GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Nov 2018, 11:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat

November 20, 2018

November 20, 2018

09:00 AM PST

10:00 AM PST

The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
• ### The winning strategy for 700+ on the GMAT

November 20, 2018

November 20, 2018

06:00 PM EST

07:00 PM EST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

# Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6517
GMAT 1: 760 Q51 V42
GPA: 3.82
Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

21 Mar 2016, 03:31
1
6
00:00

Difficulty:

75% (hard)

Question Stats:

52% (02:17) correct 48% (02:09) wrong based on 141 sessions

### HideShow timer Statistics

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

1) x>0
2) x<1

* A solution will be posted in two days.

_________________

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" "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: 7036 Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 [#permalink] ### Show Tags 21 Mar 2016, 05:35 1 1 MathRevolution wrote: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 * A solution will be posted in two days. INFO from Q stem $$1+x+x^2+x^3+x^4<\frac{1}{(1-x)}$$.. $$1+x+x^2+x^3+x^4-\frac{1}{(1-x)}<0$$.. $$\frac{{(1+x+x^2+x^3+x^4)(1-x)-1}}{(1-x)}<0$$.. $$\frac{{1+x+x^2+x^3+x^4-x-x^2-x^3-x^4-x^5)-1}}{(1-x)}<0$$.. $$\frac{-x^5}{(1-x)}<0$$.. $$\frac{x^5}{(x-1)}<0$$.. so two conditions 1) if x>0, x<1 2) if x<0, x>1 ...... Not possible lets check the statements 1) x>0 if x<1.. ans is YES if x>1.. ans is NO Insuff 2) x<1 if 0<x<1.. ans is YES if x<0.. ans is NO Insuff combined.. Suff as per point 1, ans is always YES 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 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)? 1) x>0 2) x<1 [#permalink] ### Show Tags 22 Mar 2016, 16:59 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer. _________________ 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"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Manager
Joined: 24 May 2013
Posts: 79
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

26 Mar 2016, 20:18
Is 1+x+x^2+x^3+x^4<1/(1-x)?

Although the series on L.H.S. is small otherwise we culd hav used G.P. for the sum ..
1(x^5-1)/(x-1) <1/(1-x)
x^5/(x-1)<0

----(+ve)----0-------(-ve)----------1--------(+ve)-----
So if x lies between 0 and 1 then we have YES for the above equation else NO.

1) x>0. Insufficient.
2) x<1. Insufficient.

Combining Sufficient. C is the answer.
Current Student
Status: It`s Just a pirates life !
Joined: 21 Mar 2014
Posts: 233
Location: India
Concentration: Strategy, Operations
GMAT 1: 690 Q48 V36
GPA: 4
WE: Consulting (Manufacturing)
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

25 Apr 2016, 00:16
MathRevolution wrote:
Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0
2) x<1

* A solution will be posted in two days.

I took x=1,2 and x=1/2,0

C
_________________

Aiming for a 3 digit number with 7 as hundredths Digit

Senior Manager
Joined: 29 Jun 2017
Posts: 468
GPA: 4
WE: Engineering (Transportation)
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

05 Sep 2017, 05:24
Ans is C
is
$$1+x+x^2+x^3+x^4$$ < $$\frac{1}{(1-x)}$$

Multiply and divide by 1-x on both numerator and denominator on L.H.S.

equation becomes
is
$$\frac{(1-x^5)}{(1-x)}$$ < $$\frac{1}{(1-x)}$$
or
$$\frac{x^5}{(1-x)}$$ > 0

Taking 3 points -1, 0 and 1 on number line
value above 1 i.e. x>1 will not hold this equation
x<1 but >0 will hold this equation
x<0 but > -1 will again not hold equation
x< -1 will again satisfy equation

so soln is x< -1 and x should be from (0,1) open brackets i.e. 0 and 1 not included.

1) x>0 is insufficient as it includes x>1 also which is not part of solution. A,D eliminated
2) x<0 is insufficient as it includes -1<x<0 which is not part of solution B eliminated

combine both
0<x<1
is our solution and Ans is C
_________________

Give Kudos for correct answer and/or if you like the solution.

Senior Manager
Joined: 15 Jan 2017
Posts: 359
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

17 Sep 2017, 23:14
I have a query - If I take (C)
--> it means x = 0.5, 0.3, 0.2
With x = 0.5
1+ x + x sq + x^3 + x^4 < 1/1-x
1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/1-0.5
1.93< 2 SUFF
if x= 0.3
1+0.3+0.09 + 0.027 + 0.0081 = < 1/1 - 0.3
1.42 no less than 0.7
if x = 0.2
1+0.2+0.04 + 0.008 + 0.0016 <1/ 1-0.2
1.24 not less than 0.8
So I chose E.
Can anyone explain the issue in my reasoning here?
Math Expert
Joined: 02 Sep 2009
Posts: 50627
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

17 Sep 2017, 23:26
1
I have a query - If I take (C)
--> it means x = 0.5, 0.3, 0.2
With x = 0.5
1+ x + x sq + x^3 + x^4 < 1/1-x
1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/1-0.5
1.93< 2 SUFF
if x= 0.3
1+0.3+0.09 + 0.027 + 0.0081 = < 1/1 - 0.3
1.42 no less than 0.7
if x = 0.2
1+0.2+0.04 + 0.008 + 0.0016 <1/ 1-0.2
1.24 not less than 0.8
So I chose E.
Can anyone explain the issue in my reasoning here?

It's $$\frac{1}{1 - 0.3}$$ NOT $$\frac{1}{1}- 0.3$$.

$$(1 + 0.3 +0.3^2 + 0.3^3 + 0.3^4 = 1.4251)< (\approx 1.428=\frac{1}{1 - 0.3})$$

$$(1 + 0.2 +0.2^2 + 0.2^3 + 0.2^4 = 1.2496)< (1.25=\frac{1}{1 - 0.3})$$
_________________
Senior Manager
Joined: 15 Jan 2017
Posts: 359
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

17 Sep 2017, 23:41
hi, sorry I wasn not clear on my post earlier.
So after x = 0.2
I got 1.42 < 1/0.7 --> 1.42 < 10/7 which is approx 1.4
Thus
1.42 < 1.44 (TRUE)

1.24 < 1/0.8
1.24 < 5/4
1.24 < 1.25 (TRUE)

if x = 0.6
then 2.3 < 2.5 (TRUE)

Thank you, the above cleared it up for me.
Ans C
Intern
Joined: 09 Dec 2013
Posts: 27
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

03 Dec 2017, 15:05
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

1) x>0
2) x<1

The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.

Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?
DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1370
Location: India
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1  [#permalink]

### Show Tags

04 Dec 2017, 10:22
Buttercup3 wrote:
MathRevolution wrote:
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

1) x>0
2) x<1

The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.

Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?

Here algebra is slightly better but yes, it can be done by plugging in numbers. Let me try:

(1) x > 0

Now RHS of the question stem is 1/(1-x). Since its 1-x in the denominator, and this statement says x > 0, we should immediately look for two cases: one where 0<x<1 and one where x>1 (x cannot be 1 as for that RHS would be undefined).

So lets take a value of x between 0 and 1, say 0.5. RHS becomes 1/(1-0.5) = 2 LHS becomes 1+0.5+0.25+0.125+0.0625=1.9375. LHS is less than RHS
If we try any other value in the same range say 0.2 or 0.9, still RHS would be greater.

Now lets take another case where x > 1, in this case clearly we can see that 1-x would be negative, so RHS would be negative while LHS positive. So LHS would become greater.

So not sufficient.

(2) x < 1

We have got to take cases where 0<x<1, where -1<x<0, and where x<-1

If 0<x<1, RHS would be greater as discussed in first statement.

If -1<x<0, lets say x=-0.5. RHS = 1/(1+0.5) = 0.666.. and LHS = 1-0.5+0.25-0.125+0.0625 = 0.6875. LHS is greater

We dont even need to look at third case because we can see that this statement is not sufficient.

Combining the two statements: 0<x<1. In that case RHS would be greater as discussed. Sufficient.
Re: Is 1+x+x^2+x^3+x^4<1/(1-x)? 1) x>0 2) x<1 &nbs [#permalink] 04 Dec 2017, 10:22
Display posts from previous: Sort by