Find all School-related info fast with the new School-Specific MBA Forum

It is currently 20 May 2013, 09:12
Customize  |  Hide

Points to remember : 1. if x > y then (1/x) < (1/y) (

  Question banks Downloads My Bookmarks Reviews  
Author Message
TAGS:
10 KUDOS received
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1050
Followers: 8

Kudos [?]: 214 [10] , given: 1

GMAT Tests User
Points to remember : 1. if x > y then (1/x) < (1/y) ( [#permalink] New post 18 Aug 2008, 02:44
10
This post received
KUDOS
Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

--------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

-----------------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

----------------------------------------------------------------------------------------------------------------------------------

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Check out my GMAT blog - GMAT Tips and Strategies


Last edited by amitdgr on 19 Sep 2008, 06:02, edited 3 times in total.
SVP
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7

Kudos [?]: 131 [0], given: 0

GMAT Tests User
Re: Inequalities Review [#permalink] New post 19 Aug 2008, 03:10
amitdgr, an excellent compilation.
Senior Manager
Senior Manager
User avatar
Joined: 29 Mar 2008
Posts: 351
Followers: 2

Kudos [?]: 33 [0], given: 0

GMAT Tests User
Re: Inequalities Review [#permalink] New post 19 Aug 2008, 09:46
amitdgr wrote:
Points to remember :

1. if x > y then (1/x) < (1/y)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.
--------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

-----------------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

----------------------------------------------------------------------------------------------------------------------------------

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------

1 )if x^2 > k^2 and x> 0 then this implies k >0

2) If k > 1 , then

a) k^x > 1 when x>0
b) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------

For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------


Please add if i missed something.
_________________
Imagination is more powerful than knowledge - Albert Einstein



Thanks for posting the summary.

How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

SVP
SVP
User avatar
Joined: 30 Apr 2008
Posts: 1900
Location: Oklahoma City
Schools: Hard Knocks
Followers: 25

Kudos [?]: 339 [0], given: 32

GMAT Tests User
Re: Inequalities Review [#permalink] New post 19 Aug 2008, 10:12
leonidas wrote:
How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.


I don't think this is always true, especially since this doesn't limit the rule to positive integers.

x = -2
k = 1/2

x^2 = 4
k^2 = 1/4

x !> k
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Find out what's new at GMAT Club - latest features and updates

Senior Manager
Senior Manager
User avatar
Joined: 29 Mar 2008
Posts: 351
Followers: 2

Kudos [?]: 33 [0], given: 0

GMAT Tests User
Re: Inequalities Review [#permalink] New post 18 Sep 2008, 09:41
jallenmorris wrote:
leonidas wrote:
How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.


I don't think this is always true, especially since this doesn't limit the rule to positive integers.

x = -2
k = 1/2

x^2 = 4
k^2 = 1/4

x !> k


Nice catch....... you got me there!!!!! I agree, I assumed them to be integers (though not given), got side tracked..
The above should hold good, if they were integers.
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

GMAT Instructor
Joined: 24 Jun 2008
Posts: 973
Location: Toronto
Followers: 167

Kudos [?]: 443 [0], given: 3

GMAT Tests User
Re: Inequalities Review [#permalink] New post 18 Sep 2008, 11:47
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y)
Please add if i missed something.


I didn't go through this post thoroughly, but 1 above is not quite right. Notice that if x = 3 and y = -2, for example, then x > y and 1/x is also greater than 1/y.

1 above is only true if x and y share the same sign, as one can quite easily demonstrate (start from x > y, then divide both sides by x and by y).
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Manager
Manager
Joined: 13 Jan 2009
Posts: 180
Schools: Harvard Business School, Stanford
Followers: 3

Kudos [?]: 14 [0], given: 9

GMAT Tests User
Re: Inequalities Review [#permalink] New post 22 Jul 2009, 22:11
Great compilation. Thanks!!!
Manager
Manager
User avatar
Joined: 27 Jun 2008
Posts: 163
Followers: 1

Kudos [?]: 14 [0], given: 11

GMAT Tests User
Re: Inequalities Review [#permalink] New post 22 Jul 2009, 23:55
Nice compilation, Thanks.
Manager
Manager
User avatar
Joined: 29 Oct 2009
Posts: 212
Followers: 49

Kudos [?]: 417 [0], given: 18

GMAT Tests User
Re: Inequalities Review [#permalink] New post 10 Nov 2009, 15:29
amitdgr wrote:
----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------


Hi. Appreciate the effort! Good Job!

Just a minor correction though. Point 2 and 3 (in red above) are not valid for k = 1.

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html

Manager
Manager
User avatar
Joined: 23 Jun 2009
Posts: 156
Followers: 1

Kudos [?]: 6 [0], given: 9

GMAT Tests User
Re: Inequalities Review [#permalink] New post 11 Nov 2009, 20:59
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

--------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

-----------------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

----------------------------------------------------------------------------------------------------------------------------------

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.


Thanks @amitdgr +1 from me. Great collection
Manager
Manager
Joined: 25 Dec 2009
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 3

Re: Inequalities Review [#permalink] New post 20 Jan 2010, 02:11
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

--------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

-----------------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
Quote:
a < b is equivalent to a/c < b/c;

a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

----------------------------------------------------------------------------------------------------------------------------------

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.


This wont be true if c is between 0 and 1. a < b is equivalent to a/c < b/c
Director
Director
User avatar
Joined: 21 Dec 2010
Posts: 657
Followers: 9

Kudos [?]: 58 [0], given: 51

GMAT Tests User
Re: Inequalities Review [#permalink] New post 31 Jan 2011, 00:47
@ above post ,

a/c < b/c , if a < b and 0 < c < 1 , of course even in the given constrains its true.

thanks for the help
_________________

What is of supreme importance in war is to attack the enemy's strategy.

Intern
Intern
Joined: 28 Mar 2011
Posts: 34
Followers: 0

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

Re: Inequalities Review [#permalink] New post 06 Jun 2011, 20:51
Hi,

can someone please explain this :

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r


How do we open the modulus here?

I came across this example :

If |5x + 2| – 3 < 3x, what is the value of x?
|5x+2|< 3x+3 |5x+2| < -3x-3

2x < 1 5x+2 < -3x - 3
x< 1/2 8x < -5 ---> x < -5/8


so,if we open |x-a| < r as the above:

|x-a| < r |x-a| < -r
x < r+a x < a-r


What am i missing :?



regards,
Manager
Manager
Joined: 08 Sep 2010
Posts: 176
Followers: 0

Kudos [?]: 9 [0], given: 18

GMAT Tests User
Re: Inequalities Review [#permalink] New post 10 Jun 2011, 20:22
Excellent compilation...good refresher
_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

If you like my explanations award kudos.

Re: Inequalities Review   [#permalink] 10 Jun 2011, 20:22
    Similar topics Author Replies Last post
Similar
Topics:
New posts If X>1 and Y>1, is X<Y? (1) X?/(XY+X)<1 (2) jcgoodchild 8 21 Mar 2006, 11:09
Popular new posts If X > 1 and Y > 1, is X < Y A. X^2/(XY+X) < 1 bewakoof 13 25 Jul 2007, 22:03
New posts If x > 1 and y > 1, is X < Y 1) x^2/(xy + x) < 1 x97agarwal 6 31 Jul 2008, 16:36
New posts if X>1 and Y>1 is X<y? 1)x^2/(xy+X)<1 shobuj 1 27 Sep 2008, 12:37
New posts 1 EXPERTS_POSTS_IN_THIS_TOPIC If x>1 and y>1, is x<y? jamifahad 7 02 May 2011, 05:29
Display posts from previous: Sort by

Points to remember : 1. if x > y then (1/x) < (1/y) (

  Question banks Downloads My Bookmarks Reviews  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.