Discussion
If \(\frac{x}{y} = \left(\frac{-1}{3}\right)^{-3} \div \left(\frac{2}{3}\right)^{-4}\), then what is the value of \(\left(\frac{x}{y} + \frac{y}{x}\right)^{-1}\) ?
Answer: D
The value is \(-\frac{48}{265}\).
A negative exponent inverts the fraction:
\[ \begin{aligned} \left(\tfrac{-1}{3}\right)^{-3} &= (-3)^{3} = -27 \\ \left(\tfrac{2}{3}\right)^{-4} &= \left(\tfrac{3}{2}\right)^{4} = \tfrac{81}{16} \\ \tfrac{x}{y} &= -27 \times \tfrac{16}{81} = -\tfrac{16}{3} \end{aligned} \]
\[ \begin{aligned} \tfrac{x}{y} + \tfrac{y}{x} &= -\tfrac{16}{3} - \tfrac{3}{16} = -\tfrac{265}{48} \\ \left(-\tfrac{265}{48}\right)^{-1} &= -\tfrac{48}{265} \end{aligned} \]
\(-\frac{3}{16}\) is only y/x, and the positive options miss the negative sign carried by the odd power of \(-\frac{1}{3}\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development