mathematics ( Ganita Prakash ) for Class 8

Q1: A quadrilateral has three acute angles, each measure 80°. What is the measure of the fourth angle?

Solution:

Let x be the measure of the fourth angle of a quadrilateral.

Sum of the four angles of a quadrilateral = 360°

80° + 80° + 80° + x = 360°

x = 360° – (80° + 80° + 80°)

x = 360° – 240°

x = 120°

Hence, the fourth angle is 120°.

Q2: In a quadrilateral ABCD, the measure of the three angles A, B and C of the quadrilateral is 110°, 70° and 80°, respectively. Find the measure of the fourth angle.

Solution: Let,

∠A = 110°

∠B = 70°

∠C = 80°

∠D = x

We know that the sum of all internal angles of quadrilateral ABCD is 360°.

∠A + ∠B+ ∠C+∠D = 360°

110° + 70° + 80° + x = 360°

260° + x = 360°

x = 360° – 260°

x = 100°

Therefore, the fourth angle is 100°.

Q3: The opposite angles of a parallelogram are (3x + 5)° and (61 – x)°. Find the measure of four angles.

Solution:
Given,
(3x + 5)° and (61 – x)° are the opposite angles of a parallelogram.
We know that the opposite angles of a parallelogram are equal.
Therefore,
(3x + 5)° = (61 – x)°
3x + x = 61° – 5°
4x = 56°
x = 56°/4
x = 14°
⇒ 3x + 5 = 3(14) + 5 = 42 + 5 = 47
61 – x = 61 – 14 = 47
The measure of angles adjacent to the given angles = 180° – 47° = 133°
Hence, the measure of four angles of the parallelogram are 47°, 133°, 47°, and 133°.

Q4: A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.

Solution:
Let ABCD be the rhombus.
Thus, AB = BC = CD = DA

Given that a side and diagonal are equal.
AB = BD (say)
Therefore, AB = BC = CD = DA = BD
Now, all the sides of a triangle ABD are equal.
Therefore, ΔABD is an equilateral triangle.
Similarly,
ΔBCD is also an equilateral triangle.
Thus, ∠A = ∠ABD = ∠ADB = ∠DBC = ∠C = ∠CDB = 60°
∠B = ∠ABD + ∠DBC = 60° + 60° = 120°
And
∠D = ∠ADB + ∠CDB = 60° + 60° = 120°
Hence, the angles of the rhombus are 60°, 120°, 60° and 120°.

Q5: In a trapezium PQRS, PQ || RS, ∠P : ∠S = 3 : 2 and ∠Q : ∠R = 4 : 5. Find the angles of the trapezium.

Solution:

Given,

∠P : ∠S = 3 : 2

∠Q : ∠R = 4 : 5

Let 3x and 2x be the measures of angles P and S.

And let 4x and 5x be the measures of angle Q and R.

As we know, the sum of adjacent angles at the corners of non-parallel sides of a trapezium = 360°

∠P + ∠S = 180°

3x + 2x = 360°

5x = 180°

x = 180°/5 = 36°

So, 3x = 3(36°) = 108°

2x = 2(36°) = 72°

Similarly,

4x + 5x = 180°

9x = 180°

x = 180°/9 = 20°

So, 4x = 4(20°) = 80°

5x = 5(20°) = 100°

Therefore, the angles of the trapezium are ∠P = 108°, ∠Q = 80, ∠R = 100 and ∠S = 72.

Q6: The diagonals of a rhombus are 12 cm and 7.5 cm. Find the area of a rhombus.

Solution:

Given: Length of diagonal 1 = 12 cm

Length of diagonal 2 = 7.5 cm

We know that,

Area of a rhombus = (1/2) × Diagonal 1× Diagonal 2 square units

A = (½)×12×7.5

A = 6×7.5

A = 45 cm²

Hence, the area of a rhombus is 45 cm².

Q7: ABCD is a quadrilateral, whose angles are ∠A = 5(a+2)°, ∠B = 2(2a+7)°, ∠C = 64°, ∠D = ∠C-8°. Determine the value of ∠A.

Solution:

Given that, ∠A = 5(a+2)°, ∠B = 2(2a+7)°, ∠C = 64°, ∠D = ∠C-8°

Hence, ∠D = 64° – 8°

∠D = 56°

As we know,

∠A+∠B+∠C+∠D = 360°

Now, substitute the values, we get

5(a+2)° + 2(2a+7)° + 64°+56° = 360°

5a°+10°+4a°+14° +64° +56° = 360°

9a° + 144° = 360°

9a° = 360° – 144°

9a° = 216°

a° = 216°/9

a° = 24°

Hence, the value of ∠A is:

∠A = 5(a+2)° = 5(24°+2°) = 5 (26°) = 130°

Therefore, ∠A = 130°.

Q8: The three angles of a quadrilateral are 60°, 90°, 110°. Determine the fourth angle.

Solution:

We know that the sum of interior angles of a quadrilateral is 360°.

Given three angles are 60°, 90° and 110°.

Let the unknown angle be “x”.

By using the property of quadrilateral,

60° + 90° + 110° + x = 360°

260° + x = 360°

x = 360° – 260°

x = 100°.

Hence, the fourth angle of a quadrilateral is 100°.

Q9: If the diagonals of a rhombus are 12 cm and 7.5 cm, what is the area of the rhombus?

Solution:

Given: Length of diagonal 1 = 12 cm

Length of diagonal 2 = 7.5 cm

We know that,

Area of a rhombus = (1/2) × Diagonal 1× Diagonal 2 square units

A = (½)×12×7.5

A = 6×7.5

A = 45 cm ²

Hence, the area of a rhombus is 45 cm ² .

Q10: If two angles of a quadrilateral are 76° and 68°, and the other two angles are in the ratio of 5: 7, what are the measures of these two angles?

Solution:

Given that two angles are 76° and 68°.

Let’s denote the other two angles as 5x and 7x.

As we know, the sum of interior angles of a quadrilateral is 360°.

Therefore, 76°+68°+5x + 7x = 360°

144° + 12x = 360°

12x = 360° – 144°

12x = 216°

x = 216°/12

x = 18°

Hence, the other two angles are:

5x = 5(18)° = 90°

7x = 7(18°) = 126°.

Reema’s Curiosity

Ancient civilizations needed numbers to:

• Count food, livestock, trade goods, ritual offerings.
• Track days and seasons.

Modern numbers evolved from ancient Indian concepts:

• Yajurveda Samhita lists numbers by powers of 10:
  1 (eka), 10 (dasha), 100 (shata), 1000 (sahasra), 10,000 (āyuta), … up to 10¹².
• Indian numerals (0–9) developed ~2000 years ago.
• First instance in Bakhshali manuscript (3rd century CE) with 0 as a dot.
• Aryabhata (499 CE) explained calculations using the Indian system.

Transmission of Indian numerals:

• To Arab world (~800 CE) via Al-Khwārizmī (On the Calculation with Hindu Numerals, 825 CE) and Al-Kindi (On the Use of Hindu Numerals, 830 CE).
• To Europe (~1100 CE) and popularised by Fibonacci (~1200 CE).
• Called Arabic numerals in Europe; properly Hindu or Hindu-Arabic numerals.

Key Quote:
“The ingenious method of expressing every number using ten symbols with place value emerged in India.” — LaplaceMechanism of Counting

Purpose: To determine the size of a collection of objects (e.g., counting cows).Methods of Counting

1. Objects
   • Use sticks, pebbles, or other tokens.
   • Each object represents one counted item (one-to-one mapping).
   • Example: 5 cows → 5 sticks.
2. Sounds / Names
   • Map objects to spoken words, letters, or sounds.
   • Limited by the number of available symbols.
   • Example: “a, b, c…”
3. Written Symbols
   • Use a standard sequence of symbols to represent numbers.
   • This is an early form of written numbers.

Early Number SystemsI. Body Parts

• Hands, fingers, body parts used to count (e.g., Papua New Guinea).

II. Tally Marks

• Marks on bones/walls (e.g., Ishango bone, 20,000–35,000 years ago; Lebombo bone, 44,000 years ago).

III. Counting in Groups

• Example: Gumulgal (Australia) counted in 2s:
  • 1 = urapon, 2 = ukasar, 3 = ukasar-urapon, 4 = ukasar-ukasar, …
• Similar systems in South America & South Africa.
• Advantage: Counting in groups reduces effort; precursor to grouping in larger systems (5s, 10s, 20s).

IV. Roman Numerals

• Symbols: I (1), V (5), X (10), L (50), C (100), D (500), M (1000).
• Numbers represented by sums of landmark numbers (e.g., 27 = XXVII).
• Efficiency: Better than tally marks but difficult for arithmetic operations.
• Use of abacus for calculations.

Concept of BaseI. Egyptian Number System

• Landmark numbers: 1, 10, 100, … (powers of 10, base-10 system).
• Representation: Group numbers by landmark numbers starting from the largest.
• Shortcomings: Requires new symbols for higher powers of 10.

II. Base-n Systems

• Landmark numbers: 1, n, n², n³, …
• Example: Base-5 system → 1, 5, 25, 125, …
• Advantages: Simplifies addition and multiplication; product of landmark numbers is another landmark number.

III. Abacus

• Early calculation tool using base-10 grouping of numbers.
• Counters on lines representing powers of 10.

Place Value RepresentationI. Mesopotamian Number System

• Initially used symbols for landmark numbers.
• Developed base-60 (sexagesimal) system.
• Symbols: 1 and 10.
• Numbers grouped by powers of 60 (e.g., 7530 = 2×3600 + 5×60 + 30).
• Positional system: Place of symbol determines value.
• Placeholder symbol (like 0) introduced for blanks.
• Influence seen in modern time measurement (60 min = 1 hour, 60 sec = 1 min).

II. Mayan Number System

• Central America, 3rd–10th centuries CE.
• Place value system with base-20, used placeholder for 0 (shell symbol).
• Dots = 1, Bars = 5; numbers written vertically.

III. Chinese Number System

1. Type of System
   • Used rod numerals.
   • Decimal system (base-10).
   • In use from 3rd century AD.
2. Symbols
   • Separate symbols for 1–9.
   • Vertical position of the symbol indicates powers of 10.
3. Placeholders
   • Blank spaces used as placeholders to represent missing values.
   • This is similar to the Hindu number system.

Significance

• Early use of positional notation.
• Enabled representation of large numbers efficiently.

IV. Hindu Number System

1. Base-10 system
   • Also called the decimal system.
   • Uses place value to represent numbers.
2. Symbols / Digits
   • 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
3. Use of Zero (0)
   • Used both as a digit and as a number.
   • Introduced around 200 BCE.
   • Codified by Aryabhata and Brahmagupta.
4. Advantages
   • Allows unambiguous representation of numbers.
   • Facilitates arithmetic operations like addition, subtraction, multiplication, and division.
5. Significance
   • Foundation for modern mathematics including algebra, analysis, and computational methods.

Need to Count (Stone Age)

Purposes: Food, animals, trade, rituals, calendars.

No number names/symbols → Used one-to-one mapping with:

• Sticks/Pebbles/Seeds (Method 1)
• Sounds/Names (Method 2) → Limited by available sounds/letters.
• Written symbols (Method 3) → e.g., Roman numerals.

Key Concepts

• One-to-One Mapping: Each object ↔ one counting unit.
• Numerals = Written symbols in a number system.
• Landmark Numbers: Special values to build other numbers (e.g., 1, 5, 10…).

Early Number Systems

(A)Tally Marks

• Simple notches/lines.
• Ishango Bone (20k–35k yrs old) – possibly calendar.
• Lebombo Bone (44k yrs old) – lunar calendar.

(B) Counting in Groups

• Gumulgal (Australia): Count in 2’s → Numbers = combinations of 2’s & 1’s.
• Also used by Bakairi (S. America) & Bushmen (S. Africa).
• Common groups in history: 2, 5, 10, 20.

(C) Roman Numerals

• Symbols: 
• Rules:
  1. If a smaller numeral is placed before a larger one, subtract it.
     Example: IV = 5 − 1 = 4.
  2. If a smaller numeral is placed after a larger one, add it.
     Example: VI = 5 + 1 = 6.
• Advantage: Shorter than tally marks.
• Limitations:
  1. No zero.
  2. Difficult for large numbers.
  3. Cannot perform complex arithmetic easily.

Idea of Base-n

• Base-n system: Landmark numbers = powers of n
  Example: Base-5 → 1, 5, 25, 125…
• Advantages: Consistent grouping, easier addition & multiplication.

Egyptian System

• Base-10, symbols for 1, 10, 100, 1000, 10,000…
• Build numbers by repeating symbols. For example 324 which equals 100 + 100 + 100 + 10 + 10 + 4 is written as
• Limit: Needs infinite symbols for very large numbers.

Abacus

• Decimal-based calculating tool.
• Each line = power of 10.
• Counters above line = value of 5× that landmark.

Mesopotamian System

• Location: Ancient civilisation in present-day Iraq and nearby regions.
• Time Period: Around 4000 years ago.
• Base: Base-60 (sexagesimal system).
• Symbols:
  • Two main wedge-shaped symbols (cuneiform writing) for numbers.
  • Numbers formed by repeating and combining these symbols.
• Special Use:
  • Still used today in measuring time (60 seconds in a minute, 60 minutes in an hour) and angles (360° circle).

Mayan Number System Basics

• Base: Modified base-20.
  1. 1st place: 1’s (units)
  2. 2nd place: 20’s
  3. 3rd place: 360’s (not 400, due to calendar reasons)
  4. 4th place: 7200’s, etc.
• Symbols:
  1. Dot (•) = 1
  2. Bar (—) = 5
  3. Shell = 0 (placeholder)
• Numbers are written vertically, lowest value at bottom.

Chinese Rod Numeral System – Key Points1. Purpose

Two systems existed:

• Written system – for recording quantities.
• Rod numeral system – for performing calculations efficiently.

2. Rod Numerals

• Base: Decimal (base-10), like our modern system.
• Digits 1–9: Represented using vertical or horizontal rods (small sticks or lines).
• Place value:
  1. Vertical rods → used for units and hundreds places.
  2. Horizontal rods → used for tens and thousands places.
     (This alternation prevented confusion between adjacent digits.)

3. Zero Representation

• Like the Mesopotamians: used a blank space to indicate an empty place value.
• Advantage: Due to uniform rod sizes, the blank space was easier to identify.
• Note: If they had an actual symbol for zero, it would have been a fully developed place value system like the Hindu–Arabic numerals.

Spread of Hindu–Arabic Numerals

• Base: Base-10 (decimal system).
• Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
• Place Value System: Value of a digit depends on its position.
  Example: In 375,
  3 → Hundreds place = 3 × 100 = 300
  7 → Tens place = 7 × 10 = 70
  5 → Ones place = 5 × 1 = 5
• Use of Zero: A major contribution by Indian mathematicians (Aryabhata, Brahmagupta).
• Spread: Carried to Europe by Arab traders → became the Hindu–Arabic numerals we use today.

Comparing Systems

Exponential Notation

Definition: Shorthand for repeated multiplication of the same number.

Examples:

• n² = n × n
• n³ = n × n × n
• n⁴ = n × n × n × n

Algebra:

• a³ × b² = a × a × a × b × b
• a² × b⁴ = a × a × b × b × b × b

Important Note: Addition is not exponent:

• 4 + 4 + 4 = 3 × 4 = 12
• 4 × 4 × 4 = 4³ = 64

Prime Factorization and Exponential Form

Prime factorization is expressing a number as a product of its prime numbers.

Exponential form is writing repeated prime factors using powers.

Steps to Express in Exponential Form

1. Find the prime factors of the number.
2. Group the same factors together.
3. Write each group as a power.
4. Combine to get the exponential form.

Example: Express 32,400 in exponential form

1. Prime Factorization:
32,400 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

2. Group factors:

• 2 × 2 × 2 × 2 = 2⁴
• 3 × 3 × 3 × 3 = 3⁴
• 5 × 5 = 5²

3. Exponential Form:
32,400 = 2⁴ × 3⁴ × 5²

Quick Tip: Prime factorization is useful for finding HCF, LCM, and simplifying roots of numbers.

Laws of Exponents: Multiplication and Division of Powers

1. Multiplying Same Bases

• Rule: Add the exponents
• Formula: nᵃ × nᵇ = nᵃ⁺ᵇ
• Example: p⁴ × p⁶ = p¹⁰

2. Power of a Power

• Rule: Multiply the exponents
• Formula: (nᵃ)ᵇ = nᵃ×ᵇ
• Example: (2⁵)² = 2¹⁰

3. Dividing Powers with Same Base

• Rule: Subtract the exponents (denominator from numerator)
• Formula: nᵃ ÷ nᵇ = nᵃ⁻ᵇ, n ≠ 0
• Example: 2⁴ ÷ 2³ = 2¹

4. Negative Powers

• Rule: Reciprocal of the positive power
• Formula: n⁻ᵃ = 1 ÷ nᵃ
• Example: 3⁻² = 1 ÷ 3² = 1/9

5. Zero Exponent

• Rule: Any non-zero number to the power 0 is 1
• Formula: x⁰ = 1, x ≠ 0
• Example: 7⁰ = 1

6. Multiplying Different Bases with Same Exponent

• Rule: Multiply the bases, keep the exponent
• Formula: mᵃ × nᵃ = (m × n)ᵃ
• Example: 2³ × 5³ = (2 × 5)³ = 10³

7. Dividing Different Bases with Same Exponent

• Rule: Divide the bases, keep the exponent
• Formula: mᵃ ÷ nᵃ = (m ÷ n)ᵃ
• Example: 8² ÷ 2² = (8 ÷ 2)² = 4²

 Linear vs. Exponential Growth

1. Linear Growth

Description: Adds a fixed amount per step.

Example:

• Distance to the Moon: 384,400 km = 384,400,000 m
• Step size: 20 cm = 0.2 m
• Number of steps:
  384,400,000 / 0.2 = 1,922,000,000 steps = 1.922 × 10⁹

2. Exponential Growth

Description: Multiplies by a fixed factor per step.

Example: Paper folding to the Moon:

• Initial thickness: 0.001 cm
• Number of folds: 46
• Thickness:
  T = 0.001 × 2⁴⁶ ≈ 7,036,874,841,600 cm ≈ 703,687.48 km

Powers of 10 and Scientific Notation

1. Expanded Form Using Powers of 10

For Whole Numbers:
Example: 47,561

• Expanded form:
  4 × 10⁴ + 7 × 10³ + 5 × 10² + 6 × 10¹ + 1 × 10⁰

For Decimals:
Example: 561.903

• Expanded form:
  5 × 10² + 6 × 10¹ + 1 × 10⁰ + 9 × 10⁻¹ + 0 × 10⁻² + 3 × 10⁻³

2. Scientific Notation

Any number can be written as: x × 10ʸ, where:

• x = coefficient (usually between 1 and 10)
• y = exponent (shows the scale of the number)

Examples:

• 5,900 → 5.9 × 10³
• 8,000,000 → 8 × 10⁶

Importance of Exponent: Determines scale; coefficient adjusts precision.

Importance of the Exponent

• Exponent (y) determines the scale or magnitude of the number.
• Coefficient (x) adjusts precision for significant digits.

Large Numbers in Nature

• Human population (2025) = 8 × 10⁹
• African elephants = 4 × 10⁵ → ~20,000 people per elephant
• Grains of sand on Earth ≈ 10²¹
• Stars in observable universe ≈ 2 × 10²³
• Drops of water on Earth ≈ 2 × 10²⁵

Fun Fact:

• 10⁶ seconds ≈ 11.6 days
• 10⁹ seconds ≈ 31.7 years

Q1: A new sapling starts with a height of 2 cm. How tall might the plant be after 4 years if its height doubles each year?

Solution: Initial height = 2 cm

The height doubles each year, so after 4 years:
= 2¹ × 2⁴
= (2)¹⁺⁴
= 32 cm

The plant will be 32 cm tall after 4 years.

Q2: The number of books in a library increases by 5 times every 2 years. A library starts with 100 books. How many books will there be in the library after:

(a) 6 years

(b) 10 years

Solution:  Initial number of books = 100

The number of books increases by a factor of 5 every 2 years.

(a) After 6 years (3 periods of 2 years):

Number of books after 6 years = 100 × 5³ = 100 × 125 = 12500 books

• After 6 years: 12500 books

(b) After 10 years (5 periods of 2 years):

Number of books after 10 years = 100 × 5⁵ = 100 × 3125 = 312500 books

• After 10 years: 312500 books

Q3: The number of bacteria in a culture increases 3 times every hour. A culture starts with 1 bacterium. How many bacteria will be in the culture after 5 hours?

Solution:

Initial bacteria = 1

The bacteria triple every hour, so after 5 hours:
Number of bacteria after 5 hours = 1 × 3⁵ = 1 × 243 = 243 bacteria

The number of bacteria will be 243 after 5 hours.

Q4: The planet Uranus is approximately 2,896,819,200,000 metres awayfrom the Sun. What is this distance in standard form?

Solution:

The distance of Uranus from the Sun is given as 2,896,819,200,000 meters.

To express this in standard form:

2,896,819,200,000 = 2.8968192 × 10¹²

So, the distance of Uranus from the Sun in standard form is:

2.8968192 × 10¹² meters

Q5: An inch is approximately equal to 0.02543 metres. Write this distance in standard form.

Solution:

We are given that 1 inch ≈ 0.02543 meters.

To express this in standard form:

0.02543 = 2.543 × 10⁻²

So, the distance in standard form is:

2.543 × 10⁻² meters

Q6: A particular star is at a distance of about 8.1 × 10¹³ km from the Earth. Assuring that light travels at 3 × 10⁸ m per second, find how long does light takes from that star to reach the Earth.

Solution:

Given, a particular star is at a distance of about 8.1 × 10¹³ km from the Earth.

Assuring that light travels at 3 × 10⁸ m per second.

We have to find the time the light takes from that star to reach the Earth.

We know, speed = distance / time

Given, speed = 3 × 10⁸ m/s

Distance = 8.1 × 10¹³ km

We know, 1 km = 1000 m

= 8.1 × 10¹³ × 10³

Using the law of exponents,

a^m × aⁿ = a^{m + n}

= 8.1 × 10^{13 + 3}

= 8.1 × 10¹⁶ m

Time = distance / speed

= 8.1 × 10¹⁶ / 3 × 10⁸

= (8.1 / 3) × (10¹⁶/10⁸)

= 2.7 × (10¹⁶/10⁸)

Using the law of exponents,

a^m ÷ aⁿ = a^{m – n}

= 2.7 × 10^{16 – 8}

= 2.7 × 10⁸ seconds

Therefore, the required time is 2.7 × 10⁸ seconds.

Q7: In a stack, there are 4 books, each of thickness 15mm, and 6 paper sheets, each of thickness 0.010mm. What is the total thickness of the stack?

Solution:

Thickness of each book = 15mm

Number of books in the stack = 4

Thickness of 4 books = 4 × 15 = 60mm

Thickness of each paper sheet = 0.010mm

Thickness of 6 paper sheets = 6 × 0.010 = 0.060mm

Total thickness of the stack = 60mm + 0.060mm = 60.060mm

Q8: A number when divides ( –15) ^–1 results ( –5) ^–1. Find the number.

Solution:

Let x be the number such that

( –15) ^–1 ÷ x = ( –5) ^–1

⇒ –1/15 ÷ x = –⅕

⇒ –1/15 × 1/x = –⅕

⇒ –1/15x = –⅕

⇒ 15x = 5

⇒ x = ⅓ or 3 ^–1

Q9: A savings account balance quadruples every 3 years. The initial balance in a savings account is 1500 rupees. How much will the balance be after 9 years?

Solution:

​Initial balance = 1500 rupees

The balance quadruples every 3 years, so after 9 years:

Balance after 9 years = 1500 × 4³ = 1500 × 64 = 96000 rupees

The balance will be 96000 rupees after 9 years

Q10: The volume of the Earth is approximately 7.67 × 10^–7 times thevolume of the Sun. Express this figure in usual form.

Solution:

The volume of Earth is approximately 7.67 × 10^–7 times the volume of the Sun. We are asked to express this in usual form.

To convert from scientific notation to usual form, we move the decimal point to the left by 7 places (since the exponent is -7):

7.67 × 10^–7 = 0.000000767

So, the volume of Earth as a fraction of the volume of the Sun is:

0.000000767

1. Exponential Notation

Formula: A number n multiplied by itself a times is expressed as: nᵃ

Example:
2 × 2 × 2 × 2 = 2⁴ = 16

2. Laws of Exponents

Here are the laws of exponents when a and b are non-zero integers and m, n are any integers.

a. Product Rule

Formula:
nᵃ × nᵇ = nᵃ⁺ᵇ

Example:
2³ × 2⁴ = 2³⁺⁴ = 2⁷ = 128

b. Power of a Power Rule

Formula:
(nᵃ)ᵇ = nᵃ×ᵇ = (nᵇ)ᵃ

Example:
(2³)² = 2³×² = 2⁶ = 64

c. Quotient Rule

Formula:
nᵃ ÷ nᵇ = nᵃ⁻ᵇ

Example:
2⁵ ÷ 2² = 2⁵⁻² = 2³ = 8

d. Product of Different Bases 

Formula:
nᵃ × mᵃ = (n × m)ᵃ

Example:
2³ × 3³ = (2 × 3)³ = 6³ = 216

e. Zero Exponent Rule

Formula:
n⁰ = 1 (n ≠ 0)

Example:
5⁰ = 1

f. Negative Exponent Rule

Formula:
n⁻ᵃ = 1/nᵃ

Example:
2⁻³ = 1/2³ = 1/8

3. Scientific Notation

Formula: A number is written as:
x × 10ᵃ where 1 ≤ x < 10, a is an integer

Example:

• Number: 59,853
• Scientific notation:
  59,853 = 5.9853 × 10⁴

4. Prime Factorization in Exponential Form

Formula: Express a number as a product of prime factors in exponential form.

Example:

• Number: 32,400
• Prime factorization:
  32,400 = 2⁴ × 3⁴ × 5²

5. Combinations

Formula: For k items, each with n choices, the total number of combinations is: nᵏ

Example:

A 5-digit password using digits 0–9:
10⁵ = 100,000 combinations

6. Linear vs. Exponential Growth

a. Linear Growth

Description: Adds a fixed amount per step.

Example:

• Distance to the Moon: 384,400 km = 384,400,000 m
• Step size: 20 cm = 0.2 m
• Number of steps:
  384,400,000 / 0.2 = 1,922,000,000 steps = 1.922 × 10⁹

b. Exponential Growth

Description: Multiplies by a fixed factor per step.

Example: Paper folding to the Moon:

• Initial thickness: 0.001 cm
• Number of folds: 46
• Thickness:
  T = 0.001 × 2⁴⁶ ≈ 7,036,874,841,600 cm ≈ 703,687.48 km

Q1: Give a reason to show that the number given below is a perfect square: 5963 
Sol: The unit digit of the square numbers will be 0, 1, 4, 5, 6, or 9 if we examine the squares of numbers from 1 to 10. Thus, the unit digit for all perfect squares will be 0, 1, 4, 5, 6, or 9, and none of the square numbers will end in 2, 3, 7, or 8.
Given 5963
We have the property of a perfect square, i.e. a number ending in 3 is never a perfect square.
Therefore the given number 5963 is not a perfect square.

Q2: 2025 plants are to be planted in a garden in a way that each of the rows contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Sol:
Let the number of rows be x.
Thus, the number of plants in each row = x.
Total many contributed by all the students = x  ×  x = x²
Given, x² = Rs.2025
x² = 3 × 3 × 3 × 3 × 5 × 5
⇒ x² = (3 × 3) × (3 × 3) × (5 × 5)
⇒ x² = (3 × 3 × 5) × (3 × 3 × 5)
⇒ x² = 45 × 45
⇒ x = √(45 × 45)
⇒ x = 45
Therefore,
Number of rows = 45
Number of plants in each row = 45

Q3: Find out the cube root of 13824 by the prime factorisation method.
Sol: First, let us prime factorise 13824:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2 ³ × 2 ³ × 2 ³ × 3 ³
3√13824 = 2 × 2 × 2 × 3 = 24

Q4: (13/10) ³ 
Sol: The cube of a rational number is the result of multiplying a number by itself three times.
To evaluate the cube of (13/10) ³
Firstly we need to convert into proper fractions, i.e.(13/10) ³
We need to multiply the given number three times, i.e. (13/10) × (13/10) × (13/10) = (2197/1000)
∴ the cube of (1 3/10) is (2197/1000)

​Q5: By what least number should the number be divided to obtain a number with a perfect square? In this, in each case, find the number whose square is the new number 4851.
Sol: The number is a perfect square if and only if the prime factorization creates pairs; it is not exactly a perfect square if it is not paired up.
Given 4851,
Resolving 4851 into prime factors, we obtain
4851 = 3 X 3 X 7 X 7 X 11
= (32 X 72 X 11)
To obtain a perfect square, we need to divide the above equation by 11
we obtain, 9075 = 3 X 3 X 7 X 7
The new number = (9 X 49)
= (3² X 7² )
Taking squares on both sides of the above equation, we obtain
∴ The new number = (3 X 7)²
= (21)²
Therefore, the new number is a square of 21

Q6: Find the cube root of 10648 by the prime factorisation method.
Sol:10648 = 2 × 2 × 2 × 11 × 11 × 11
Grouping the factors in triplets of number equal factors,
10648 = (2 × 2 × 2) × (11 × 11 × 11)
Here, 10648 can be grouped into triplets of number equal factors,
∴ 10648 = 2 × 11 = 22
Therefore, the cube root of 10648 is 22.

Q7: Without adding, find the sum of the following:
(1+3+5+7+9+11+13+15+17+19+21+23)
Sol:   (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
As per the given property of perfect square, for any natural number n, we have some of the first n odd natural numbers = n²
But here n = 12
By applying the above the law, we get
thus, (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)
= 12² 
= 144

Q8: By what least number should the given number be divided to get a perfect square number? In each of the following cases, find the number whose square is the new number 1575.
Sol: A method for determining the prime factors of a given number, such as a composite number, is known as prime factorisation.
Given 1575,
Resolve 1575 into prime factors, we get
1575 = 3 X 3 X 5 X 5 X 7 = (3² X 5² X 7)
To obtain a perfect square, we have to divide the above equation by 7
Then we get, 3380 = 3 X 3 X 5 X 5
New number = (9 X 25) = (3² X 5² )
Taking squares on both sides of the above equation, we get
∴ New number = (3 X 5)² = (15)²

​Q9:If m is the required square of a natural number given by n, then n is
(a) the square of m
(b) greater than m
(c) equal to m
(d) √m
Ans: (d)
Sol: n² = m
Then,
= n = √m
Q10: The cube of 100 will have _________ zeroes.
Sol: The cube of 100 will have  six zeroes.
= 1003
= 100 × 100 × 100
= 1000000

Q11: Use the following identity and find the square of 189.
(a – b)² = a² – 2ab + b²
Sol: 189 = (200 – 11) ²
= 40000 – 2 x 200 x 11 + 112
= 40000 – 4400 + 121
= 35721

Q12: What would be the square root of the number 625 using the identity
(a +b)² = a² + b² + 2ab?
Sol: (625)²
= (600 + 25)²
= 600² + 2 x 600 x 25 +25²
= 360000 + 30000 + 625
= 390625

Q13:Show that the sum of two consecutive natural numbers is 13².
Sol: Let 2n + 1 = 13
So, n = 6
So, ( 2n + 1)² = 4n² + 4n + 1
= (2n² + 2n) + (2n² + 2n + 1)
Substitute n = 6,
(13)² = ( 2 x 6² + 2 x 6) + (2 x 6² + 2 x 6 + 1)
= (72 + 12) + (72 + 12 + 1)
= 84 + 85

​
Q14: (1.2) ³ = _________.
Sol: (1.2) ³  = 12/10
= (12/10) × (12/10) × (12/10)
= 1728/1000
= 1.728

Q15: Find the cube root of 91125 by the prime factorisation method.
Sol: 91125 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
By grouping the factors in triplets of equal factors, 91125 = (3 × 3 × 3) × (3 × 3 × 3) × (5 × 5 × 5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3 × 3 × 5) = 45
Thus , 45 is the cube root of 91125.

Q16: A cuboid of plasticine made by Parikshit with sides 5 cm, 2 cm, and 5 cm. How many such cuboids will be needed to form a cube?
Sol: The given side of the cube is 5 cm, 2 cm and 5 cm.
Therefore, volume of cube = 5 × 2 × 5 = 50
The prime factorisation of 50 = 2 × 5 × 5
Here, 2, 5 and 5 cannot be grouped into triples of equal factors.
Therefore, we will multiply 50 by 2 × 2 × 5 = 20 to get the perfect square.
Hence, 20 cuboids are needed to form a cube.

Q17: State true or false.
(i) The cube of any odd number is even
(ii) A perfect cube never ends with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two-digit number may be a three-digit number.
(vi) The cube of a two-digit number may have seven or more digits.
(vii) The cube of a single-digit number may be a single-digit number.
Sol:
(i) This statement is false.
Taking a cube of any required odd numbers
3³= 3 x 3 x 3 = 27
7³=7 x 7 x 7= 343
5³=5 x 5 x 5=125
All the required cubes of any given odd number will always be odd.
(ii) This statement is true.
10³= 10 x 10 x 10= 1000
20³ = 20 x 20 x 20 = 2000
150³ =150 x150 x150 = 3375000
Hence a perfect cube will never end with two zeros.
(iii) This statement is false.
15²= 15 x15= 225
15³= 15 x 15 x 15= 3375
Thus, the square of any given number ends with 5; then the cube ends with the number 25 is an incorrect statement.
(iv) This statement is false.
2³= 2x2x2= 8
12³ = 12 x 12 x 12= 1728
Accordingly, There are perfect cubes ending with the number 8
(v) This statement is false.
The minimum two digits number is 10
And 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never be a three-digit number.
(vi) This statement is false 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never have seven or more digits.
(vii) This statement is true
1³ = 1 x 1 x 1= 1
2³ = 2 x 2 x 2= 8
According to the cube, a single-digit can be a single-digit number.

Q18: Find the cube of 3.5.
Sol: 3.53 = 3.5 x 3.5 x 3.5
= 12.25 x 3.5
= 42.875

Q19: Find the smallest whole number from which 1008 should be multiplied to obtain a perfect square number. Also, find out the square root of the square number so obtained.
Sol:
Let us factorise the number 1008.1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 
= ( 2 × 2 ) × ( 2 × 2 ) × ( 3 × 3 ) × 7
Here, 7 cannot be paired.
Therefore, we will multiply 1008 by 7 to get a perfect square.
New number so obtained = 1008 ×7 = 7056
Now, let us find the square root of 70567056 = 2 × 2 × 2 × 2 × 3 × 3× 7 × 7
7056 = (2 ×  2 ) × ( 2 × 2 ) × ( 3 × 3 ) ×( 7 × 7 )
7056 = 2² × 2² × 3² × 7²
7056 = (2 × 2 × 3 × 7)²
Therefore;
√7056 = 2 × 2 × 3 × 7
= 84

Q20: √(1.96) = _________.
Sol: We have,
= √(1.96)
= √(196/100)
= √((14 × 14)/(10 × 10))
= √(14² / 10²)
= 14/10
= 1.4


Q21: There are _________ perfect cubes between 1 and 1000.
Sol: 
There are 8 perfect cubes between 1 and 1000.
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
5 × 5 × 5 = 125
6 × 6 × 6 = 216
7 × 7 × 7 = 343
8 × 8 × 8 = 512
9 × 9 × 9 = 729

Q22: Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product is a perfect cube.
Sol: The prime factorisation of 392 gives:
392 = 2 x 2 x 2 x 7 x 7
As we can see, number 7 cannot be paired in a group of three. Therefore, 392 is not a perfect cube.
We must multiply the 7 by the original number to make it a perfect cube.
Thus,
2 x 2 x 2 x 7 x 7 x 7 = 2744, which is a perfect cube, such as 23 x 73 or 143.
Hence, the smallest natural number, which should be multiplied by 392 to make a perfect cube, is 7.

Q23: Which of the following numbers are in perfect cubes? In the case of a perfect cube, find the number whose cube is the given number 256 
Sol: A perfect cube can be expressed as a product of three numbers of equal factors
Resolving the given number into prime factors, we obtain
256 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2
Since the number 256 has more than three factors
∴ 256 is not a perfect cube.

Q24: Find the smallest number by which 128 must be divided to get a perfect cube.
Sol:  The prime factorisation of 128 is given by:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
By grouping the factors in triplets of equal factors,
128 = (2 × 2 × 2) × (2 × 2 × 2) × 2
Here, 2 cannot be grouped into triples of equal factors.
Therefore, to obtain a perfect cube, we will divide 128 by 2.

​Q25: There are _________ perfect squares between 1 and 100.
Sol: There are 8 perfect squares between 1 and 100.
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 36
7 × 7 = 49
8 × 8 = 64
9 × 9 = 81

Q26: Show that each of the numbers is a perfect square. In each case, find the number whose square is the given number:
7056
Sol: 7056,
A perfect square is always expressed as a product of pairs of prime factors.
Resolving 7056 into prime factors, we obtain
7056 = 11 X 539 
= 12 X 588 
= 12 X 7 X 84 
= 84 X 84 
= (84)²
Thus, 84 is the number whose square is 5929
Therefore,7056 is a perfect square.

Square Numbers1. Understanding Squares

• Definition: Area of a square = Side × Side.
• Examples: 1² = 1, 2² = 4, 3² = 9, 4² = 16.
• Squares of natural numbers = perfect squares.

2. Notation

• Square of n = n² (“n squared”).
• Example: 2² = 4, 3² = 9.

Square of Fractions and Decimal Numbers

Fractions/decimals can also be squared: (2.5)² = 6.25.

1. Square of Fractions 

The square of a fraction means multiplying the fraction by itself:

Steps to Square a Fraction

Step 1: Square the numerator – multiply the top number by itself.

Step 2: Square the denominator – multiply the bottom number by itself.

Step 3: Simplify if possible.

Examples

2. Square of Decimal Numbers 

The square of a number is obtained by multiplying the number by itself.

Steps to Square a Decimal Number

Step 1: Ignore the decimal point

• Treat the decimal number as a whole number (remove the decimal point temporarily).

Step 2: Square the number

• Multiply the whole number by itself.

Step 3: Place the decimal

• Count the total number of decimal places in the original number.
• In the square, place the decimal point so that the number of decimal places is twice the original.

Examples

A) 0.3²

• Ignore decimal and square → 3² = 9
• Original decimal places = 1 → Square decimal places = 2
• Answer → 0.09

B) 1.2²

• Ignore decimal and square → 12² = 144
• Original decimal places = 1 → Square decimal places = 2
• Answer → 1.44

3. Patterns and Properties of Perfect Squares

(a) Units Digit Rule

• A square ends only in: 0, 1, 4, 5, 6, 9.
• Never ends in: 2, 3, 7, 8.
• Example: 16, 25, 49 are squares; but 23, 47 are not.

(b) Digits Pattern

• Numbers ending in 1 or 9 → square ends in 1.
• Numbers ending in 4 or 6 → square ends in 6.
• Example: 19² = 361 (ends in 1).

(c) Zeros Rule

• If number ends in n zeros, square ends in 2n zeros.
• Example: 100² = 10,000 (2 zeros → 4 zeros).

(d) Parity (Even/Odd)

• Square of even number = even.
• Square of odd number = odd.
• Example: 12² = 144 (even), 25² = 625 (odd).

(e) Odd Number Differences

Difference of consecutive squares = odd number.

• 2² – 1² = 3, 3² – 2² = 5, 4² – 3² = 7.

Sum of first n odd numbers = n².

(f) Perfect Square Test (Subtraction Rule)

• Subtract consecutive odd numbers from n.
• If result becomes 0 → number is a perfect square.
• Example: 25 – 1 – 3 – 5 – 7 – 9 = 0 → 25 is a square.

(g) Finding Next Square

• (n+1)² = n² + (2n+1).
• Example: 35² = 1225 → 36² = 1225 + 71 = 1296.

(h) Numbers Between Squares

• Between n² and (n+1)² → always 2n numbers.
• Example: Between 4² = 16 and 5² = 25 → 8 numbers.

(i) Triangular Numbers Relation

• Triangular numbers: 1, 3, 6, 10, 15, …
• Sum of two consecutive triangular numbers = perfect square.

Square Roots

Definition

• If y = x², then x = √y.
• Every square root has ± values, but usually positive root is used.
• Example: √49 = ±7.

Methods to Check/Find Square Roots

1. Listing Squares → compare with nearby perfect squares.
   List squares of natural numbers:
   1² = 1, 2² = 4, 3² = 9, 4² = 16, …
   Compare the given number with the list of squares.
   If it matches a square → the square root is the corresponding number.
   Quick Tip:
   If the number is not a perfect square, it lies between the squares of two numbers → √n is between those two numbers.
2. Successive Subtraction of Odd Numbers → subtract until 0.
   – Start with the given number.
   – Subtract 1, 3, 5, 7, 9… successively (odd numbers in order).
   – Count how many subtractions you can do until the result becomes 0.
   – The number of subtractions = the square root.Quick Tip:
   • This method works only for perfect squares.
   • The sequence of odd numbers always starts from 1.
3. Prime Factorisation → group factors in pairs.
   • Example: 256 = 2⁸ 
     → √256 = 2⁴ = 16.
4. Estimation → use nearby perfect squares.
   – Identify perfect squares closest to the given number – one smaller, one larger.
   – Conclude that the square root lies between the roots of these perfect squares.
   – Refine by checking multiples to find the exact root (if it is a perfect square).
   Quick Tip:
   • For numbers not perfect squares, this method gives a good approximate value.
   • Works well with decimal approximations too.

For Non-Perfect Squares

When a number is not a perfect square, its square root can be estimated by comparing it with nearby perfect squares.

Steps:

1. Identify the perfect squares just below and above the number.
2. Conclude that the square root lies between the roots of these perfect squares.
3. Refine the estimate using linear approximation or trial and error.

Quick Tip:

• This method gives a quick approximation.
• For more precision, use long division or a calculator.

Cubic Numbers

Definition and Notation

• Cube = n³ = n × n × n.
• Represents volume of cube of side n.
• Examples: 2³ = 8, 3³ = 27, 4³ = 64.

Properties

• Cubes grow faster than squares.
• Possible last digits of cubes → any digit (0–9).

Relation with Odd Numbers

• n³ = sum of n consecutive odd numbers.
• Example: 4³ = 13+15+17+19 = 64.

Taxicab Numbers

• First discovered by Ramanujan (famous Hardy–Ramanujan number).
• 1729 = 1³+12³ = 9³+10³.
• Smallest number expressible as sum of two cubes in two ways.
• Other examples: 4104, 13832.

Cube Roots

Definition

• If y = x³ → x = ³√y.
• Example: ³√8 = 2.

Finding Cube Roots (Prime Factorisation)

• Group factors into triplets.
• Example: 3375 = (3×3×3)(5×5×5) → ³√3375 = 15.
• Example: ³√8000 = 20.8000 = (2×2×2) (2×2×2) (5×5×5)= 2×2×2 = 20.

Successive Differences of Powers

When we list powers of natural numbers in order and calculate the successive differences, a pattern appears:

• For squares (n²) → the second differences are constant.
• For cubes (n³) → the third differences are constant.
• General Rule: For nth powers, the differences become constant at the nth level.

1. Squares (n²)

Numbers:
1, 4, 9, 16, 25

First differences:
3, 5, 7, 9

Second differences:
2, 2, 2  (constant)

2. Cubes (n³)

Numbers:
1, 8, 27, 64, 125

First differences:
7, 19, 37, 61

Second differences:
12, 18, 24

Third differences:
6, 6  (constant)

General Rule

• For numbers to the power 1 (linear) → first differences constant
• For squares → second differences constant
• For cubes → third differences constant
• For nth powers → constant differences appear at nth level

What’s a Perfect Square?

A natural number n is a perfect square if n = m², where m is a natural number. So, when a number is the square of another number, it’s called a perfect square!
Example:
9 = 3² (The square of 3 is 9)
25 = 5² (The square of 5 is 25)

Digits of a Perfect Square:

• If a number ends in 2, 3, 7, or 8, it can never be a perfect square.
• The squares of even numbers are even, and the squares of odd numbers are odd.
• A number ending in an odd number of zeros cannot be a perfect square.

Between Two Squares:

• Between n² and (n+1)², there are exactly 2n non-perfect square numbers!

The Pythagorean Triplet Connection:

• For any natural number n greater than 1, the numbers 2n, (n² – 1), and (n² + 1) form a Pythagorean triplet. 

Square Roots

• The square root is the opposite of squaring a number!
  Example: If you square 4, you get 16. If you take the square root of 16, you get 4.

Number of Digits in Square Numbers

If a number n has n digits, then the number of digits in its square root is:

• n/2 if n is even.
• (n+1)/2 if n is odd.

[Question: 694414]

Square of a Number: 

• When you multiply a number by itself, it’s called squaring that number.
  Example:
  3 × 3 = 9 (This means 9 is the square of 3!)
  5 × 5 = 25 (This means 25 is the square of 5!)

Perfect Square: 

• A number like 16, which can be expressed as 4², is called a perfect square! But remember, not every number is a perfect square. For example, 32 is not a square number. So, always check if a number is the square of another number.

Remember

All-natural numbers are not perfect squares or square numbers, 32 is not a square number. In general, if a natural number ‘m’ can be expressed as n², where n is also a natural number, then ‘m’ is the perfect square. The numbers like 1, 4, 9, 16, 25, and 36 are called square numbers. 

Table: Square of numbers from 1 and 10. 

Properties of Square Number

Table: Let us consider the square of all natural numbers from 1 to 20. 

From the table, we conclude that:

Property 1: “The ending digits (the digits in the one’s place) of a square number is 0, 1, 4, 5, 6 or 9 only.”

[Question: 694416]

Some Interesting Patterns

• Triangular numbers are: 1, 3, 6, 10, 15, 21, etc. If we combine two consecutive triangular numbers, we get a square number.
  1 + 3 = 4, ‘4’ is a square number
  3 + 6 = 9, ‘9’ is a square number
  6 + 10 = 16, ‘16’ is a square number
  and so on.

• 1² =1
  11² = 121
  111² = 12321
  1111² = 1234321
• 7² = 49
  67² = 4489
  667² = 444889
  6667² = 44448889 and so on.

Cube of a number 

A natural number multiplied by itself three times gives a cube of that number, e.g.
1 × 1 × 1 = 1
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
The numbers 1, 8, 27, 64, … are called cube numbers or perfect cubes.

Perfect Cube: A number is a perfect cube if it can be expressed as n³ for some integer n.
Prime Factor Test: In the prime factorization of a perfect cube, every prime factor appears in groups of three.

Cube Root of a Number The cube root of a number is the side length of a cube whose volume is that number.
 is the inverse operation of cubing x.For example :

This means the cube root of 8 is 2. 

Prime Factorization Method: Factorize the number, group identical factors in threes, and multiply one factor from each triplet to get the cube root.

Steps to Calculate Cube Root

1. Prime Factorize the Number
Break the number down into its prime factors.

For example: Prime Factorize 8000
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5

2. Group the Factors in Threes
Arrange identical factors into sets of three.

8000 = (2 × 2 × 2) (2 × 2 × 2) (5 × 5 × 5)

3. Multiply One Factor from Each Triplet
From each group of three identical primes, take one prime and multiply them together.

8000 = (2 × 2 × 2) (2 × 2 × 2) (5 × 5 × 5) 
Picking  one prime from each triplet: 2 × 2 × 5 = 20

4. Result
The product you get in Step 3 is the cube root of the original number.

Therefore, cube root of 8000 is 

[Question: 1284091]

Properties of Perfect Cubes

(a) Property 1
If the digit in the one’s place of a number is 0, 1, 4, 5, 6 or 9, then the digit in the one’s place of its cube will also be the same digit.

(b) Property 2

If the digit in the one’s place of a number is 2, the digit in the one’s place of its cube is 8, and vice-versa.

(c) Property 3

If the digit in the one’s place of a number is 3, the digit in the one’s place of its cube is 7 and vice-versa.

Examples of Properties 1, 2 & 3

(d) Property 4 

Cubes of even natural numbers are even.

Examples of Property 4

(e) Property 5

Cubes of odd natural numbers are odd.

Examples of Property 5

(f) Property 6

Cubes of negative integers are negative.

Examples of Property 6

Some Interesting Patterns in Cubes

1. Adding consecutive odd numbers

Note that we start with [n * (n – 1) + 1] odd number.

2. Difference of two consecutive cubes:

2³ – 1³ = 1 + 2 * 1 * 3

3³ – 2³ = 1 + 3 * 2 * 3

4³ – 3³ = 1 + 4 * 3 * 3

5³ – 4³ = 1 + 5 * 4 * 3

3. Cubes and their prime factor
Each prime factor of the number appears three times in its cube.

Facts That Matter

• If we multiply a number by itself three times, the product so obtained is called the perfect cube of that number.
• There are only 10 perfect cubes from 1 to 1000.
• Cubes of even numbers are even and those of odd numbers are odd.
• The cube of a negative number is always negative.
• If the prime factors of a number cannot be made into groups of 3, it is not a perfect cube.

[Question: 1284092]

Q1: Is 176 a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.

Solution: Prime factorization of 176:
176 = 2⁴ × 11.To make it a perfect square, multiply by 11.
176 × 11 = 1936, which is a perfect square.
√1936 = 44.

Q2: Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Solution: Prime factorization of 9720:
9720 = 2³ × 3⁵ × 5 To make it a perfect cube, divide by 3² × 5 = 45
Divide 9720 by 21725 to get a perfect cube.

Q3: By what smallest number should 216 be divided so that the quotient is a perfect square? Also, find the square root of the quotient.

Solution:
Prime factorization of 216:
216 = 2³ × 3³.
To make the quotient a perfect square, divide by 2 and 3.
216 ÷ 6 =  36
√36 = 6

Q4: By what smallest number should 3600 be multiplied so that the quotient is a perfect cube? Also, find the cube root of the quotient.

Solution:
Prime factorization of 3600:
3600 = 2⁴ × 3² × 5².
To make it a perfect cube, multiply by 2² × 3 × 5 = 4 × 3 × 5 = 60.
3600 × 60 = 216000, and the cube root of 216000 is 60.

Q5: A farmer wants to plough his square field of side 150m. How much area will he have to plough?

Solution:
The area of a square field is given by the formula:
Area = (side)²
Here, the side of the square field is 150 meters.
Area = (150)² = 22,500 m²
So, the farmer will have to plough 22,500 square meters.

Q6: What will be the number of unit squares on each side of a square graph paper if the total number of unit squares is 256?

Solution:
The total number of unit squares on a square graph paper is the square of the number of unit squares on each side.
Let the number of unit squares on each side be x.
So, x² = 256.
Taking the square root of both sides:
x = √256 = 16
Therefore, there are 16 unit squares on each side of the square graph paper.

Q7: If one side of a cube is 15m in length, find its volume.

Solution:
The volume of a cube is given by the formula:
Volume = side³
Here, the side length is 15 meters.
Volume = 15³  = 3375 m³
So, the volume of the cube is 3375 cubic meters.

Q8: Find the number of plants in each row if 1024 plants are arranged so that the number of plants in a row is the same as the number of rows.

Solution:
The total number of plants is 1024, and we are told that the number of plants in each row is the same as the number of rows.
Let the number of rows (and plants per row) be x.
The total number of plants is the product of the number of rows and the number of plants per row:
x × x = 1024
x² = 1024
Taking the square root of both sides:
x = √1024 = 32.
Therefore, the number of plants in each row is 32.

Q11: Difference of two perfect cubes is 189. If the cube root of the smaller of the two numbers is 3, find the cube root of the larger number.

Solution:
Let the two perfect cubes be x³ and y³, where x is the cube root of the smaller number and y is the cube root of the larger number.
We are given:
y³ – x³ = 189 and x = 3.
So, 3³ = 27.
Now, substitute into the equation:
y³ – 27 = 189
y³ = 189 + 27 = 216
So, y = ∛216 = 6.
Therefore, the cube root of the larger number is 6.

Q10: A hall has a capacity of 2704 seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.

Solution:
Let the number of rows and the number of seats in each row be x.
The total number of seats is given by:
x × x = 2704
x² = 2704
Taking the square root of both sides:
x = √2704 = 52.
Therefore, the number of seats in each row is 52.