Hermione Posted October 4, 2005 Author Share Posted October 4, 2005 Hi people! After I got a PM from a TMian with non-math background, that the statistics formulae (with sigma 'n all) are diffucult to use, i thought i would rewrite it... in the mean time i found a post by Econ (one of TM's 'the best' math gurus, i think :)) in which he/she had given a very lucid explanation for finding sd, variance etc... i am giving it as it is here: Say that x1, x2, x3, x4, x5, ...., xn are n draws from a (random) sample. Then: Step 1: Compute the mean, i.e. m =[ Sum xi (i=1,..., n) ] / n Step 2: Compute the squared deviation of each observation from its mean, i.e. For x1 --------> (x1-m)^2 For x2---------> (x2-m)^2 ..... For xn---------> (xn-m)^2 Step 3: The variance is V= [(x1-m)^2 + (x2-m)^2 + .... + (xn-m)^2 ] / n Step 4: The s.d. is s.d. = V^(1/2) Example: Let x1=10, x2= 20 and x3=30 Then: (1) m=20 (3) V = [ (10-20)^2 + 0 + (30-20)^2] / 3 = 200/3 (4) s.d. = (20/3) ^ (1/2) Quote Link to comment Share on other sites More sharing options...

SLACIBangladesh Posted October 4, 2005 Share Posted October 4, 2005 Mixtures first.... When three different ingredients are mixed then the ratio in which they have to be mixed in order to get a final strength of vm is: n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (v2 - vm)(vm - v1) (All these formulae have been taken from a test prep book issued by a private coaching center) ;) How this formula comes? Explanation needed. :rolleyes: :rolleyes: :rolleyes: Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 5, 2005 Author Share Posted October 5, 2005 How this formula comes? Explanation needed. :rolleyes: :rolleyes: :rolleyes: when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing will be: Vm = (v1.n1 + v2.n2) / (n1 + n2) (I assume that you understood this...:D) vm (n1 + n2) = v1 n1 + v2 n2 n1 (vm - v1) = n2 (v2 - vm) so, n1/n2 = (v2 - vm)/(vm - v1) ----> (1) similarly if you mix n2 and n3, then their ratio would be given by n2/n3 = (v3 - vm)/(vm - v2) ----> (2) now assume we mix n1, n2 and n3 of different ingredients of value v1, v2 and v3. the individual ratios (1) and (2) will still be the same. now combine these ratios to get n1:n2:n3 by making the denominators common n1/n2 = (v2 - vm)(v3 - vm)/(vm - v1)(v3 - vm) and n2/n3 = (v3 - vm)(vm - v1)/(vm - v2)(vm - v1) rearrange this and you will get the formula: n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1) Hope this is clear... Quote Link to comment Share on other sites More sharing options...

manwiththemission2005 Posted October 5, 2005 Share Posted October 5, 2005 Correction : n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1) (Posted by Suja) Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 5, 2005 Author Share Posted October 5, 2005 Correction : n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1) (Posted by Suja) yeah right! @ MWM: let me correct it in the derivation to avoid confusion... thanks for pointing it out... :) PS: i cant say that mistake was a typo.. it was a typo in the original formula (post #1), but i made the same mistake again in the derivation by cut pasting the end result...:blush: Quote Link to comment Share on other sites More sharing options...

Ibn Abbas Posted October 9, 2005 Share Posted October 9, 2005 Bump :) Quote Link to comment Share on other sites More sharing options...

kronique Posted October 9, 2005 Share Posted October 9, 2005 hey i jus saw these strings.....guys u r doin n awesome job!!! keep it up....ll try me best to put up sum formulae too......... Quote Link to comment Share on other sites More sharing options...

SagarSHAH Posted October 10, 2005 Share Posted October 10, 2005 HI,,GUYS,,I JUST SAW THIS THREAD TODAY AND WAS SHOCKED BY THE EFFORT U GUYS ARE TAKING FOR URSELF AND OTHERS...AND I ALSO I MUST APPRECIATE SUJA TO START THIS WITHOUT WHICH THIS REPLY IS INCOMPLETE....MY XAM IS ON 20OCT SO I WONT BE ABLE TO PAY MUCH CONTRIBUTION NOW,,BUT SURELY,,AFTER MY XAM I WILL ADD TO THIS...TO HAVE A SMALL CONTRIBUTION A FORMULA WHICH I THINK U SCHOLARS MUCT BE ALREADY KNOWING,,,BUT STILL I THINK SOME OF U MUST NOT HAVE NOTICED,,EVEN I DUNNO IF THIS IS AVAILABLE NYWHERE,,COZ I NOTICED THIS THING WHILE PRACTICING PROBLEMS,,AND KINDDA INVENTED A TINY-MINY FORMULA WHICH SOMETIMES SOLVES A BIG PROB.. TO FIND a% OF a (IF a IS TWO DIGIT NO):: SQUARE THE NUMBER AND DIVIDE BY 100.. 70% OF 70 IS 49...25% OF 25 IS 6.25... IT ALSO GOES FOR 1 DIGIT AND 3,4...NY NUMBER OF DIGITS,,BUT I THINK 2 DIGIT IS FREQUENTLY REQUIRED... ITS SAME AS FIRST CALCULATING 10% BY DIVIDING NNUMBER BY 10 AND THEN MULTIPLYING THAT NUMBER BY (REQUIRED PERCENTAGE /10)..LIKE VEDIC MATHS,,, ITS SAME BUT STILL MY FORMULAE REDUCES ONE STEP ( 5-15 SECS)...ITS UPTO U ,,U USE IT OR NOT,,, I JUST GAVE THIS COZ I DNT HAVEE NY OTHER FORMULAE APART FROM ALREADY POSTED...I THINK ALMOST ALL TOPICS ARE COVERD BY SUJA... HELP OTHER TO HELP U Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 10, 2005 Author Share Posted October 10, 2005 Thanks everone for all those words! 1. The area of a cyclic quad= the sqrt [(s-a) (s-b) (s-c) (s-d)],where a, b, c, and d are the sides of the quadrilateral and s=semiperimeter= a+b+c+d/2. 2. The number of diagonals in a polygon with N sides = [N*(N-3)] / 2 3. The sum of angles in a polygon with N sides = (N-2)*180 degrees 4.The distance from P1(x1, y1, z1) to P2 (x2, y2, z2) is d = sqrt [(x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2] PS: I would really like this thread to keep growing but I am not able to find time to post new formulae with all the class work, college exams, AGRE and on top of all this, applications. So i request those who are preparing for GRE to try and keep this thread active.... -sujatha Quote Link to comment Share on other sites More sharing options...

parag_bhadra Posted October 10, 2005 Share Posted October 10, 2005 I still cant see how the step was reduced. 49% of 49 = 49*49/100 - is what we would do in any case so where is the step reduction or am OO missing something. Suja - Do you have something on Inequalities ? HI,,GUYS,,I JUST SAW THIS THREAD TODAY AND WAS SHOCKED BY THE EFFORT U GUYS ARE TAKING FOR URSELF AND OTHERS...AND I ALSO I MUST APPRECIATE SUJA TO START THIS WITHOUT WHICH THIS REPLY IS INCOMPLETE....MY XAM IS ON 20OCT SO I WONT BE ABLE TO PAY MUCH CONTRIBUTION NOW,,BUT SURELY,,AFTER MY XAM I WILL ADD TO THIS...TO HAVE A SMALL CONTRIBUTION A FORMULA WHICH I THINK U SCHOLARS MUCT BE ALREADY KNOWING,,,BUT STILL I THINK SOME OF U MUST NOT HAVE NOTICED,,EVEN I DUNNO IF THIS IS AVAILABLE NYWHERE,,COZ I NOTICED THIS THING WHILE PRACTICING PROBLEMS,,AND KINDDA INVENTED A TINY-MINY FORMULA WHICH SOMETIMES SOLVES A BIG PROB.. TO FIND a% OF a (IF a IS TWO DIGIT NO):: SQUARE THE NUMBER AND DIVIDE BY 100.. 70% OF 70 IS 49...25% OF 25 IS 6.25... IT ALSO GOES FOR 1 DIGIT AND 3,4...NY NUMBER OF DIGITS,,BUT I THINK 2 DIGIT IS FREQUENTLY REQUIRED... ITS SAME AS FIRST CALCULATING 10% BY DIVIDING NNUMBER BY 10 AND THEN MULTIPLYING THAT NUMBER BY (REQUIRED PERCENTAGE /10)..LIKE VEDIC MATHS,,, ITS SAME BUT STILL MY FORMULAE REDUCES ONE STEP ( 5-15 SECS)...ITS UPTO U ,,U USE IT OR NOT,,, I JUST GAVE THIS COZ I DNT HAVEE NY OTHER FORMULAE APART FROM ALREADY POSTED...I THINK ALMOST ALL TOPICS ARE COVERD BY SUJA... HELP OTHER TO HELP U I was wondering about a probability What is the probability that If I tick answers randomly in a Multiple choice Test (say) with 5 choices in each question and 20 Questions - I will get all the answers right. I can figure out that one. = 1 Correct set / 5^20 sets But what about say "If I need to get (say) 40% (ie 8) answers right" Yours Anxiously Parag Does [(No of Choices - 1*)^(Total No of Questions - No. of Correct Questions)] / ( No of Choices ^ Total No of Questions) Look right ? Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 10, 2005 Author Share Posted October 10, 2005 Suja - Do you have something on Inequalities ? check this: http://tutorial.math.lamar.edu/AllBrowsers/1314/SolveLinearInequalities.asp ----------------------------------------------------------------------------------------------------- Hmm... This is difficult. Ok guys, i first thought I wouldnt say anything. but let me tell this for the first and last time: (I know this is a cliche but still: NO OFFENCE MEANT) RULES FOR CONTRIBUTING TO THIS THREAD: 1. No more thank-you posts... it just increases the length of the thread and makes it difficult for the users to go through this as last minute revision. 2. No doubts 'whatsoever' other than those about the formulae posted here. Do not post problems to be solved in this thread. make them into new threads. This thread was meant for formulae and let it remain so... 3. When you post a formula make an effort to make it distinct among the other lines.... I mean, dont make others search for the formula among the text. also, try to see if a formula has already been posted or not before posting it. sorry, if i sound like a nag... the point is I am one. :) -suja Quote Link to comment Share on other sites More sharing options...

parag_bhadra Posted October 10, 2005 Share Posted October 10, 2005 Point taken !:blush: Here is a Contribution to redeem myself.:notworthy Some useful things to remember for combinations If nCr = M when M is known If n is also known then r can have 2 values (r1+r2 = n) If nCr = M when M is known If r is also known then n can have 1 value Let M = 495, If n = 12 then r = 4, 8 (4+8 = 12) If r = 4 or if r = 8 then n = 12 If M is known to be a multiple Combination value (all number can have the value nC1 but if r is not one then there are only specific numbers which are possible) then it can have few options which are M = MC1 = MC(M-1) & a specific nCr, nC(n-r) In above eg apart from 12C4 = 12C8 = 495 also remember to consider 495C1=495C494 = 495 If nCr = M & M = r+1 then n = r+1 if nC12 = 13 then n = 13 ******************** End of Formulae ******************* 2 reasons for putting that Probability thing here 1. Was only vaguely recalling the formula - so wanted to check up fast. 2. Was hoping that the stalwarts:wizard: here will quickly establish its veracity. My original comment about 49% of 49 still remains unanswered - am I still entitled to that ? Sorry if I sound stubbborn ...... I am:boxing: ! Sincere thanks for the Inequality thread. ***************************************************** check this: http://tutorial.math.lamar.edu/AllBrowsers/1314/SolveLinearInequalities.asp ----------------------------------------------------------------------------------------------------- Hmm... This is difficult. Ok guys, i first thought I wouldnt say anything. but let me tell this for the first and last time: (I know this is a cliche but still: NO OFFENCE MEANT) RULES FOR CONTRIBUTING TO THIS THREAD: 1. No more thank-you posts... it just increases the length of the thread and makes it difficult for the users to go through this as last minute revision. 2. No doubts 'whatsoever' other than those about the formulae posted here. Do not post problems to be solved in this thread. make them into new threads. This thread was meant for formulae and let it remain so... 3. When you post a formula make an effort to make it distinct among the other lines.... I mean, dont make others search for the formula among the text. also, try to see if a formula has already been posted or not before posting it. sorry, if i sound like a nag... the point is I am one. :) -suja Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 11, 2005 Author Share Posted October 11, 2005 @parag... sent you a PM @everyone...i generally do not say things that would hurt others' ego (this should explain why i didnt reply to the comment by parag on sagarSHAH's post). it is out of sincere interest that i kinda moderate this thread. i stand by what i said. doubts regarding ANYTHING other than previously posted formulae WILL NOT be answered HERE. moreover, starting a new thread is the fastest way to get answers to your doubts... A REQUEST: I do not want any more discussions abt this to continue in this thread... if it does, i will not involve myself in that. i am off guys... :( Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 12, 2005 Author Share Posted October 12, 2005 check this link: could be useful... :hmm: http://mathforum.org/dr.math/faq/faq.divisibleto50.html -sujatha Quote Link to comment Share on other sites More sharing options...

SagarSHAH Posted October 14, 2005 Share Posted October 14, 2005 Harmonic Progression A series of non-zero numbers is said to be harmonic progression (abbreviated H.P.) if the series obtained by taking reciprocals of the corresponding terms of the given series is an arithmetic progression. For example, the series 1 +1/4 +1/7 +1/10 +..... is an H.P. since the series obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10 +... is an A.P. A general H.P. is 1/a + 1/(a + d) + 1(a + 2d) + ... nth term of an H.P. = 1/[a +(n -1)d] Three numbers a, b, c are in H.P. iff 1/a, 1/b, 1/c are in A.P. i.e. iff 1/a + 1/c = 2/b i.e. iff b= 2ac/(a + c) Thus the H.M. between a and b is H = 2ac/(a + c) ----- If A, G, H are arithmetic, geometric and harmonic means between two distinct, positive real numbers a and b, THEN G² = AH i.e. A, G, H are in G.P. A, G, H are in descending order of magnitude i.e. A > G > H. Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 17, 2005 Author Share Posted October 17, 2005 http://library.thinkquest.org/20991/gather/formula/index.html http://www.ilovemaths.com/3.htm http://mathforum.org/dr.math/faq/ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm http://www.math.hmc.edu/calculus/tutorials/algebrareview/ http://www.math.com/tables/index.html http://www.mathpower.com/tutorial.htm hope this helps! -suja Quote Link to comment Share on other sites More sharing options...

seagull Posted October 17, 2005 Share Posted October 17, 2005 thanks a l---o---t :) Quote Link to comment Share on other sites More sharing options...

manwiththemission2005 Posted October 18, 2005 Share Posted October 18, 2005 Important Things to Note: When we take square root of a positive number, only +ve values are considered. For eg: root(16) is only +4 and not -4 When we are asked to find x.. for eg: x^2 = 16.. then x= +/-4 When you have square root on 1 side or on both sides in a quantitative comparison q, square the terms on both the sides. That will make it easier to solve the problem. 0 is an even number. 1 Quote Link to comment Share on other sites More sharing options...

bscout Posted October 18, 2005 Share Posted October 18, 2005 Pick's theorem (Taken from this site) Pick's theorem provides an elegant formula for the area of a simple lattice polygon: a lattice polygon whose boundary consists of a sequence of connected nonintersecting straight-line segments. http://www.mcs.drexel.edu/%7Ecrorres/Archimedes/Stomachion/Pick.gif The formula is Area = I +B/2 – 1where I = number of interior lattice points (http://www.mcs.drexel.edu/%7Ecrorres/Archimedes/Stomachion/BlueDot.gif) and B = number of boundary lattice points (http://www.mcs.drexel.edu/%7Ecrorres/Archimedes/Stomachion/RedDot.gif) For example, the area of the simple lattice polygon in the figure is 31 + 15 /2 – 1 = 37.5. http://www.mcs.drexel.edu/%7Ecrorres/Archimedes/Stomachion/StomaGrid.gif The interior and boundary lattice points of the fourteen pieces of the Stomachion are indicated on the second figure. Using Pick's theorem the areas of the fourteen pieces can be determined as in the above example; e.g., the blue piece in the upper right-hand corner has area 18 + 14 /2 – 1 = 24 Quote Link to comment Share on other sites More sharing options...

manwiththemission2005 Posted October 18, 2005 Share Posted October 18, 2005 A remainder rule to remember: If a product of 2 integers, x and y is divided by an integer n, then the remainder that you get will be the product of the remainders when x is divided by n and y is divided by n. R[] ---> remainder function R[(1046*1047*1048)/11] = R[1046/11]*R[1047/11]*[1048/11] = 1*2*3 = 6 Note: Sometimes the product of the remainders will be greater than the original divisor. In this case you'll have to repeat the process. [postED BY ARJMEN IN GMAT MATH FORUM] 1 Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 18, 2005 Author Share Posted October 18, 2005 Good job Karthik and bscout! :tup: 1 Quote Link to comment Share on other sites More sharing options...

psychodementia Posted October 21, 2005 Share Posted October 21, 2005 Are such arcane and esoteric formulae ever asked in the GRE? IMO this is the resort for the formula dependent testers, who wish every problem in the GRE can be solved by merely using a formula. Unfortunately there are no cookie cutter solutions to get that elusive 800 in quant. An application of concepts and mind would yield the answer without having the delve into the depths of such formulae. Arun Quote Link to comment Share on other sites More sharing options...

Hermione Posted October 21, 2005 Author Share Posted October 21, 2005 Hi arun! I partly agree with what you say! Nor do i think one can ace GRE/GMAT just by memorizing these formulae. I am quoting myself: Dont let these formulae intimidate you greenhorn. They are all 'just in case' stuff. I personally believe that you dont need to know any formula to do GRE. All this is to save time for cross checking and DI problems. So dont worry if you cannot read up all these formulae (especially if you dont have much time). This thread is meant to be an archive of formulae, either directly or even remotely helpful to solve problems that might come up in GRE. This is for everyone....I wanted to tell something else also here, which i think is very important. While all these formulae are meant to help you do problems quickly, never ever substitute values blindly into a formula. You can go terribly wrong if you dont plug-in data properly. And you will never know where you went wrong coz whenever we get an answer wrong, we all tend to check up our calculations rather than substitution of data. So i take this oppurtunity to just tell all you guys to use these formulae carefully. Please dont mistake me if i sound patronizing. I am telling this out of the responsibility i feel.. If any of you guys go wrong coz you used a formula that was put up in this thread, i would really feel guilty. so be careful. :)so it is true that one can do well even without knowing any of these formulae and moreover formulae can be terribly misleading if you plug in wrong data. but as i see it, the time spent in studying all these forumulae is justified if you can solve a problem in your test faster by using one of these formulae. afterall, dont we study 3500 words inorder to know the 30 odd words that are gonna come in the test? If something 'might' help, why not learn it too? hope you get my idea! Quote Link to comment Share on other sites More sharing options...

priyankagre2005 Posted November 15, 2005 Share Posted November 15, 2005 Addition/Subtraction Property for Inequalities If a If a Multiplication/Division Properties for Inequalities · when multiplying/dividing by a positive value If a If a · when multiplying/dividing by a negative value If a bc If a b/c Natural (or Counting) Numbers : N = {1, 2, 3, 4, 5, ...} Whole Numbers : {0, 1, 2, 3, 4, 5, ...} Integers : Z = {..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...} Real Numbers : R = {x | x corresponds to point on the number line} Quote Link to comment Share on other sites More sharing options...

priyankagre2005 Posted November 15, 2005 Share Posted November 15, 2005 If a cube of sides n*n*n is painted and then divided into 1*1*1 size cubes then number cubes with NO face painted is given by (n-2)^2 1 Quote Link to comment Share on other sites More sharing options...

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