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A Square and a Cube Class 8 Maths Worksheet with Answers Chapter 1

Updated:5 Mins Read

Ganita Prakash Part-I, Grade 8, Textbook of Mathematics – Here we present the 1st chapter from the NCERT Ganita Prakash Part-I book. We want the students to attempt these questions after completing the chapter. The worksheet is divided into 4 parts, namely – Basic, Standard, Advance, HOTS. Answers are provided at the bottom for each.

Class 8 Chapter 1 Maths Worksheet: A Square and A Cube

Basic

  1. In the 100-locker puzzle (lockers numbered 1 to 100), which locker numbers remain open at the end?
  2. What is a square number? Write any three square numbers.
  3. Find:

    1. 12²
    2. 15²
    3. 18²
  4. Which of these are perfect squares?

    121, 125, 144, 150, 169, 171
  5. Find the (positive) square root:

    1. √64
    2. √49
    3. √100
  6. Find:

    1. 4³
    2. 6³
    3. 10³

Standard

  1. Without calculating the full square, decide which numbers are definitely not perfect squares:

    327, 462, 2032, 1089
  2. The squares 1², 9², 11², 19², 21², 29² all end with 1.

    Write the next two such squares (after 29²).
  3. Using prime factorisation, decide and write the answer:

    1. Is 156 a perfect square?
    2. Is 324 a perfect square? If yes, find √324.
  4. How many square numbers are there:

    1. between 1 and 100 (inclusive)?
    2. between 101 and 200 (inclusive)?
  5. Find √576 without a calculator.
  6. Find the cube roots:

    1. ∛64
    2. ∛512
    3. ∛729

Advance

  1. Estimate √250 to the nearest integer.
  2. Find the smallest square number that is divisible by 4, 9, and 10.
  3. Find the smallest number by which 9408 must be multiplied so that the product becomes a perfect square.

    Also find the square root of the product.
  4. 2800 is not a perfect square.

    Find the smallest number by which 2800 must be multiplied to make it a perfect square, and find the square root of the new number.
  5. Find:

    1. ∛27000
    2. ∛10648
  6. Find the smallest number by which 1323 must be multiplied so that the product becomes a perfect cube.

    Also find the cube root of the product.

HOTS

  1. Explain (in 2–3 lines) why only perfect squares have an odd number of factors.
  2. The “passcode” is made from the first five locker numbers that were touched exactly twice.

    What is the passcode? (Write the five numbers in order.)
  3. Prove quickly: A perfect square can never end with 2, 3, 7, or 8.
  4. Write 4104 and 13832 as the sum of two cubes in two different ways each.
  5. Arrange the numbers 1 to 17 (without repetition) in a row such that the sum of every adjacent pair is a perfect square.

    Write one valid arrangement.
  6. If a number ends with exactly 3 zeros, how many zeros will its square end with?

    Also write the general rule for a number ending with k zeros.

Answer Key

Basic Answers

  1. Open lockers are the perfect squares:
    1, 4, 9, 16, 25, 36, 49, 64, 81, 100
  2. A square number is a number of the form n².
    Examples: 1, 4, 9 (any three squares are fine).
    1. 12² = 144
    2. 15² = 225
    3. 18² = 324
  4. Perfect squares: 121, 144, 169
    1. √64 = 8
    2. √49 = 7
    3. √100 = 10
    1. 4³ = 64
    2. 6³ = 216
    3. 10³ = 1000

Standard Answers

  1. Definitely not perfect squares: 327, 462, 2032
    (Because they end in 7 or 2. 1089 is a perfect square: 33².)
  2. Next two such squares:
    31² = 961 and 39² = 1521
    1. 156 is not a perfect square (prime factors cannot be paired).
    2. 324 is a perfect square, and √324 = 18.
    1. Between 1 and 100: 10 squares (1² to 10²).
    2. Between 101 and 200: 4 squares (11², 12², 13², 14²).
  5. √576 = 24 (since 24² = 576).
    1. ∛64 = 4
    2. ∛512 = 8
    3. ∛729 = 9

Advance Answers

  1. √250 is between 15 and 16, and closer to 16 (since 16² = 256).
    Nearest integer: 16
  2. LCM(4, 9, 10) = 180, smallest square multiple is 180 × 5 = 900.
    Answer: 900
  3. 9408 = 2⁶ × 3 × 7².
    Multiply by 3 to make powers even: 9408 × 3 = 28224.
    √28224 = 168
  4. 2800 = 2⁴ × 5² × 7.
    Multiply by 7: 2800 × 7 = 19600.
    √19600 = 140
    1. ∛27000 = 30 (because 30³ = 27000)
    2. ∛10648 = 22 (because 22³ = 10648)
  6. 1323 = 3³ × 7².
    Multiply by 7 to make 7 a triplet: 1323 × 7 = 9261 = 21³.
    Cube root = 21

HOTS Answers

  1. Factors usually come in pairs (a, b) with a × b = n, so they pair up evenly.
    Only for squares, one pair becomes (k, k), which is a single unpaired factor, so squares have an odd number of factors.
  2. Lockers touched exactly twice are prime numbers.
    First five primes: 2, 3, 5, 7, 11
  3. The last digit of a square can only be 0, 1, 4, 5, 6, or 9.
    So a number ending in 2, 3, 7, or 8 cannot be a perfect square.
  4. 4104 = 2³
    13832 = 2³ + 24³ = 18³ + 20³
  5. One valid arrangement:
    16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17
  6. If a number ends with 3 zeros, its square ends with 6 zeros.
    General rule: if a number ends with k zeros, its square ends with 2k zeros.

Worksheet From Other Chapters

Ganita Prakash Part-I

Ganita Prakash Part-II

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

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