MANAGEMENT: EXAM PREPARATIONS CAT,MAT,BANK FREE MATERIAL PART-1

Number Systems

Number 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 Number

Starting 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 a unique product of prime numbers.

Examples:

${550 = 2 \times 5^2 \times 11}$

${1200 = 2^4 \times 3 \times 5^2}$

Thus, prime numbers are the basic building blocks of any positive integer. This factorization will also help in finding GCD and LCM quickly.

Perfect Numbers

A 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 Numbers

Two numbers are co-prime to each other, if they do not have any common factor except 1.

Example: 25 and 9, since they don’t have a common factor other than 1

Points to Remember

The number 1 is neither prime nor composite.
The number 2 is the only even number which is prime.
(xⁿ + yⁿ) is divisible by (x + y), when n is an odd number.
(xⁿ – yⁿ) is divisible by (x + y), when n is an even number.
(xⁿ – yⁿ) is divisible by (x – y), when n is an odd or an even number.

Number Systems - II

Factors of a Number

Representing a number as prime factors helps in analyzing problems.

$N = p^a + q^b + r^c$ 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

${{(a+1)(b+1)(c+1)} \over 2}$

Sum of Factors = ${({a^{p+1} - 1})({b^{q+1} - 1})({c^{r+1} - 1}) \over {(a-1)(b-1)(c-1)}}$

Concept of Cyclicity

Concept of cyclicity is used to find unit's digit in case the numbers are occuring in powers.

Cyclicity of 1, 5, 6 - 1

Cyclicity of 4, 9 - 2

Cyclicity of 2, 3, 7, 8 - 4

Maximum Power of p (prime nubmer) in n! (n factorial)

${n \over p} + {n \over p^2} + {n \over p^3} + \dots$

Basic Formulae of Algebra

(a + b)² = a²+ b² + 2ab
(a - b)²= a²+ b² - 2ab
(a + b)² - (a - b)² = 4ab
(a + b)²+ (a - b)² = 2 (a² + b²)
(a²- b²) = (a + b) (a - b)
(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)
(a³ + b³) = (a +b) (a² - ab + b²)
(a³ - b³) = (a - b) (a² + ab + b²)
(a³ + b³ + c³ - 3abc) = (a + b + c) (a² + b² + c² - ab - bc - ca)
If a + b + c = 0, then a³ + b³+ c³ = 3abc

Divisibility Rules

Divisibility by 2

The last digit is even (0, 2, 4, 6, or 8).

Divisibility by 3

The sum of the digits is divisible by 3.

Divisibility by 4

The last two digits divisible by 4.

Divisibility by 5

The last digit is 0 or 5.

Divisibility by 6

The sum of the digits is divisible by 3 and the number itself is divisible by 2.

Divisibility by 7

Subtract 2 times the last digit from the rest.

Divisibility by 8

If 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.

Divisibility by 9

The sum of the digits is divisible by 9.

Divisibility by 10

The last digit is 0.

Divisibility by 11

Add the digits in blocks of two from right to left. Example - 627: 6 + 27 = 33

Divisibility by 12

It is divisible by 3 and by 4.

LCM and HCF

Factors and Multiples

Suppose there are two numbers - a and b. If a number a divides another number b exactly, we say that a is a factor of b and b is called a multiple of a.

Highest Common Factor (HCF) or Greatest Common Divisor (GCD)

The greatest common divisor (gcd), also known as the greatest common denominator, greatest common factor (gcf), or highest 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)

The lowest common multiple or (LCM) least common multiple or smallest common multiple of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. 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

Co-primes: Two numbers are said to be co-primes if their HCF is 1. HCF of two co-prime numbers is 1.

HCF and LCM of Fractions

$HCF= {\mbox{HCF of Numerators} \over \mbox{LCM of Denominator}}$

$LCM = {\mbox{LCM of Numerators} \over \mbox{HCF of Denominator}}$

$\operatorname{gcd}(a,b)=\frac{a\cdot b}{\operatorname{lcm}(a,b)}$

Percentages

Percentage is a way of expressing a number as a fraction of 100 (per cent means per hundred). It is denoted using the percent sign, %.

Example: $25\% = {25 \over 100}$

What 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%.

Interests

The 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.

$Amount = Principal + Interest$

The rate of interest is as per annum (unless indicated).

Simple Interest

Simple interest is simply calculated on principal amount using the following formula:

${SI = {P \times R \times T \over 100}}$

where, P = principal, R = rate per annum, T = time in years

Amount can be calculated by adding interest to principal.

Compound Interest

When the borrower and the lender agree to fix up a certain unit of time (say yearly or half-yearly or quarterly) to settle the previous account. In such cases ,the amount after the first unit of time becomes the principal for the 2nd unit. The amount after second unit becomes the principal for the 3rd unit and so on. After a specified period, the difference between the amount and the money borrowed 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.11000

After 2 years: Interest = Rs.1100; New Principal = 12100

After 3 years: Interest = Rs.1210; You get Rs.13310. So, there is gain of Rs.310 if you lend at compound interest!

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Speed, Time and Distance

The questions on speed, time and distance are based on one general formula:

${Speed \times Time = Distance}$

The Concept

There 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 Streams

In 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 V km/hr, then

speed downstream = (U + V) km/hr
speed upstream = (U - V) km/hr

If 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/hr

Trains

A train has a definite length. The distance covered by the train depends on the length of the train.

Time taken by a train x mt 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 cover x mt.
Time taken by a train x mt long in passing a stationary object of length y mt is equal to time taken by the train to cover (x+y) mt.
Suppose two trains are moving in the same direction at u kmph and v kmph such that u > v, then their relative speed is u-v kmph.
Suppose two trains are moving in opposite direction at u kmph and v kmph then, their relative speed is equal to (u+v) kmph.
If two trains of length x km and y km are moving in opposite diredtions at u kmph and v kmph, then time taken by the train to cross each other is equal to (x+y)/(u+v) hr.
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 Work

Work 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

$Work = Man \times Days$

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 1/10^th of total work.

Pipes and Cisterns

Inlet: 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.

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).
If pipe A is ‘x’ times bigger than pipe B, then pipe A will take 1/x^th (less time) of the time taken by pipe B to fill the cistern.
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.
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.
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).

Averages

Average is defined as the ratio of sum of the quantities to the number of quantities.

$Average = {{x_1 + x_2 + \dots + x_n} \over n}}$

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 Allegations

Mixtures are formed when two or more quantities of different values are mixed together.

Alligation is a practical method of solving arithmetic problems related to mixtures of ingredients. There are two kinds of problems:

To find the quantity of a mixture given the quantities of its ingredients. (Alligation Medial)
To find the amount of each ingredient needed to make a mixture of a given quantity. (Alligation Alternate)

Important Facts and Formula

Allegation: It is the rule that enables us to find the ratio in which two of more ingredients 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 mixture is 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 units are taken out and replaced by water. After n operations the quantity of pure liquid = x (1 - y/x)n units.

Linear Equations

A Linear Equation is 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, x³ or x^1/2.

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 = o

Example 3: 2(x+1) = 6(x-4) is also a linear equation.

Solving Linear Equations

Any 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 - 24

2x - 6x = - 24 - 2

- 4x = - 26

4x = 26

x = 26 / 4

x = 13 / 2 or 6.5

MANAGEMENT

Monday, June 4, 2012

EXAM PREPARATIONS CAT,MAT,BANK FREE MATERIAL PART-1