RD Sharma Solutions for Class 8 Chapter 4 - Cubes and Cube Roots ... - Free Printable
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Step-by-step solution for: RD Sharma Solutions for Class 8 Chapter 4 - Cubes and Cube Roots ...
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Step-by-step solution for: RD Sharma Solutions for Class 8 Chapter 4 - Cubes and Cube Roots ...
Problem Analysis and Solution Explanation
The image contains three problems related to cubes and cube roots. Below, I will explain each problem and its solution step by step.
---
#### Problem 7:
Question:
What is the smallest number by which 8192 must be divided so that the quotient is a perfect cube? Also, find the cube root of the quotient so obtained.
Solution:
1. Prime Factorization of 8192:
\[
8192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{13}
\]
2. Condition for a Perfect Cube:
A number is a perfect cube if all the exponents in its prime factorization are multiples of 3. Here, the exponent of 2 is 13, which is not a multiple of 3.
3. Making the Quotient a Perfect Cube:
To make \( 8192 \) a perfect cube, we need to reduce the exponent of 2 to the nearest multiple of 3. The closest multiple of 3 less than 13 is 12.
\[
2^{13} \div 2 = 2^{12}
\]
So, dividing 8192 by 2 gives:
\[
8192 \div 2 = 4096
\]
4. Prime Factorization of 4096:
\[
4096 = 2^3 \times 2^3 \times 2^3 \times 2^3 = (2^3)^4
\]
This shows that 4096 is a perfect cube.
5. Cube Root of 4096:
\[
\sqrt[3]{4096} = \sqrt[3]{(2^3)^4} = 2^4 = 16
\]
Final Answer for Problem 7:
\[
\boxed{2, 16}
\]
---
#### Problem 8:
Question:
Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
1. Let the Numbers Be:
Let the numbers be \( x \), \( 2x \), and \( 3x \).
2. Sum of Their Cubes:
According to the problem:
\[
x^3 + (2x)^3 + (3x)^3 = 98784
\]
3. Simplify the Equation:
\[
x^3 + 8x^3 + 27x^3 = 98784
\]
\[
36x^3 = 98784
\]
4. Solve for \( x^3 \):
\[
x^3 = \frac{98784}{36} = 2744
\]
5. Find \( x \):
\[
x = \sqrt[3]{2744}
\]
Prime factorize 2744:
\[
2744 = 2 \times 2 \times 2 \times 7 \times 7 \times 7 = 2^3 \times 7^3
\]
\[
\sqrt[3]{2744} = \sqrt[3]{2^3 \times 7^3} = 2 \times 7 = 14
\]
6. Find the Numbers:
\[
x = 14, \quad 2x = 2 \times 14 = 28, \quad 3x = 3 \times 14 = 42
\]
Final Answer for Problem 8:
\[
\boxed{14, 28, 42}
\]
---
#### Problem 9:
Question:
The volume of a cube is 9261000 m³. Find the side of the cube.
Solution:
1. Volume of a Cube:
The volume \( V \) of a cube with side length \( a \) is given by:
\[
V = a^3
\]
Here, \( V = 9261000 \) m³.
2. Find the Side Length \( a \):
\[
a = \sqrt[3]{9261000}
\]
3. Prime Factorization of 9261000:
\[
9261000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7
\]
Group the factors into cubes:
\[
9261000 = (2^3) \times (3^3) \times (5^3) \times (7^3)
\]
4. Cube Root:
\[
\sqrt[3]{9261000} = \sqrt[3]{2^3 \times 3^3 \times 5^3 \times 7^3} = 2 \times 3 \times 5 \times 7 = 210
\]
Final Answer for Problem 9:
\[
\boxed{210}
\]
---
Final Answers:
1. Problem 7: \(\boxed{2, 16}\)
2. Problem 8: \(\boxed{14, 28, 42}\)
3. Problem 9: \(\boxed{210}\)
Parent Tip: Review the logic above to help your child master the concept of finding cube roots worksheet.