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CTET Paper II (Mathematics and Science) 4 Jan 2022 Solved Paper

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Question : 33 of 150

Marks: +1, -0

The value of

[(\frac{4}{3})^{2}]^{3}×(\frac{2}{3})^{-4}×(\frac{27}{16})^{2}

÷ 3^{3} is

3
$\frac{3}{4}$
9
$\frac{9}{16}$

Solution:

Concept:

Use the laws of exponents:

(a^m)^n = a^{mn}

,

(a/b)^{-n} = (b/a)^n

, and

a^m \times a^n = a^{m+n}

. Simplify the expression by converting all numbers to prime bases.

Explanation:

First, simplify

[(\frac{4}{3})^{2}]^{3} = (\frac{4}{3})^{6}

.

Next,

(\frac{2}{3})^{-4} = (\frac{3}{2})^{4}

.

Then,

(\frac{27}{16})^{2} = \frac{27^2}{16^2} = \frac{(3^3)^2}{(2^4)^2} = \frac{3^6}{2^8}

.

Division by

3^3

means multiply by

\frac{1}{3^3}

.

Now rewrite

(\frac{4}{3})^{6}

as

(2^2)^6 / 3^6 = 2^{12}/3^6

, and

(\frac{3}{2})^{4} = 3^4/2^4

.

Multiply all terms:

\frac{2^{12}}{3^6} \times \frac{3^4}{2^4} \times \frac{3^6}{2^8} \times \frac{1}{3^3}

.

Combine powers of 2:

2^{12-4-8} = 2^0 = 1

.

Combine powers of 3:

3^{-6+4+6-3} = 3^{1} = 3

.

Thus, the simplified value is

3

.

Answer:

3

(Option A)

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