the smallest natural number by which $256$ must be multiplied so that the product becomes a perfect cube. #### Solution: 1. **Find the prime factorisation of $256$:** $$\begin{aligned} 256 \div 2 &= 128 \\ 128 \div 2 &= 64 \\ 64 \div 2 &= 32 \\ 32 \div 2 &= 16 \\ 16 \div 2 &= 8 \\ 8 \div 2 &= 4 \\ 4 \div 2 &= 2 \\ 2 \div 2 &= 1 \end{aligned}$$ 2. **Express $256$ as a product of prime factors:** $$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$$ 3. **Group the factors into triplets:** $$256 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2)$$ 4. **Analyse the remaining factors:** The prime factor $2$ appears in two complete triplets, but the last group contains only two $2$s ($(2 \times 2)$). To make it a complete triplet of three $2$s, we need **one more $2$**. 5. **Conclusion:** Multiply $256$ by $2$: $$\text{New Number} = 256 \times 2 = 512$$ Since $512 = 8^3$, $512$ is a perfect cube. $$\mathbf{Answer: \text{The smallest natural number is } 2.}$$ --- ### Example 2: Prime Factorisation Method **Problem:** Find the cube root of $13824$ using the prime factorisation method. #### Solution: 1. **Find the prime factorisation of $13824$:** $$\begin{aligned} 13824 \div 2 &= 6912 \\ 6912 \div 2 &= 3456 \\ 3456 \div 2 &= 1728 \\ 1728 \div 2 &= 864 \\ 864 \div 2 &= 432 \\ 432 \div 2 &= 216 \\ 216 \div 2 &= 108 \\ 108 \div 2 &= 54 \\ 54 \div 2 &= 27 \\ 27 \div 3 &= 9 \\ 9 \div 3 &= 3 \\ 3 \div 3 &= 1 \end{aligned}$$ 2. **Write $13824$ as a product of prime factors:** $$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$$ 3. **Group factors into triplets:** $$13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$$ 4. **Take one factor from each triplet to find the cube root:** $$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3$$ $$\sqrt[3]{13824} = 24$$ $$\mathbf{Answer: \sqrt[3]{13824} = 24}$$ --- ### Example 3: Estimation Method **Problem:** Estimate the cube root of $857375$. #### Solution: 1. **Step 1: Form groups of three digits from right to left:** $$\underbrace{857}_{\text{Group 2}} \quad \underbrace{375}_{\text{Group 1}}$$ 2. **Step 2: Determine the Units Digit from Group 1 ($375$):** * The number $375$ ends in the digit $5$. * We know that a number ending in $5$ always has a cube ending in $5$. * Therefore, the **units digit** of the cube root is **$5$**. 3. **Step 3: Determine the Tens Digit from Group 2 ($857$):** * Find the two consecutive perfect cubes between which $857$ lies: $$9^3 = 729 \quad \text{and} \quad 10^3 = 1000$$ $$729 < 857 < 1000 \implies 9^3 < 857 < 10^3$$ * The smaller integer whose cube is less than $857$ is $9$. * Therefore, the **tens digit** of the cube root is **$9$**. 4. **Step 4: Combine the digits:** $$\text{Tens digit} = 9, \quad \text{Units digit} = 5 \implies 95$$ $$\mathbf{Answer: \sqrt[3]{857375} = 95}$$ --- ### Example 4: Real-World Word Problem **Problem:** A cubic water reservoir has a volume of $17.576\text{ m}^3$. Find the side length of the reservoir. #### Solution: 1. **Define the formula:** $$\text{Volume of cube} = \text{side}^3 \implies s^3 = 17.576$$ $$s = \sqrt[3]{17.576} = \sqrt[3]{\frac{17576}{1000}} = \frac{\sqrt[3]{17576}}{\sqrt[3]{1000}}$$ 2. **Find $\sqrt[3]{1000}$:** $$\sqrt[3]{1000} = 10$$ 3. **Find $\sqrt[3]{17576}$ using Estimation:** * Group the number: $\underbrace{17}_{\text{Group 2}} \quad \underbrace{576}_{\text{Group 1}}$ * Group 1 ($576$) ends in $6$, so the units digit is **$6$**. * Group 2 ($17$) lies between $2^3 = 8$ and $3^3 = 27$ ($8 < 17 < 27$). The smaller cube root is $2$, so the tens digit is **$2$**. * Therefore, $\sqrt[3]{17576} = 26$. 4. **Calculate final side length:** $$s = \frac{26}{10} = 2.6\text{ metres}$$ $$\mathbf{Answer: \text{The side length of the reservoir is } 2.6\text{ m.}}$$ --- ## 4. Common Student Mistakes to Avoid ### Mistake 1: Confusing Cubing with Multiplying by 3 * **Incorrect:** $4^3 = 4 \times 3 = 12$ * **Correct:** $4^3 = 4 \times 4 \times 4 = 64$ * **Correction Strategy:** Remember that the exponent $3$ tells you how many times the base appears in a multiplication chain, not what to multiply the base by. ### Mistake 2: Writing Square Root Symbol Instead of Cube Root * **Incorrect:** $\sqrt{27} = 3$ (This is technically reading "square root of 27") * **Correct:** $\sqrt[3]{27} = 3$ * **Correction Strategy:** Always include the small index $3$ inside the V-notch of the radical sign ($\sqrt[3]{\phantom{x}}$). Leaving it blank implies an index of $2$ (square root). ### Mistake 3: Grouping Factors in Pairs Instead of Triplets * **Incorrect:** When computing $\sqrt[3]{216}$, grouping as $(2 \times 2) \times (2 \times 3) \dots$ like in square roots. * **Correct:** Cubes require **triplets** (groups of three identical factors): $$216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \implies \sqrt[3]{216} = 2 \times 3 = 6$$ ### Mistake 4: Grouping Digits Left-to-Right in Estimation * **Incorrect:** Grouping $17576$ as $\underbrace{175}_{\text{Group 1}} \quad \underbrace{76}_{\text{Group 2}}$ from left to right. * **Correct:** Grouping must **always start from the rightmost digit** (units place): $$\underbrace{17}_{\text{Group 2}} \quad \underbrace{576}_{\text{Group 1}}$$ --- ## 5. Practice Questions for Self-Assessment ### Question 1 Find the smallest number by which $675$ must be divided so that the quotient becomes a perfect cube. <details> <summary><b>Click to view Step-by-Step Solution</b></summary> 1. **Prime factorise $675$:** $$675 \div 3 = 225$$ $$225 \div 3 = 75$$ $$75 \div 3 = 25$$ $$25 \div 5 = 5$$ $$5 \div 5 = 1$$ $$675 = 3 \times 3 \times 3 \times 5 \times 5 = (3 \times 3 \times 3) \times (5 \times 5)$$ 2. **Analyse triplets:** The factor $3$ forms a complete triplet $(3 \times 3 \times 3)$. The factor $5$ forms only a pair $(5 \times 5 = 25)$. 3. **Determine divisor:** To eliminate the extra factors, we must divide by $5 \times 5 = 25$. 4. **Verify:** $$\text{Quotient} = \frac{675}{25} = 27 = 3^3$$ **Final Answer:** The smallest number by which $675$ must be divided is **$25$**. </details> --- ### Question 2 Find the cube root of $91125$ using the prime factorisation method. <details> <summary><b>Click to view Step-by-Step Solution</b></summary> 1. **Prime factorise $91125$:** $$\begin{aligned} 91125 \div 3 &= 30375 \\ 30375 \div 3 &= 10125 \\ 10125 \div 3 &= 3375 \\ 3375 \div 3 &= 1125 \\ 1125 \div 3 &= 375 \\ 375 \div 3 &= 125 \\ 125 \div 5 &= 25 \\ 25 \div 5 &= 5 \\ 5 \div 5 &= 1 \end{aligned}$$ 2. **Group into triplets:** $$91125 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5)$$ 3. **Calculate cube root:** $$\sqrt[3]{91125} = 3 \times 3 \times 5 = 45$$ **Final Answer:** $\mathbf{\sqrt[3]{91125} = 45}$ </details> --- ### Question 3 Estimate the cube root of $110592$. <details> <summary><b>Click to view Step-by-Step Solution</b></summary> 1. **Group digits from right to left:** $$\underbrace{110}_{\text{Group 2}} \quad \underbrace{592}_{\text{Group 1}}$$ 2. **Units digit from Group 1 ($592$):** * Ends in $2$. Numbers whose cubes end in $2$ have a units digit of $8$. * **Units digit = $8$**. 3. **Tens digit from Group 2 ($110$):** * Compare with known cubes: $4^3 = 64$ and $5^3 = 125$. * $64 < 110 < 125 \implies 4^3 < 110 < 5^3$. * Take the smaller value ($4$). * **Tens digit = $4$**. 4. **Combine:** $$\text{Cube Root} = 48$$ **Final Answer:** $\mathbf{\sqrt[3]{110592} = 48}$ </details> --- ## 6. Exam Revision & FAQs ### Q1: Is the cube of a negative number always negative? Explain why. **Answer:** Yes, the cube of any negative number is always negative. When you cube a negative number, you multiply three negative terms together: $$(-a) \times (-a) \times (-a)$$ Multiplying the first two negatives gives a positive value ($(-a) \times (-a) = +a^2$). Multiplying this positive value by the third negative term gives a negative result ($(+a^2) \times (-a) = -a^3$). --- ### Q2: How many zeros appear at the end of a cube number if the original number ends in two zeros? **Answer:** If a number ends in $k$ zeros, its cube will end in $3k$ zeros. * If a number ends in **$2$ zeros** (e.g., $100$), its cube will end in **$2 \times 3 = 6$ zeros**: $$100^3 = (10^2)^3 = 10^6 = 1,000,000 \quad \text{(6 zeros)}$$ --- ### Q3: What is the main difference between finding square roots and cube roots using prime factorisation? **Answer:** * **Square Root ($\sqrt{n}$):** Prime factors are grouped into **pairs** (groups of two identical factors), and one factor is extracted from each pair. * **Cube Root ($\sqrt[3]{n}$):** Prime factors are grouped into **triplets** (groups of three identical factors), and one factor is extracted from each triplet. --- ### Q4: Can we use the estimation method to find the cube root of any random number? **Answer:** No. The standard shortcut estimation method taught in Class 8 works **only for perfect cubes**. If applied to a non-perfect cube, it will output an incorrect integer result because it assumes the input can be written as an exact integer cubed.
Verified NCERT & Board Exam Aligned Material
Ravindra Higher Secondary School, Waidhan
Previous GuideDirect and Inverse Proportions - Identifying direct and inverse variation between quantities and solving real-world word problems using proportion formulasNext GuideImprovement in Food Resources - Methods of crop yield improvement, including crop variety improvement, nutrient management, cropping patterns, and crop protection

Related Study Notes

MathematicsClass 6

Integers

Integers - Understanding negative numbers, representation of integers on a number line, and performing addition and subtraction of integers

Read Article
MathematicsClass 6

Basic Geometrical Ideas

Basic Geometrical Ideas - Understanding points, line segments, rays, intersecting and parallel lines, curves, polygons, and basic elements of a circle

Read Article
MathematicsClass 8

Mensuration

Mensuration - Calculating the area of trapeziums, general quadrilaterals, and polygons, alongside surface area and volume of cubes, cuboids, and cylinders

Read Article

NCERT Study Guide Directory

Textbook solutions, chapter notes & practice worksheets by grade

Interlinked Syllabus
Class 10 NCERT Guides27 chapters
  • Database Management System
  • Areas Related to Circles
  • Our Environment
  • Coordinate Geometry
  • Pair of Linear Equations in Two Variables
  • Probability
  • Statistics
  • Heredity and Evolution
  • How do Organisms Reproduce?
  • Control and Coordination
  • Some Applications of Trigonometry
  • Surface Areas and Volumes
  • Metals and Non-metals
  • Triangles
  • Circles
  • The Human Eye and the Colourful World
  • Carbon and its Compounds
  • Magnetic Effects of Electric Current
  • Arithmetic Progressions
  • Electricity
  • Light - Reflection and Refraction
  • Life Processes
  • Acids, Bases and Salts
  • Chemical Reactions and Equations
  • Introduction to Trigonometry
  • Quadratic Equations
  • Real Numbers
Class 9 NCERT Guides20 chapters
  • Quadrilaterals
  • Linear Equations in Two Variables
  • Circles
  • Improvement in Food Resources
  • Triangles
  • Lines and Angles
  • Is Matter Around Us Pure?
  • Heron's Formula
  • Tissues
  • Structure of the Atom
  • Atoms and Molecules
  • Work and Energy
  • Gravitation
  • Force and Laws of Motion
  • Motion
  • The Fundamental Unit of Life
  • Matter in Our Surroundings
  • Coordinate Geometry
  • Number Systems
  • Polynomials
Class 8 NCERT Guides23 chapters
  • Mensuration
  • Direct and Inverse Proportions
  • → Cubes and Cube Roots (Mathematics)
  • Light
  • Reaching the Age of Adolescence
  • Factorisation
  • Exponents and Powers
  • Chemical Effects of Electric Current
  • Force and Pressure
  • Practical Geometry
  • Practical Geometry
  • Crop Production and Management
  • Comparing Quantities
  • Algebraic Expressions and Identities
  • Friction
  • Squares and Square Roots
  • Sound
  • Combustion and Flame
  • Coal and Petroleum
  • Microorganisms: Friend and Foe
  • Linear Equations in One Variable
  • Understanding Quadrilaterals
  • Rational Numbers
Class 7 NCERT Guides16 chapters
  • Data Handling
  • Reproduction in Plants
  • Transportation in Animals and Plants
  • The Triangle and Its Properties
  • Electric Current and Its Effects
  • Respiration in Organisms
  • Physical and Chemical Changes
  • Motion and Time
  • Acids, Bases and Salts
  • Heat
  • Nutrition in Animals
  • Nutrition in Plants
  • Perimeter and Area
  • Integers
  • Rational Numbers
  • Simple Equations
Class 6 NCERT Guides15 chapters
  • Integers
  • Basic Geometrical Ideas
  • Body Movements
  • Ratio and Proportion
  • Fun with Magnets
  • Light, Shadows and Reflections
  • Playing With Numbers
  • Algebra
  • Decimals
  • Fractions
  • Knowing Our Numbers
  • Electricity and Circuits
  • Components of Food
  • Getting to Know Plants
  • Separation of Substances
Ravindra Higher Secondary School Logo

Ravindra Higher Secondary School

Waidhan, Singrauli (M.P.)

We Serve Society By Serving People

Established in 1988, Ravindra Higher Secondary School (RHS Waidhan) is dedicated to delivering excellence in education, character building, and holistic growth for students in Waidhan, Singrauli (MP).

Quick Links

  • Home Page
  • About RHS & Leadership
  • Academic Programs & Curriculum
  • Admissions Process 2026-27
  • Campus & Facilities
  • Faculty & Staff Members
  • Photo & Video Gallery
  • Notice Board & Announcements
  • Contact & Location

Shift & Office Hours

KG to Class 5th (Morning Shift)

07:30 AM – 11:30 AM

Class 6th to 12th (Afternoon Shift)

12:00 PM – 05:00 PM

Administrative Office Hours

Mon – Sat: 09:00 AM – 04:00 PM

Address & Location

  • Ravindra Higher Secondary School, Main Campus, Waidhan, Singrauli, Madhya Pradesh – 486886
  • +91 9826986106
  • [email protected]

© 2026 Ravindra Higher Secondary School, Waidhan, Singrauli. All rights reserved.

Privacy Policy•Contact Us•Student Portal