Saturday, August 16, 2025

Grade 8 Ganita Prakash style case-based MCQs

Case-Based MCQs: Exponential Growth — Folding Paper & King's Diamonds (Grade 8)
Grade 8 • Case-Based MCQs Exponential Growth

Power Play: Folding Paper & The King’s Diamonds

Two real-life style passages with MCQs testing understanding of doubling, powers, and geometric growth. Includes answers with explanations.

Case Study 1 — The Magical Folding Paper

A sheet of paper has an initial thickness of 0.001 cm. When you fold it once, its thickness doubles. This doubling continues with every fold. This is a classic example of exponential growth.

After 10 folds: 1.024 cm (≈ 1 cm)
After 17 folds: ≈ 131 cm (≈ 1.31 m)
After 26 folds: ≈ 670 m
After 30 folds: ≈ 10.7 km (typical airplane altitude)
After 46 folds: > 700,000 km (Earth–Moon scale)

General rule: after n folds, thickness t(n) = 0.001 × 2n cm.

Q1. The thickness after 1 fold is
  1. 0.001 cm
  2. 0.002 cm
  3. 0.01 cm
  4. 0.1 cm
Show Answer & Explanation
Answer: (b) 0.002 cm. One fold doubles the thickness.
Q2. After 10 folds, the thickness is about
  1. 1 cm
  2. 10 cm
  3. 100 cm
  4. 0.1 cm
Show Answer & Explanation
Answer: (a) 1 cm. Exactly 1.024 cm since 0.001 × 210 = 1.024 cm.
Q3. 131 cm (after 17 folds) is closest to
  1. length of a pencil
  2. height of a chair
  3. height of a person
  4. height of a door
Show Answer & Explanation
Answer: (c) height of a person (~1.3 m).
Q4. After 26 folds ≈ 670 m. Which is closest?
  1. Eiffel Tower (330 m)
  2. Burj Khalifa (830 m)
  3. Qutub Minar (73 m)
  4. Empire State Building (443 m)
Show Answer & Explanation
Answer: (b) Burj Khalifa (830 m) is closest to 670 m among the options.
Q5. 10.7 km (after 30 folds) is comparable to
  1. Mariana Trench depth
  2. Height at which airplanes fly
  3. Height of Mount Everest
  4. Depth of a swimming pool
Show Answer & Explanation
Answer: (b) airplane cruising altitude (~10–12 km).
Q6. After 20 folds, thickness is approximately
  1. 1 m
  2. 10 m
  3. 100 m
  4. 1 km
Show Answer & Explanation
Answer: (b) 10 m. 0.001 × 220 = 1048.576 cm ≈ 10.49 m.
Q7. The thickness increases after every 3 folds by
  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times
Show Answer & Explanation
Answer: (d) 8 times, since 23 = 8.
Q8. Exponential growth here means
  1. increase by a fixed number
  2. double every step
  3. decrease every step
  4. remain constant
Show Answer & Explanation
Answer: (b) doubles every fold.
Q9. After 46 folds the thickness reaches about 700,000 km. This can reach
  1. the Sun
  2. the Moon
  3. the Mariana Trench
  4. Mount Everest
Show Answer & Explanation
Answer: (b) the Moon (Earth–Moon distance scale).
Q10. This growth pattern follows
  1. Arithmetic progression
  2. Linear growth
  3. Geometric progression
  4. Subtraction pattern
Show Answer & Explanation
Answer: (c) Geometric progression (common ratio 2).

Case Study 2 — The King’s Diamonds

A king has 3 daughters. Each daughter has 3 baskets. Each basket has 3 keys. Each key opens 3 rooms. Continuing this multiplicative pattern through seven stages gives the total diamonds:

Total diamonds = 37 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187

Computed as 37 = (34) × (33) = 81 × 27 = 2187. Here, 34 corresponds to rooms, and multiplying by 33 accounts for the remaining levels to reach diamonds.

Q1. Total baskets
  1. 3
  2. 6
  3. 9
  4. 27
Show Answer & Explanation
Answer: (c) 9, since 3 daughters × 3 baskets each.
Q2. Total keys
  1. 9
  2. 18
  3. 27
  4. 81
Show Answer & Explanation
Answer: (c) 27, because 9 baskets × 3 keys each.
Q3. Total rooms
  1. 27
  2. 34
  3. 81
  4. 243
Show Answer & Explanation
Answer: (c) 81 = 34 (27 keys × 3 rooms each).
Q4. Total number of diamonds
  1. 243
  2. 729
  3. 2187
  4. 6561
Show Answer & Explanation
Answer: (c) 2187. The passage uses seven factors of 3: 37 = 2187.
Note: If you counted only “rooms × 3 diamonds per room,” you’d get 243, but the text clearly frames a 7-stage multiplicative pattern.
Q5. The total diamonds can be expressed as
  1. 34
  2. 35
  3. 36
  4. 37
Show Answer & Explanation
Answer: (d) 37. Seven successive “×3” stages.
Q6. If 34 = 81 (rooms), to get 37 you multiply 81 by
  1. 3
  2. 9
  3. 27
  4. 243
Show Answer & Explanation
Answer: (c) 27, since 37 = 34 × 33 = 81 × 27.
Q7. The concept illustrated is
  1. Arithmetic progression
  2. Powers & exponents
  3. Fractions
  4. Decimals
Show Answer & Explanation
Answer: (b) Powers & exponents (repeated multiplication by 3).

Teacher Tips

  • Estimate vs exact: Have students compute a few exact powers, then use order-of-magnitude comparisons (m, km).
  • Structure trees: Draw branching trees for the King’s problem to make exponents countable.
  • Discussion prompt: Why does doubling (or multiplying by 3) explode so fast compared to adding?

Friday, August 15, 2025

Squares and Cubes – Complete Notes

Squares and Cubes – Complete Notes


1. Square of a Number

  • Definition: The square of a number nn is n×nn \times n, written as n2n^2.

  • Read as: “n squared”

  • Example: =25

  • Inverse Operation: Square root (   √).


2. Perfect Squares

  • Definition: A number whose square root is a whole number.

  • List: 1,4,9,16,25,36,49,64,81,100,

  • Properties:

    1. If a number ends in 2, 3, 7, or 8 → Not a perfect square.

    2. Perfect squares end in 0, 1, 4, 5, 6, or 9.

    3. Perfect squares can only have an even number of zeros at the end.

    4. Square of even numbers → even; square of odd numbers → odd.

    5. A number ending in an odd number of zeros is never a perfect square.

    6. Difference between squares of two consecutive numbers:

      (n+1)2n2=2n+1(n+1)^2 - n^2 = 2n + 1
    7. Pythagorean Triplet: m2+n2=p2m^2 + n^2 = p^2. Example: 32+42=523^2 + 4^2 = 5^2.


3. Patterns in Perfect Squares

  • Last digit patterns:

    • Ends in 1 or 9 → Square ends in 1

    • Ends in 4 or 6 → Square ends in 6

    • Ends in 5 → Square ends in 5

  • Sum of first nn odd numbers:

    1+3+5++(2n1)=n21 + 3 + 5 + \dots + (2n - 1) = n^2

    Example: 1+3+5=9=321 + 3 + 5 = 9 = 3^2

  • Numbers between two consecutive perfect squares:

    2mwhere m and m+1 are consecutive numbers2m \quad \text{where } m \text{ and } m+1 \text{ are consecutive numbers}

4. Square Root

  • Definition: If y=x2y = x^2, then xx is the square root of yy.

  • Notation: y\sqrt{y}

  • Example: 49=±7\sqrt{49} = \pm 7 (positive and negative roots for integers).

  • Perfect Square Test: A number is a perfect square if prime factors can be paired into two identical groups.


5. Cube of a Number

  • Definition: The cube of a number nn is n×n×nn \times n \times n, written as n3n^3.

  • Example: 33=273^3 = 27

  • Can be positive or negative: (2)3=8(-2)^3 = -8.


6. Perfect Cubes

  • Definition: A number whose cube root is a whole number.

  • Example: 273=3\sqrt[3]{27} = 3 → 27 is a perfect cube.

  • List: 1,8,27,64,125,216,343,512,729,1000,1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, \dots

  • Properties:

    1. Cube of even number → even; cube of odd number → odd.

    2. Cube of a number ending in:

      • 0 → ends in 0

      • 1 → ends in 1

      • 2 → ends in 8

      • 3 → ends in 7

      • 4 → ends in 4

      • 5 → ends in 5

      • 6 → ends in 6

      • 7 → ends in 3

      • 8 → ends in 2

      • 9 → ends in 9

    3. Sum of cubes of first nn natural numbers:

      13+23++n3=(n(n+1)2)21^3 + 2^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2
    4. In prime factorisation, if each prime factor occurs three times, the number is a perfect cube.


7. Cube Root

  • Definition: If y=x3y = x^3, then xx is the cube root of yy.

  • Notation: y3\sqrt[3]{y}

  • Example: 83=2\sqrt[3]{-8} = -2.

  • Perfect Cube Test: Prime factors can be split into three identical groups.


8. Special Numbers – Hardy–Ramanujan (Taxicab) Numbers

  • Definition: Numbers that can be expressed as the sum of two cubes in two different ways.

  • Example:

    1729=13+123=93+1031729 = 1^3 + 12^3 = 9^3 + 10^3
  • First few taxicab numbers: 1729, 4104, 13832, ...


9. Key Patterns to Remember

Concept Pattern / Formula
Difference between consecutive squares 2n+12n+1
Sum of first nn odd numbers n2n^2
Sum of first nn cubes (n(n+1)2)2\left( \frac{n(n+1)}{2} \right)^2
Last digit pattern for cubes (0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9)


Tuesday, August 12, 2025

Chapter 1: A Square and A Cube

Chapter 1: A Square and A Cube

Study Material

  • Square of a number: The square of a number n is n × n, written as n2. Example: 52 = 25. (Competency: Understanding and computing squares)
  • Cube of a number: The cube of a number n is n × n × n, written as n3. Example: 33 = 27. (Competency: Understanding and computing cubes)
  • Properties of squares:
    • All squares are positive numbers.
    • Square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...
  • Properties of cubes:
    • Cubes can be positive or negative depending on the number.
    • Cube numbers: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, ...
  • Perfect square: A number whose square root is a whole number. Example: 81 is a perfect square because √81 = 9. (Competency: Identifying perfect squares)
  • Perfect cube: A number whose cube root is a whole number. Example: 27 is a perfect cube because ∛27 = 3. (Competency: Identifying perfect cubes)
  • Patterns: The difference between consecutive square numbers increases by 2 each time: 1, 4, 9, 16 → differences 3, 5, 7, ...

Insert NCERT diagram of squares and cubes from textbook page 2.

Worksheet

I. Multiple Choice Questions (MCQs)

  1. The square of 12 is:
    1. 124
    2. 144
    3. 122
    4. None of these
    (Competency: Calculating squares)
  2. The cube of -4 is:
    1. -64
    2. 64
    3. -124
    4. None of these
    (Competency: Calculating cubes with negative numbers)
  3. Which of the following is NOT a perfect square?
    1. 49
    2. 36
    3. 50
    4. 64
    (Competency: Identifying perfect squares)

II. Assertion–Reasoning

  1. Assertion (A): The square of any odd number is odd. Reason (R): Odd × Odd = Even.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false and R is true. (Competency: Logical reasoning with properties of numbers)
  2. Assertion (A): The cube of any even number is even. Reason (R): Even × Even × Even = Even.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false and R is true. (Competency: Logical reasoning with cubes)

III. True or False

  1. All prime numbers are perfect squares. (Competency: Understanding perfect squares)
  2. 125 is a perfect cube. (Competency: Identifying perfect cubes)
  3. The square root of 81 is 9. (Competency: Computing square roots)

IV. Short Answer (1 mark)

  1. Find the square of 25. (Competency: Calculating squares)
  2. Find the cube of 6. (Competency: Calculating cubes)

V. 2-Mark Questions

  1. Write the first five square numbers. (Competency: Listing square numbers)
  2. Write the first five cube numbers. (Competency: Listing cube numbers)

VI. 3-Mark Questions

  1. Find the difference between the square of 15 and the cube of 5. (Competency: Application of squares and cubes)
  2. If x2 = 196, find x. (Competency: Finding square roots)

VII. Long Answer (5 Marks)

  1. A box contains small cubes of side 2 cm. How many such cubes can be placed in a cubical box of side 10 cm? (Competency: Application of cube and volume concepts)

VIII. Case-Based Questions (CBQ – Competency Based Questions)

Read the following and answer the questions:

A builder is making a square floor of side 12 m and wants to decorate it with cubic tiles of side 1 m. He also needs to paint a cubic water tank of side 4 m.

  1. Area of the floor is:
    1. 144 m2
    2. 120 m2
    3. 100 m2
    4. None of these
    (Competency: Calculating area of a square)
  2. Number of tiles needed:
    1. 144
    2. 120
    3. 100
    4. 96
    (Competency: Application of square measurement in real life)
  3. Volume of the water tank is:
    1. 64 m3
    2. 48 m3
    3. 32 m3
    4. None of these
    (Competency: Calculating volume of a cube)
  4. If 1 m2 requires 2 liters of paint, total paint needed for the tank is:
    1. 192 liters
    2. 160 liters
    3. 128 liters
    4. None of these
    (Competency: Application of surface area of a cube)

Insert NCERT image of cube and square real-life examples from textbook page 6.

Sunday, August 10, 2025

Case-Based Study Questions ch1 a square and a cube class 8

 

Case-Based Study Questions (4 MCQs)

Case Study:
Khoisnam was part of a puzzle challenge involving 100 lockers. Each person toggled lockers according to their assigned number (Person 1 opened every locker, Person 2 toggled every second locker, and so on). At the end, Khoisnam noticed only certain lockers remained open. He also found that some lockers were toggled exactly twice and these helped him form a passcode.


Q1. Which locker numbers will remain open at the end of the toggling process?
(a) All even-numbered lockers
(b) All prime-numbered lockers
(c) All perfect square lockers
(d) All lockers with even number of factors
Answer: (c)

Q2. If the first five locker numbers toggled exactly twice form a code, what will be the code?
(a) 2, 3, 4, 5, 6
(b) 2, 3, 5, 7, 11
(c) 2, 4, 6, 8, 10
(d) 3, 5, 7, 11, 13
Answer: (b)

Q3. Which statement best explains why perfect square lockers remain open?
(a) Perfect squares have an even number of factors.
(b) Perfect squares have an odd number of factors due to a repeated factor.
(c) Perfect squares are always prime numbers.
(d) Perfect squares are always even numbers.
Answer: (b)

Q4. If locker number 49 remains open, which statement is correct?
(a) 49 has an even number of factors.
(b) 49 is a cube number.
(c) 49 is a perfect square.
(d) 49 is a prime number.
Answer: (c)

Case Study:
Akhil is designing square and cube-shaped tiles for a math exhibition. The side lengths of the tiles are whole numbers, and he labels each tile with its area (for squares) or volume (for cubes). While checking his labels, he notices patterns in the numbers and decides to play a guessing game with his friends.


Q1. Akhil has a square tile with area 256 cm². What is the length of its side?
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 16 cm
Answer: (d)


Q2. One of the cube-shaped tiles has a volume of 512 cm³. What is the length of its side?
(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 6 cm
Answer: (b)


Q3. Akhil notices that some numbers cannot be perfect squares because of their unit digit. Which of the following cannot be a perfect square?
(a) 441
(b) 676
(c) 538
(d) 784
Answer: (c)


Q4. The largest square tile Akhil can make with an integer side length from a cloth of area 300 cm² has a side of:
(a) 16 cm
(b) 17 cm
(c) 18 cm
(d) 15 cm
Answer: (d)

(Because 15² = 225 < 300 and 16² = 256 < 300, but 17² = 289 is also possible — however, for maximum integer side length without exceeding the cloth area, we pick the largest perfect square ≤ 300, which is 289 → 17 cm)


Q5. Akhil labels a cube tile with volume 1331 cm³. Which of these is correct?
(a) It is not a perfect cube.
(b) Its side length is 11 cm.
(c) Its side length is 13 cm.
(d) Its side length is 15 cm.
Answer: (b)

Case Study 1 – The Locker Puzzle

Queen Ratnamanjuri’s 100-locker puzzle required each person to toggle lockers based on their number. At the end, only certain lockers stayed open, and some lockers were toggled exactly twice, forming a passcode.

Q1. Which lockers remain open at the end?
(a) All even-numbered lockers
(b) All prime-numbered lockers
(c) All perfect square lockers
(d) All lockers with even number of factors
Answer: (c)

Q2. If the first five lockers toggled exactly twice form the code, what is it?
(a) 2, 3, 4, 5, 6
(b) 2, 3, 5, 7, 11
(c) 2, 4, 6, 8, 10
(d) 3, 5, 7, 9, 11
Answer: (b)

Q3. Why do perfect square lockers remain open?
(a) They have an odd number of factors due to a repeated factor.
(b) They are all prime numbers.
(c) They are all even numbers.
(d) They have an even number of factors.
Answer: (a)

Q4. If locker #81 is open, which is true?
(a) 81 is a cube number.
(b) 81 is a perfect square.
(c) 81 is a prime number.
(d) 81 has an even number of factors.
Answer: (b)


Case Study 2 – Square Tile Designs

Akhil designs square tiles for a math exhibition, each with integer side lengths. He writes their areas and notices patterns in the last digits.

Q1. Which of these numbers cannot be a perfect square?
(a) 784
(b) 625
(c) 676
(d) 538
Answer: (d)

Q2. A tile has an area of 441 cm². What is the side length?
(a) 20 cm
(b) 21 cm
(c) 22 cm
(d) 23 cm
Answer: (b)

Q3. The largest square tile that can be cut from a cloth of area 300 cm² has side length:
(a) 15 cm
(b) 16 cm
(c) 17 cm
(d) 18 cm
Answer: (c) (Because 17² = 289 ≤ 300)

Q4. Which of these areas ends in 6 and is a perfect square?
(a) 256
(b) 216
(c) 676
(d) 196
Answer: (c) (676 = 26²)


Case Study 3 – Cube Blocks

Ravi stacks wooden cubes to make cube-shaped blocks. Each cube has equal side length in cm.

Q1. A block has volume 512 cm³. What is the side length?
(a) 6 cm
(b) 7 cm
(c) 8 cm
(d) 9 cm
Answer: (c)

Q2. Which of these is not a perfect cube?
(a) 729
(b) 125
(c) 216
(d) 500
Answer: (d)

Q3. The cube of a number ending in 5 will have a units digit:
(a) 5
(b) 0
(c) 1
(d) 6
Answer: (a)

Q4. Which cube has side length 11 cm?
(a) 1331 cm³
(b) 1728 cm³
(c) 2197 cm³
(d) 2744 cm³
Answer: (a)


Case Study 4 – The Hardy–Ramanujan Number

When Hardy visited Ramanujan, he mentioned taxi number 1729. Ramanujan said it was special — the smallest number expressible as the sum of two cubes in two different ways.

Q1. 1729 can be written as:
(a) 1³ + 12³ = 9³ + 10³
(b) 1³ + 12³ = 8³ + 11³
(c) 1³ + 11³ = 8³ + 10³
(d) 1³ + 13³ = 8³ + 10³
Answer: (a)

Q2. Numbers like 1729 are called:
(a) Square numbers
(b) Perfect cubes
(c) Taxicab numbers
(d) Prime numbers
Answer: (c)

Q3. The next taxicab number after 1729 is:
(a) 1331
(b) 4104
(c) 4913
(d) 32768
Answer: (b)

Q4. Which of these is a perfect cube from the list above?
(a) 1331
(b) 4104
(c) 1729
(d) 32768
Answer: (a) (1331 = 11³)


Case Study 5 – Odd Number Patterns

Manoj explores a pattern:
1 = 1²,
1 + 3 = 4 = 2²,
1 + 3 + 5 = 9 = 3²,
and so on.

Q1. The sum of the first 10 odd numbers is:
(a) 10² = 100
(b) 10³ = 1000
(c) 20² = 400
(d) 5² = 25
Answer: (a)

Q2. The 15th odd number is:
(a) 27
(b) 29
(c) 31
(d) 33
Answer: (c) (Formula: 2n – 1)

Q3. The sum of the first n odd numbers equals:
(a) n³
(b) 2n
(c) n²
(d) n(n + 1)/2
Answer: (c)

Q4. If 1225 is the sum of the first 35 odd numbers, then adding the 36th odd number gives:
(a) 1250
(b) 1261
(c) 1296
(d) 1369
Answer: (c)

CLASS 6 CH-4 DATA HANDLING AND PRESENTATION NCERT SOLUTIONS GANITA PRAKASH

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