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If $0<x<y$, which of the following must be negative?
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05 May 2025, 14:29
05 May 2025, 14:29
Expert Reply
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Question Stats:
33% (02:27) correct
66% (02:24) wrong based on 3 sessions
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If $\(0<x<y\)$, which of the following must be negative?
A. $\(x^2-y\)$
B. $\(\frac{1}{y}-\frac{1}{x}\)$
C. $\(\frac{x^2-y^2}{y-1}\)$
D. $\(\frac{x^2}{y}-y\)$
E. $\(\frac{1}{y^2}-\frac{1}{y}\)$
A. $\(x^2-y\)$
B. $\(\frac{1}{y}-\frac{1}{x}\)$
C. $\(\frac{x^2-y^2}{y-1}\)$
D. $\(\frac{x^2}{y}-y\)$
E. $\(\frac{1}{y^2}-\frac{1}{y}\)$
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Re: If $0<x<y$, which of the following must be negative?
[#permalink]
09 May 2025, 04:30
09 May 2025, 04:30
Expert Reply
Option B: $\(\frac{1}{y}-\frac{1}{x}\)$
- Since $\(x<y, \frac{1}{y}<\frac{1}{x}\)$ (reciprocals reverse inequalities for positives).
- Thus, $\(\frac{1}{y}-\frac{1}{x}<0\)$. Always negative.
$\(\checkmark\)$ Must be negative
Option D: $\(\frac{x^2}{y}-y\)$
- Rewrite: $\(\frac{x^2-y^2}{y}\)$.
- Since $\(x<y, x^2<y^2\)$ (both positive), so numerator is negative. Denominator $\(y>0\)$.
- Thus, $\(\frac{x^2-y^2}{y}<0\)$. Always negative.
$\checkmark$ Must be negative
Why Others Fail
- A: $\(x^2-y\)$ can be positive (e.g., $\(x=2, y=3 \rightarrow 4-3=1>0\)$ ).
- C: Positive if $\(y<1\)$ (e.g., $\(x=0.5, y=0.8 \rightarrow \frac{0.25-0.64}{0.8-1}=\frac{-0.39}{-0.2}>0\)$ ).
- E : Positive if $\(y<1\)$ (e.g., $\(y=0.5 \rightarrow \frac{1}{0.25}-\frac{1}{0.5}=4-2=2>0\)$ ).
Final Answer
Options B and D must be negative.
- Since $\(x<y, \frac{1}{y}<\frac{1}{x}\)$ (reciprocals reverse inequalities for positives).
- Thus, $\(\frac{1}{y}-\frac{1}{x}<0\)$. Always negative.
$\(\checkmark\)$ Must be negative
Option D: $\(\frac{x^2}{y}-y\)$
- Rewrite: $\(\frac{x^2-y^2}{y}\)$.
- Since $\(x<y, x^2<y^2\)$ (both positive), so numerator is negative. Denominator $\(y>0\)$.
- Thus, $\(\frac{x^2-y^2}{y}<0\)$. Always negative.
$\checkmark$ Must be negative
Why Others Fail
- A: $\(x^2-y\)$ can be positive (e.g., $\(x=2, y=3 \rightarrow 4-3=1>0\)$ ).
- C: Positive if $\(y<1\)$ (e.g., $\(x=0.5, y=0.8 \rightarrow \frac{0.25-0.64}{0.8-1}=\frac{-0.39}{-0.2}>0\)$ ).
- E : Positive if $\(y<1\)$ (e.g., $\(y=0.5 \rightarrow \frac{1}{0.25}-\frac{1}{0.5}=4-2=2>0\)$ ).
Final Answer
Options B and D must be negative.
gmatclubot
Re: If $0<x<y$, which of the following must be negative? [#permalink]
09 May 2025, 04:30
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