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How many cubes of side 1 cm will make a cube of side 3 cm?
The volume of a cube is side × side × side. For a cube with side 3 cm, Volume = 3³ = 27 cm³. Each small cube has volume 1 cm³. So, number of 1 cm³ cubes needed = Total volume ÷ 1 = 27 ÷ 1 = 27 cubes. Thus, 27 small cubes of 1 cm³ are required to form a 3 cm cube. For more NCERT Solutions forRead more
The volume of a cube is side × side × side. For a cube with side 3 cm,
Volume = 3³ = 27 cm³.
Each small cube has volume 1 cm³.
So, number of 1 cm³ cubes needed = Total volume ÷ 1 = 27 ÷ 1 = 27 cubes.
Thus, 27 small cubes of 1 cm³ are required to form a 3 cm cube.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessHow many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
The total number of tiny squares in the image is 144, arranged in a 12 × 12 square. The prime factorisation of 144 is: 144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3² All exponents are even, so it confirms 144 is a perfect square. So, number of tiny squares = 144 Prime factorisation: 2⁴ × 3² Square root: 12Read more
The total number of tiny squares in the image is 144, arranged in a 12 × 12 square. The prime factorisation of 144 is:
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
All exponents are even, so it confirms 144 is a perfect square.
So, number of tiny squares = 144
Prime factorisation: 2⁴ × 3²
Square root: 12
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessHow many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32. (ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198. In general, numbers between n² and (n+1)² are 2n. Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers. For more NCERT SRead more
(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32.
(ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198.
In general, numbers between n² and (n+1)² are 2n.
Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168. So, the smallest number to multiply is 3 and the square root of the result iRead more
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168.
So, the smallest number to multiply is 3 and the square root of the result is 168.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the smallest square number that is divisible by each of the following numbers: 4, 9 and 10.
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM: 4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180. Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.Read more
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM:
4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180.
Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.
Answer: 900.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See less