CBSE Class 8 Mathematics A Square and A Cube Practice Paper Worksheet Set 01 with Answers

Class 8 Maths Ganita Prakash Chapter 1 A Square and A Cube Practice Paper 05 is a 40-mark test of 20 questions in Sections A to E, given here with clear step-by-step answers. It checks how well you can spot perfect squares and cubes, solve the locker puzzle, make a number a square or a cube by multiplying or dividing, find square and cube roots by prime factorisation, and use the pattern that every square is a sum of consecutive odd numbers, along with the Hardy–Ramanujan number 1729.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the smallest square number which is exactly divisible by 12, 15, and 20.
Answer:
The smallest number divisible by 12, 15 and 20 is their LCM.
12 = 2 × 2 × 3, 15 = 3 × 5, 20 = 2 × 2 × 5, so LCM = 2 × 2 × 3 × 5 = 60.
In 60 = (2 × 2) × 3 × 5, the factors 3 and 5 are unpaired.
Multiply by 3 × 5 = 15: 60 × 15 = 900 = (2 × 3 × 5)² = 30².
The smallest such square number is 900.
In simple words: Take the LCM 60 and multiply by the unpaired factors 3 and 5 to get 900.

Teacher's Note:
1. Any number divisible by all three must be a multiple of the LCM.
2. Check: 900 ÷ 12 = 75, 900 ÷ 15 = 60, 900 ÷ 20 = 45.

Question 12. Find the square root of 2401 by prime factorisation method.
Answer:
Divide by the smallest prime that works: 2401 ÷ 7 = 343, 343 ÷ 7 = 49, 49 ÷ 7 = 7, 7 ÷ 7 = 1.
2401 = 7 × 7 × 7 × 7 = (7 × 7) × (7 × 7).
Take one factor from each pair: √2401 = 7 × 7 = 49.
In simple words: 2401 is four 7s multiplied together, so its square root is 7 × 7 = 49.

Teacher's Note:
1. 2401 is not divisible by 2, 3 or 5, so try 7 next.
2. Check: 49 × 49 = 2401.

Question 13. Find the cube root of 2744, by the prime factorisation method.
Answer:
2744 ÷ 2 = 1372, ÷ 2 = 686, ÷ 2 = 343, ÷ 7 = 49, ÷ 7 = 7, ÷ 7 = 1.
2744 = 2 × 2 × 2 × 7 × 7 × 7 = (2 × 2 × 2) × (7 × 7 × 7).
Take one factor from each triplet: ∛2744 = 2 × 7 = 14.
In simple words: 2744 is three 2s and three 7s, so its cube root is 2 × 7 = 14.

Teacher's Note:
1. For a cube root, group the prime factors in threes, not in pairs.
2. Check: 14 × 14 × 14 = 196 × 14 = 2744.

Question 14. Find the smallest number by which 200 must be multiplied so that the product becomes a perfect cube.
Answer:
200 = 2 × 2 × 2 × 5 × 5.
The 2s form a complete triplet, but there are only two 5s.
One more 5 completes the triplet: 200 × 5 = 1000 = 10³.
The smallest multiplier is 5.
In simple words: 200 has three 2s but only two 5s, so multiply by one more 5.

Teacher's Note:
1. Multiply by whatever is missing to complete every triplet.
2. The product 1000 = 10³ confirms the answer.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the smallest number by which 405 must be divided to obtain a perfect square. Also, find the square root of the quotient.
Answer:
405 = 3 × 3 × 3 × 3 × 5 = (3 × 3) × (3 × 3) × 5.
The 3s are in pairs; the factor 5 is unpaired.
Divide by 5 to remove it: 405 ÷ 5 = 81.
81 = (3 × 3) × (3 × 3), so √81 = 3 × 3 = 9.
The smallest divisor is 5 and the square root of the quotient is 9.
In simple words: Remove the lonely factor 5 by dividing; 81 is left, and √81 = 9.

Teacher's Note:
1. When dividing, remove the unpaired factors; when multiplying, supply their partners.
2. Check: 9 × 9 = 81 and 81 × 5 = 405.

Question 16. Find the smallest number by which 1352 must be multiplied so that the product has a cube root. Also, find the cube root of the product.
Answer:
1352 = 2 × 2 × 2 × 13 × 13.
The 2s form a triplet, but 13 appears only twice.
Multiply by one more 13: 1352 × 13 = 17576.
17576 = (2 × 2 × 2) × (13 × 13 × 13), so ∛17576 = 2 × 13 = 26.
The smallest multiplier is 13 and the cube root of the product is 26.
In simple words: Add one more 13 to complete the triplet; then the cube root is 2 × 13 = 26.

Teacher's Note:
1. "Has a cube root" here means the product is a perfect cube.
2. Check: 26 × 26 × 26 = 676 × 26 = 17576.

Question 17. Find the smallest number by which 1536 must be divided so that the quotient has a cube root. Also, find the cube root of the quotient.
Answer:
1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 (nine 2s and one 3).
The nine 2s make three complete triplets; the single 3 is left over.
Divide by 3: 1536 ÷ 3 = 512.
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), so ∛512 = 2 × 2 × 2 = 8.
The smallest divisor is 3 and the cube root of the quotient is 8.
In simple words: Remove the extra 3 by dividing; 512 = 8³ is left.

Teacher's Note:
1. Count the prime factors carefully; nine 2s divide exactly into three groups of three.
2. Check: 8³ = 512 and 512 × 3 = 1536.

Section D: Long Answer Question (5 Marks)

Question 18. A school has 1475 students. What is the minimum number of students that need to be added so that the total number of students can be arranged in a perfect square grid?
Answer:
A perfect square grid has as many rows as columns, so the total must be a perfect square.
1475 = 5 × 5 × 59; the factor 59 is unpaired, so 1475 is not a perfect square.
We want the smallest perfect square greater than 1475 (adding students can only increase the total).
38² = 1444, which is less than 1475; 39² = 1521, which is more than 1475.
So the next perfect square is 1521.
Students to be added = 1521 − 1475 = 46.
The minimum number of students to be added is 46, giving a grid of 39 rows × 39 columns.
In simple words: The next square after 1475 is 39² = 1521, so 46 more students are needed.

Teacher's Note:
1. Multiplying by the unpaired factor 59 would also give a square, but that means adding far more students; the question asks for the fewest to add.
2. Estimate square roots using known squares: 40² = 1600 is just above, so try 38 and 39.
3. Always check that the number found is the nearest square above, not just any square.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1 (Squares and Odd Numbers): Riya was reading about ancient Indian mathematicians. She discovered that square numbers can be expressed as the sum of consecutive odd numbers. She tried with 36:
36 = 1 + 3 + 5 + 7 + 9 + 11.
She wondered if she could find other squares this way. She also noticed that the number of odd numbers added equals the square root of the number.
Answer the following:
(a) Express 49 as the sum of consecutive odd numbers.
(b) How many odd numbers are required to express 121 as a sum?
(c) Find whether 50 can be expressed as the sum of consecutive odd numbers.
(d) If 2025 is a perfect square, how many odd numbers will be required to express it?
Answer:
The sum of the first n odd numbers is n², as the picture of a 7 × 7 square built from L-shaped layers of 1, 3, 5, …, 13 squares shows:

(a) √49 = 7, so add the first 7 odd numbers: 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13.
(b) √121 = 11, so 11 odd numbers (1 + 3 + 5 + … + 21) are needed.
(c) 7² = 49 and 8² = 64, so 50 lies between two consecutive squares and is not a perfect square.
Therefore 50 cannot be expressed as the sum of consecutive odd numbers starting from 1.
(d) 2025 = 3 × 3 × 3 × 3 × 5 × 5 = (3 × 3) × (3 × 3) × (5 × 5), so √2025 = 3 × 3 × 5 = 45.
So 45 odd numbers are required.
In simple words: A square n² is the sum of the first n odd numbers, so the count of odd numbers is just the square root.

Teacher's Note:
1. The last odd number in the sum for n² is 2n − 1, for example 13 for 49.
2. In fact 50 cannot be such a sum even without starting from 1: an odd count of odd numbers gives an odd total, and an even count gives a multiple of 4.
3. Prime factorisation is the quickest way to find √2025.

Question 20. Case Study 2 (Cubes and Taxicab Numbers): During a math club discussion, Arjun told the famous story of Hardy and Ramanujan. Hardy thought 1729 was dull, but Ramanujan replied it was special since 1729 = 1³ + 12³ = 9³ + 10³. This made Arjun curious about cubes and their properties. He started exploring other such numbers called “taxicab numbers.”
Answer the following:
(a) Find the units digit of the cube of 25.
(b) What is the smallest number by which 250 must be multiplied to make it a perfect cube?
(c) What is the cube root of 2744?
(d) Can a perfect cube end with exactly two zeros (00)? Justify your answer.
Answer:
(a) The units digit of a cube depends only on the units digit of the number. 25 ends in 5 and 5³ = 125 ends in 5.
So the units digit of 25³ is 5 (25³ = 15625).
(b) 250 = 2 × 5 × 5 × 5. The 5s form a triplet, but there is only one 2.
Two more 2s are needed, so multiply by 2 × 2 = 4 (250 × 4 = 1000 = 10³).
(c) 2744 = (2 × 2 × 2) × (7 × 7 × 7), so ∛2744 = 2 × 7 = 14.
(d) No. Zeros at the end come from factors of 10 = 2 × 5. In a cube, each prime factor appears in groups of three, so the factors of 10 also come in groups of three.
Hence a cube ends in 0, 3, 6, … zeros (10³ = 1000, 20³ = 8000, 100³ = 1000000), never exactly two zeros.
In simple words: Cube of a number ending in 5 ends in 5; 250 needs two more 2s; ∛2744 = 14; cubes end in zeros in groups of three.

Teacher's Note:
1. The question has been made precise as "exactly two zeros", because 1000 does end in the digits 00.
2. 1729 is the smallest number that is the sum of two cubes in two different ways.
3. Units digits of cubes: 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9, 0→0.