\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} ?

CTET 2 Math and Science 21 Jan 2024 Paper

Question : 49 of 150

Marks: +1, -0

\;{x}/{y} = (\;{−1}/{3}) ^{−3} ÷ (\;{2}/{3}) ^{−4}

then what is the value of

(\;{x}/{y} + \;{y}/{x}) ^{−1}

Solution:

\;\;{x}/{y}= (\;{-1}/{3}) ^{-3} ÷ (\;{2}/{3}) ^{-4}

\;x \/ y=(-3)^3 \/(3 \/ 2)^4=(-1)(3)^3 × 2^4 \/(3)^4=-2^4 \/ 3=-16 \/ 3

Now

(\;{x}/{y}+\;{y}/{x}) ^{-1}=(-16 \/ 3-3 \/ 16)^{-1}=(-(256+9) \/ 48)^{-1}=(-265 \/ 48)^{-1}=-48 \/ 265

Go to Question:

Prev Question

Next Question