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805+ (Hard)|   Inequalities|      
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Bunuel
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If x 0, is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? 04 May 2021, 06:10
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95% (hard)

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34% (02:23) correct 66% (01:56) wrong based on 102 sessions

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If x ≠ 0, is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}\) ?


(1) \(x > 0\)

(2) \(x < 1\)


M36-73

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Bunuel
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If x 0, is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? 09 Nov 2021, 09:18
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Bunuel
If x ≠ 0, is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}\) ?


(1) \(x > 0\)

(2) \(x < 1\)


M36-73
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Official Solution:


If \(x \neq 0\), is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}?\)

Let's simplify the question:

Is \(1+x+x^2+x^3+x^4+x^5-\frac{1}{1-x} < 0?\)

Is \(\frac{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}{1-x} < 0?\)

Is \(\frac{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}{1-x} < 0?\)

Is \(\frac{-x^6}{1-x} < 0?\)

Is \(\frac{x^6}{x-1} < 0?\)

For \(\frac{x^6}{x-1}\) to be negative \(x\) must not be 0 (because if \(x=0\), then \(\frac{x^6}{x-1}\) would be 0) AND \(x-1\) must be negative. In this case we'd have \(\frac{x^6}{x-1}=\frac{positive}{negative}=negative\). We are given in the stem that \(x \neq 0\), so, the question basically asks whether \(x < 1\) (\(x -1 < 0\) means \(x < 1 \)).

(1) \(x > 0\). Not sufficient.

(2) \(x < 1\). Sufficient.


Answer: B
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If x 0, is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? 04 May 2021, 06:20
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Bunuel
If x ≠ 0, is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}\) ?


(1) \(x > 0\)

(2) \(x < 1\)
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Similar questions to practice:
https://gmatclub.com/forum/is-1-x-x-2-x ... 15368.html
https://gmatclub.com/forum/is-1-x-x-2-x ... 56226.html
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GMAT Focus 1: 735 Q90 V89 DI81
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If x 0, is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? 04 May 2021, 06:54
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Bunuel
If x ≠ 0, is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}\) ?


(1) \(x > 0\)

(2) \(x < 1\)
Show more


\(1+x+x^2+x^3+x^4+x^5<\frac{1}{(1-x)}\)..

\(1+x+x^2+x^3+x^4+x^5-\frac{1}{(1-x)}<0\)..

\(\frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0\)

\(\frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0\)

\(\frac{{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}}{(1-x)}<0\)

\(\frac{-x^6}{1-x}<0\)

\(\frac{x^6}{x-1}<0\)

x^6 is always non negative, so the question becomes : Is x-1<0? OR Is x<1?

lets check the statements
1) x>0
if x<1.. ans is YES
if x>1.. ans is NO
Insuff

2) x<1
Exactly what we were looking for
suff

B
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If x 0, is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? 19 Feb 2023, 09:16
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chetan2u
Bunuel
If x ≠ 0, is \(1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}\) ?


(1) \(x > 0\)

(2) \(x < 1\)


\(1+x+x^2+x^3+x^4+x^5<\frac{1}{(1-x)}\)..

\(1+x+x^2+x^3+x^4+x^5-\frac{1}{(1-x)}<0\)..

\(\frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0\)

\(\frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0\)

\(\frac{{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}}{(1-x)}<0\)

\(\frac{-x^6}{1-x}<0\)

\(\frac{x^6}{x-1}<0\)

x^6 is always non negative, so the question becomes : Is x-1<0? OR Is x<1?

lets check the statements
1) x>0
if x<1.. ans is YES
if x>1.. ans is NO
Insuff

2) x<1
Exactly what we were looking for
suff

B
Show more

In the last step before evaluating the statements, when we multiply an inequality with a -1, the inequality should flip right? How or why have you kept the inequality sign same? Help me understand, and please correct me if I'm wrong. Thank you for working out the answer. Appreciate it!
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