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

Redeem

Inequality Question MGMAT

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Wed Feb 02, 2011 5:17 am
Thanked: 1 times

Inequality Question MGMAT

by krnverma » Tue Aug 16, 2011 9:45 am
Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?
Top
Source: — Data Sufficiency |
User avatar
Senior | Next Rank: 100 Posts
Posts: 64
Joined: Sun Jul 24, 2011 9:11 am
Thanked: 13 times
GMAT Score:610

by sumgb » Tue Aug 16, 2011 10:31 am
Negative solution
-2x<-2
x>2
This is incorrect. This translates into x > 1. (although this doesn't change the outcome, one must be careful :-) )

stmnt 2:

As you have said, stmnt 2 translates into
(x-7)(x-3) >= 0
"these are boundary points, on a number line."
These are indeed boundary conditions. The boundary conditions are -
either 1. x >= 7 or 2. x <= 3

If you notice, when x >= 7 the inequality holds true (if x = 7 then 0 and if x > 7 both values positive so result > 0)

when x <= 3 again the inequality holds tru (if x = 3 then 0 and if x < 3 both values are negative and the result > 0)

hence from stmnt 2 it follows x can take negative value.

try x = -1 in inequality (x-7)(x-3) >= 0 you get
(-8) * (-4) > 0 true. so x can take negative values so you cant conclude if x definitely greater than 0. Hence insuff.

Thus, answer A

Hope this helps.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 312
Joined: Tue Aug 02, 2011 3:16 pm
Location: New York City
Thanked: 130 times
Followed by:33 members
GMAT Score:780

by gmatboost » Wed Aug 17, 2011 8:04 pm
You can visualize this with three number lines

One shows where x-7 is positive and negative

----------------------7++++++++

One shows where x-3 is positive and negative

--------------3++++++++++++++++

One shows the product of these two, which is determined by the signs on the number lines above

++++++++++++++3-------7+++++++

When x < 3, neg * neg = pos
When 3 < x < 7, pos * neg = neg
When x > 7, pos * pos = pos

This approach can be helpful anytime you have an inequality with several expressions on one side and zero on the other.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Wed Feb 02, 2011 5:17 am
Thanked: 1 times

by krnverma » Fri Aug 19, 2011 6:44 am
Thank you everyone.
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Mon Jul 11, 2011 11:47 am
Thanked: 1 times
Followed by:1 members

by AVbyT » Thu Aug 25, 2011 11:07 am
krnverma - My answer is the for

2) (x^2) - 10x >= -21 => (x^2) -10x + 21 >=0 => (x-3)(x-7)=>0 => x >= 3 OR x>=7 which brings us to x >=7

AND Also, (7-x)(3-x) > 0 => 7 >= x OR 3 >=x Which brings us to 3 >=x

Finally, x >= 7 OR 3 >=x which means the stem is insufficient
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 398
Joined: Tue Jul 26, 2011 11:39 pm
Location: India
Thanked: 41 times
Followed by:6 members

by prateek_guy2004 » Thu Aug 25, 2011 11:20 am
krnverma wrote:Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?
Well to correct answer aint D.

Because Statement 1 is sufficient X<11.

statement 2 Gives 2 answers + and - so its may be hence insufficient.

Hence Answer is A.
Top
Senior | Next Rank: 100 Posts
Posts: 87
Joined: Sun Jul 24, 2011 10:36 am

by [email protected] » Wed Sep 07, 2011 12:15 pm
krnverma wrote:Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?
please tell why statement 2 is not sufficient. HELP HELP
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Wed Feb 02, 2011 5:17 am
Thanked: 1 times

by krnverma » Thu Sep 08, 2011 12:10 am
3 >=x gives us 3 positive values (3, 2, 1) and infinity negative values. So statement 2 insufficient. Read earlier posts under this thread for more details.

Thanks
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Sep 08, 2011 4:13 am
krnverma wrote:Q: Is x>0?

1) |2x-12| < 10
2) (x^2) - 10x >= -21

OA : A

But when I solve it I get D.

My solution (Please correct me):

Statement 1: |2x-12|<10

Positive solution
2x<22
x<11

Negative solution
-2x<-2
x>2

(1) SUFFICIENT

Statement 2: (x^2) - 10x >= -21

x^2 -10x +21 >= 0
(x-7)(x-3) >= 0

So x >= 7 or 3

Both greater than 0 so Statement (2), also, SUFFICIENT.

My answer D.

MGMAT has the solution of second statement as below:
After doing the quadratic and reaching the values x = 7 or 3, they say "these are boundary points, on a number line." And then show a diagram, basically, showing how 3>x>7, which means it can be any number greater than 7 and less than 3. As this includes negative numbers, so we do not have a definite "yes" for the question. And therefore, OA: A

Please help me understand how is (x-7)(x-3) >= 0 lead to a negative value for x?
Statement 1: |2x-12| < 10.
-10 < 2x-12 < 10
2 < 2x < 22
1 < x < 11.
Sufficient.

Statement 2: x²-10x ≥ -21.
It's possible that x=0:
(0)²-10(0)≥-21.
0≥-21.

It's possible that x=1:
1² - 10(1)≥-21.
-9≥-21.

Thus, we cannot determine whether x>0.
Insufficient.

The correct answer is A.

Here's an algebraic way to evaluate statement 2:

x²-10x+21 ≥ 0.
(x-3)(x-7)≥0.

The critical points are x=3 and x=7.
These are the only values where x²-10x+21=0.
When x is any other value, x²-10x+21>0 or x²-10x+21<0.
To determine the range of x, test one value to the left and right of each critical point.

x<3:
Plug x=0 into x²-10x+21 ≥ 0:
0²-10(0)+21≥ 0
21≥0.
This works.
x<3 is part of the range.

3<x<7:
Plug x=5 into x²-10x+21 ≥ 0:
5²-10(5)+21 ≥ 0.
-4 ≥ 0.
Doesn't work.
3<x<7 is not part of the range.

x>7:
Plug x=10 into x²-10x+21≥0:
10²-10(10)+21≥0
21≥0.
This works.
x>7 is part of the range.

Thus, x≤3 OR x≥7.

Using algebra for statement 2 makes the problem HARDER.
I would use algebra to evaluate statement 1 but would plug in values to evaluate statement 2.
Don't limit yourself to algebra.
Many DS questions can be solved more easily -- and more QUICKLY -- by using a COMBINATION of techniques.

Other posts about critical points:

https://www.beatthegmat.com/inequality-c ... 89518.html

https://www.beatthegmat.com/inequalities ... 89788.html

https://www.beatthegmat.com/knewton-q-t89317.html

https://www.beatthegmat.com/which-is-true-t89111.html

Determining the critical points to solve a problem from GMATPrep:
https://www.beatthegmat.com/need-explanation-t85192.html
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
User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

by sl750 » Thu Sep 08, 2011 6:08 am
Statement 1 can be written as,

-10 < 2x -12 < 10, after rearranging the terms, we have 1<x<11. Sufficient

In statement 2, we have a quadratic equation

(x-3)(x-7)>=0, we have two solutions to this, either x>=7 or x<=3. The latter doesn't satisfy x>0. So it is insufficient
Top
Post new topic Post Reply
• Page 1 of 1