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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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



Director
Joined: 21 Sep 2004
Posts: 607

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



Director
Joined: 21 Sep 2004
Posts: 607

can we have some basic sets and venn diagrams principles here?



Manager
Joined: 28 Aug 2004
Posts: 205

If x > y, then x  y divides x^n  y^n.



VP
Joined: 18 Nov 2004
Posts: 1433

Honghu,
I think ur first example on inequalities is wrong, for x4<9...solution shud be 5<x<13



VP
Joined: 18 Nov 2004
Posts: 1433

Dan wrote: If x > y, then x  y divides x^n  y^n.
interesting....can someone provide a proof for this...thx



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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 ) otherwise i can't show some of the workings here.



Director
Joined: 21 Sep 2004
Posts: 607

oh. I was wondering that we were able to post attachments for a while.. hmmm looks like its gone..



Senior Manager
Joined: 19 Nov 2004
Posts: 281
Location: Germany

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.



Intern
Joined: 04 Oct 2004
Posts: 10

ywilfred wrote: alright, some more as I've promised....
Here's one way to solve absolue inequalities:
x4 < 9 We need to solve both positive and negative x
Solving for positive: x4 < 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,
x4 < 9
1: In order to get rid of absolute value we need to consider + and ve.
=> X4 < 9 => X < 9 + 4 => X < 13 > (1)
and second,
=> X4 > 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.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

brravindra wrote: ywilfred wrote: alright, some more as I've promised....
Here's one way to solve absolue inequalities:
x4 < 9 We need to solve both positive and negative x
Solving for positive: x4 < 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, x4 < 9 1: In order to get rid of absolute value we need to consider + and ve. => X4 < 9 => X < 9 + 4 => X < 13 > (1) and second, => X4 > 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.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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



SVP
Joined: 03 Jan 2005
Posts: 2233

Good job ywilfred! We'll definitely have to include this in the Ebook.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

Glad it's of use !



Director
Joined: 27 Dec 2004
Posts: 897

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 ) use pointers on advanced rectangular coordinates as well.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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.



Director
Joined: 27 Dec 2004
Posts: 897

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



SVP
Joined: 03 Jan 2005
Posts: 2233

Working with Ratios [#permalink]
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30 Mar 2005, 09:00
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/(ab)=c/(cd)
(a+b)/(ab)=(c+d)/(cd)
(a+c)/(b+d)=c/d
(ac)/(bd)=c/d
etc
Basically, you can do all kinds of additions and subtractions.
Example:
a/b=3/5 (1)
2ab=4 (2)
What is a?
From (1) we get a/(2ab)=3/1, so a=3*4=12
Explanation: a is 3 share, b is 5 share. Two a is 6 share, 2ab 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)
2a5/3a=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.



Manager
Joined: 28 Jan 2004
Posts: 198
Location: Ghana

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
_________________
It's not over until it's OVER!



Intern
Joined: 17 Mar 2005
Posts: 14
Location: Philadelphia

Divisibility by 7 [#permalink]
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07 Apr 2005, 11:08
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.







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