# Simplify: (i) ((2^{5})^{2} × 7^{3}) / (8^{3} × 7) (ii) (25 × 5^{2} × t^{8}) / (10^{3} × t^{4})

(iii) (3^{5} ×10^{5} × 25) / (5^{7} × 6^{5})

**Solution:**

To solve this question, we will be using the laws of exponents.

(i) ((2^{5})^{2} × 7^{3}) / (8^{3} × 7)

= (2^{5 × }^{2} × 7^{3}) / ((2^{3})^{3} × 7) [(a^{m})^{n} = a^{m}^{n}]

= (2^{10} × 7^{3 - 1}) / 2^{9} [a^{m} ÷ a^{n} = a^{m - n}]

= 2^{1} × 7^{2}

= 98

(ii) (25 × 5^{2} × t^{8}) / (10^{3} × t^{4})

= (25 × 5^{2} × t^{8}) / ((2 × 5)^{3} × t^{4})

= ((5)^{2 + 2} × t^{8}) / (2^{3} × 5^{3} × t^{4}) [a^{m} × a^{n} = a^{m + n}]

= (5^{4} × t^{8}) / (2^{3} × 5^{3} × t^{4})

= (5^{4 - 3} × t^{8 - 4}) / 8 [a^{m} ÷ a^{n} = a^{m - n}]

= (5 × t ^{4}) / 8

(iii) (3^{5} × 10^{5} × 25) / (5^{7} × 6^{5})

= (3^{5} × (2 × 5)^{5} × (5 × 5)) / (5^{7} × (2 × 3)^{5})

= (3^{5} × 2^{5} × 5^{5} × 5^{2}) / (5^{7} × 2^{5} × 3^{5})

= (3^{5} × 2^{5} × 5^{7}) / (5^{7} × 2^{5} × 3^{5}) [a^{m} × a^{n} = a^{m + n}]

= 5^{7 - 7} × 2^{5 - 5} × 3^{5 - 5} [a^{m} ÷ a^{n} = a^{m - n}]

= 3^{0} × 2^{0}× 7^{0 }

= 1 [a^{0} = 1]

**☛ Check: **NCERT Solutions for Class 7 Maths Chapter 13

**Video Solution:**

## Simplify: (i) ((2⁵)² × 7³) / (8³ × 7) (ii) (25 × 5² × t⁸) / (10³ × t⁴) (iii) (3⁵ ×10⁵ × 25) / (5⁷ × 6⁵)

Maths NCERT Solutions Class 7 Chapter 13 Exercise 13.2 Question 5

**Summary:**

We have simplified the following: (i) ((2^{5})^{2} × 7^{3}) / (8^{3} × 7) (ii) (25 × 5^{2} × t^{8}) / (10^{3} × t ^{4}) (iii) (3^{5} ×10^{5} × 25) / (5^{7} × 6^{5}) as follows (i) 98 , (ii) (5 × t^{4})/8 , (iii) 1.

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