Maharashtra Board Class 7 Chapter 6 Set 29 Indices Solutions

Question 1. Simplify:
(i) \( \left[ \left(\frac{15}{12}\right)^3 \right]^4 \)
(ii) \( (3^4)^{-2} \)
(iii) \( \left[ \left(\frac{1}{7}\right)^{-3} \right]^4 \)
(iv) \( \left[ \left(\frac{2}{5}\right)^{-2} \right]^{-3} \)
(v) \( (6^5)^4 \)
(vi) \( \left[ \left(\frac{6}{7}\right)^5 \right]^2 \)
(vii) \( \left[ \left(\frac{2}{3}\right)^{-4} \right]^5 \)
(viii) \( \left[ \left(\frac{5}{8}\right)^3 \right]^{-2} \)
(ix) \( \left[ \left(\frac{3}{4}\right)^6 \right]^7 \)
(x) \( \left[ \left(\frac{2}{5}\right)^{-3} \right]^2 \)
Answer:
Solution:
(i) \( \left[ \left(\frac{15}{12}\right)^3 \right]^4 \)
\( = \left(\frac{15}{12}\right)^{3 \times 4} = \left(\frac{15}{12}\right)^{12} \)
(ii) \( (3^4)^{-2} \)
\( = 3^{4 \times (-2)} \)
\( = 3^{-8} \)
(iii) \( \left[ \left(\frac{1}{7}\right)^{-3} \right]^4 \)
\( = \left(\frac{1}{7}\right)^{(-3) \times 4} = \left(\frac{1}{7}\right)^{-12} \)
(iv) \( \left[ \left(\frac{2}{5}\right)^{-2} \right]^{-3} \)
\( = \left(\frac{2}{5}\right)^{(-2) \times (-3)} = \left(\frac{2}{5}\right)^6 \)
(v) \( (6^5)^4 \)
\( = 6^{5 \times 4} \)
\( = 6^{20} \)
(vi) \( \left[ \left(\frac{6}{7}\right)^5 \right]^2 \)
\( = \left(\frac{6}{7}\right)^{5 \times 2} = \left(\frac{6}{7}\right)^{10} \)
(vii) \( \left[ \left(\frac{2}{3}\right)^{-4} \right]^5 \)
\( = \left(\frac{2}{3}\right)^{(-4) \times 5} = \left(\frac{2}{3}\right)^{-20} \)
(viii) \( \left[ \left(\frac{5}{8}\right)^3 \right]^{-2} \)
\( = \left(\frac{5}{8}\right)^{3 \times (-2)} = \left(\frac{5}{8}\right)^{-6} \)
(ix) \( \left[ \left(\frac{3}{4}\right)^6 \right]^7 \)
\( = \left(\frac{3}{4}\right)^{6 \times 1} = \left(\frac{3}{4}\right)^6 \)
(x) \( \left[ \left(\frac{2}{5}\right)^{-3} \right]^2 \)
\( = \left(\frac{2}{5}\right)^{(-3) \times 2} = \left(\frac{2}{5}\right)^{-6} \)
In simple words: This question demonstrates the law of indices \((a^m)^n = a^{m \times n}\), where an index raised to another index results in the multiplication of the exponents. For negative exponents, the base is inverted, and the exponent becomes positive.

🎯 Exam Tip: Remember the power of a power rule for indices; multiplying the exponents is key. Pay close attention to negative signs in the exponents as they indicate reciprocals.

Question 2. Write the following numbers using positive indices:
(i) \( \left(\frac{2}{3}\right)^{-2} \)
(ii) \( \left(\frac{11}{3}\right)^{-5} \)
(iii) \( \left(\frac{1}{6}\right)^{-3} \)
(iv) \( (y)^{-4} \)
Answer:
Solution:
(i) \( \left(\frac{2}{3}\right)^{-2} \)
\( = \left(\frac{3}{2}\right)^2 \)
(ii) \( \left(\frac{11}{3}\right)^{-5} \)
\( = \left(\frac{3}{11}\right)^5 \)
(iii) \( \left(\frac{1}{6}\right)^{-3} \)
\( = 6^3 \)
(iv) \( (y)^{-4} \)
\( = \frac{1}{y^4} \)
In simple words: To convert a negative index to a positive one, you simply take the reciprocal of the base and change the sign of the exponent. For fractions, this means flipping the numerator and denominator.

🎯 Exam Tip: A common mistake is to only change the sign of the exponent without taking the reciprocal of the base. Ensure you perform both operations to correctly convert to a positive index.