Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 15:01

### 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 June
Open Detailed Calendar

# Inequality questions

Author Message
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

### Show Tags

11 Feb 2008, 05:52
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi Guys,

I am studying topic Inequalities and have been facing few problems. I will be glad if anyone can tell me what fundamental mistake am I making.

Q1. What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

Correct ans is (d)
I am getting (a)

what am I doing wrong ?

Q2. Solve x^2 + 8x + 7 < 0

OA is -7< x < -1

When I try to solve I do -
(x+1) (x+7) < 0
x + 1 < 0
x < -1
And
x + 7 < 0
x < -7.

something wrong??

Similarly -

Q3 Solution of inequality -x^2 + 5x - 6 < 0

ans???

Thanks,

Cheers,
JG
Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [0], given: 1

### Show Tags

11 Feb 2008, 07:31
greatchap wrote:
Hi Guys,

I am studying topic Inequalities and have been facing few problems. I will be glad if anyone can tell me what fundamental mistake am I making.

Q1. What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

Correct ans is (d)
I am getting (a)

what am I doing wrong ?

Q2. Solve x^2 + 8x + 7 < 0

OA is -7< x < -1

When I try to solve I do -
(x+1) (x+7) < 0
x + 1 < 0
x < -1
And
x + 7 < 0
x < -7.

something wrong??

Similarly -

Q3 Solution of inequality -x^2 + 5x - 6 < 0

ans???

Thanks,

Cheers,
JG

here is what's important: solution of (x-a)*(x-b) > 0 (where a < b) is {x < a} U {x > b} , consequently solution of < 0 is a < x < b.

let's look at the first problem 2/(x-2) < 1 //multiply both sides by [(x-2)^2]/2 - positive number - inequality sign doesn't change

2*(x-2) < x^2-4*x + 4 -> x^2-6*x + 8 > 0 -> a = 2, b = 4
solution: { x < 2} U {x > 4} -> D

Second problem:

x^2 + 8*x + 7 < 0 -> a = -7, b = -1 -> -7 < x < -1

Third problem:
-x^2 + 5*x - 6 < 0
x^2 - 5*x + 6 > 0
a = 2, b = 3 -> {x < 2} U {x > 3}
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

### Show Tags

11 Feb 2008, 08:20
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?
Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [0], given: 1

### Show Tags

11 Feb 2008, 08:44
greatchap wrote:
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?

When you solve a problem, you can replace it with equivalent problems and solve those. However, if you multiply both sides by x-2 the new problem is not going to be equivalent to the initial one because there are going to be two situations:

4 < 2*(x-2) if x-2 > 0
4 > 2*(x-2) if x-2 < 0

because when you multiply both sides by the same positive number, inequality sign doesn't change, and the sign changes when you multiply both sides by the same negative number (for example if (-2) < (-1) and you multiply both sides by (-1) it would be incorrect to say: (-1)*(-2) < (-1)*(-1) or 2 < 1)

In order to avoid this problem, I mutliply both sides by a positive number (x-2)^2 and keep the inequality sign unchanged. I hope that helps.
CEO
Joined: 21 Jan 2007
Posts: 2742
Location: New York City
Followers: 11

Kudos [?]: 942 [0], given: 4

### Show Tags

11 Feb 2008, 08:52
greatchap wrote:
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?

because x is a variable and we have an inequality here. u cant cross multiply unles u know the sign of the variable. a negative flips the inequality sign
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

SVP
Joined: 28 Dec 2005
Posts: 1561
Followers: 3

Kudos [?]: 159 [0], given: 2

### Show Tags

11 Feb 2008, 09:55
maratikus wrote:
greatchap wrote:
Thanks for the solutions maratikus.
4 < 2*(x-2) if x-2 > 0
4 > 2*(x-2) if x-2 < 0

because when you multiply both sides by the same positive number, inequality sign doesn't change, and the sign changes when you multiply

here is my question: when you solve both of those inequalities, you get that either x is less than 4, or that x is greater than 4. how do you get to the final answer of x<2 and x>4 ?
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

### Show Tags

11 Feb 2008, 10:31
I still have a confusion so am writing again...

I came across a ques:

Q-Z Find the value of x in (3x^2 + x) / (x^2 + 2) < 3.

A- here we did a normal cross multiply -
3x^2 + x < 3(x^2 + 2)
3x^2 + x < 3x^2 + 6

so whats the difference in that ques & (solution set of inequality 4/x-2 < 2 ) ??

cant we do -> 4 < 2(x-2)
4 < 2x - 4
8 < 2x
x > 4. (wrong)

If we can't then, why cross multiplication in Q-Z. moreover with what no u multiplied in 4/x-2 < 2. why not x-2 to both sides. ??
SVP
Joined: 28 Dec 2005
Posts: 1561
Followers: 3

Kudos [?]: 159 [0], given: 2

### Show Tags

11 Feb 2008, 10:50
i think the difference here is that in Q-Z, you are dealing with x^2, and so those terms will always be positive no matter what.
Manager
Joined: 14 May 2007
Posts: 182
Location: India
Followers: 2

Kudos [?]: 74 [0], given: 11

### Show Tags

13 Feb 2008, 09:51
CEO
Joined: 17 Nov 2007
Posts: 3585
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3984 [0], given: 360

### Show Tags

13 Feb 2008, 11:23
$$\frac{4}{(x-2)} < 2$$

my logic:

1. it is true for all negative (x-2) or for x<2
2. the function $$f=\frac{4}{(x-2)}$$ is $$\infty$$ at x close to 2 at the greater side.
3. When x increases from 2, the function f always decreases and equal to 2 at x=4. Therefore, f<2 at x>4
4. Combine two conditions: x<2 & x>4
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

SVP
Joined: 28 Dec 2005
Posts: 1561
Followers: 3

Kudos [?]: 159 [0], given: 2

### Show Tags

13 Feb 2008, 20:57
walker wrote:
$$\frac{4}{(x-2)} < 2$$

my logic:

1. it is true for all negative (x-2) or for x<2

how can you show this algebraically ?

im getting 4 >x....
CEO
Joined: 17 Nov 2007
Posts: 3585
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3984 [0], given: 360

### Show Tags

13 Feb 2008, 23:50
pmenon wrote:
walker wrote:
$$\frac{4}{(x-2)} < 2$$

my logic:

1. it is true for all negative (x-2) or for x<2

how can you show this algebraically ?

im getting 4 >x....

1. the inequality is true when 4/(x-2) is negative.
2. 4/(x-2) is negative when (x-2) is negative
3. (x-2) is negative when x<2 (x-2<0 --> x<2)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Re: Inequality questions   [#permalink] 13 Feb 2008, 23:50
Display posts from previous: Sort by