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x<2x<x^2
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05 May 2025, 03:18
05 May 2025, 03:18
Expert Reply
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Question Stats:
0% (00:00) correct
100% (01:01) wrong based on 2 sessions
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\(
x<2x<x^2\)
x<2x<x^2\)
Quantity A |
Quantity B |
\(x\) |
\(\frac{x}{2} + 1\) |
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Re: x<2x<x^2
[#permalink]
08 May 2025, 04:30
08 May 2025, 04:30
Expert Reply
Step 1: Solve Compound Inequality
1. First Part $\((x<2 x)\)$ :
- Subtract $\(x: 0<x \rightarrow x>0\)$.
2. Second Part ( $\(2 x<x^2\)$ ):
- Rewrite: $\(x^2-2 x>0 \rightarrow x(x-2)>0\)$.
- Solution: $x<0$ or $x>2$.
3. Combine with $x>0$ :
- Only $x>2$ satisfies both.
Step 2: Compare Quantities for $x>2$
- Simplify Comparison:
$x$ vs. $\(\frac{x}{2}+1 \rightarrow\)$ Subtract $\(\frac{x}{2}\)$ :
$\(\frac{x}{2}\)$ vs. $\(1 \rightarrow x\)$ vs. 2 .
- Since $\(x>2, \frac{x}{2}>1\)$, so:
$\(x>\frac{x}{2}+1\)$.
Step 3: Verify with Example
- Let $x=3$ :
- Quantity A: 3
- Quantity B: $\(\frac{3}{2}+1=2.5\)$
- $\(3>2.5\)$ -
Conclusion
For all valid $x>2$, Quantity A $>$ Quantity $B$.
1. First Part $\((x<2 x)\)$ :
- Subtract $\(x: 0<x \rightarrow x>0\)$.
2. Second Part ( $\(2 x<x^2\)$ ):
- Rewrite: $\(x^2-2 x>0 \rightarrow x(x-2)>0\)$.
- Solution: $x<0$ or $x>2$.
3. Combine with $x>0$ :
- Only $x>2$ satisfies both.
Step 2: Compare Quantities for $x>2$
- Simplify Comparison:
$x$ vs. $\(\frac{x}{2}+1 \rightarrow\)$ Subtract $\(\frac{x}{2}\)$ :
$\(\frac{x}{2}\)$ vs. $\(1 \rightarrow x\)$ vs. 2 .
- Since $\(x>2, \frac{x}{2}>1\)$, so:
$\(x>\frac{x}{2}+1\)$.
Step 3: Verify with Example
- Let $x=3$ :
- Quantity A: 3
- Quantity B: $\(\frac{3}{2}+1=2.5\)$
- $\(3>2.5\)$ -
Conclusion
For all valid $x>2$, Quantity A $>$ Quantity $B$.
gmatclubot
Re: x<2x<x^2 [#permalink]
08 May 2025, 04:30
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