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CTET Paper II (Mathematics and Science) 24 Dec 2021 Solved Paper

Question : 32 of 150

Marks: +1, -0

(\frac{-3}{2})^{-3}\div x=(\frac{9}{4})^{-2}

, then the value of (2x + 5)^{-1} is

Solution:

Concept:

Use the laws of exponents to simplify powers and solve for

x

. Then substitute to find the value of the given expression.

Explanation:

First, simplify

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

. Invert the fraction and cube it:

(\frac{-2}{3})^3 = \frac{-8}{27}

Next, simplify

(\frac{9}{4})^{-2}

. Invert and square:

(\frac{4}{9})^2 = \frac{16}{81}

The equation becomes

\frac{-8}{27} \div x = \frac{16}{81}

, which is

\frac{-8}{27} \times \frac{1}{x} = \frac{16}{81}

Multiply both sides by

x

\frac{-8}{27} = \frac{16}{81} x

. Then

x = \frac{-8}{27} \times \frac{81}{16}

Simplify:

\frac{-8 \times 81}{27 \times 16} = \frac{-8 \times 3}{16} = \frac{-24}{16} = \frac{-3}{2}

Now find

(2x+5)^{-1}

: substitute

x = \frac{-3}{2}

to get

2 \times \frac{-3}{2} + 5 = -3 + 5 = 2

. So

(2)^{-1} = \frac{1}{2}

Answer:

\frac{1}{2}

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