But before we engage ourselves into taking over this section completely, first let’s have a perspective on what QA is all about and how it is so much important for a paper like CAT. Syllabus:- There is no fix syllabus but the following sections are considered as the most relevant by the experts in this section who followed the pattern over the years.
HCF / LCM
Quadratic Equations Inequalities and Modulus Special Equations Functions and Graphs Maxima and Minima Progressions Logarithms Indices and Surds
The below 2 sections are the most important as they compose the largest part among the total number of questions. a. Number SystemNumber System problems are usually very simple but at times they become quite typical if approached wrongly. They require certain tricks that you can pick up from any good textbook. Number Base System also deserves the mention here. This contributes 3-4 questions to every CAT, and so it is a very important topic. You should be comfortable writing numbers in their algebraic form. You should also learn about divisibility tests and the other points that will the handling of numbers pretty easy. b. AlgebraAlgebra itself comprises of multiple sections which are equally important and a serious aspirant should get his hand into carving all of them. The other topics that you need to look at are Permutations and Combinations, Probability (very basic, including die and card problems and perhaps Bayes’ theorem; basic concepts must be clear), Progressions (A.P, G.P. and A.G.P), Logarithms, Indices and Surds. c. Arithmetic Arithmetic is usually vast but it compose a lesser part and only a few topics make it to the final but since one can never be sure, hence you should have the insight in all the topics. Major topics that you need to cover are Set Theory (especially Venn diagrams) and problems on Time, Speed and Distance, Percentage, Profit and Loss and Simple and Compound Interest. All these topics are covered as part of the school syllabus, but may need some brushing up on as the difficulty somewhat increases. d. Geometry These topics are diagram based hence they are combined together. Maximum weightage is given to geometry among the three of them, although every CAT paper will have 3-4 questions on mensuration, as well as a couple of questions on coordinate geometry. Topics that need to be covered in geometry are basic theorems involving triangles, circles, parallel lines and other shapes. Similarity and Congruency of triangles is an important property to look after for as they shape the pattern and type of the questions. In coordinate geometry, straight lines and circles hold their due importance. Given the equation of a circle, you should be able to comprehend the centre and radius of the circle and plot it with respective intercepts. Tangents and other lines can add up to the complexity. Similarly, you should know what the slope and y-intercept of a given straight line equation is, and be able to draw the line on a piece of graph paper. This will help in forming its equation and solving the problem further. As far as mensuration is concerned, formulae on lateral and total surface areas and volumes of triangles, circles, cylinders, cones, cuboids and spheres are to be learned Some derived formulae are to be taken care off too like finding the slant height in a cone. Mensuration problems require lots of practice as the calculation in finding the surface areas and volumes are pretty difficult and intensive. Before going deep into the knowledge part, first consider some basic points which should be kept in mind while preparing and practising for QA. 1. The QA Questions featured in the CAT are mostly are mostly application oriented and test one’s grasp of fundamentals.So what one really should do is to get all the fundamentals straight and it would be an easy cakewalk thereafter. 2. Overall Approach and AttitudeYour overall frame of mind should always be positive about solving up the questions. Negligence or giving up will only hamper things. 3. Develop A Sense Of Purpose Anyone wanting to do well in QA should develop a great sense of patience and insight in that direction. If any question is not opening up may be it is made up for leaving. But yes giving up must be the last option on the table. 4. The real difficulty does not lie in the task itself; rather it lies in your lack of motivation and clarity as to what the task really means to you. 5. Let Commitment And Hard Work Be Your Fuel. Give your best shot at all the questions. Acquaint yourself with all the fundamentals and then attempt the questions and above all you are your own strength so make up your own tactics and shortcuts. 6. Depending on where your skill levels in QA stand, you will need to proportionately plan for anywhere between one to two hours per day toward QA preparation. 7. Be Confident And Preserve. Being confident will always help you no matter what. You should have faith in yourself that you will crack the question wide upon how much difficult it might be but yes do not let that fool yourself. 8. Prefer Quality Over Quantity Common sense must be understood as the quality of your effort. Quality is sometimes underrated by students. One should not make this fault. The quality aspect of your preparation is so critical that it may in fact make or break your chances of success. 9. Pacing your Preparation Take your preparation at a correct pace. Do not move too fast or too slow practising the exercises. 10. Test Taking Taking tests at adequate time and intervals is very much important as it defines how well or bad you will score or the least it can tell you where you are weak and vice-versa. Make all your plans of attempting the exam and implement them. It will help you find which one you are good at.Now we can move onto some tricks, tips and techniques that are being used in cracking QA section. 1. 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). 2. 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. 3. (m + n)! is divisible by m! * n!. 4. 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. 5. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y). 6. To Find Square of a 3-Digit Number. Let the number be XYZ Steps. a. Last digit = Last digit of Sq(Z). b. Second last digit = 2*Y*Z + any carryover from STEP 1. c. Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2. d. Fourth last digit is 2*X*Y + any carryover from STEP 3. e. Beginning of result will be Sq(X) + any carryover from Step 4. Eg) Let us find the square of 431. a. Last digit = Last digit of Sq(1) = 1. b. Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6. c. Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1. d. Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2. e. Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18 THUS SQ(431) = 185761. 7. If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only. -> The sum of first n natural numbers = n(n+1)/2. -> The sum of squares of first n natural numbers is n(n+1)(2n+1)/6. -> The sum of cubes of first n natural numbers is (n(n+1)/2)2/4. -> The sum of first n even numbers= n (n+1). -> The sum of first n odd numbers= n2. 8. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then -> the total number of factors is (x+1)(y+1)(z+1) .... -> the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c).... -> the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c).... -> the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...) -> Total no. of prime numbers between 1 and 50 is 15. -> Total no. of prime numbers between 51 and 100 is 10. -> Total no. of prime numbers between 101 and 200 is 21. -> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6. -> The number of rectangles in n*m board is given by n+1C2 * m+1C2. 9. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational. 10. The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6 -> The number of rectangles in n*m board is given by n+1C2 * m+1C2. 11. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational. 12. Certain nos. to be remembered. -> 2^10 = 4^5 = 32^2 = 1024. -> 3^8 = 9^4 = 81^2 = 6561. -> 7 * 11 * 13 = 1001. -> 11 * 13 * 17 = 2431. -> 13 * 17 * 19 = 4199. -> 19 * 21 * 23 = 9177. -> 19 * 23 * 29 = 12673. 13. 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. 14. 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. 15. 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. 16. 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. 17. 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. 18. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e. 145 = 1! + 4! + 5!. 19. When 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. 20. When 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. 21. The relationship between base 10 and base ‘e’ in log is given by log10N = 0.434 logeN. 22. WINE and WATER Mixture 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. 23. 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 24. 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 25. When 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. 26. When ‘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. The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003. 27. -> 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. -> 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. -> 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. -> 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). 28. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5. 29. -> 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. -> When an unbiased coin is tossed even no. (2n) of times, then, P (no. of heads=no. of tails) = 1-(2nCn/22n). 30. 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 31. CALENDAR -> Calendar repeats after every 400 years. -> Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400. -> Century has 5 odd days and leap century has 6 odd days. -> 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. -> January 1, 1901 was a Tuesday. 32. -> 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) -> For any regular polygon, the sum of interior angles =(n-2)*180 degrees So measure of one angle is (n-2)/n *180 -> If any parallelogram can be inscribed in a circle, it must be a rectangle. -> If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal). 33. 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). 34. -> 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. -> 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. -> Area of a Rhombus = Product of Diagonals/2. 35. 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]. 36. Area of a triangle -> 1/2*base*altitude -> 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B -> root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2 -> a*b*c/(4*R) where R is the circumradius of the triangle -> r*s ,where r is the inradius of the triangle 37. In any triangle -> a=b*cos C + c*cos B -> b=c*cos A + a*cos C -> c=a*cos B + b*cos A -> a/sin A=b/sin B=c/sin C=2R, where R is the circumradius -> cos C = (a^2 + b^2 - c^2)/2ab -> sin 2A = 2 sin A * cos A -> cos 2A = cos^2 (A) - sin^2 (A) 38. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1. 39. Appollonius Theorem In a triangle ABC, if AD is the median to side BC, then AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2). 40. -> In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base. -> In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides. 41. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle. 42. Let W be any point inside a rectangle ABCD, then, WD2 + WB2 = WC2 + WA2. 43. 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)). 44. -> 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) -> Distance between 2 points (x1, y1) and (x2, y2) is given by Sq((x1-x2)2+ (y1-y2)2). 45. 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. Moving on to the divisibility rules for numbers and examples- Divisibility by 2 -> A number is divisible by 2 if and only if the last digit is divisible by 2. Example: Not required :) Divisibility by 3 -> A number is divisible by 3 if and only if the sum of the digits is divisible by 3. Example: Which of the following two numbers is/are fully divisible by 3: 97533222 or 97533322? Answer: 97533222 Reason: Sum of digits for 97533222 = 33 / 3 = 11 Sum of digits for 97533322 = 34 / 3 = 11.33 Divisibility by 4 -> A number is divisible by 4 if and only if the last 2 digits is a number divisible by 4. Example: Which of the following two number is fully divisible by 4: 5555444446 or 5555444448 ? Answer: 5555444448 Reason: Last 2 digits for 5555444446 are 46, 46 is not fully divisible by 4 (46/4 = 11.5) Last 2 digits for 5555444448 are 48, 48 is fully divisible by 4 (48/4 = 12) Divisibility by 5 -> A number is divisible by 5 if and only if the last digit is divisible by 5. Example: If a number ends in 5 or 0 then it would be divisible by 5. Such as 777995, 13170 Divisibility by 6 -> A number is divisible by 6 if and only if it is divisible by 2 and 3. Example: Which of the following numbers is/are fully divisible by 3: 97533222, 97533322, 97533225, 97533228? Answer: 97533222 and 97533228 Reason: For a number to be fully divisible by 6 it should satisfy following two conditions: 1. It must be a even number (which means out of 4 options we can decide that 97533225 is surely not divisible by 6 as it is not an even number) 2. It must be divisible by 3(sum of all digits of particular even number should be divisible by 3) Sum of digits for 97533222 = 33 (fully divisible by 3) Sum of digits for 97533322 = 34 (not fully divisible by 3) Sum of digits for 97533228 = 39 (fully divisible by 3) Divisibility by 7 -> To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number, if the resulting number is divisible by 7 then original number is divisible by 7. If you don't know the new number's divisibility, you can apply the rule again. Example: Which of the following numbers is/are fully divisible by 7: 203, 3192, 3197, 38241? Answer : 203, 3192, 38241 Reason: For 203, you would double the last digit to get six, and subtract that from 20 to get 14. Hence 203 is divisible by 7. Applying the same rule to check if 3192 is divisible by 7: 3192 => 319 - 2*2 = 315 => 31 - 2*5 = 21, 21 is divisible by 7 hence 3192 is divisible by 7 Applying the same rule to check if 3197 is divisible by 7: 3197 => 319-2*7 = 305 => 30 -2*5 = 20, 20 is not divisible by 7 hence 3197 is not divisible by 7 Applying the same rule to check if 38241 is divisible by 7: 38241 => 3824 -2*1 = 3822 => 382 - 2*2 = 378 => 37 -2*8 = 21, 21 is divisible by 7 hence 38241 is divisible by 7 Divisibility by 8 -> A number is divisible by 8 if and only if the last 3 digits of a number divisible by 8. Example: Which of the following numbers is/are fully divisible by 8: 97533224, 97533328, 97533222, 97533228? Answer: 97533224 and 97533328 Reason: 224 and 328 are divisible by 8 whereas 222 and 228 are not divisible by 8. Divisibility by 9 -> A number is divisible by 9 if and only if the sum of the digits is divisible by 9. Example: Which of the following numbers is/are fully divisible by 9: 111111111, 111222, 2222244, 66669 699999? Answer: 111111111, 111222 and 2222244 Reason: Sum of digits for 111111111, 111222 and 2222244 are 9, 9 and 18 respectively all of which are divisible by 9. Whereas sum of digits for 66669 & 699999 are 33 & 51 respectively both of which are not divisible by 9. Divisibility by 10 -> A number is divisible by 10 if and only if the number ends in n zeros. Example: If a number ends in 0 then it would be divisible by 5. Such as 111110, 876543200 Divisibility by 11 -> A number is divisible by 11 if the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11. Example: Which of the following numbers is/are fully divisible by 11: 6050, 3035362, 90002, 900002? Answer: 6050, 3035362, 90002 Reason: Sum of digits at even place in 6050 = 11 and sum of digits at odd place is 0. Difference between odd and even = 11. Hence 6050 is divisible by 11. Similarly sum of digits at even place in 3035362 = 11 and sum of digits at odd place is 11. Difference between odd and even = 0. Hence 3035362 is divisible by 11. Similarly Difference between odd and even digits for 90002 is 11 whereas for 900002 difference between odd and even digits is 7. Hence 90002 is divisible by 11 but 900002 is not. Divisibility by 12-> A number is divisible by 12 if the number is divisible by both 3 and 4 Example: Which of the following two numbers is/are divisible by 12: 22584 and 22756? Answer: 22584 Reason: 22584 is divisible by 3 because sum of all digits =21 which is divisible by 3 and 22584 is also divisible by 4 because last 2 digits of 22584 i.e. 84 is divisible by 4. Hence 22584 is divisible by 12 22756 is not divisible 3 as sum of digits are 22 which is not divisible by 3. Divisibility by 13 -> To find out if a number is divisible by 13 take the last digit of number and multiple it by 4 and add it to number formed with remaining digits check if the resultant number is divisible by 13. In case resultant number is big repeat process of multiplying last digit by 4 and adding it to number formed by remaining digits. Example: Which of the following numbers is/are divisible by 13: 11013, 110006, 110026 ? Answer: 110006 Reason: Applying the rule above of multiplying last digits by 4 and adding it to number formed by remaining digit we get : 1100+4*4= 11024 Continuing the same way with resultant number we get : 1102+4*4= 1118 => 1118 = 111+8*4 = 143 => 14+3*4 = 26. 26 is divisible by 13. Hence 110006 is divisible by 13 11013 is not divisible by 13 as per results below: 11013= 1101+3*4 = 1113 => 1113 = 111+3*4 = 123 => 123 = 12+3*4 =24 24 is not divisible 13 hence 11013 is not divisible by 13. -> If n is even , n(n+1)(n+2) is divisible by 24. Some more points to remember:
Now based on previous year CAT Question Papers we are giving some illustrations:
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Wednesday, May 8, 2013
Quantitative Ability for CAT,MAT,XAT
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