Maharashtra Board Class 7 Chapter 6 Set 27 Indices Solutions

Question 1. Simplify:
(i) \( 7^4 \times 7^2 \)
(ii) \( (-11)^5 \times (-11)^2 \)
(iii) \( \left(\frac{6}{7}\right)^3 \times \left(\frac{6}{7}\right)^5 \)
(iv) \( \left(-\frac{3}{2}\right)^5 \times \left(-\frac{3}{2}\right)^3 \)
(v) \( a^{16} \times a^7 \)
(vi) \( \left(\frac{P}{5}\right)^3 \times \left(\frac{P}{5}\right)^7 \)
Answer:
(i) \( 7^4 \times 7^2 = 7^{4+2} = 7^6 \)
(ii) \( (-11)^5 \times (-11)^2 = (-11)^{5+2} = (-11)^7 \)
(iii) \( \left(\frac{6}{7}\right)^3 \times \left(\frac{6}{7}\right)^5 = \left(\frac{6}{7}\right)^{3+5} = \left(\frac{6}{7}\right)^8 \)
(iv) \( \left(-\frac{3}{2}\right)^5 \times \left(-\frac{3}{2}\right)^3 = \left(-\frac{3}{2}\right)^{5+3} = \left(-\frac{3}{2}\right)^8 \)
(v) \( a^{16} \times a^7 = a^{16+7} = a^{23} \)
(vi) \( \left(\frac{P}{5}\right)^3 \times \left(\frac{P}{5}\right)^7 = \left(\frac{P}{5}\right)^{3+7} = \left(\frac{P}{5}\right)^{10} \)
In simple words: This question demonstrates the product rule for exponents, where \( a^m \times a^n = a^{m+n} \). When multiplying powers with the same base, we simply add their exponents to find the simplified form.

🎯 Exam Tip: Always identify the base first. If the bases are the same, you can directly add the exponents. Pay close attention to negative bases and fractional bases to ensure accuracy in the final result.