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I'm going to make a sticky thread where the very very basic

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 [#permalink] New post 15 Mar 2005, 11:07
Here's antmavel's question where you can try out the method I posted to find the sum of a terms in a series

http://www.gmatclub.com/phpbb/viewtopic.php?t=14796
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 [#permalink] New post 15 Mar 2005, 13:39
on the same note.

If d is POSITIVE and |x| < d, then -d < x < d
If d is NEGATIVE and |x| < d, then there is no solution

If d is POSITIVE and |x| > d, then x < -d OR x > d
If d is NEGATIVE and |x| > d, then x is all real numbers
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 [#permalink] New post 17 Mar 2005, 08:10
can we have some basic sets and venn diagrams principles here?
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 [#permalink] New post 18 Mar 2005, 02:15
If x > y, then x - y divides x^n - y^n.
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 [#permalink] New post 18 Mar 2005, 07:54
Honghu,

I think ur first example on inequalities is wrong, for |x-4|<9...solution shud be -5<x<13
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 [#permalink] New post 18 Mar 2005, 07:55
Dan wrote:
If x > y, then x - y divides x^n - y^n.


interesting....can someone provide a proof for this...thx
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 [#permalink] New post 22 Mar 2005, 00:24
vprabhala wrote:
can we have some basic sets and venn diagrams principles here?


I can add them in, but soemone needs to solve my problem with attachements (somehow, I do not have the option to attach documents anymore, must have been a bad boy :-D ) otherwise i can't show some of the workings here.
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 [#permalink] New post 22 Mar 2005, 06:11
oh. I was wondering that we were able to post attachments for a while.. hmmm looks like its gone.. :roll:
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 [#permalink] New post 22 Mar 2005, 06:14
I've been meaning to post a couple of documents with tips on AWA and verbal, but no possibility of creating an attachment.

If anyone figures out how to go about attachments, please let me know.
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 [#permalink] New post 22 Mar 2005, 12:12
ywilfred wrote:
alright, some more as I've promised....

Here's one way to solve absolue inequalities:

|x-4| < 9
We need to solve both positive and negative x

Solving for positive: x-4 < 9 --> x <13
Solving for negative:
-x+4 < 9 --> - x < 5 --> x>5 (dividing by negative number, switch the sign)

So now we have 5<x<13


Hello Ywilfred,
I am not sure if "- x < 5 --> x>5" is the correct !
Since -X < 5 => X > -5.
I think for absolute value problems, we need to do as follows,

|x-4| < 9
1: In order to get rid of absolute value we need to consider + and -ve.
=> X-4 < 9 => X < 9 + 4 => X < 13 ----> (1)
and second,
=> X-4 > -9 => X > -5 ------> (2)
(note that when we take the -ve value we flip the operator)
Combining (1) and (2) we get
-5 < X < 13.
Kindly let me know if I am wrong....
Cheers,
Ravindra.
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 [#permalink] New post 23 Mar 2005, 19:13
brravindra wrote:
ywilfred wrote:
alright, some more as I've promised....

Here's one way to solve absolue inequalities:

|x-4| < 9
We need to solve both positive and negative x

Solving for positive: x-4 < 9 --> x <13
Solving for negative:
-x+4 < 9 --> - x < 5 --> x>5 (dividing by negative number, switch the sign)

So now we have 5<x<13


Hello Ywilfred,
I am not sure if "- x < 5 --> x>5" is the correct !
Since -X < 5 => X > -5.
I think for absolute value problems, we need to do as follows,

|x-4| < 9
1: In order to get rid of absolute value we need to consider + and -ve.
=> X-4 < 9 => X < 9 + 4 => X < 13 ----> (1)
and second,
=> X-4 > -9 => X > -5 ------> (2)
(note that when we take the -ve value we flip the operator)
Combining (1) and (2) we get
-5 < X < 13.
Kindly let me know if I am wrong....
Cheers,
Ravindra.



yes, you're right. Its a typographical error. I'll go edit the post later on. Thanks for bringing it up. :-D
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 [#permalink] New post 25 Mar 2005, 07:28
vprabhala wrote:
can we have some basic sets and venn diagrams principles here?


Hi vprabhala, I've promised to post something regarding basic sets and venn diagram principles here and so here it is, finally.

I've written a document explaining two concepts you can use to solve problems invovlving sets. In this document, I've taken out some examples posted on GMATClub quite some time ago and apply the concepts explained.

Hope this document is useful for members here.

As usual, if you spot any mistakes in my working, please let me know and I'll have them changed. Thanks. :-D

(Note: I've converted my document to a PDF format so it takes up less space)
Attachments

Working with Sets.pdf [74.48 KiB]
Downloaded 1838 times

To download please login or register as a user

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 [#permalink] New post 25 Mar 2005, 09:07
Good job ywilfred! We'll definitely have to include this in the Ebook. :)
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 [#permalink] New post 25 Mar 2005, 09:44
Glad it's of use ! :-D
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 [#permalink] New post 25 Mar 2005, 19:53
You guys are great.
Can someone please provide some tips/hints on tackling mixture problems? And i mean hard core mixture problems with 2 variables, 2 equations etc. Some of us (i can :lol: :lol: ) use pointers on advanced rectangular coordinates as well.
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 [#permalink] New post 26 Mar 2005, 08:09
sure thing folaa3. Give me some time to compile it, and i'll see if i can add in anything about rectangular coordinates too. By the way, what do you mean by advanced pointers ? The ones tested seem pretty fundamental.
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 [#permalink] New post 26 Mar 2005, 20:17
ywilfred wrote:
sure thing folaa3. Give me some time to compile it, and i'll see if i can add in anything about rectangular coordinates too. By the way, what do you mean by advanced pointers ? The ones tested seem pretty fundamental.


Well, if the ones tested are pretty fundamental like you said, then i guess anything you or other members have to offer as far as tips would be great. Thanks :lol:
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Working with Ratios [#permalink] New post 30 Mar 2005, 08:00
Working with Ratios

Ratio questions are very easy to solve if you have mastered the way of thinking.

Basically if you have
a/b=c/d (or a:b=c:d)
then you can immediately derive a variaty of correlated ratios, such as:
a/(a+b)=c/(c+d)
a/(a-b)=c/(c-d)
(a+b)/(a-b)=(c+d)/(c-d)
(a+c)/(b+d)=c/d
(a-c)/(b-d)=c/d
etc

Basically, you can do all kinds of additions and subtractions.

Example:

a/b=3/5 (1)
2a-b=4 (2)
What is a?
From (1) we get a/(2a-b)=3/1, so a=3*4=12
Explanation: a is 3 share, b is 5 share. Two a is 6 share, 2a-b is one share. If one share is 4, then 3 share is 12.

Of course this question can be solved using the more traditional algebra approach:
b=5/3a
substitute in (2)
2a-5/3a=4
1/3a=4
a=12

You can see the two approaches are really the same in nature. However the first approach is very straight forward and does not involve calculation in fractions. Sometimes it can save you lots of time, especially when using this method with word problems such as mixture problems.

Mixture Problems

Example:

A fruit mixture is made up by 25% fruit A and 75% fruit B. Now if the amount of fruit A is doubled, what is their relative share in the new mixture?

A:B=25:75
2A:B=50:75=2:3
The new mixture total quantity is 2A+B
2A:(2A+B)=2:5
B:(2A+B)=3:5
Therefore the new shares are fruit A 40%, fruit B 60%.

Example 2:

In a picnic 60% people ate two hotdogs, 30% people ate one hamburger, and 10% people ate one hotdog. The total number of hotdog and hamburgers consumed is 80. How many hamburgers and hotdogs are consumed?

People:
T:H:O=6:3:1 (1)
Food:
2T+H+O=80
From (1)
T:H:O:(2T+H+O)=6:3:1:16
Therefore
T:(2T+H+O)=3:8=30:80 30 people ate two hotdogs
H:T=3:6=1:2=15:30 15 people ate one hamburger
O:T=1:6=5:30 5 people ate one hotdogs
Total people 50, total hotdogs 65, total hamburgers 15.
Verify, total hotdogs and hamburgers=65+15=80.
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 [#permalink] New post 01 Apr 2005, 11:28
Very useful guys. Don't think anyone can crack the GMAT when you still have problems with no. properties and the like.

i sometimes can't believe the ones i miss but that should be in the past now. cheers
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Divisibility by 7 [#permalink] New post 07 Apr 2005, 10:08
Just learned this today so thought i would add it here -

A simple test to find out if a number is divisible by 7 -

"Double the units and subtract from the tens. Keep going in the chain till you cannot get any further. If the end result is a zero or a multiple of 7 then the entire number is divisible by 7"

e.g - 1365 -> 136-(5*2) -> 126 -> 12-(6*2) -> 0. Hence 1365 is divisible by 7.

e.g - 1367 -> 136 - (2*7) -> 122 -> 12 - (4*2) -> 4. Hence 1367 is not divisible by 7.
Divisibility by 7   [#permalink] 07 Apr 2005, 10:08
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