bscout Posted March 4, 2006 Share Posted March 4, 2006 I don't know if you already read all the thread but as Suja said before: 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. ../images/smilies/smile.gif 1 Quote Link to comment Share on other sites More sharing options...

yogesh_kkk Posted March 14, 2006 Share Posted March 14, 2006 Awsome post i just loved it anyways best of luck for all. Quote Link to comment Share on other sites More sharing options...

bscout Posted March 22, 2006 Share Posted March 22, 2006 Posted by TwinnSplitter in this thread. If there are three sets A, B, and C, then P(AuBuC) = P(A) + P(B) + P© – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) Number of people in exactly one set = P(A) + P(B) + P© – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC) Number of people in exactly two of the sets = P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC) Number of people in exactly three of the sets = P(AnBnC) Number of people in two or more sets = P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC) I found this formulae very useful. Quote Link to comment Share on other sites More sharing options...

varun_gre2005 Posted April 28, 2006 Share Posted April 28, 2006 i might sound a bit silly but if one those all of these formulae is that enough ? :) Quote Link to comment Share on other sites More sharing options...

eye Posted April 28, 2006 Share Posted April 28, 2006 nice post, thanks. Quote Link to comment Share on other sites More sharing options...

rufus16 Posted July 14, 2006 Share Posted July 14, 2006 Friends, lets not lose this post.. this should be of great use to us ... Thanks to Sujatha, who started this thread and helped me in finding this thread . Quote Link to comment Share on other sites More sharing options...

coolgeek Posted July 14, 2006 Share Posted July 14, 2006 i have attached a pdf which has a very good info about the properties if SD (very useful!) [ATTACH]2585[/ATTACH] Quote Link to comment Share on other sites More sharing options...

rufus16 Posted July 14, 2006 Share Posted July 14, 2006 thanks coolgeek :) the pdf is exhaustive Quote Link to comment Share on other sites More sharing options...

Mystery Man Posted July 14, 2006 Share Posted July 14, 2006 Amke this a stickie Quote Link to comment Share on other sites More sharing options...

mishu007 Posted July 14, 2006 Share Posted July 14, 2006 thanks man a nice post............ Quote Link to comment Share on other sites More sharing options...

bscout Posted November 4, 2006 Share Posted November 4, 2006 *Bumped* Many TManiacs, will find this thread very useful. Quote Link to comment Share on other sites More sharing options...

jinics Posted November 5, 2006 Share Posted November 5, 2006 POWERS AND INDICES To find the unit digit of p^n If there is an odd no. in the unit place of p eg 741,843 etc Multiply the unit digit by itself until u get 1. Example: If u need to find the unit digit of (743)^38: Multiply 3 four times to get 81. (743)^38=(743)^36 X (743)^2 36 is a multiple of 4, and 3 when multiplied 4 times gives 1 in the unit digit.Therefore, when multiplied 9 x 4 times, it will still give 1 in the unit digit. the unit digit of (743)^38, hence will be 1 x 9 =9 In short (…..1)^n =(…..1) (….3)^4n=(…..1) (….7)^4n=(…..1) (….9)^2n=(….1) If the unit digit of p is even and u need to find the unit digit of (p)^n Multiply the unit digit of p by itself until a 6 is in the unit place (…2)^4n=(….6) (….4)^2n=(….6) (….6)^n=(….6) (….8)^4n=(….6) For numbers ending with 1,5,6, after any times of multiplication, you get only 1, 5, 6 respectively. Can somebody please elaborate this method? Quote Link to comment Share on other sites More sharing options...

rezbipul Posted February 13, 2007 Share Posted February 13, 2007 Restricted – Combinations (a) Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = n-pCr-p. (b) Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = n-pCr Example: In how many ways can a cricket-eleven be chosen out of 15 players? if (i) A particular player is always chosen, (ii) A particular is never chosen. Ans: (i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players. =. Required number of ways = 14C10 = 14C4 = 14!/4!x19! = 1365 (ii) A particular players is never chosen, it means that 11 players are selected out of 14 players. => Required number of ways = 14C11 = 14!/11!x3! = 364 (iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:- 2n-1 Proof: Number of ways of selecting one thing, out of n-things = nC1 Number of selecting two things, out of n-things =nC2 Number of ways of selecting three things, out of n-things =nC3 Number of ways of selecting ‘n’ things out of ‘n’ things = nCn =>Total number of ways of selecting one or more things out of n different things = nC1 + nC2 + nC3 + ------------- + nCn = (nC0 + nC1 + -----------------nCn) - nC0 = 2n – 1 [ nC0=1] Example: John has 8 friends. In how many ways can he invite one or more of them to dinner? Ans. John can select one or more than one of his 8 friends. => Required number of ways = 28 – 1= 255. (iv) Number of ways of selecting zero or more things from ‘n’ identical things is given by :- n+1 Example: In how many ways, can zero or more letters be selected form the letters AAAAA? Ans. Number of ways of : Selecting zero 'A's = 1 Selecting one 'A's = 1 Selecting two 'A's = 1 Selecting three 'A's = 1 Selecting four 'A's = 1 Selecting five 'A's = 1 => Required number of ways = 6 [5+1] (V) Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by :- (p+1) (q+1) (r+1)2n – 1 Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen. Ans: Number of ways of selecting apples = (3+1) = 4 ways. Number of ways of selecting bananas = (4+1) = 5 ways. Number of ways of selecting mangoes = (5+1) = 6 ways. Total number of ways of selecting fruits = 4 x 5 x 6 But this includes, when no fruits i.e. zero fruits is selected => Number of ways of selecting at least one fruit = (4x5x6) -1 = 119 Note :- There was no fruit of a different type, hence here n=o => 2n = 20=1 (VI) Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’. Example: In how many ways 5 balls can be selected from ‘12’ identical red balls? Ans. The balls are identical, total number of ways of selecting 5 balls = 1. Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5? Ans. Here n = 5 [Number of digits] And r = 4 [ Number of places to be filled-up] Required number is 5P4 = 5!/1! = 5 x 4 x 3 x 2 x 1 1 Quote Link to comment Share on other sites More sharing options...

rezbipul Posted March 11, 2007 Share Posted March 11, 2007 Here is venn diagram related thread: http://www.www.urch.com/forums/gre-math/56879-formula.html Quote Link to comment Share on other sites More sharing options...

rezbipul Posted April 6, 2007 Share Posted April 6, 2007 1) A fraction in lowest terms with a prime denominator other than 2 or 5 (i.e. coprime to 10) always produces a recurring decimal. 2) If the repeating decimal is between 0.1 and 1, and the repeating block is n digits long occurring right at the decimal point, then the fraction (not necessarily reduced) will be the n-digit block over n digits of 9. For example, 0.444444... = 4/9 since the repeating block is 4 (a 1-digit block), 0.565656... = 56/99 since the repeating block is 56 (a 2-digit block), 0.789789... = 789/999 since the repeating block is 789 (a 3-digit block), etc. If the repeating decimal is between 0 and 0.1, and the repeating n-digit block is preceded only by k digits of 0 (all of which are to the right of the decimal point), then the fraction (not necessarily reduced) will be the n-digit block over the integer consists of n digits of 9 followed by k digits of 0. For example, 0.000444... = 4/9000 since the repeating block is 4 and this block is preceded by 3 zeros, 0.005656... = 56/9900 since the repeating block is 56 and it is preceded by 2 zeros, 0.0789789... = 789/9990 since the repeating block is 789 and it is preceded by 1 zero. For any repeating decimal not prescribed above, it can be written as a sum of a terminating decimal and a repeating decimal of one of the two above types. For example, 1.23444... = 1.23 + 0.00444... = 123/100 + 4/900 = 1107/900 + 4/900 = 1111/900 0.3789789 ... = 0.3 + 0.0789789... = 3/10 + 789/9990 = 2997/9990 + 789/9990 = 3786/9990 = 631/1665 Quote Link to comment Share on other sites More sharing options...

rezbipul Posted April 15, 2007 Share Posted April 15, 2007 When number is of the form a^n - b^n -> it is always divisible by a-b. -> it is divisible by a+b if n is even When number is of the form a^n + b^n -> it is divisible by a+b if n is odd Quote Link to comment Share on other sites More sharing options...

rezbipul Posted April 21, 2007 Share Posted April 21, 2007 In a more general case of a committee of n people from a group of mcouples with this restriction (no couple in a committee), the number of ways to select the committee would be: m! ----------- * 2^n (m - n)!*n! Quote Link to comment Share on other sites More sharing options...

KBTA Posted June 14, 2007 Share Posted June 14, 2007 Thanks a lot Suja. I hope this thread goes ROCKING jus like the PROBABILITY thread which I started b4. :) Hi manwiththemission,Could you help me with your probability thread through mailing me. My mail adress: shilajatu@yahoo.comWith Regards Jakir Quote Link to comment Share on other sites More sharing options...

bscout Posted June 15, 2007 Share Posted June 15, 2007 Hi manwiththemission,Could you help me with your probability thread through mailing me. My mail adress: shilajatu@yahoo.comWith Regards Jakir MWM wrote his last post a long time ago. Here you have the thread he created. Quote Link to comment Share on other sites More sharing options...

rezbipul Posted July 17, 2007 Share Posted July 17, 2007 x/Lnx > y/lny if x > y > 2.71 which is e x/Lnx x > y > 0 x/Lnx x > y > 1 Quote Link to comment Share on other sites More sharing options...

mewidyu Posted August 10, 2008 Share Posted August 10, 2008 so bad that this gr8 thread is dying.. ok as i cant add any more formula here... i will repost all formulas in here only.... in one post... so that one can copy it easily.... Quote Link to comment Share on other sites More sharing options...

mewidyu Posted August 10, 2008 Share Posted August 10, 2008 Formulae and Shortcuts - making life easier!!! Mixtures first.... 1. 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) you can use this to find the final price of say two types of rice being mixed or final strength of acids of different concentration being mixed etc.... the ratio in which they have to be mixed in order to get a mean value of vm can be given as: n1/n2 = (v2 - vm)/(vm - v1) 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) 2. If from a vessel containing M units of mixtures of A & B, x units of the mixture is taken out & replaced by an equal amount of B only .And If this process of taking out & replacement by B is repeated n times , then after n operations, Amount of A left/ Amount of A originally present = (1-x/M)^n 3. If the vessel contains M units of A only and from this x units of A is taken out and replaced by x units of B. if this process is repeated n times, then: Amount of A left = M [(1 - x/M)^n] Ths formula can be applied to problem involving dilution of milk with water, etc... (All these formulae have been taken from a test prep book issued by a private coaching center) file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image001.gif More formulae: PROGRESSION: Sum of first n natural numbers: 1 +2 +3 + .... + n = [n(n+1)]/2 Sum of first n odd numbers: 1 + 3 + 5 + .... upto n terms = n^2 Sum of first n even numbers: 2 + 4 + 6 + ... upto n terms = n(n+1) ARITHMETIC PROGRESSION nth term of an Arithmetic progression = a + (n-1)d Sum of n terms in an AP = s = n/2 [2a + (n-1)d] where, a is the first term and d is the common differnce. If a, b and c are any three consequtive terms in an AP, then 2b = a + c GEOMETRIC PROGRESSION nth term of a GP is = a[r^(n-1)] sum of n terms of a GP: s = a [(r^n - 1)/(r-1)] if r > 1 s = a [(1 - r^n)/(r-1)] if r sum of an infinite number of terms of a GP is s(approx.) = a/ (1-r) if r If a, b and c are any three consequtive terms in a GP, then b^2 = a*c PS: Everybody!!! please contribute or this thread will never grow. Problems on trains a km/hr = (a* (5/18)) m/s a m/s = (a* (18/5)) km/hr Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres. Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l+b) metres. Suppose 2 trains or 2 bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relative speed = (u-v) m/s Suppose 2 trains or 2 bodies are moving in the opposite direction at u m/s and v m/s, where u>v, then their relative speed = (u+v) m/s If 2 trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then the time taken by the trains to cross each other = (a+b)/(u+v) sec If 2 trains of length a metres and b metres are moving in same directions at u m/s and v m/s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec If 2 trains(or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then (A's speed)file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gifB's speed) = (root(B):root(A)) This one is a nice formula for finding the number of unique divisors for any number and also the sum of those divisors.... such questions are there in powerprep and so you might also get it in your real GRE. If N is a number such that N = (a^m) (b^n) (c^p).... where, a, b, c, ... are prime numbers, then the number of divisors of N, including N itself is equal to: (m+1) (n+1) (p+1).... and the sum of the divisors of N is given by: S = [(a^m+1) - 1]/[a - 1] * [(b^n+1) - 1]/[b - 1] * [(c^p+1) - 1]/[c- 1]..... Example: for say N = 90, on factorizing you get 90 = 3*3*5*2= (3^2)*(5^1)*(2^1) then the number of divisors of 90 are (2+1)(1+1)(1+1) = 12 the 12 divisors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 And the sum of the divisors would be [(3^3) - 1]/[3 - 1] * [(5^2) - 1]/[5 - 1] * [(2^2) - 1]/[2 - 1] = (26/2) (24/4) (3/1) = 234 Though this method looks more complicated than listing the factors and adding them, once you get used to this formula, it saves lot of time.. Let the coordinates of P1 be (x1,y1) and of P2 be (x2,y2) - The distance from P1 to P2 is: d = sqrt[(x1-x2)2+ (y1-y2)2] - The coordinates of the point dividing the line segment P1P2 in the ratio r/s are: ([r x2+s x1]/[r+s], [r y2+s y1]/[r+s]) - As a special case, when r = s, the midpoint of the line segment has coordinates: ([x2+x1]/2,[y2+y1]/2) - The slope m of a non-vertical line passing through the points P1 and P2: slope = m = (y2 -y1)/(x2 -x1) Two (non-vertical) lines are parallel if their slopes are equal. Two (non-vertical) lines are perpendicular if the product of their slopes = -1. Slope of a perpendicular line is the negative reciprocal of the slope of the given line. file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image003.gifHi The population of a town decreases by 'x%' during the first year, decreases by 'y%' during the second year and again decreases by 'z%' during the third year. If the present population of the town is 'P', then the population of the town three years ago was:: P*100*100*100 ----------------------- (100-x)(100-y)(100-z). The population of a town is 'P'.It decreases by 'x%' during the first year, decreases by 'y%' during the second year and again deceases by 'z%' during the third year. The population after 3 years will be: P*(100-x)(100-y)(100-z) -------------------------- 100*100*100. If 'X' litres of oil was poured into a tank and it was still 'x%' empty, then the quantity of oil that must be poured into the tank in order to fill it to the brim is: X*x ------- litres. 100 - x If 'X' liters of oil was poured into a tank and it was still 'x%' empty, then the capacity of the tank is: X*100 ---------- litres. 100 - x A candidate scoring x% in an examination fails by "a" marks, while another candidate who scores y% marks gets "b" marks more than the minimum required pass marks. Then the maximum marks for that exam = 100(a+b) ---------- y-x . The pass marks in an examination is x%. If a candidate who secures y marks fails by z marks, then the maximum marks is given by 100(y+z) ----------- x. Permutations And Combinations 1. If one operation can be performed in m ways and another operation in n ways, then the two operations in succesion can be done in m*n ways 2. The linear permutation of n distinct objects (that is, the number of ways in which these n objects can be arranged is n! and the circular permutation of n distinct objects is (n-1)! But if the clockwise and anticlockwise directions are indistinguishable then the circular permutations of n different things taken at a time is (n-1)!/2 3. But out of these n objects, if there are n1 objects of a certain type, n2 of another type and n3 of another, and so on, Then the number of arrangements (linear permutations) possible is n!/n1!n2!...nz! 4. The total number of ways of arranging r things from n things is given by nPr = n!/(n-r)! 5. The number of ways to select r things out of n things is given by nCr = n!/(r!*(n-r)!) 6. nPr = r! * nCr Numbers And Percentages COUNTING SUM OF FIRST “n” NATURAL NUMBERS = n(n+1)/2 Sum of first “n” ODD integers = n*n Sum of first “n” EVEN integers = n(n+1) Sum of the squares of the first n integers = n(n+1)(2n+1)/6 Sum of the cubes of first n integers =(n(n+1)/2)^2 IF n is even, then No. of odd no.s from 1 to n is n/2 No. of even no.s from 1 to n is n/2 If n is odd then, No. of odd no.s from 1 to n is (n+1)/2 No. of even no.s from 1 to n is (n-1)/2 POWERS AND INDICES To find the unit digit of p^n If there is an odd no. in the unit place of p eg 741,843 etc Multiply the unit digit by itself until u get 1. Example: If u need to find the unit digit of (743)^38: Multiply 3 four times to get 81. (743)^38=(743)^36 X (743)^2 36 is a multiple of 4, and 3 when multiplied 4 times gives 1 in the unit digit.Therefore, when multiplied 9 x 4 times, it will still give 1 in the unit digit. the unit digit of (743)^38, hence will be 1 x 9 =9 In short (…..1)^n =(…..1) (….3)^4n=(…..1) (….7)^4n=(…..1) (….9)^2n=(….1) If the unit digit of p is even and u need to find the unit digit of (p)^n Multiply the unit digit of p by itself until a 6 is in the unit place (…2)^4n=(….6) (….4)^2n=(….6) (….6)^n=(….6) (….8)^4n=(….6) For numbers ending with 1,5,6, after any times of multiplication, you get only 1, 5, 6 respectively. Number of numbers divisible by a certain integer: How many numbers upto 100 are divisible by 6? Soln: Divide 100 by 6, the resulting quotient is the required answer Here, 100/6 = 16x6+4 16 is the quotient and 6 is the remainder. Therefore, there are 16 numbers within 100 which are divisible by 6. PERCENTAGES If the value of a number is first increased by x% and later decreased by x%, the net change is always A DECREASE= (x^2)/100 if the value of a number is first increased by x% and then decreased by y%, then there is (x-y-(xy/100))% increase if positive , and decrease if negative If the order of increase or decrease is changed, THE RESULT IS UNAFFECTED If the value is increased successively by x% and y%,then final increase is given by x+y+(xy/100) % Suppose you have a name with n letters, and there are k1 of one letter, k2 of another letter, and so on, up to kz. For example, in ELLEN, n = 5, k1 = 2 [two E's], k2 = 2 [two L's], k3 = 1 [one N]). Then the number of rearrangements is n!/k1!k2!...kz! Taking from http://mathforum.org/library/drmath/view/56109.html. This formula was provided by mauveÄster in another thread. 1.Area of a triangle with base b and height h = (1/2)*b*h 2. The area of an equilateral triangle with side a is [sqrt(3)/4]*a^2 3. The area of any triangle given the length of its 3 sides a, b and c:is sqrt[s(s-a)(s-b)(s-c)] where s= (a+b+c)/2 Time And Work TIME AND WORK If A can do a piece of work in x days, then A’s one day’s work=1/x If the ratio of time taken by A and B in doing a work is x:y, then, ratio of work done is 1/x :1/y=y:x. And the ratio in which the wages is to be distributed is y:x If A can do a work in x days and B can do the same work in y days, then A and B can together do the work in (xy)/(x+y) days If “a” men or “b” women can do a piece of work in x days, then “m” men and “n” women can together finish the work in (abx)/(an+bm) days If A is x times efficient than B, and working together, they finish the work in y days, then Time taken by A=y(x+1)/(x), Time taken by B=y(x+1) If A and B can finish a work in “x” and “ax” days respectively, that is if A is “a” times efficient than B, then working together, they can finish the work in (ax)/(a+1) days If A and B working together can complete a work in x days, whereas B working alone can do the same work in y days, ten, A alone will complete the work in (xy)/(y-x) days. Pipe A can fill a tank in x hrs and B can empty a tank in y hrs.If both pipes are opened together, the tank will be filled in (xy)/(y-x) hrs A pipe can fill a cistern in x hrs but due to leakage in the bottom, it is filled in y hrs, then the time taken by the leak to empty the cistern is (xy)/(y-x) hrs Area, Surface area and Volume - 1 Most of you will know all the formulae given below. This is just to give a comprehensive list (all these can be found in Math tutorials, resources, help, resources and Math Worksheets ) Rectangle: file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image004.gif Area = lwPerimeter = 2l + 2wParallelogram: file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image005.gif Area = bhTriangle file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image006.gif Area = 1/2 of the base X the height = 1/2 bhPerimeter = a + b + cTrapezoid file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image007.gif file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image008.gif Perimeter = P = a + b1 + b2 + c Circle: file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image009.gifThe distance around the circle is a circumference. The distance across the circle is the diameter (d). The radius ® is the distance from the center to a point on the circle. (Pi = 3.14) d = 2r c = pd = 2 pr A = pr2 (p=3.14) Area, Surface area and Volume -2 Rectangular Solid file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image010.gif Volume = lwh Surface = 2lw + 2lh + 2whPrisms file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image011.gif Volume = Base area X Height Surface = 2b + Ph (b is the area of the base P is the perimeter of the base)Cylinder file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image012.gif Volume = pr2 h Surface = 2prh Area, Surface area and Volume - 3 Pyramid file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image013.gif V = 1/3 bh b is the area of the base Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. The areas of the triangular faces will have different formulas for different shaped bases. Cones file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image014.gif Volume = 1/3 pr2 x height = 1/3 pr2h Surface = pr2 + prs = pr2 + pr file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image015.gif Sphere file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image016.gif Volume = 4/3 pr^3 Surface area = 4pr^2 Distance of a Point from a Line Distance of a Point from a Line The perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by: d =| ax1 +by1 +c|/[file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image017.gif(a² +b²)] Simple And Compound Interest 1. Simple Interest = PNR/100 where, P --> Principal amount N --> time in years R --> rate of interest for one year 2. Compound interest (abbreviated C.I.) = A -P = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image018.gif where A is the final amount, P is the principal, r is the rate of interest compounded yearly and n is the number of years 3. When the interest rates for the successive fixed periods are r1 %, r2 %, r3 %, ..., then the final amount A is given by A = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image019.gif 4. S.I. (simple interest) and C.I. are equal for the first year (or the first term of the interest period) on the same sum and at the same rate. 5. C.I. of 2nd year (or the second term of the interest period) is more than the C.I. of Ist year (or the first term of the interest period), and C.I. of 2nd year -C.I. of Ist year = S.I. on the interest of the first year. Mean. (i) Mean (for ungrouped data) = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image020.gifwhere x1, x2, x3, ..., xn are the observations and n is the total no. of observations. (ii) Mean (for grouped data) = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image021.gif, where x1, x2, x3, ..., xn are different variates with frequencies f1, f2, f3, ..., fn respectively. (iii) Mean for continuous distribution. Let there be n continuous classes, yi be the class mark and fi be the frequency of the ith class, then mean = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image022.gif(Direct method) Let A be the assumed mean, then mean = A +file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image023.gif, where di = yi -A (Short cut method) If the classes are of equal size, say c, then mean = A +c x file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image024.gif, where ui = file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image025.gif(Step deviation method) Median. (i) Median is the central value (or middle observation) of a statistical data if it is arranged in ascending or descending order. (ii) Let n be the total number of observations, then Median =file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image026.gif Quartiles (i) Lower Quartile =file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image027.gif (ii) Upper Quartile =file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image028.gif (iii) Inter quartile-range = upper quartile -lower quartile Mode. (i) Mode (or modal value) of a statistical data is the variate which has the maximum frequency. (ii) The class with maximum frequency is called the modal-class. Analytical Geometry In A Nutshell LINES - BASICS: 1. The equation of X axis: y =0 2. The equation of Y axis: x = 0 3. Equation of straight line parallel to X axis: y =a, where a is any constant 4. Equation of straight line parallel to Y axis: x =a, where a is any constant 5. Equation of a straight line through a given point (x1, y1) and having a given slope m: y -y1 = m (x - x1) 6. Equation of a straight line through a given point (0, 0) and having a given slope m: y = m x 7. Equation of a straight line with a slope m and y-intercept c is: y = mx + c 8. Equation of a straight line passing through two points (x1, y1) and (x2, y2) is: (y -y1)/(y2 - y1) = (x -x1)/(x2 -x1) 9. Equation of a straight line whose x and y intercepts are a and b is: x/a + y/b = 1 10. The length of the perpendicular drawn from origin (0,0) to the line Ax + By + C =0 is : C/ sqrt(A^2 + B^2) 11. Length of the perpendicular from (x1, y1) to the line Ax + By + C =0 is: Ax1 + By1 +C / sqrt(A^2 + B^2) 12. The point of intersection of two lines a1x + b1y +c1 = 0 and a2x + b2y +c2 = 0 is : ([b1*c2 - b2*c1]/[a1*b2 - a2*b1], [c1*a2 - c2*a1]/[a1*b2 - a2*b1]) 13. The condition for concurrency of three lines a1x + b1y +c1 = 0, a2x + b2y +c2 = 0 and a3x + b3y +c3 = 0 is (in determinant form) | a1 b1 c1 | | a2 b2 c2 | = 0 | a3 b3 c3 | 14. The angle between two lines y = m1x + c1 and y = m2x + c2 is tan inverse of the modulus of : [(m1 - m2)/(1 + m1*m2)] 15. Condition for parallelism of two lines with slopes m1 and m2 is m1 = m2 16. Condition for perpendicularity of two lines with slopes m1 and m2 is m1*m2 =-1 CIRCLES: 17. General equation of a circle with centre (x1, y1) and radius r is: (x - x1)^2 + (y - y1)^2 = r^2 18. The equation of a circle whose diameter is the line joining the points (x1, y1) and (x2, y2) is : (x - x1)(x - x2) + (y - y1)(y - y2) = 0 19. The equation of the tangent to the circle x^2 + y^2 = a^2 (where a is the radius of the circle) at the point (x1, y1) on it is : x*x1 + y*y1 + a^2 20. The condition for y = mx + c to be a tangent to the circle x^2 + y^2 = a^2 is : c^2 = a^2 (1 + m^2) Multiplication of 2digit by 2 digit number ab x cd ------ pqrs 1. first multiply bd - write down the unit fig at s carry over the tens fig. 2. Multiply axd & bxc add them together and also add the carry over from step 1 write down the units fig at r and carry over the tens fig. 3. Multiply axc and add the carry over from step 2. write down at pq. Its a bit difficult to explain in text - just do it a copule of times and you will get a hang of it. Trignometry - only the most relevant part (IMO) I found this 'Mnemonic for All Special Angles' in Oak Road Systems -- Software Since 1984 this might be useful to solve some geometry problems For angle A = 0, 30° (π/6), 45° (π/4), 60° (π/3), 90° (π/2): sin A = (sqrt0)/2, (sqrt1)/2, (sqrt2)/2, (sqrt3)/2, (sqrt4)/2 cos A = (sqrt4)/2, (sqrt3)/2, (sqrt2)/2, (sqrt1)/2, (sqrt0)/2 tan A = 0, (sqrt3)/3, 1, sqrt3, undefined In any triangle: sine = (opposite side) / hypotenuse cosine = (adjacent side) / hypotenuse tan = (opposite side)/(adjacent side) = (sine/cosine) Probability - 'The Rules' 1. If two events are mutually exclusive (i.e. they cannot occur at the same time), then the probability of them both occurring at the same time is 0. then: P(A and B) = 0 and P(A or B) = P(A) + P(B) 2. if two events are not-mutually exclusive (i.e. there is some overlap)then: P(A or B) = P(A) + P(B) - P(A and B) 3. If events are independent (i.e. the occurrence of one does not change the probability of the other occurring), then the probability of them both occurring is the product of the probabilities of each occurring. Then: P(A and B) = P(A) * P(B) 4. If A, B and C are not mutually exclusive events, then P(A or B or C) = P(A) + P(B) + P© - P(A and B) - P(B and C) - P(C and A) + P(A and B and C) and = union or = intersection sorry for making the formulae clusmsy using this 'and' and 'or'. Harmonic Mean The harmonic mean of x1,...,xn is n / (1/x1 + ... + 1/xn) As the name implies, it's a mean (between the smallest and largest values). An example of the use of the harmonic mean: Suppose we're driving a car from Amherst (A) to Boston (B) at a constant speed of 60 miles per hour. On the way back from B to A, we drive a constant speed of 30 miles per hour (damn Turnpike). What is the average speed for the round trip? We would be inclined to use the arithmetic mean; (60+30)/2 = 45 miles per hour. However, this is incorrect, since we have driven for a longer time on the return leg. Let's assume the distance between A and B is n miles. The first leg will take us n/60 hours, and the return leg will take us n/30 hours. Thus, the total round trip will take us (n/60) + (n/30) hours to cover a distance of 2n miles. The average speed (distance per time) is thus: 2n / {(n/60) + (n/30)} = 2 / (1/20) = 40 miles per hour. The reason that the harmonic mean is the correct average here is that the numerators of the original ratios to be averaged were equal (i.e. n miles at 60 miles/hour versus n miles at 30 miles/hour). In cases where the denominators of two ratios are averaged, we can use the arithmetic mean. This is another good site on harmonic mean : * Product of 2 numbers is the produst of their LCM & HCF. * LCM of a fraction = LCM of numerator/HCF 0f denominator. *HCF of a fraction = HCF of numer./LCM of denom. Ratio & Proportion: * if a/b = c/d = e/f = ..... then, a/b = c/d = e/f =(a+c+e+...)/(b+d+f+...) * If a/b = c/d, Then, i) b/a = d/c ii) a/c = b/d iii) (a+b)/ b = (c+d)/d iv) (a-b)/b = (c-d)/d v) (a+b)/(a-b) = (c+d)/(c-d) file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image029.gif file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image030.gif Join Date: Sep 2005 Posts: 71 file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image031.gif Money in Compound Interest gets doubled in 70/r years (approximately) ie. P(1+r/100)^N = 2P when N=70/r Got this in OG11 for GMAT. Checked it for full range of N & r in Excel. Seems to work fine. Would be great if someone can give a derivation for the calculation. Divisibility Rules Divisibility by:2If the last digit is even, the number is divisible by 2.3If the sum of the digits is divisible by 3, the number is also. 4If the last two digits form a number divisible by 4, the number is also. 5If the last digit is a 5 or a 0, the number is divisible by 5. 6If the number is divisible by both 3 and 2, it is also divisible by 6. 7 Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also. 8 If the last three digits form a number divisible by 8, then so is the whole number. 9If the sum of the digits is divisible by 9, the number is also. 10If the number ends in 0, it is divisible by 10. 11Alternately add and subtract the digits from left to right. If the result (including 0) is divisible by 11, the number is also. Example: to see whether 365167484 is divisible by 11, start by subtracting: 3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11. 12If the number is divisible by both 3 and 4, it is also divisible by 12. 13 Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13,then so is the original number Statistics 1. Mean of a distribution x1, x2, x3, ......, xn is given by the formula: file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image032.gif where n is the number of terms in the given set. 2. Median value of an ordered distribution y1, y2, y3, ......., yn-1, yn can be given as: file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image032.gif 3. Mode is the most common value obtained (or value that occurs at highest frequency) in a set of observations. 4. The sample variance may be computed as file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image032.gif where file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image032.gifis the sample mean. 5. The square root of the sample variance of a set of N values is the sample standard deviation 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 file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image033.gif) 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) 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...file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image034.gif) 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... 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... 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.... Here is a Contribution to redeem myself.file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image035.gif 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 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. 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 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. file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image036.gif The formula is Area = I +B/2 – 1where I = number of interior lattice points (file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image037.gif) and B = number of boundary lattice points (file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image038.gif) For example, the area of the simple lattice polygon in the figure is 31 + 15 /2 – 1 = 37.5. file:///E:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image039.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 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] Facts -- 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} 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 Some more Facts: If SD of x1, x2, x3, ... xn is sigma then SD of x1+k , x2+k, x3+k ... Xn+k is also sigma If SD of x1, x2, x3, ... xn is sigma then SD of x1*k , x2*k, x3*k ... Xn*k is k*sigma Variance (kx) = k^2 Variance(x) Binomial probability mass function: P(x) = nCx * p^x * q ^ (n-x) where x is happening event, n is total number of event, p is probability of happening of event and q is probability of not happening. Standard deviation- normal distribution z values that you need to memorize: PICTURE VIEW - Area within .5 Standard Deviation above and below the mean is 38% - Area within 1 Standard Deviation above and below the mean is 68% - Area within 2 Standard Deviation above and below the mean is 95% - Area within 3 Standard Deviation above and below the mean is 99.7 - Area below 1 standard deviation is 84% - Area below 2 standard deviation is 97.7% - Area above 1 standard deviation is 15.8% - Area above 2 standard deviation is 2.27% - If perimeters of a square and parallelogram are equal, then area of a square is always greater than area of a parallelogram. - Similarly, if perimeters of a square and circle are same, then area of a circle is greater than area of a square. ONE VARIABLE INCREASED/ DECREASED PROBLEMS: PRICE INCREASED AND REDUCTION OF THE CONSUMPTION: 1. Price of sugar is increased 25%. How much percent must a house hold must reduce his consumption of sugar so as not to increase his expenditure? how much time u require to this problem? just try this short cut less then 5 sec u will get the answer % REDUCTION= (INCREASE/100+INCREASE)* 100 lets try this with short cut increase = 25% so reduction = (25/ 100+25 ) * 100 = (25/125) * 100 = 1/5* 100 = 20 % so house hold have to decrease 20% of their consuption to keep constant . PRICE DECREASED INCREASE IN CONSUPTION: PROBLEM : 2 Certain familyhave fixed budget for ice cream purchase for year . but,Ice cream price decreased by 20% due to winter season. find by how much % a consumer must increase his consumpion of ice creame so as not to decrease his expenditure. Here the short cut %INCREASE IN CONSUMPTION = (REDUCTION/ 100- REDUCTION) *100 just as mentioned above decrease icecreame price= 20 = 20/(100-20) *100 =20/80 * 100 =1/4 *100 = 25% BOTH VARIABLES INCREASED/ DECREASED PROBLEMS TYPE3: Petrol tax is increased by 20% and the costumer comsumption also increased by 20%.Find the % increase or decrease in the expenditure OR Water tax is increased by 20% and consumption also increased by 20% find what is the net effect in change? the short cut: [(A+B) + AB]/100 increase A: 20 increase B: 20 = (20+20) + (20*20)/100 = 40+ 400/100 = 40 +4 = 44 % net increase ONE INCREASED ONTHER DECREASED PROBLEMS: Shop keeper decreased the price of a article by 20% and then increased the artical by 30% what is the net effect of the artical is it increased or decreased? SHORT CUT IS SAME AS ABOVE so first decreased the price so we have to take as negative value for A decrease A : -20% increase A : 30% = (-20 + 20)+ (-20)*20/100 = (0) +(-400)/100 = -4% so net effect is 4% loss to the shop keeper. Quantitative Ability – POINTS TO REMEMBER 1. If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots. Eg: x3+3x2+2x+6=0 has no positive roots 2. For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots. 3. For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) 4. Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum. 5. ü For a cubic equation ax3+bx2+cx+d=o · Sum of the roots = - b/a · Sum of the product of the roots taken two at a time = c/a · Product of the roots = -d/a ü For a bi-quadratic equation ax4+bx3+cx2+dx+e = 0 · Sum of the roots = - b/a · Sum of the product of the roots taken three at a time = c/a · Sum of the product of the roots taken two at a time = -d/a · Product of the roots = e/a 6. If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case) 7. Consider the two equations a1x+b1y=c1 a2x+b2y=c2 Then, ü If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations. ü If a1/a2 = b1/b2 c1/c2, then we have no solution. ü If a1/a2 b1/b2, then we have a unique solution. 8. Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1 9. |a| + |b| = |a + b| if a*b>=0 else, |a| + |b| >= |a + b| 10. The equation ax2+bx+c=0 will have max. value when a0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a 11. ü If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4. ü If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k). 12. Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other. 13. For any 2 numbers a, b where a>b ü a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively) ü (GM)^2 = AM * HM 14. For three positive numbers a, b, c ü (a + b + c) * (1/a + 1/b + 1/c)>=9 15. For any positive integer n ü 2 16. a2 + b2 + c2 >= ab + bc + ca If a=b=c, then the case of equality holds good. 17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1) 18. (n!)2 > nn 19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s 20. If n is even, n(n+1)(n+2) is divisible by 24 21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3) 22. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity Note: 2 23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e] 24. (m + n)! is divisible by m! * n! 25. When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9. 26. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) 27. ü The sum of first n natural numbers = n(n+1)/2 ü The sum of squares of first n natural numbers is n(n+1)(2n+1)/6 ü The sum of cubes of first n natural numbers is (n(n+1)/2)2/4 ü The sum of first n even numbers= n (n+1) ü The sum of first n odd numbers= n2 28. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then ü the total number of factors is (x+1)(y+1)(z+1) .... ü the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c).... ü the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c).... ü the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...) 29. ü Total no. of prime numbers between 1 and 50 is 15 ü Total no. of prime numbers between 51 and 100 is 10 ü Total no. of prime numbers between 101 and 200 is 21 30. ü The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6 ü The number of rectangles in n*m board is given by n+1C2 * m+1C2 31. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational. 32. Certain nos. to be remembered ü 210 = 45 = 322 = 1024 ü 38 = 94 = 812 = 6561 ü 7 * 11 * 13 = 1001 ü 11 * 13 * 17 = 2431 ü 13 * 17 * 19 = 4199 ü 19 * 21 * 23 = 9177 ü 19 * 23 * 29 = 12673 33. Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square. 34. If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no. 35. To find out the sum of 3-digit nos. formed with a set of given digits This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits) Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8. Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there) = 25 * 24 * 11111 =6666600 36. Consider the equation x^n + y^n = z^n As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3. 37. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then, N^(p-1) – 1 is always divisible by p. 38. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e. 145 = 1! + 4! + 5! 39. ü Where a no. is of the form a^n – b^n, then, · The no. is always divisible by a - b · Further, the no. is divisible by a + b when n is even and not divisible by a + b when n is odd ü Where a no. is of the form a^n + b^n, then, · The no. is usually not divisible by a - b · However, the no. is divisible by a + b when n is odd and not divisible by a + b when n is even 40. The relationship between base 10 and base ‘e’ in log is given by log10N = 0.434 logeN 41. WINE and WATER formula Let Q - volume of a vessel, q - qty of a mixture of water and wine be removed each time from a mixture, n - number of times this operation is done and A - final qty of wine in the mixture, then, A/Q = (1-q / Q)^n 42. Pascal’s Triangle for computing Compound Interest (CI) The traditional formula for computing CI is CI = P*(1+R/100)^N – P Using Pascal’s Triangle, Number of Years (N) ------------------- 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 … 1 .... .... ... ... ..1 Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount? Step 1: Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331 The coefficients - 1,3,3,1 are lifted from the Pascal's triangle above. Step 2: CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1) If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210 CI = 2 * 100 + 1* 10 = Rs.210 43. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by: Final Difference% = X - Y - XY/100 Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold? Applying the formula, Final difference% = 40 – 25 - (40*25/100) = 5 %. So if 5 % = 1,000 Then, 100 % = 20,000. Hence, C.P = 20,000 & S.P = 20,000+ 1000= 21,000 44. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100. 45. ü Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)2 ü The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003 46. ü If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time. ü If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time. ü If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time ü If A can finish a work in X time and B in Y time and A, B & C together in S time then · C can finish that work alone in (XYS)/ (XY-SX-SY) · B+C can finish in (SX)/(X-S); and · A+C can finish in (SY)/(Y-S) 47. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5 48. ü When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0 ü When an unbiased coin is tossed even no. (2n) of times, then, P (no. of heads=no. of tails) = 1-(2nCn/22n) 49. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m! Eg)1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C? Here n=10, m=3 (i.e. A, B, C) Hence, P (A>B>C) = 1/3! = 1/6 Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning? P (M>S) = 1/2! = 1/2 50. CALENDAR ü Calendar repeats after every 400 years. ü Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400. ü Century has 5 odd days and leap century has 6 odd days. ü In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day. ü January 1, 1901 was a Tuesday. 51. ü For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides) ü For any regular polygon, the sum of interior angles =(n-2)*180 degrees So measure of one angle is (n-2)/n *180 ü If any parallelogram can be inscribed in a circle, it must be a rectangle. ü If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal). 52. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order) 53. ü For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is 0.5*d1*d2, where d1, d2 are the length of the diagonals. ü For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where s=(a + b + c + d)/2 Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle. ü Area of a Rhombus = Product of Diagonals/2 54. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for [(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2] 55. Area of a triangle ü 1/2*base*altitude ü 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B ü root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2 ü a*b*c/(4*R) where R is the circumradius of the triangle ü r*s ,where r is the inradius of the triangle 56. In any triangle ü a=b*cos C + c*cos B ü b=c*cos A + a*cos C ü c=a*cos B + b*cos A ü a/sin A=b/sin B=c/sin C=2R, where R is the circumradius ü cos C = (a^2 + b^2 - c^2)/2ab ü sin 2A = 2 sin A * cos A ü cos 2A = cos^2 (A) - sin^2 (A) 57. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 58. Appollonius Theorem In a triangle ABC, if AD is the median to side BC, then AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2) 59. ü In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base. ü In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides. 60. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle. 61. Let W be any point inside a rectangle ABCD, then, WD2 + WB2 = WC2 + WA2 62. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1)) 63. ü Distance between a point (x1, y1) and a line represented by the equation ax + by + c=0 is given by |ax1+by1+c|/Sq(a2+b2) ü Distance between 2 points (x1, y1) and (x2, y2) is given by Sq((x1-x2)2+ (y1-y2)2) 64. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2. Lets say the initial distance between A and B was (x+y) kms. A and B cross each other after traveling a distance of x and y km respectively. ------------*------------------------- A---- x------------ y---------------- B Speed = distance travelled / time taken So, we can form the following equations: 1. x/ A’s speed = y/ B’s speed, as they take the same time to cross that distance. 2. A takes 'a' seconds to travel the distance y after crossing B. Hence, y/A’s speed = 'a' secs => a * A’s speed = y Similarly, x/B’ speed = 'b' secs and b * B’ speed = x Now substituting these for x and y in eqn. 1 b * B’s speed/ A’s speed = a * A’s speed/ B’s speed That is, A’s speed squared/ B’s speed squared = b/ a which gives (A's speed): (B's speed) = (root(B):root(A)) If we have to find: "How much of a v1 something should be melted into n2 of a v2 something to produce an alloy which is R something?" we can use this formula: n1=n2(R-v2)/(v1-r) As example: How much of a 75% copper alloy should be melted into 62 kg of a 35% copper alloy to produce an alloy which is 50% copper? n1=62*(0.5-0.35)/(0.75-0.5)= 37.2. If there are three sets A, B, and C, then P(AuBuC) = P(A) + P(B) + P© – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) Number of people in exactly one set = P(A) + P(B) + P© – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC) Number of people in exactly two of the sets = P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC) Number of people in exactly three of the sets = P(AnBnC) Number of people in two or more sets = P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC) Restricted – Combinations (a) Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = n-pCr-p. (b) Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = n-pCr Example: In how many ways can a cricket-eleven be chosen out of 15 players? if (i) A particular player is always chosen, (ii) A particular is never chosen. Ans: (i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players. =. Required number of ways = 14C10 = 14C4 = 14!/4!x19! = 1365 (ii) A particular players is never chosen, it means that 11 players are selected out of 14 players. => Required number of ways = 14C11 = 14!/11!x3! = 364 (iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:- 2n-1 Proof:Number of ways of selecting one thing, out of n-things = nC1 Number of selecting two things, out of n-things =nC2 Number of ways of selecting three things, out of n-things =nC3 Number of ways of selecting ‘n’ things out of ‘n’ things = nCn =>Total number of ways of selecting one or more things out of n different things = nC1 + nC2 + nC3 + ------------- + nCn = (nC0 + nC1 + -----------------nCn) - nC0 = 2n – 1 [ nC0=1] Example: John has 8 friends. In how many ways can he invite one or more of them to dinner? Ans. John can select one or more than one of his 8 friends. => Required number of ways = 28 – 1= 255. (iv) Number of ways of selecting zero or more things from ‘n’ identical things is given by :- n+1 Example: In how many ways, can zero or more letters be selected form the letters AAAAA? Ans. Number of ways of : Selecting zero 'A's = 1 Selecting one 'A's = 1 Selecting two 'A's = 1 Selecting three 'A's = 1 Selecting four 'A's = 1 Selecting five 'A's = 1 => Required number of ways = 6 [5+1] (V) Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by :- (p+1) (q+1) (r+1)2n – 1 Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen. Ans: Number of ways of selecting apples = (3+1) = 4 ways. Number of ways of selecting bananas = (4+1) = 5 ways. Number of ways of selecting mangoes = (5+1) = 6 ways. Total number of ways of selecting fruits = 4 x 5 x 6 But this includes, when no fruits i.e. zero fruits is selected => Number of ways of selecting at least one fruit = (4x5x6) -1 = 119 Note :- There was no fruit of a different type, hence here n=o => 2n = 20=1 (VI) Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’. Example: In how many ways 5 balls can be selected from ‘12’ identical red balls? Ans. The balls are identical, total number of ways of selecting 5 balls = 1. Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5? Ans. Here n = 5 [Number of digits] And r = 4 [ Number of places to be filled-up] Required number is 5P4 = 5!/1! = 5 x 4 x 3 x 2 x 1 Shortcuts for fractions 1) A fraction in lowest terms with a prime denominator other than 2 or 5 (i.e. coprime to 10) always produces a recurring decimal. 2) If the repeating decimal is between 0.1 and 1, and the repeating block is n digits long occurring right at the decimal point, then the fraction (not necessarily reduced) will be the n-digit block over n digits of 9. For example, 0.444444... = 4/9 since the repeating block is 4 (a 1-digit block), 0.565656... = 56/99 since the repeating block is 56 (a 2-digit block), 0.789789... = 789/999 since the repeating block is 789 (a 3-digit block), etc. If the repeating decimal is between 0 and 0.1, and the repeating n-digit block is preceded only by k digits of 0 (all of which are to the right of the decimal point), then the fraction (not necessarily reduced) will be the n-digit block over the integer consists of n digits of 9 followed by k digits of 0. For example, 0.000444... = 4/9000 since the repeating block is 4 and this block is preceded by 3 zeros, 0.005656... = 56/9900 since the repeating block is 56 and it is preceded by 2 zeros, 0.0789789... = 789/9990 since the repeating block is 789 and it is preceded by 1 zero. For any repeating decimal not prescribed above, it can be written as a sum of a terminating decimal and a repeating decimal of one of the two above types. For example, 1.23444... = 1.23 + 0.00444... = 123/100 + 4/900 = 1107/900 + 4/900 = 1111/900 0.3789789 ... = 0.3 + 0.0789789... = 3/10 + 789/9990 = 2997/9990 + 789/9990 = 3786/9990 = 631/1665 When number is of the form a^n - b^n -> it is always divisible by a-b. -> it is divisible by a+b if n is even When number is of the form a^n + b^n -> it is divisible by a+b if n is odd Quote Link to comment Share on other sites More sharing options...

rina777 Posted March 28, 2010 Share Posted March 28, 2010 This is from lost thread: For equal perimeter: Area of a circle>Area of a Square>Area of Parallelogram>Area of a Triangle For equal Area: Perimeter of a Triangle>Perimeter of a Parallelogram>Perimeter of a Square>Perimeter of a Circle Quote Link to comment Share on other sites More sharing options...

rina777 Posted March 28, 2010 Share Posted March 28, 2010 For a Parallelogram, sum of the square of all sides = sum of the square of two diagonals, If ABCD is a parallelogram then, AB^2+BC^2+ CD^2+ DA^2=AC^2+ BD^2 Quote Link to comment Share on other sites More sharing options...

rina777 Posted March 28, 2010 Share Posted March 28, 2010 case 1: a is odd and b is even or a is even and b is odd the number of odd integer=(b-a+1)/2 the number of even integer=(b-a+1)/2 case 2: a, b are odd the number of odd integer=[(b-a)/2]+1 the number of even integer=(b-a)/2 case 3: a, b are even the number of odd integer=[(b-a)/2] the number of even integer=[(b-a)/2]+1 Quote Link to comment Share on other sites More sharing options...

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