CAT Preparation Tips

Important Note:-CAT 2009 is going to be online so prepare at least 10 mock tests on computer.The Indian Institutes of Management (IIMs) are India's premier management institutes. IIM, not to mention is the deity to CAT aspirants. Last year one is added to the list i.e IIM Shillong which increases the probability of getting into IIM, af course if number of aspirants remain the same; but still this number is very bleak. So what matters the most in this world of competition is how you prepare for this tough and 'smart' exam, yes, smart is the key word because the person who prepares smartly is the winner of this game as there is a famous adage: Don't work hard work smart. So overall I will be discussing about some smart techniques for killing this meow (CAT).What you need to firstly do is to see where you stand currently i.e what are your strengths and weaknesses, and you can do this by taking one or two mock tests. Never fear if you find that you are weak nearly in every area because converting your weakness to your strength doesn't need more time, it needs smart time. After deciphering your strengths and weaknesses, it's time either to collect the reading material or to join a coaching class. Many students ask why do we need coaching classes; well then the answer to this question is: coaching institutes will organize your study habit; off course they aren't alchemist who will turn you into gold. But if you are a working official and think that you are a good manager in organising your study time then nothing is better than that. Therefore, organized study is the first main key point to this smart journey of preparation; after all you are going to become a manager so everything should be organized and managed.Second key point is 'Stick to the Rules'; though the irony is CAT never does but by not sticking to the rules they are checking whether you does. So; make a timetable of your study plan and keep it in your pocket or paste it on your notice board of office or home. Timetable of every aspirant will be different so keep in mind your weaknesses while making the itinerary. Give more time to your weaknesses and practice as much as you can. Main problem with many students is that they organize their studies for few days, and after that they procrastinate the assigned targets or don't give the required time to their studies. But perseverance and patience are the two traits which every aspirant should have as CAT will check your patience by giving you a whole together a different paper; so you should maintain your calmness and don't get flummoxed.Third point is to keep track of your performance and your mistakes. After all you are preparing to improve and if you are not tracking your performance then how will you get to know whether you are improving in your weak areas or not. Never repeat your mistakes. Keep your mind fit by eating healthy food and by reading some good novels and newspapers. Good is a subjective term but here what I imply is read those novels or articles in newspapers which you hate reading in normal circumstances; CAT is not good in giving your favorite RC passages. After all for ESLs (English as a Second Language) reading is the strongest tool to hone its English skills.Lastly, never think that your aim is to clear written exam. You are not preparing for getting a call, you are preparing for getting into IIMs, so participate in GD sessions and keep yourself updated with current affairs. If you haven't join any coaching institute then you can join any public forum online or offline for practicing GD and Interview sessions.At the end I will summarize my rules:a) Know your strengths and weaknesses.b) Organize your studies.c) Keep track of your performance and mistakes.d) Read healthy.e) Keep preparing for GD & Interview.As it is not in my scope to give you each and every aspect of preparation but I have tried to write some of the key points which I have noticed in every IIM student. So, don't just toil, have some smart oil.Preparation StrategyI would like to recommend you to spend at least 5 hours a day for the ultimate preparation.

Concerned tasks for CAT preparation

* Preparing Maths* Preparing Data Interpretation* Preparing Analytical reasoning* Preparing Reading comprehension* Preparing Verbal ability* word list* Reading

There are two phases of preparation

First phase:

In this phase you should get good grip on fundamentals and get familiar with all chapters and problems. Do not look for speed solving or feel any sort of pressure while preparing fundamentals. Learn with your own efficiency, without getting into fundamentals clearly speed solving is not useful. Please manage to allocate at least one hour a day for reading newspapers, magazines. It is better to subscribe for any good magazines during the preparation. Get in touch with current affairs.

I request you to schedule your weeks to prepare accordingly because haphazardness will waste lot of time. If you have been preparing without any study structure then whatever you may achieve would have been increased up to 50%. Try to shuffle the subjects also, so that you do not get bored with one subject. A good time table for 1 days look like below:

Day 1:

1 hour reading

2 hours of math

1 hour of analytical reasoning / data interpretation

1 hour of reading comprehension exercises / verbal ability

Second phase(self analyzing and improving):

By the time you finish the fundamentals which is about second week of august, you will get a fare idea on the area of your weakness and strengths. Make a tabular form of topics categorizing strengths and weakness to have a clear picture on it. Keep weak areas in mind and try to improve yourself by focusing them more.

From second week of August you can take section tests along with revision of topics in alternate day, here if you think you have to allocate more time you should do.. You must start taking comprehensive tests by the beginning of September.

Things you look to learn in comprehensive tests of 2-3 hour long, is how to manage paper so the attitude here should be the same like a batsman who wants to score as many runs as possible in the last 10 overs in onedayers. You should think of scoring as much as possible, but most people will get emotional to have a challenging problem and stuck up solving only that problem.

After you tested yourself spend time to analyze it. The key points to analyze your test: Speed:see if your speed is better than previous tests, notice if you do not understand basics you cannot apply them to tougher problems. Both these attributes are a prerequisite for speed solving; Concentration:Most of we do not use two hours sitting at a stretch; a high degree of concentration is also not easy for first time test takers. Comprehensive tests make us to sit for two hours at a stretch and we increase our attentive span. Time Management:the most important value provided by the comprehensive test is an idea of our strengths and weakness. So you should manage time and allocate certain time to each section depending on your strength and weakness Accuracy:accuracy is the number of correct questions out of attempted questions. if this is low you need to go and revise the chapters which are reducing your accuracy because marking wrong option will impose negative marks in cat Finding traps:you should avoid false positives or false negatives i.e., you shouldnt leave easy questions thinking they are hard and attempt hard questions thinking they are easy. Analyzing your progress on the above four: compare with the previous attempted paper on above all points

Take comprehensive test for yourself on every alternate days with analyzing and revision of topics on another day,

Some institutes also conduct all India mock test on weekends please try to give as many mocks as possible you can.

Try to attempt at least 30 comprehensive tests (including all India mocks)

Finally give yourself a break on every weekend to relax and start fresh from the following Monday.Best of Luck

You can buy books for CAT preparation like "How to prepare for Quantitative Aptitude for the CAT" by Arun Sharma.You can even join Mindworkzz-CAT Online Classes by Arun Sharma.

Quantitative AbilityQuantitative Ability Syllabus1. Number Systems2. LCM and HCF3. Percentages4. Profit, Loss and Discount5. Interest (Simple and Compound)6. Speed, Time and Distance7. Time and Work8. Averages9. Ratio and Proportion10. Linear Equations11. Quadratic Equations12. Complex Numbers13. Logarithm14. Progressions (Sequences & Series)15. Binomial Theorem16. Surds and Indices17. Inequalities18. Permutation and Combination19. Probability20. Functions21. Set Theory22. Mixtures and Alligations23. Geometry24. Co-ordinate Geometry25. Trigonometry26. MensurationData InterpretationData is given in form of tables, charts and graphs. In this section it is tested that how can you interpret the given data and answer the questions based on it.1. Tables2. Column Graphs3. Bar Graphs4. Line Charts5. Pie Chart6. Venn Diagrams7. CaseletsCombination of two or more types linked to each other.Logical Reasoning1. Number and Letter Series2. Calendars3. Clocks4. Cubes5. Venn Diagrams6. Binary Logic7. Seating Arrangement8. Logical Sequence9. Logical Matching10. Logical Connectives11. Syllogism12. Blood RelationsVerbal AbilityTypes of Questions

Vocabulary Based (Synonyms Antonyms) English Usage or Grammar Sentence Correction Fill in the blanks Cloze Passage Analogies or Reverse Analogies Jumbled Paragraph Meaning-Usage Match Summary Questions Verbal Reasoning Facts / Inferences / Judgements Reading ComprehensionVocabulary:Vocabulary questions test the candidates knowledge of the primary meanings of words, secondary shades of meaning, usage, idioms and phrases, antonyms, related words, etc.Grammar:Grammar-based questions test the candidates ability to spot and correct grammatical errors. CAT generally tests knowledge of high school level grammar and includes areas like subject-verb agreement, use of modifiers, parellel construction, redundancy, phrasal verbs, use of articles, prepositions, etc.Verbal Reasoning:Verbal reasoning questions are designed to test the candidates ability to identify relationships or patterns within groups of words or sentences.Study material of Quant ability-Number SystemsNumber Systems forms the base for quant ability and clearing of concepts is important for CAT and other related exams. Following table gives a brief introduction to system of numbers.

Prime NumberStarting from the basic knowledge, a prime number is a natural number which has only two distinct divisors: 1 and itself.The number 1 is not a prime number.There are 25 prime numbers under 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.Prime Factorization Theorem: This is the area where prime numbers are used. This theorem states that any integer greater than 1 can be written as auniqueproduct of prime numbers.Examples:{tex}{550 = 2 \times 5^2 \times 11}{/tex}{tex}{1200 = 2^4 \times 3 \times 5^2}{/tex}Thus, prime numbers are the basic building blocks of any positive integer. This factorization will also help in finding GCD and LCM quickly.Perfect NumbersA number is a perfect number if the sum of its factors, excluding itself and but including 1, is equal to the number itself.Example: 6 (1 + 2 + 3 = 6), 28 (1 + 2 + 4 + 7 +14 = 28)Co-Prime NumbersTwo numbers are co-prime to each other, if they do not have any common factor except 1.Example: 25 and 9, since they dont have a common factor other than 1Points to Remember1. The number 1 is neither prime nor composite.2. The number 2 is the only even number which is prime.3. (xn+ yn) is divisible by (x + y), when n is an odd number. 4. (xn yn) is divisible by (x + y), when n is an even number. 5. (xn yn) is divisible by (x y), when n is an odd or an even number.

Number Systems - IIFactors of a NumberRepresenting a number as prime factors helps in analyzing problems.{tex}N = p^a + q^b + r^c{/tex} Where p, q, r are prime numbers and a, b, c are the number of times each prime number occurs.Number of Factors = (a + 1)(b + 1)(c + 1)Number of Ways of Expressing a Given Number as a Product of Two Factors{tex}{{(a+1)(b+1)(c+1)} \over 2}{/tex}Sum of Factors = {tex}{({a^{p+1} - 1})({b^{q+1} - 1})({c^{r+1} - 1}) \over {(a-1)(b-1)(c-1)}}{/tex}Concept of Cyclicity (Power Cycle)Concept of cyclicity is used to find unit's digit in case the numbers are occuring in powers.The last digit of a number of the form abfalls in a particular sequence that depends on the unit digit of thenumber (a) and the power the number is raised to (b).Consider the power cycle of 2: 21= 2 22= 4 23= 8 24= 16 25= 32 26= 64You can see that the unit digit gets repeated after every fourth power of 2. Hence, you can say that 2 has a power cycle of 2, 4, 8, 6 with cyclicity 4.This is applicable for all numbers ending in 2.Unit DigitPower CycleCyclicity

001

111

22, 4, 8, 64

33, 9, 7, 14

44, 62

551

661

77, 9, 3, 14

88, 4, 2, 64

99, 12

Applications of FactorialMeaning of Factorial N! = 1 x 2 x 3 x ... x (N-1) x N Example 3! = 6; 5! = 120Maximum power of p (prime nubmer) in n! (n factorial)To find the highest power of a prime number (p) in a factorial (n!), keep dividing n by p and add all the quotients. Alternatively, use the formula:{tex}{n \over p} + {n \over p^2} + {n \over p^3} + \dots{/tex}Basic Formulae of Algebra1. (a + b)2= a2+ b2+ 2ab2. (a - b)2= a2+ b2- 2ab3. (a + b)2- (a - b)2= 4ab4. (a + b)2+ (a - b)2= 2 (a2+ b2)5. (a2- b2) = (a + b) (a - b)6. (a + b + c)2= a2+ b2+ c2+ 2 (ab + bc + ca)7. (a3+ b3) = (a +b) (a2- ab + b2)8. (a3- b3) = (a - b) (a2+ ab + b2)9. (a3+ b3+ c3- 3abc) = (a + b + c) (a2+ b2+ c2- ab - bc - ca)10. If a + b + c = 0, then a3+ b3+ c3= 3abcDivisibility RulesDivisibility by 2The last digit is even (0, 2, 4, 6, or 8).Divisibilityby 3The sum of the digits is divisible by 3.Divisibilityby 4The last two digits divisible by 4.Divisibilityby 5The last digit is 0 or 5.Divisibilityby 6The sum of the digits is divisible by 3 and the number itself is divisible by 2.Divisibilityby 7Subtract 2 times the last digit from the rest.Divisibilityby 8If the hundreds digit is even, examine the number formed by the last two digits. If the hundreds digit is odd, examine the number obtained by the last two digits plus 4.Divisibilityby 9The sum of the digits is divisible by 9.Divisibilityby 10The last digit is 0.Divisibilityby 11Add the digits in blocks of two from right to left. Example - 627: 6 + 27 = 33Divisibilityby 12It is divisible by 3 and by 4.LCM and HCFFactors and MultiplesSuppose there are two numbers - a and b. If a numberadivides another numberbexactly, we say thatais a factor ofbandbis called a multiple ofa.Highest Common Factor (HCF) or Greatest Common Divisor (GCD)Thegreatest common divisor(gcd), also known as the greatest common denominator, greatest common factor (gcf), orhighest common factor(hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. For example,the GCD of 8 and 12 is 4.The HCF of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the HCF of a given set of numbers:1. Factorization Method:In this method, express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives HCF.2. Division Method:Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required HCF.Finding the HCF of more than two numbers:H.C.F. of [(H.C.F. of any two) and (the third number)] gives the HCF of three given numbers.Least Common Multiple (LCM)Thelowest common multipleor (LCM) least common multiple or smallest common multiple of two rational numbersaandbis the smallest positive rational number that is an integer multiple of bothaandb. The definition can be generalised for more than two numbers.The least number which is exactly divisible by each one of the given numbers is called their LCM.1. Factorization Method of Finding LCM:Resolve each one of the given numbers into a product of prime factors. Then, LCM is the product of highest powers of all the factors.2. Common Division Method (Short-cut Method) of Finding LCM:Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required LCM of the given numbers.Product of two numbers = Product of their HCF and LCM{tex}\operatorname{gcd}(a,b)=\frac{a\cdot b}{\operatorname{lcm}(a,b)}{/tex}Co-primesTwo numbers are said to be co-primes if their HCF is 1. HCF of two co-prime numbers is 1. To fing LCM of co-prime numbers, just multiply them. No need to find factors.HCF and LCM of Fractions{tex}HCF= {\mbox{HCF of Numerators} \over \mbox{LCM of Denominator}}{/tex} {tex}LCM = {\mbox{LCM of Numerators} \over \mbox{HCF of Denominator}}{/tex}PercentagesPercentageis a way of expressing a number as a fraction of 100 (per cent meansper hundred). It is denoted by using the percent sign,%.Example:{tex inline}25\% = {25 \over 100}{/tex}Practice QuestionsWhat is 200% of 30? Answer: 200% 30 = (200 / 100) 30 = 60.What is 13% of 98? Answer: 13% 98 = (13 / 100) 98 = 12.74.60% of all university students are male. There are 2400 male students. How many students are in the university? Answer: 2400 = 60% X, therefore X = (2400 / (60 / 100)) = 4000.There are 300 cats in the village, and 75 of them are black. What is the percentage of black cats in that village? Answer: 75 = X% 300 = (X / 100) 300, so X = (75 / 300) 100 = 25, and therefore X% = 25%.The number of students at the university increased to 4620, compared to last year's 4125, an absolute increase of 495 students. What is the percentual increase? Answer: 495 = X% 4125 = (X / 100) 4125, so X = (495 / 4125) 100 = 12, and therefore X% = 12%.Constant Product RuleYou can apply this rule when you have two parameters whose product is constant. In other words, when two parameters are inversely proportional to each other. For examples, Time x Speed = Distance Price x Consumption = Expenditure Length x Breadth = AreaThe rule states that 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter.Let's understand with the help of example. Suppose speed increases by 25% (or 1/4) and distance is constant, time required will decrease by 1/(4+1) or 1/5 or 20%.Profit and LossProfit and Loss is mainly used in finance and business transactions.Cost Price (CP):Expenses occured in producing a product or service.Selling Price (SP):The price at which goods or services are sold.Marked Price (MP):Price printed on the product for sale. If seller gives any discount, selling price will be different from the marked price.Profit and Loss Formulae1.Profit (or Loss) = SP CP (profit is made only when SP is greater then CP)2.Profit % (or Loss %) = (Actual Profit/Loss CP) 100%3.CP = (SP 100 ) (100 + profit %)4.SP = (100 + profit %) C.P 1005.Actual Discount = MP SP6.Discount % = (Actual Discount MP) 100%7.SP = (MP Discount %) of MPInterestsThe lending and borrowing of money involves the concept of simple interest and compound interest. If you borrow money for certain period of time, you would have to return the this sum of money (Principal) with some extra money. this extra money is called Interest.The money borrowed is called principal. The sum of interest and principal is called the amount. The time for which money is borrowed is called period.{tex}Amount = Principal + Interest{/tex}The rate of interest is as per annum (unless indicated).Simple InterestSimple interest is simply calculated on principal amount using the following formula:{tex}{SI = {P \times R \times T \over 100}}{/tex}where, P = principal, R = rate per annum, T = time in years

Amount can be calculated by adding interest to principal.Compound InterestWhen the borrower andthe lender agree to fix up a certain unit of time (say yearly or half-yearlyor quarterly) to settle the previous account.In such cases ,the amount after the first unit of time becomes the principalfor the 2nd unit. The amount after second unit becomes the principal for the3rd unit and so on.After a specified period, the difference between the amount and the moneyborrowed is called Compound Interest for that period.Suppose you lend Rs.10000 (principal) for 3 years at 10% per annum. So, you will get Rs.1000 as interest per annum (simple interest) For three years, interest will be Rs.3000 and thus you will get total amount of Rs.13000.Compound interest involves interest on interest too, thus will give you better amount after 3 years. While calculating the compound interest, the principal amount keeps changing year after year (if the interest is compounded annually).After 1 year: Interest = Rs.1000; New Principal = Rs.11000After 2 years: Interest = Rs.1100; New Principal = 12100After 3 years: Interest = Rs.1210; You get Rs.13310. So, there is gain of Rs.310 if you lend at compound interest!{tex}A = P \left(1 + \frac{r}{n}\right)^{nt}{/tex}Trigonometry

Trigonometric Identities

Speed, Time and DistanceThe questions on speed, time and distance are based on one general formula:Distance = Speed X Time

The ConceptThere are three parameters: Speed, Time and Distance. Keeping one parameter constant and changing another, the third parameter also gets changed.Example: Suppose you travel from your office (A) to your home (B). The distance between A and B is 30 km. You travel at the speed of 40 km/hr and it takes 45 minutes.Boats and StreamsIn water or river, the direction along the stream is called downstream.Direction against the stream is called upstream.If the speed of boat in still water is U km/hr and the speed of stream is Vkm/hr, then speed downstream = (U + V) km/hr speed upstream = (U - V) km/hrIf the speed downstream is A km/hr and the speed upstream is B km/hr, then Speed of boat in still water = 1/2(A+B) km/hr Rate of stream or river = 1/2(A-B) km/hrTrainsA train has a definite length. The distance covered by the train depends on the length of the train.1. Time taken by a trainxmt long in passing a stationary point (it can be a signal post or a pole or a standing man) is equal to time taken by the train to coverxmt.2. Time taken by a trainxmt long in passing a stationary object of lengthymt is equal to time taken by the train to cover (x+y) mt.3. Suppose two trains are moving in the same direction atu kmphandv kmphsuch that u > v, then their relative speed isu-v kmph.4. Suppose two trains are moving in opposite direction atu kmphandv kmphthen, their relative speed is equal to(u+v) kmph.5. If two trains of lengthx kmandy kmare moving in opposite diredtions atu kmphandv kmph, then time taken by the train to cross each other is equal to(x+y)/(u+v) hr.6. If two trains start at the same time from 2 points A and B towards each other and after crossing they take a and b hours in reaching B and A respectively, then ratio of A's speed : B's speed = (b^1/2 : a^1/2).Time and WorkWork is the job or task completed (Distance) in a specified time. Time and Work are in direct proportion. If the amount of work increases, time to complete the work also increases.Work as Man Days{tex}Work = Man \times Days{/tex}If a man does a work in 10 days, total work does is 10 man days. Two men will take 5 days (10 man days). The work done by one man in one day is 1/10thof total work.Pipes and CisternsInlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is known as an outlet.1. If an inlet pipe can fill a cistern in A hours, the part filled in 1 hour = 1/A (same principle as time and work).2. If pipe A is x times bigger than pipe B, then pipe A will take 1/xth(less time) of the time taken by pipe B to fill the cistern.3. If an inlet pipe can fill a tank in A hours and an outlet pipe empties the full tank in B hours (B>A), then the net part filled in 1 hour when both the pipes are opened will be 1/A - 1/B.4. If X and Y fill a cistern in m and n hours respectively, then together they will take (1/m + 1/n) hours to fill the cistern.5. If an inlet pipe fills a cistern in a hours and takes x minutes longer to fill the cistern due to a leak in the cistern, then the time in which the leak will empty the cistern is given by a(1+a/x).AveragesAverage is defined as the ratio of sum of the quantities to the number of quantities.{tex}Average = {{x_1 + x_2 + \dots + x_n} \over n}}{/tex} If each number is increased or decreased by a certain quantity, then the average also increases or decreases by the same quantity. If each number is multiplied or divided by a certain quantity, then the average also gets multiplied or divided by the same quantity.Tip:Suppose a man covers a certain distance at x kmph and an equal distance at y kmph, then the average speed during the whole journey is (2xy/x+y) kmph.Mixtures and AlligationsMixturesare formed when two or more quantities of different values are mixed together.Alligationis a practical method of solving arithmetic problems related to mixtures of ingredients. There are two kinds of problems:1. To find the quantity of a mixture given the quantities of its ingredients. (Alligation Medial)2. To find the amount of each ingredient needed to make a mixture of a given quantity. (Alligation Alternate)Important Facts and FormulaAllegation: It is the rule that enables us to find the ratio in which two of moreingredients at the given price must be mixed to produce a mixture of a desired price.Mean Price: The cost price of a unit quantity of the mixtureis called the mean price.

Rule of Allegation: If two ingredients are mixed then Quantity of Cheaper / Quantity of Dearer = (C.P. of Dearer - Mean Price) / (Mean Price - C.P. of Cheaper)

Cheaper quantity : Dearer quantity = (d - m) : (m - c)Suppose a container contains x units of liquid from which y unitsare taken out and replaced by water. After n operations thequantity of pure liquid = x (1 - y/x)n units.Ratio and ProportionRatioRatiois a relation between two quantities or numbers.A ratio ofaandbis denoted bya:band is read as:a is to b.In a ratio, the first part (a) is called Antecedent and second part (b) is called Consequent.ProportionProportionis a statement that two ratios are equal.When two ratios are equal, the four terms involved, taken in order are called proportional, and they are said tobe in proportion. a/b = c/dContinued ProportionThree quantities are said to be in continued proportion, if the ratio of the first to the second is same as theratio of the second to the third. a/b = b/c; b is called mean proportion.Compounded Ratioof two ratios a/b and c/d is ac/bd or ac : bd.Invertendo:If a : b :: c : d then b : a :: d : cAlternendo:If a : b :: c : d then a : c :: b : dComponendo:If a : b :: c : d then (a +b) : b :: (c +d) : dDividendo:If a : b :: c : d then (a - b) : b :: (c - d) : dComponendo and Dividendo:If a : b :: c : d then (a +b) : (a - b) :: (c +d) : (c - d)Linear EquationsALinear Equationis an equation whose graph is a straight line. Each term has a degree of atmost 1. Each term can have degree 0 (constant term) or degree 1. A linear equation in one variable is an equation that involves only one variable x. Geeral form of linear equation can be written as ax + b = 0. There are no higher or lower order terms such as x2, x3or x1/2.Example 1:6x + 5 = 0 is a linear equation. Note that it is in the form ax + b = 0, where a and b are constants. In this case a = 6 and b = 2. Any equation which can be reduced to form ax + b is also called linear eqation.Example 2:7x + 3 = 9, this can be written as 7x - 6 = 0Example 3:2(x+1) = 6(x-4) is also a linear equation.Solving Linear EquationsAny linear equation can have only one solution. If you solve a linear equation, you will get one value of x. Solving linear equations is very simple. First, open all brackets and take all terms involving x to Left Hand Side (LHS) and constant terms to Right Hand Side (RHS). Then, using simple additions and subtractions, you will get the value of x.Example 4:2(x+1) = 6(x-4)2x + 2 = 6x - 242x - 6x = - 24 - 2- 4x = - 264x = 26x = 26 / 4x = 13 / 2 or 6.5Quadratic Equations{tex}ax^2+bx+c=0{/tex}Linear Equation can have degree of atmost 1 and has only one solution.Quadratic Equationcan have degree of atmost 2 and has two solutions.General form of quadratic equation: ax2+ bx + c = 0, where a, b and c are constants. Note that maximum degree of x is 2.Solving Quadratic EquationsUnlike linear equations, any quadratic equation always has two solutions called roots of quadratic equation. After solving quadratic equation, you will get two values of x. To solve quadratic equation, you can use directly quadratic formula.{tex}x={-b\pm\sqrt{b^2-4ac} \over 2a}{/tex}DiscriminantIn the above quadratic formula, the expression underneath the square root sign is called thediscriminantof the quadratic equation. Discriminant is used to find the nature of roots.{tex}\Delta = b^2 - 4ac{/tex}Case 1: {tex}\Delta > 0{/tex}Real and distinct rootsCase 2: {tex}\Delta = 0{/tex}Real and one distict root (two same roots)Case 3: {tex}\Delta < 0{/tex}Roots are imaginary and occur as complex conjugates of each otherSum and Product of RootsLet {tex}\alpha{/tex} and {tex}\beta{/tex} be the roots of quadratic equation {tex inline}x^2+px+q=0{/tex}{tex}x^2+px+q=(x-\alpha)(x-\beta){/tex}{tex}x^2+px+q=x^2-(\alpha+\beta)x+\alpha \beta{/tex}{tex}p=-(\alpha+\beta){/tex}{tex}q=\alpha \beta{/tex}{tex}\text{Sum of the roots} = -p{/tex}{tex}\text{Product of roots} = q{/tex}ProgressionsA Progression is a sequence of numbers which have some kinf of relation. This relation determines what kind of a progression is. Generally, there are two types of progressions:1. Arithmetic Progression (AP)2. Geometric Progression (GP)Any progression (AP or GP) can be generally expressed as{tex}a_1 + a_2 + a_3 +\dots+ a_{n-1} + a_n{/tex}Total Terms: {tex}n{/tex}First Term: {tex}a_1{/tex}Last Term: {tex}a_n{/tex}Arithmetic ProgressionIn AP, the relation amoung sequence of numbers is that the difference between any two successive numbers is same.Example: 3, 5, 7, 9, 11, 13, ... is an AP with difference 2. This difference is called common difference.{tex}a_n = a_1 + (n - 1)d{/tex}{tex}S_n=\frac{n}{2}( a_1 + a_n)=\frac{n}{2}[ 2a_1 + (n-1)d]{/tex}Geometric ProgressionIn GP, each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.Example: 2, 6, 18, 54, ...{tex}a + ar + ar^2 + ar^3 + ar^4 + \cdots{/tex}{tex}a_n = a\,r^{n-1}{/tex}FunctionsA function is a rule which indicates an operation to perform.Graph Transformations1. y = f(x) + a is the same as the graph y = f(x), shifted upwards by a units.2. y = f(x - a) shifts the graph a units to the right.3. y = f(ax) is a stretch with scale factor 1/a parallel to the x-axis.4. y = a.f(x) is a stretch with scale factor a parallel to the y-axis.Binomial TheoremBinomial Expression: An algebraic expression consisting of two terms with a positive or negative sign between them.Example: (x+y)The expansion of binomial expression raised to power n is calledBinomial Theorem.{tex}{(x + y)^n = x^n + ^nC_1x^{n-1}y + ^nC_2x^{n-2}y^2 + \dots + y^n}{/tex}{tex}^nC_1, ^nC_2, \dots , ^nC_n{/tex} areBinomial Coefficients.Points to Note:1. There are total of (n+1) terms in the expansion.In each term, sum Indices and SurdsLaws of Indices{tex}a^m \times a^n = a^{m+n}{/tex}{tex}a^m \div a^n = a^{m-n}{/tex}{tex}(a^m)^n = a^{mn}{/tex}{tex}a^{1 \over m} = \sqrt[m]{a}{/tex}{tex}a^{-m} = \frac{1}{a^m}{/tex}{tex}a^{\frac{m}{n}} = \sqrt[n]a^m{/tex}{tex}a^0 = 1{/tex}{tex}a^1 = a{/tex}2. of the indices of x and y is equal to n.3. InequalitiesAcomparison relationshipbetween two algebraic expressions or quantities is known as an Inequalities. There are two main types of inequalities:1. Greater than or greater than equal to:{tex inline}> or \geq{/tex}2. Less than or less than equal to:{tex inline}< or \leq{/tex}Rules for solving inequalities1. Rule of Addition and Subtraction:Adding or subtracting a fixed number to each side of an inequality produces an equivalent inequality.2. Rule of multiplication/division by a positive number:All terms on both sides of an inequality can bemultiplied or divided by a positive number.3. Rule of multiplication/division by a negative number:If all terms on both sides of an inequality aremultiplied or divided by a negative number, the sign of the inequality will be reversed.Logarithms{tex}x = b^y{/tex}{tex}\therefore \log x = \log b^y{/tex}{tex} \log x = y \log b{/tex}{tex}{{\log x} \over {\log b}} = y{/tex}{tex}\log_b x = y{/tex}Therefore,{tex}\text{ if }x = b^y,\text{ then }y = \log_b (x){/tex}Logarithmic Identities1. {tex}\log_b(xy) = \log_b(x) + \log_b(y){/tex}2. {tex}\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y){/tex}3. {tex}\log_b(x^d) = d \log_b(x){/tex}4. {tex}\log_b\!\left(\!\sqrt[y]{x}\right) = \begin{matrix}\frac{\log_b(x)}{y}\end{matrix}{/tex}5. {tex}x^{\log_b(y)} = y^{\log_b(x)}{/tex}6. {tex}c\log_b(x)+d\log_b(y) = \log_b(x^c y^d){/tex}7. {tex}\log_b(1) = 0{/tex}8. {tex}\log_b(b) = 1{/tex}9. {tex}b^{\log_b(x)} = x{/tex}10. {tex}\log_b(b^x) = x{/tex}11. {tex}\log_a b = {\log_c b \over \log_c a}{/tex}Permutation & CombinationCombination & Permutation deals with arrangement of thing. If the order doesn't matter, then it is calledCombination. If the order does matter, then it is aPermutation.In other words,Permutation is an ordered Combination.Permutation{tex}{^nP_r} = {n! \over {(n-r)!}{/tex}

There are basically two types of permutation:1. When repetition is allowed2. No repetition1. Permutations with RepetitionTo chooserthings from n when repetition is allowed, the permutations are:n n ... (r times) = nr(Because there arenpossibilities for the first choice, then there arenpossibilites for the second choice, and so on.)CombinationsNumber of ways objects can be selected from a group.{tex}{^nC_r} = {{^nP_r} \over r!}{/tex}Circular PermutationsCircular TableA circular table has no fixed starting or ending point. If n persons are to be arranged in a straight line, there are n! unique ways. When n persons are to sit around a circular table, each arrangement will be repeated n times, There will be (n-1)! different arrangements.Circular WireArrangement of beads (which are all different) around a circular wire diifers from table. Why? Because when you turn it over, you can see other side of it like a mirror image. So, total number of different arrangements decreases by half. Thus, n beads on a circular wire can be arranged in (n-1)!/2 ways.ProbabilityProbabilityis the likelihood or chance of an event occurring.Some Concepts When we toss a coin, then either a Head (H) or a Tail (T) appears. A dice is a solid cube ,having 6 faces,marked 1, 2, 3, 4, 5, 6respectively. When we throw a die, the outcome is the numberthat appears on its upper face. A pack of cards has 52 cards.It has 13 cards of each suit, namely spades, clubs, hearts anddiamonds. Cards of spades and clubs are balck cards.Cards of hearts and diamonds are red cards.There are four honours of each suit.These are Aces, Kings, Queens and Jacks.These are called Face cards.The probability of a certain event occurring can be represented by P(A). The probability of a different event occurring can be written as P(B). Therefore, for two events A and B,{tex}\displaystyle P(A) + P(B) - P(A\cap B) = P(A\cup B){/tex}Mutually Exclusive EventsEvents A and B are mutually exclusive if they have no events in common. If two events are mutually exclusive,{tex}\displaystyle P(A) + P(B) = P(A\cup B){/tex}Independent EventsTwo events are independent if (and only if){tex}\displaystyle P(A\cap B) = P(A)P(B){/tex}Conditional ProbabilityConditional probability is the probability of an event occurring, given that another event has occurred.{tex}\displaystyle P(A|B){/tex} means the probability of A occurring, given that B has occurred.For two events A and B,{tex}\displaystyle P(A\cap B) = P(A|B)P(B){/tex}{tex}\displaystyle P(A\cap B) = P(B|A)P(A){/tex}Geometry BasicsGeometry is branch of mathematics concerned with shapes, sizes and properties of figures. Geometry includes knowledge of angles, lines, triangles, quadrilaterals, circles and polygons.Triangles Sum of the angles of a triangle is 180 degrees. The sum of any two sides of a triangle is greaterthan third side. Pythagoras Theorem:In a right angled triangle (Hypotenuse)2= (Base)2+ (Height)2 The line joining the mid point of a side of a triangleto the opposite vertex is called the Median. The point where the three medians of a triangle meet,is called Centroid. The centroid divides each of themedians in the ratio 2:1. In an isosceles triangle, the altitude from thevertex bisects the base. The median of a triangle divides it into two trianglesof the same area. The area of the triangle formed by joining the mid pointsof the sides of a given triangle is one-fourth of the areaof the given triangle.Similar TrianglesIf the angles of one triangle are equal to the angles of another triangle, then the triangles are said to beEquiangular. Equiangular triangles have the same shape but may have different sizes. Therefore, equiangular triangles are also calledSimilar Triangles.Two triangles are similar if their corresponding angles are equal and corresponding sides are proportional.

{tex}{AB \over DE} = {BC \over EF} = {CA \over FD}{/tex}{tex}\angle A = \angle D{/tex}{tex}\angle B = \angle E{/tex}{tex}\angle C = \angle F{/tex}Properties1. The ratio of area of similar triangles is equal to the square of ratio of sides.2. If the ratio of corresponding sides equals to 1, then triangles become congruent.QuadrilateralA quadrilateral is a polygon with four sides (edges) and four vertices (corners). The word quadrilateral is made of the words quad (meaning four) and lateral (meaning sides). So, quadrilateral is simply a four sided figure.The sum of interior angles for all quadrilaterals must be equal to 360 degrees.Types of Quadrilaterals1. Square2. Rectangle3. Parallelogram4. Rhombus5. Trapezoid3.

Quadrilateral: Square & RectangleSquare:A square has four equal sides and four equal angles (90-degree angles or right angles).

Perimeter of SquarePerimeter of the square is 4 times the length of the side. For a square of length L, perimeter = 4 * LArea of SquareArea of square is the square of length of the side. Area = L * LProperties of Square Thediagonals of a squarebisect each other and meet at right angles (90 degrees) The diagonals of a square bisect its angles The diagonals of a square areperpendicular Opposite sides of a square are bothparallel and equal in length All four angles of a square are equal. (every angle of a square is a right angle) The diagonals of a square are equal.Rectangle:A rectangle is a quadrilateral with 4 right angles. It is similar to square except that its sides are not equal.

Perimeter of RectanglePerimeter of the rectangle is sum of the length of its sides. Perimeter = 2 * (Length + Breadth)Area of RectangleArea of rectangle is multiplication of its length and breadth. Area = Length * BreadthProperties of Rectangle All angles are right angles (90 degrees) Opposite sides are parallel and equal. Diagonals bisect each other.Quadrilateral: ParallelogramAparallelogramis a quadrilateral with two pairs of parallel sides.

Properties of parallelogram Eachdiagonal divides the quadrilateral into twocongruent triangles with the same orientation. The opposite sides are parallel and equal in length. The diagonals bisect each other. The opposite angles are equal in measure. The sum of the squares of the sides equals the sum of the squares of the diagonals. (Parallelogram Law) Adjacent angles aresupplementary (180 degrees). Square, Rectangle and Rhobus are special types of parallelograms.Perimeter & Area of parallelogramThe perimeter of parallelogram is sum of the length of its sides. Perimeter = 2(L + B)Area of parallelogram is multiplication of its longer side and distance between them. Area = L * HParallelogram LawIt states that the sum of the squares of the lengths of the four sides of aparallelogram equals the sum of the squares of the lengths of the two diagonals. If ABCD is parallelogram, then{tex}2(AB)^2+2(BC)^2=(AC)^2+(BD)^2{/tex}In case of rectangle and square, this law reduces to pythagoras theorem.CirclesA circle is a locus of all points which are equidistant from a point.

Circumference{tex}C = 2\pi r = \pi d{/tex}Area of Circle{tex}\mathrm{Area} = \pi r^2{/tex}MensurationMensurationis the branch of mathematics which deals with the study of geometric shapes, their area, volume and different parameters in geometric objects.Important Mensuration Formulae1.Area of rectangle (A) = length(l) * Breath(b) {tex}A = l \times b{/tex}2.Perimeter of a rectangle (P) = 2 * (Length(l) + Breath(b)) {tex}P = 2 \times(l + b){/tex}3.Area of a square (A) = Length (l) * Length (l)A = {tex}l \times l{/tex}4.Perimeter of a square (P) = 4 * Length (l){tex}P = 4 \times l{/tex}5.Area of a parallelogram(A) = Length(l) * Height(h){tex}A = l \times h{/tex}6.Perimeter of a parallelogram (P) = 2 * (length(l) + Breadth(b)) {tex}P = 2 \times (l + b){/tex}7.Area of a triangle (A) = (Base(b) * Height(b)) / 2{tex}A = \frac{1}{2} \times b \times h{/tex}Perimeter = (a + b + c)Area of triangle ={tex inline}A = \sqrt{s(s-a)(s-b)(s-c)}{/tex} [Heros formula]8.Area of triangle (A) = {tex inline}\frac{1}{2} a \times b \times \angle C = \frac{1}{2} b \times c \times \angle A = \frac{1}{2} a \times c \times \angle B{/tex}9.Area of isosceles triangle = {tex inline}\frac{b}{4}\sqrt{4a^2 - b^2}{/tex}10.Area of trapezium (A) = {tex inline}\frac{1}{2} (a+b) \times h{/tex}11.Perimeter of a trapezium (P) = sum of all sides12.Area of rhombus (A) = Product of diagonals / 213.Perimeter of a rhombus (P) = 4 * length14.Area of quadrilateral (A) = 1/2 * Diagonal * (Sum of offsets)15. Area of a Kite (A) = 1/2 * product of its diagonals16.Perimeter of a Kite (A) = 2 * Sum on non-adjacent sides17. Area of a Circle (A) = {tex inline}\pi r^2 = \frac{\pi d^2}{4}{/tex}18.Circumference of a Circle ={tex inline}2 \pi r = \pi d{/tex}19.Total surface area ofcuboid ={tex inline}2 (lb + bh + lh){/tex}20.Total surface area ofcuboid ={tex inline}6 l^2{/tex}21.length of diagonal ofcuboid ={tex inline}\sqrt{l^2+b^2+h^2}{/tex}22.length of diagonal ofcube ={tex inline}\sqrt{3 l}{/tex}23.Volume of cuboid = l * b * h24.Volume of cube = l * l * l25.Area of base of a cone = {tex inline}\pi r^2{/tex}26. Curved surface area of a cone = C = {tex inline}\pi r l{tex}27.Total surface area of a cone ={tex inline}\pi r (r+l){/tex}28.Volume of right circular cone ={tex inline}\frac{1}{3} \pi r^2 h{/tex}29.Surface area of triangular prism = (P * height) + (2 * area of triangle)30.Surface area of polygonal prism = (Perimeter of base * height ) + (Polygonal base area * 2)31.Lateral surface area of prism = Perimeter of base * height32.Volume of Triangular prism = Area of the triangular base * height33.Curved surface area of a cylinder ={tex inline}2 \pi r h{/tex}34.Total surface area of a cylinder ={tex inline}2 \pi r(r + h){/tex}35.Volume of a cylinder ={tex inline}\pi r^2 h{/tex}36.Surface area of sphere ={tex inline}4 \pi r^2 = \pi d^2{/tex}37.Volume of a sphere ={tex inline}\frac{4}{3} \pi r^3 = \frac{1}{6} \pi d^3{/tex}38.Volume of hollow cylinder ={tex inline}\pi r h(R^2-r^2){/tex}39.Surface area of a right square pyramid ={tex inline}a \sqrt{4b^2 - a^2}{/tex}40.Volume of a right square pyramid ={tex inline}\frac{1}{2} \times base \, \, area \times height{/tex}41.Area of a regular hexagon ={tex inline}\frac{3\sqrt{3}a^2}{2}{/tex}42.area of equilateral triangle ={tex inline}\frac{\sqrt{3}}{4} a^2{/tex}43.Curved surface area of a Frustums ={tex inline}\pi h (r_1 + r_2){/tex}44.Total surface area of a Frustums ={tex inline}\pi (r_1^2 + h(r_1+r_2) + r_2^2){/tex}45.Curved surface area of a Hemisphere ={tex inline}2 \pi r^2{/tex}46.Total surface area of a Hemisphere ={tex inline}3 \pi r^2{/tex}47.Volume of a Hemisphere = {tex inline}\frac{2}{3} \pi r^3 = \frac{1}{12} \pi d^3{/tex}48.Area of sector of a circle ={tex inline}\frac{\theta r^2 \pi}{360}{/tex}Vedic Maths - IntroductionVedic Mathematicsis a system of mathematics consisting of a list of 16 basicsutras, or aphorisms. They were presented by a Hindu scholar and mathematician,Bharati Krishna Tirthaji Maharaja, during the early part of the 20th century.The calculation strategies provided by Vedic mathematics are said to be creative and useful, and can be applied in a number of ways to calculation methods in arithmetic and algebra.16 Sutras translated in English (from Sanskrit) are:1. By one more than the one before2. All from 9 and the last from 103. Vertically and Cross-wise4. Transpose and Apply5. If the Samuccaya is the Same it is Zero6. If One is in Ratio the Other is Zero7. By Addition and by Subtraction8. By the Completion or Non-Completion9. Differential Calculus10. By the Deficiency11. Specific and General12. The Remainders by the Last Digit13. The Ultimate and Twice the Penultimate14. By One Less than the One Before15. The Product of the Sums16. All the MultipliersIt is amazing that with the help of Vedic Mathematics, you will be able to solve or calculate complex mathematical problems mentally.By one more than the previous one1. Square of numbers ending in 565 x 65 = (6 x (6+1) ) 25 = (6x7) 25 = 422545 x 45 = (4 x (4+1) ) 25 = (4x5) 25 = 2025105 x 105 = (10 x (10+1) 25 = (10 x 11) 25 = 110252. When sum of the last digits is 10 and previous parts are the same44 x 46 = (4 x (4+1)) (4 x 6) = (4 x 5) (4 x 6) = 202437 x 33 = (3 x (3+1)) (7 x 3) = (3 x 4) (7 x 3) = 122111 x 19 = (1 x (1+1)) (1 x 9) = (1 x 2) (1 x 9) = 209