Evaluate : [(6×10)/(2²×5³)]²×(25/27)

Solution:

Given, the expression is [(6×10)/(2²×5³)]²×(25/27)

We have to evaluate the expression.

[(6×10)/(2²×5³)]² can be written as (6×10)²/(2²×5³)²

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × b^m = (a×b)^m

(6×10)² = 6² × 10²

6² can be written as (2 × 3)²

(2 × 3)² = 2² × 3²

10² can be written as (2 × 5)²

(2 × 5)² = 2² × 5²

Now, 6² × 10² = 2² × 3² × 2² × 5²

Considering 2²×2²,

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m × aⁿ = a^m+n

Here, a = 2, m = 2 and n = 2

m + n = 2 + 2 = 4

So, 2²×2² = 2⁴

Now, (6×10)² = 2⁴ × 3² × 5²

Considering (2²×5³)²,

(2²×5³)² = (2²)²×(5³)²

For any non-zero integer ‘a’ and whole numbers m and n,

(a^m)ⁿ = (a)^mn

So, (2²)² = 2²^ײ = 2⁴

Similarly, (5³)² = 5⁶

Now, (2²×5³)² = 2⁴×5⁶

[(6×10)/(2²×5³)]²×(25/27) = [(2⁴ × 3² × 5²)/(2⁴×5⁶)] × (25/27)

25 can be written as 5²

27 can be written as 3³

So, [(2⁴ × 3² × 5²)/(2⁴×5⁶)] × (25/27) = [(2⁴ × 3² × 5²)/(2⁴×5⁶)] × (5²/3³)

= (2⁴ × 3² × 5²×5²)/(2⁴×5⁶×3³)

5²×5² can be written as 5⁴

So, (2⁴ × 3² × 5²×5²)/(2⁴×5⁶×3³) = (2⁴ × 3² × 5⁴)/(2⁴×5⁶×3³)

Considering 2⁴/2⁴,

2⁴/2⁴ = 1

For any non-zero integers ‘a’ and ‘b’ and whole numbers m and n,

a^m ÷ aⁿ = a^m-n

Considering 3²/3³,

Here, a = 3, m = 2 and n = 3

m - n = 2 - 3 = -1

So, 3²/3³ = 3⁻¹

= 1/3

Considering 5⁴/5⁶,

Here, a = 5, m = 4 and n = 6

m - n = 4 - 6 = -2

So, 5⁴/5⁶ = 5⁻²

= 1/5²

= 25

Now, (2⁴ × 3² × 5⁴)/(2⁴×5⁶×3³) = 1 × 1/3 × 1/25

= 1/(3×25)

= 1/75

Therefore, the required value is 1/75.

✦ Try This: Evaluate : [(25×9)/(2²/5)]²×(36/27)

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

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NCERT Exemplar Class 7 Maths Chapter 11 Problem 85 (e)

Evaluate : [(6×10)/(2²×5³)]²×(25/27)

Summary:

On evaluating [(6×10)/(2²×5³)]²×(25/27) we get 1/75.

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