Class 7 RS Aggarwal Chapter-5 Exponents Solutions - Free PDF Download
FAQs on RS Aggarwal Class 7 Solutions Chapter-5 Exponents
1. How do the RS Aggarwal Class 7 Solutions for Chapter 5 help in solving problems from all exercises like 5A, 5B, and 5C?
The RS Aggarwal Class 7 Solutions for Exponents provide a complete, step-by-step walkthrough for every single problem in all the exercises, including 5A, 5B, and 5C. Each solution demonstrates the correct method and application of exponent laws, ensuring you understand how to approach different types of questions, from basic simplification to more complex ones involving multiple operations.
2. What is the correct method to simplify a fraction with a negative exponent as explained in RS Aggarwal Chapter 5?
To simplify a fraction with a negative exponent, you should use the reciprocal of the base to make the exponent positive. The rule is (a/b)-n = (b/a)n. For example, to solve (2/3)-2, you would first convert it to (3/2)2 and then calculate the result, which is 9/4. This is a fundamental method used to solve many problems in this chapter.
3. How are numbers written in standard form (scientific notation) using exponents in this chapter?
The RS Aggarwal solutions explain how to express very large or small numbers in standard form (k x 10n) using a clear two-step method:
Step 1: Place a decimal point after the first non-zero digit of the number.
Step 2: Multiply this new decimal number by a power of 10. The exponent (n) will be equal to the number of places the decimal point was moved. The exponent is positive if the decimal moves left (for large numbers) and negative if it moves right (for small numbers).
For instance, 56,700,000 is written as 5.67 x 107.
4. Why is any non-zero number raised to the power of zero equal to 1?
Any non-zero number raised to the power of zero is 1 because of the quotient law of exponents (am / an = am-n). Consider the example am / am. We know that any number divided by itself is 1. Using the exponent law, this is also equal to am-m = a0. Therefore, a0 = 1. The RS Aggarwal solutions use this fundamental principle to solve various simplification problems.
5. How do I apply the laws of exponents when the bases are different but the exponents are the same?
When multiplying or dividing terms with different bases but the same exponent, you combine the bases first and keep the exponent. The rules are:
For multiplication: am × bm = (a × b)m. For example, 23 × 53 = (2 × 5)3 = 103 = 1000.
For division: am ÷ bm = (a ÷ b)m. For example, 104 ÷ 54 = (10 ÷ 5)4 = 24 = 16.
This technique is crucial for simplifying complex expressions found in the later exercises of the chapter.
6. What is a common mistake when solving problems with negative numbers as a base, like (-3)⁴ vs -3⁴?
A common mistake is confusing the placement of the negative sign. The parentheses are extremely important. The expression (-3)⁴ means (-3) is multiplied by itself four times: (-3) × (-3) × (-3) × (-3) = 81. In contrast, -3⁴ means you calculate 3⁴ first and then apply the negative sign: -(3 × 3 × 3 × 3) = -81. The RS Aggarwal solutions highlight this distinction to ensure calculations are accurate.
7. How can the step-by-step RS Aggarwal solutions help me solve complex problems with multiple exponent laws?
The solutions break down complex problems into manageable steps, showing which law to apply at each stage. For a problem like simplifying [(xᵃ/xᵇ)ᶜ], the solution would first guide you to apply the quotient rule inside the bracket to get [(xᵃ⁻ᵇ)ᶜ], and then apply the power of a power rule to get the final answer xᶜ⁽ᵃ⁻ᵇ⁾. Following this structured approach helps build the logical thinking needed for exams.
















