Wednesday, 22 October, 2014 Search  
Quantitative Ability (QA)
If there is something that can upset the strong will and confidence of even the most serious aspirants it would definitely be the Quantitative Ability. This section has always been an area of contention for students as here no guess work helps at all and neither the choice of eliminating options for that matter. It is a section which over the years have created a dubious image of itself in the minds of the students and has been one of the most important sections in deciding whether they would make it into their dream college or not.
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.

  1. NUMBER SYSTEM

   Number Base System
   HCF / LCM

  1. SET THEORY
  2. ALGEBRA

   Simple Equations
   Quadratic Equations
   Inequalities and Modulus
   Special Equations
   Functions and Graphs
   Maxima and Minima
   Progressions
   Logarithms
   Indices and Surds

  1. PERCENTAGES
  2. PROFIT AND LOSS AND PARTNERSHIPS
  3. SIMPLE INTEREST AND COMPOUND INTEREST
  4. AVERAGE, MIXTURES AND ALLIGATIONS
  5. RATIO, PROPORTION AND VARIATION
  6. GEOMETRY
  7. MENSURATION
  8. TIME AND DISTANCE
  9. TIME AND WORK
  10. PERMUTATIONS AND COMBINATIONS
  11. PROBABILITY
  12. COORDINATE GEOMETRY
  13. TRIGONOMETRY
  14. STATISTICS

All above topics can be divided into 4 main sub heads:-
The below 2 sections are the most important as they compose the largest part among the total number of questions.

a. Number System

Number 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. Algebra

Algebra 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 Attitude

Your 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:

  • 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+3×2+2x+6=0 has no positive roots.
  • 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.

 

  • 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)
  • 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. 

 

  • For a cubic equation ax3+bx2+cx+d=0
    • 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
  • 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
  • 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)

 

  • Consider the two equations, a1x+b1y=c1 and  a2x+b2y=c2 Then,
    • If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
    • If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
    • If a1/a2 <> b1/b2, then we have a unique solution.
  • Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1

 

  • |a| + |b| = |a + b| if a*b>=0 else, |a| + |b| >= |a + b|
  • The equation ax2+bx+c=0 will have max. value when a<0 and min. value when a>0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a

 

  • 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.
  • 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).

 

  • 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.
  • For any 2 numbers a, b where a>b
    • a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
    • (GM)^2 = AM * HM

 

  • For three positive numbers a, b, c
    • (a + b + c) * (1/a + 1/b + 1/c)>=9
  • For any positive integer n
    • 2<= (1 + 1/n)^n <=3

 

  • a^2 + b^2 + c^2 >= ab + bc + ca
    • If a=b=c, then the case of equality holds good.
  • a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)
  • (n!)2 > n^n
  • 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
  • If n is even, n(n+1)(n+2) is divisible by 24

 

  • 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)
  • e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity
    • Note: 2 < e < 3

 

  • 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]
  • (m + n)! is divisible by m! * n!

 

  • 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.
  • Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)

 

  • 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
  • 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
  • If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.

 

  • Certain nos. to be remembered
    • 210 = 45 = 322 = 1024
    • 38 = 94 = 812 = 6561
    • 7 * 11 * 13 = 1001
    • 11 * 13 * 17 = 2431
    • 13 * 17 * 19 = 4199
  • 21 * 23 = 9177
  • 23 * 29 = 12673
  • 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.

 

  • 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.
  • 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
  • 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.

 

  • 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
  • The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003

 

  • In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5
  • 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)
  • Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!

 

  • 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.
  • Finding number of Factors
    • To find the number of factors of a given number, express the number as a product of powers of prime numbers.

 

    • In this case, 48 can be written as 16 * 3 = (24 * 3)
    • Now, increment the power of each of the prime numbers by 1 and multiply the result.

 

    • In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)
    • Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.

 

  • Sum of n natural numbers
      • 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 first n even numbers= n (n+1)
      • The sum of first n odd numbers= n^2
  • Finding Squares of numbers
    • To find the squares of numbers near numbers of which squares are known
    • To find 41^2 , Add 40+41 to 1600 =1681
    • To find 59^2 , Subtract 60^2-(60+59) =3481

 

  • Finding number of Positive Roots
    • If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
    • Eg: x^4+3x^2+2x+6=0 has no positive roots .

 

  • Finding number of Imaginary Roots
    • 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) .
    • Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)
  • Reciprocal Roots
    • The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a

 

  • Roots
    Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1
  • Finding Sum of the roots

 

    • For a cubic equation ax^3+bx^2+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
    • For a biquadratic equation ax^4+bx^3+cx^2+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

 

  • Maximum/Minimum
    -> If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if x=y(=k/2). The maximum product is then (k^2)/4
      • If for two numbers x*y=k(=constant), then their SUM is MINIMUM if x=y(=root(k)). The minimum sum is then 2*root(k) .
  • Inequalties
    -> x + y >= x+y ( stands for absolute value or modulus ) (Useful in solving some inequations)
      • a+b=a+b if a*b>=0 else a+b >= a+b
      • 2<= (1+1/n)^n <=3 -> (1+x)^n ~ (1+nx) if x<<<1> When you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.

 

  • Product Vs HCF-LCM 
    • 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
  • AM GM HM
    • For any 2 numbers a>b a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively) (GM)^2 = AM * HM

 

  • Sum of Exterior Angles
    • 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 in
    • Square-----=90
      Pentagon--=108
      Hexagon---=120
      Heptagon--=128.5
      Octagon---=135
      Nonagon--=140
      Decagon--=144

 

  • Problems on clock
    • Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is , the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute . Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .
    • The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) .
  • Co-ordinates
    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]

 

  • Ratio
    If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to (k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to (a1+a2+a3+............./b1+b2+b3+..........)
  • Finding multiples

 

    • 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)
  • Exponents
    e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity 2 <>GP
      • In a GP the product of any two terms equidistant from a term is always constant.
      • The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
  • Mixtures
    If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture then , A/Q = (1-q/Q)^n

 

  • Some Pythagorean triplets:
    • 3,4,5----------(3^2=4+5)
      5,12,13--------(5^2=12+13)
      7,24,25--------(7^2=24+25)
      8,15,17--------(8^2 / 2 = 15+17 )
    • 9,40,41--------(9^2=40+41)
      11,60,61-------(11^2=60+61)
      12,35,37-------(12^2 / 2 = 35+37)
    • 16,63,65-------(16^2 /2 = 63+65)
    • 20,21,29-------(EXCEPTION)
  • Function
    Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y).

 

  • Finding Squares
    • To find the squares of numbers from 50 to 59
    • For 5X^2 , use the formulae

 

    • (5X)^2 = 5^2 +X / X^2
    • Eg ; (55^2) = 25+5 /25
    • =3025
      (56)^2 = 25+6/36
    • =3136
      (59)^2 = 25+9/81
    • =3481

 

  • Successive Discounts
    • Formula for successive discounts
    • a+b+(ab/100)
      This is used for succesive discounts types of sums.like 1999 population increses by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999 and if there is a decrease then it will be preceeded by a -ve sign and likewise.

 

  • Rules of Logarithms:
  • loga(M)=y if and only if M=ay
  • loga(MN)=loga(M)+loga(N)
  • loga(M/N)=loga(M)-loga(N)
  • loga(Mp)=p*loga(M)
  • loga(1)=0-> loga(ap)=p
  • 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]

Now based on previous year CAT Question Papers we are giving some illustrations:

  • The cost price of 90 apples is equal to the selling price of 60 apples. Find the Percentage profit?
    • a)  50%                  b)  45%                  c)  35%                  d)  40%
    • Answer Option: a
  • Shyam bought a total of 30 set top boxes and TV sets for Rs 500000. He then sold 75% of the set top boxes and 5 TV sets for a profit of Rs 50000. Each set top box was marked up by 20% over cost and each TV set was sold at a profit of Rs 5000. The remaining set top boxes and 5 TV sets couldn’t be sold. What is Shyam’s overall loss (approx.)?
    • a)  Rs 156550          b)  Rs 158300          c)  Rs 149000         d)  Rs 135700
    • Answer Option: b

 

  • Divide Rs.1064 into three parts in such a way that 1/2 of the first part, 1/4 of the second part and 1/8 of the third part are equal. Find value of all the parts.
  • 142, 284, 568     b)   164, 328, 656     c)   158, 316, 632     d)   152, 304, 608
    • Answer Option: d
  • A man can row a boat up the stream in 225 minutes; he can row the same boat downstream in 20 minutes less than he can row it in still water. How long would he take to row down the stream?
  • 25 or 180 minutes                             c)   28 or 202 minutes
  • 65 minutes                                       d)   19 or 140 minutes
    • Answer Option: a

 

  • Find the sum of all possible distinct remainders which are obtained when squares of prime numbers are divided by 6.
  • a)  10                                                  b)  9
  • c)  8                                                    d)  7
  • Answer Option: c
  • Which of the following must be true, if A is any composite number and B is any prime number?
  • a)  A + B is odd                                      b)  AB is odd
  • c)  (A + B) / B is odd                                         d)  none of these
  • Answer Option: d
  • Find the value of the following expression:
  • a)  17/22                b)  53/64                c)  23/25                 d)  67/71
  • Answer Option: b
  • Karunya is given a series by his younger brother to taunt him but instead he solved the problem owing to his good practice. The series is: S = 8 + 88 + 888 +…up to 20 terms.
  • a)  80/81 [1020 - 19] b)  70/81 [1020 - 46] c)  60/81 [1020 - 28]  d)  50/81 [1020 - 37]
  • Answer Option: a
  • 2 taps are opened simultaneously to fill up a water tank, their individual timings being 8 and 12 hours respectively. When the tank is supposed to be full, it is found that there is a leak at the bottom of the tank due to which only 2/3 of the tank volume is filled. How much total time will it take to fill up the tank if the leak is sealed off after the detection?
  • a)  7                      b)  33/8                  c)  5                       d)  32/5
  • Answer Option: d
  • In a coal mining company, the workmen are divided into 2 groups S and F based on their rate of working. 4 from S and 8 from F complete a specific task in 10 days. The same task is completed by 6 from S and 3 from F in 8 days. How much time will 1 from each group together take to complete the same task?
  • a)  48                     b)  45                    c)  36                     d)  none of these
  • Answer Option: b

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