Question
Easy

If $\frac{x}{y}= (\frac{-1}{3})^{-3}$ $\div(\frac{2}{3})^{-4}$, then what is the value of ($\frac{x}{y}+\frac{y}{x}$)$^{-1}$?

1
-$\frac{3}{16}$
2
$\frac{19}{48}$
3
$\frac{38}{73}$
4
-$\frac{48}{265}$
Question Details
Time to Solve: 12
Exam: CTET
Level/Paper: Paper 2
Chapter: Power & Roots
Topic: Exponents & Powers
Correct Answer
Option D
Explanation

To solve the given problem, we need to evaluate the expression \(\frac{x}{y} = \left(\frac{-1}{3}\right)^{-3} \div \left(\frac{2}{3}\right)^{-4}\) and then find the value of \(\left(\frac{x}{y} + \frac{y}{x}\right)^{-1}\). ### Step-by-Step Solution: 1. Evaluate \(\frac{x}{y}\): - First, calculate \(\left(\frac{-1}{3}\right)^{-3}\): \[ \left(\frac{-1}{3}\right)^{-3} = \left(-3\right)^3 = -27 \] - Next, calculate \(\left(\frac{2}{3}\right)^{-4}\): \[ \left(\frac{2}{3}\right)^{-4} = \left(\frac{3}{2}\right)^4 = \frac{81}{16} \] - Now, divide the two results: \[ \frac{x}{y} = \frac{-27}{1} \div \frac{81}{16} = -27 \times \frac{16}{81} =…Read More

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