BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times
by Max@Math Revolution » Thu Oct 18, 2018 11:55 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x

by fskilnik@GMATH » Fri Oct 19, 2018 7:46 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1
\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\]
Another beautiful problem, Max. Congrats!
\[\left( 1 \right)\,\,x > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{32}} + \frac{1}{{64}}\,\,\, < \,\,\,1\, + 1\,\, = 2 = \frac{1}{{1 - \frac{1}{2}}}\,\,} \right]\,\, \hfill \\
\,{\text{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + 2 + \ldots + {2^6}\,\,\,\mathop < \limits^? \,\, - 1\,\,} \right]\,\,\,\, \hfill \\
\end{gathered} \right.\]

Image

\[\left( 2 \right)\,\,x < 1\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\,1 + \underbrace {\left( { - 1} \right) + {{\left( { - 1} \right)}^2}}_{ = \,\,0} + \underbrace {{{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^4}}_{ = \,\,0} + \underbrace {{{\left( { - 1} \right)}^5} + {{\left( { - 1} \right)}^6}}_{ = \,\,0}\,\,\,\mathop < \limits^? \,\,\frac{1}{2}\,\,} \right]\,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,0 < x < 1\]
\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{1 - x\, > \,\,0} \,\,\,\,\,\,\left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\,\,\,\mathop < \limits^? \,\,\,1\,\]
\[\left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\,\,\, = \,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,\,1\,\,\,\,\,\mathop \Rightarrow \limits^{0\, < \,x\, < \,1} \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Last edited by fskilnik@GMATH on Fri Oct 19, 2018 8:10 am, edited 1 time in total.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

by fskilnik@GMATH » Fri Oct 19, 2018 8:02 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1
Alternate way: ("Algebraic only" solution)
\[1 - {x^7} = \left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\]
\[1 + x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}\,\,\mathop < \limits^? \,\,\,\frac{1}{{1 - x}}\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \,\,\left( {1 - x} \right)} \,\,\,\,\,\,\left\{ \begin{gathered}
\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,,\,\,{\text{when}}\,\,\left( {1 - x} \right) > 0\,\,,\,\,{\text{i}}{\text{.e}}{\text{.}}\,,\,\,x < 1 \hfill \\
\,1 - {x^7}\,\,\mathop > \limits^? \,\,1\,\,\,,\,\,{\text{when}}\,\,\left( {1 - x} \right) < 0\,\,,\,\,{\text{i}}{\text{.e}}{\text{.}}\,,\,\,x > 1 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\left( * \right)\]
\[\left( 1 \right)\,\,x > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,x = \frac{1}{2}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\,x = 2\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,1 - {x^7}\,\,\mathop > \limits^? \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,x < 1\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,0 < x < 1\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,1 - {x^7}\,\,\mathop < \limits^? \,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Fri Oct 19, 2018 12:35 pm
Checking if the left hand side is < right hand side
LHS<RHS
$$LHS=1+x+x^2+x^3+x^4+x^5+x^6$$
$$RHS=\frac{1}{1-x}$$
$$1+x+x^2+x^3+x^4+x^5+x^6\ <\ \frac{1}{1-x}$$
$$STATEMENT\ 1\ -->\ x>0$$
$$if\ x=2,\
$$1+2+2^2+2^3+2^4+2^5+2^{6\ }<\ \frac{1}{1-2}$$
$$1+2+4+8+16+32+64<\frac{1}{-1}$$
$$127<-1$$
$$LHS>RHS\ \&\ 127\ >-1$$
$$If\ \ x=0.5$$
$$1+0.5\ +0.5^2+0.5^3+0.5^4+0.5^4+0.5^5+0.5^6<\frac{1}{1-0.5}$$
$$1+0.5+0.25+0.125+0.0625+0.03125+0.015625<\frac{1}{0.5}$$
$$1.984375<2$$
$$LHS<RHS\ \&1.984375<2$$
$$statement\ 1\ is\ NOT\ SUFFICIENT$$
$$STATEMENT\ 2\ =x<1
$$ $$if\ \ x\ =-1$$
$$1+\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+\left(-1\right)^4+\left(-1\right)^5+\left(-1\right)^6<\frac{1}{1-\left(-1\right)}$$
$$1-1+1-1+1-1+1<\frac{1}{1-\left(-1\right)}$$
$$1-1+1-1+1-1+1\ <\frac{1}{2}$$ $$1-1+1-1+1-1+1\ <\frac{1}{2}$$
$$1<\frac{1}{2}$$
$$LHS\ >RHS\ \ and\ 1>\frac{1}{2}$$

if x=0.5
$$1+0.5+0.5^2+0.5^3+0.5^4+0.5^5+0.5^6\ <\ \frac{1}{1-0.5}$$
$$1+0.5+0.25+0.125+0.0625+0.03125+0.015625\ <\ \frac{1}{0.5}$$
1.984375 < 2

LHS<RHS and 1.984375 < 2
Statement 2 is NOT SUFFICIENT.
Combine statement 1 and 2 together ---->>>
Statement 1 = x>0
statement 2= x<0
From this, we can deduce that
0<x<1

This means that 'x' is greater than 0 and less than 1. Definitely, 'x' will always be a positive integer in the range 0.1 and 0.9.

The question stem is to find cut if
$$1+x+x^2+x^3+x^4+x^5+x^6\ <\ \frac{1}{1-x}$$
if x=0.1
$$1+0.1+0.1^2+0.1^3+0.1^4+0.1^5+0.1^6\ <\ \frac{1}{1-0.1}$$
1.111111 < 1.1111111111

LHS is less than right hand side

if x=0.9
$$1+0.9+0.9^2+0.9^3+0.9^4+0.9^5+0.9^6\ <\ \frac{1}{1-0.9}$$
4.217031 < 10

Hence, the two statement together are SUFFICIENT

OPTION C IS THEREFORE CORRECT.....
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

inequality

by GMATGuruNY » Fri Oct 19, 2018 4:43 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)?

1) x>0
2) x<1
CRITICAL POINTS in an equality occur when the two sides are EQUAL or the inequality is undefined.
Here, the two statements provide the critical points for the inequality in the prompt:
When x=0, the two sides are equal.
When x=1, the inequality is undefined.
To determine the range(s) where the left side is LESS than the right side, test one value to the left and right of each critical point.

Case 1: x > 1
If x=2, the left side is positive, while the right side is negative.
Since the right side is greater than the left side, x>0 is not a valid range.
Case 2: 0 < x < 1
If x=0.99, the left side is relatively small, while the right side = 1/(1-0.99) = 1/0.01 = 100/1 = 100.
Since the the left side is less than the right side, 0 < x < 1 is a valid range.
Case 3: x < 0
If x=-1, the left side = 1, while the right side = 1/2.
Since the left side is greater than the right side, x<0 is not a valid range.

The inequality is valid only when 0 < x < 1.
Question stem, rephrased:
Is 0 < x < 1?

Clearly, neither statement alone is sufficient.
When the statements are combined, 0 < x < 1.
SUFFICIENT.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times

by Max@Math Revolution » Sun Oct 21, 2018 5:41 pm
=>

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 the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question 1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x) is equivalent to 0 < x < 1 as shown below:

For x ≠ 1,
1+x+x^2+x^3+x^4+x^5+x^6 < 1/(1-x)
=> (1+x+x^2+x^3+x^4+x^5+x^6)(1-x)^2 < (1-x)
=> (1 - x^7)(1 - x) < 1 - x
=> 1 - x^7 - x +x^8 < 1 - x
=> - x^7 + x^8 < 0
=> x^7( x - 1 ) < 0
=> x( x - 1 ) < 0
=> 0 < x < 1

Since both conditions must be applied together to obtain this inequality, both conditions 1) & 2) are sufficient, when applied together.

Therefore, C is the answer.
Answer: C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1