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# Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)?  [#permalink]

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11 Dec 2017, 00:52
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Difficulty:

65% (hard)

Question Stats:

53% (02:12) correct 47% (01:53) wrong based on 68 sessions

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[GMAT math practice question]

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

(1) $$x>0$$
(2) $$x<1$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Expert Joined: 02 Aug 2009 Posts: 7035 Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)? [#permalink] ### Show Tags 11 Dec 2017, 02:48 MathRevolution wrote: [GMAT math practice question] Is $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$? (1) $$x>0$$ (2) $$x<1$$ Good Q MathRevolution $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$.. $$1+x+x^2+x^3+x^4=\frac{1(1-x^5)}{1-x}$$.... sum of a Geometric Progression so $$\frac{1-x^5}{1-x}<\frac{1}{1-x}..........\frac{1-x^5}{1-x}-\frac{1}{1-x}<0.........\frac{1-x^5-1}{1-x}<0.............-\frac{x^5}{1-x}<0$$ so both numerator and denominator are of opposite sign.. two cases (A) 1-x<0 or x>1, so$$-x^5 >0........... x^5<0........x<0$$ x cannot be <0 and >1 at same time Not possible (B) 1-x>0 or x<1, so$$-x^5 <0........... x^5>0........x>0$$ so x lies between 0 and 1 lets see the statements.. (1) $$x>0$$ (2) $$x<1$$ combined both tells us that x lies between 0 and 1 suff C _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 Location: India GPA: 3.82 Re: Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)? [#permalink] ### Show Tags 11 Dec 2017, 06:10 MathRevolution wrote: [GMAT math practice question] Is $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$? (1) $$x>0$$ (2) $$x<1$$ In case one doesn't know about or remember the GP series as explained by chetan2u, then you can use below approach - $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$---------------(1) Case 1: $$x>1$$, for eg. $$x=2$$, then LHS of the above inequality will be greater than $$2$$ but RHS of inequality (1) will be negative. So we have a NO for this case Case 2: $$0<x<1$$, for eg. $$x=\frac{1}{2}$$, then LHS of inequality will be $$1+0.5+0.25+0.125+0.0625<2$$ but RHS of the inequality will be $$\frac{1}{(1-0.5)}=2$$. So we have a YES for this case Case 3: $$x<0$$, for eg. $$x=-1$$, then LHS of the inequality will be $$1-1+1-1+1=1$$ but RHS of the inequality will be $$\frac{1}{(1+1)}=0.5$$. So we have a NO for this case With these understandings, we can now check the statements Statement 1: from this statement, both Case 1 & Case 2 are possible. Hence we have a Yes & a No. Insufficient Statement 2: from this statement, both Case 2 & Case 3 are possible. Hence we have a Yes & a No. Insufficient Combining 1 & 2, we have $$0<x<1$$, so only Case 2 possible, hence we have a Yes. Sufficient Option C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6517 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)? [#permalink] ### Show Tags 12 Dec 2017, 23:28 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. Modifying the question: $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$ $$⇔ (1+x+x^2+x^3+x^4)(1-x)^2<(1-x)$$ $$⇔ (1-x^5)(1-x)<(1-x)$$ $$⇔ 1-x^5-x+x^6<1-x$$ $$⇔ x^6-x^5<0$$ $$⇔ x^5(x-1)<0$$ $$⇔ x(x-1)<0$$ $$⇔ 0<x<1$$ The question asks if $$0<x<1$$. This question has the unique answer, ‘yes’, if we require both $$x>0$$ and $$x<1$$. Thus, both conditions together are sufficient. Therefore, C is the answer. In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)?  [#permalink]

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14 Dec 2017, 09:20
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

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

(1) $$x>0$$
(2) $$x<1$$

Good Q MathRevolution

$$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$..
$$1+x+x^2+x^3+x^4=\frac{1(1-x^5)}{1-x}$$.... sum of a Geometric Progression

so $$\frac{1-x^5}{1-x}<\frac{1}{1-x}..........\frac{1-x^5}{1-x}-\frac{1}{1-x}<0.........\frac{1-x^5-1}{1-x}<0.............-\frac{x^5}{1-x}<0$$
so both numerator and denominator are of opposite sign..
two cases
(A) 1-x<0 or x>1,
so$$-x^5 >0........... x^5<0........x<0$$
x cannot be <0 and >1 at same time
Not possible
(B) 1-x>0 or x<1,
so$$-x^5 <0........... x^5>0........x>0$$
so x lies between 0 and 1

lets see the statements..
(1) $$x>0$$
(2) $$x<1$$

combined both tells us that x lies between 0 and 1
suff

C

Hi chetan2u

Sum of an GP is a$$(r^n - 1)/r - 1$$ for r > 0 and for r < 0 Sum of an GP is a $$(1-r^n)/1-r$$

Now, from Stmnt 1: As x > 0, $$x^5 - 1/x-1 > 1 / 1- x$$
$$x^5 - 2/ x - 1 > 0$$

As x > 0, Can we make the Statemnt 1 insufficient based on x =1.. Please advise. Other than 1 on which values of x, statement 1 would be insufficient. I can only think 1.

Please let me know if there is anything wrong with my approach.
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Joined: 02 Aug 2009
Posts: 7035
Re: Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)?  [#permalink]

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14 Dec 2017, 17:26
1
rahul16singh28 wrote:
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

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

(1) $$x>0$$
(2) $$x<1$$

Good Q MathRevolution

$$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$..
$$1+x+x^2+x^3+x^4=\frac{1(1-x^5)}{1-x}$$.... sum of a Geometric Progression

so $$\frac{1-x^5}{1-x}<\frac{1}{1-x}..........\frac{1-x^5}{1-x}-\frac{1}{1-x}<0.........\frac{1-x^5-1}{1-x}<0.............-\frac{x^5}{1-x}<0$$
so both numerator and denominator are of opposite sign..
two cases
(A) 1-x<0 or x>1,
so$$-x^5 >0........... x^5<0........x<0$$
x cannot be <0 and >1 at same time
Not possible
(B) 1-x>0 or x<1,
so$$-x^5 <0........... x^5>0........x>0$$
so x lies between 0 and 1

lets see the statements..
(1) $$x>0$$
(2) $$x<1$$

combined both tells us that x lies between 0 and 1
suff

C

Hi chetan2u

Sum of an GP is a$$(r^n - 1)/r - 1$$ for r > 0 and for r < 0 Sum of an GP is a $$(1-r^n)/1-r$$

Now, from Stmnt 1: As x > 0, $$x^5 - 1/x-1 > 1 / 1- x$$
$$x^5 - 2/ x - 1 > 0$$

As x > 0, Can we make the Statemnt 1 insufficient based on x =1.. Please advise. Other than 1 on which values of x, statement 1 would be insufficient. I can only think 1.

Please let me know if there is anything wrong with my approach.

approach is correct but you have made two errors..
1) while solving $$\frac{x^5 - 1}{x-1} > \frac{1}{1- x}$$, you have not made th edenominator equal you have taken x-1 same as 1-x and got answer $$\frac{x^5 - 2}{x - 1} > 0$$..
$$\frac{x^5 - 1}{x-1} > \frac{1}{1- x}........\frac{x^5 - 1}{x-1} - \frac{1}{1- x}>0.............\frac{x^5 - 1}{x-1} -(- \frac{1}{x-1})>0.........\frac{x^5 - 1}{x-1} + \frac{1}{x-1}>0.............\frac{x^5}{x-1}>0$$

2) even if you have an equation $$\frac{x^5 - 2}{x - 1}> 0$$, the values will be x^5>2, so 1 or 1.05 etc will not hold true as x need not be integer
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)?  [#permalink]

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14 Dec 2017, 18:22
Hello,

As an engineer, I really aprecciate well strucutured resolution. But, unless well trained in this mathematical developments, one have to struggle using these method to answer this type of question in about 2 minutes.

My reasoning in this case of question is 1) find a value of x that is an exception or 2) give a 'good guess' for the possibly values of x.

Note: a 'good guess' is not guessing literally. There is study area called Numerical Analysis that uses iterative methods to find solutions for some types of equations. Although there are different methods, they have in common the need for a initial value to begin iterations. This is the 'guess'. There are even equations with the aim to calculate what would be a good 'guess' value.

1) X > 0, so try values like 0.5, 1, 10. You will see that is insufficient to address the inequality
2) X < 1, so try 0.5, 0, -1, -10. You will see that is insufficient to address the inequality
1/2) 0 < X < 1, so try 0.1 (low), 0.5 (mid) and 0.9 (high). You will see that is sufficient

Why use those numbers? Because they make sense:
-> Numbers between 0 and 1: always try them when there is no mention to integers;
-> -1, 0, 1: 0 could be an inflection point (where the possible solutions of x change signal; think about a parabola and its most upper/lower point);
-> -10, 10: numbers well above -1 and 1

Best,
Is 1 + x + x^2 + x^3 + x^4 < 1/(1 - x)? &nbs [#permalink] 14 Dec 2017, 18:22
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