Finding the X

Preliminaries

This section is an extension of the “Linear equations” which most of you might have studied in your high school math class. Therefore, we can start with a discussion of two of the basic algebraic methods for solving a pair of linear equations.

Algebraic methods of solving a pair of linear equations

1. Substitution method:

Illustration 1:

Solve the following pair of equations

7x – 5y = 2                 (1)

x + 2y = 3                  (2)

Solution:

Step 1: We pick either of the equations and write one variable in terms of the other.

It is easier if we first consider the equation in which the coefficient of x or y is 1.

Here in equation (2), the coefficient of x is 1.

Therefore, consider equation (2)

x + 2y = 3

• x = 3 – 2y ——-(3)

Step 2: Substitute the above value of x in equation (1). We get

7(3 – 2y) – 5y = 2

• 21 – 14y – 5y = 2
• 19 = 19y
• y = 1

Step 3: Substituting this value of y in equation (3)

x = 3 – 2 = 1

Thus, x = 1, y = 1 is the solution of the given pair of equations.

1. Elimination method:

Illustration 2:

The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save Rs.2000 per month, find their monthly incomes.

Solution:

Let us denote the incomes of the two persons by 9x and 7x and their expenditures by 4y and 3y respectively. Then the equations formed in the situation is given by:

9x – 4y = 2000 ————(1)

7x – 3y = 2000 ————(2)

Step 1: Multiply equation (1) by 3 and equation (2) by 4 to make the coefficients of y equal.

27x – 12y = 6000 ———(3)

28x – 12y = 8000 ———(4)

Step 2: Subtract equation (3) from equation (4) to eliminate y, because the coefficients of y are the same.

(28x – 27x) – (0) = 8000 – 6000

• x = 2000

Step 3: Substituting this value of x in equation (1), we get

9(2000) – 4y = 2000

• 4y = 18000 – 2000
• 4y = 16000
• y = 4000

Thus, the solution of the equations is x = 2000, y = 4000.

Therefore, the monthly incomes of the persons are Rs.18,000 and Rs.14,000 respectively.

Daily life problems

These are the problems which depicts everyday life situations. You should analyse the situation and express it in equations to get the required answer. All these questions demand you to express them in one or two variables, say x and y. A naive outline for approaching these problems is given below.

Step 1: Analyse the question and decide whether to use one variable or two variables (“x” or “x” and “y”)

Step 2: Express the situation in terms of the above variable(s).

Step 3: Solve the equation to find the value of the variable(s).

Tip: Always assign variable “x” for what you require as the final answer.

Age problem

Age Problem is one of the most important type in the daily life problems.

In the age problem, statements regarding age of one or more people will be given. You should analyse those statements to find the age of a particular person asked in the question.

“For detailed theory, refer the book “CSIR-NET General Aptitude – A New Outlook”

Distances and Directions

Direction Sense Test

Points to remember

• A person facing towards North, on taking left turn will face towards West and on taking the right turn towards the East. To make it simple, remember that right turn means clockwise rotation and left turn means anticlockwise rotation.
• Each turn mentioned in the question is equivalent to an angle rotation of 90o.
• In order to determine the shortest straight distance between the two given points the Pythagoras theorem should be used.

Pythagoras Theorem

In a right triangle, (Hypotenuse)2 = (Base)2 + (Height)2

The numbers which satisfy this relation, are called Pythagorean triplets.
Example:
(3, 4, 5), (5, 12, 13), (6, 8, 10), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (15, 20, 25) etc.

Note: If you see a triangle with sides exactly as any of these Pythagorean triplets, you can be sure that the triangle is a right triangle.

Heights and Distances

Problems under this section usually involves the use of trigonometry in finding the height of a building or object, mostly will involve a triangle, with at least an angle given. There can be questions in which some right-angled triangles are involved and applying Pythagoras theorem will do the trick in most of those problems. A good hold of solving the quadratic equations is also required to solve some of the problems. The primary step to solve any of this question is to picture the given situation.

Weight is measured in Kilograms and Pounds. Likewise, there are two units to measure angles: degrees and radians.

In almost all cases, especially cases involving  (and if the answer options are not in degrees), it is better to use radians for calculation.

“For detailed theory, refer the book “CSIR-NET General Aptitude – A New Outlook”

How to approach questions on Probability?

The first objective while trying to solve any question in probability is to define the EVENT. The event whose probability is to be determined is described in the question and your task in trying to solve the problem is to define it.

You can define the event either narrowly or broadly. Narrow definitions of events are the building blocks of any probability problem and whenever there is a doubt about a problem, it is better to get into the narrowest form of the event definition.

The difference between the narrow and broad definition of event is explained through the following illustration.

Illustration :

What is the probability of getting a number greater than 3, in a throw of an unbiased dice having 6 faces?

(a) 1/6                          (b) 1/2                         (c) 1/3                          (d) 2/3

The broad definition of the event here is getting a number greater than 3 and this probability is 3/6. However, this event can also be broken down into its basic definitions as:

The event is defined as getting 4 or 5 or 6. The individual probabilities of each of these are 1/6, 1/6 and 1/6 respectively.

Hence the required probability is 1/6 + 1/6 + 1/6 = 3/6 =1/2. Option (b).

Although in this example it seems highly trivial, the narrow event-definition approach is very effective in solving difficult problems on probability.

In general, event definition means breaking up the event to the most basic building blocks, which have to be connected together through the two English conjunctions – AND and OR.

The use of the conjunction AND

Whenever you see AND as the natural conjunction joining two separate parts of the event definition, you can replace the AND with the multiplication sign.

Thus if events A AND B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A AND B occur is got by connecting P(A) AND P(B). Replacing the AND with multiplication sign, we get the required probability as: P(A) × P(B)

Example

If the probability of student A being in a particular team is 1/4 and the probability of student B being in the same team is 1/3, then the probability that both the students are in the same particular team is got by

Event Definition: A being in the team AND B being in the team.

That is, P(A) × P(B) = 1/4 × 1/3 = 1/12

Note that since we use the conjunction AND in the event definition here, we multiply the individual probabilities that are connected by the conjunction AND.

The use of the conjunction OR

Whenever you see OR as the natural conjunction joining two separate parts of the event definition, you can replace the OR by the addition sign.

Thus if events A OR B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A OR B occur is got by connecting P(A) OR P(B). Replacing the OR by addition sign, we get the required probability as:
P(A) + P(B).

Example

If the probability of student A being in a particular team is 1/4 and the probability of student B being in the same team is 1/3, then the probability that either of the two students are in the particular team is got by

Event Definition: A being in the team OR B being in the team.

That is, P(A) + P(B) = 1/4 + 1/3 = 7/12

Note that since we use the conjunction OR in the event definition here, we add the individual probabilities that are connected by the conjunction OR.

Combination of AND and OR

In examination, questions may not be simple enough to solve directly using any of the two conjunction AND or OR. However, no matter how complicated the problem on probability is, it can be broken down into its narrower parts, which can be connected by ANDs and ORs to get the event definition. Once the event is defined, the probability of each narrow event within the broad event is calculated and all the narrow events are connected by Multiplication (for AND) or by Addition (for OR) to get the final solution.

Example: What is the probability that the sum of the numbers is not less than 11, when two dice are thrown simultaneously?

Event definition: The sum of the numbers is not less than 11 if it is either 11 or 12.

That is, (6 AND 5) OR (5 AND 6) OR (6 AND 6)

= (1/6×1/6) + (1/6×1/6) + (1/6×1/6) = 1/36 + 1/36 + 1/36 = 3/36 = 1/12

Another approach of Probability

The probability of an event is defined as

P(Event) =

This means that the probability of any event can be obtained by counting the numerator and the denominator independently.

The counting can be done using different methods like physical counting, use of concepts of permutations and combinations etc.

Example: When we roll dice, we could define event A as the occurrence of even numbers.
Here, S = {1, 2, 3, 4, 5, 6}, n(S) = 6, A ={2, 4, 6}, n(A)=Number of outcomes = 3

P(Occurrence of even numbers) =  =  =  =

• The probability of any event ranges from zero to one. That is, 0 ≤ P(Event) ≤ 1.
• The sum of probabilities of all the events in a sample space is 1. That is,

P(E1) + P(E2) + P(E3) + … = 1.

• For any two events A and B, we have

P(A ᴜ B) = P(A) + P(B) – P(A ∩ B).

• For any two mutually exclusive events A and B, we have

P(A ᴜ B) = P(A) + P(B).

Permutations:

A permutation is an arrangement of all or part of a set of objects with regard to the order of the arrangement. Note that here the order of arrangements matters.

Example 1: Suppose we have a set of three letters: A, B, and C. Consider the different ways by which we can arrange two letters from that set. Each possible arrangement would be an example of a permutation. The complete list of possible permutations would be AB, BA, AC, CA, BC and CB.

Example 2: All permutations made with the letters A, B, C taking all at a time are {ABC, BCA, CAB, CBA, BAC, ACB}

Combinations:

A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected.

Example: Suppose we have a set of three letters: A, B, and C. Consider the different ways by which we can select 2 letters from that set. Each possible selection would be an example of a combination. The complete list of possible selections would be: AB, AC, and BC. Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter (unlike permutations).

Tip to understand whether permutation or combination to be used:

A permutation focuses on the arrangement of objects with regard to the order in which they are arranged. A combination in contrast, focuses on the selection of objects without regard to the order in which they are selected.

Hence seeing a question, first check whether the order matters. If the order matters, it is permutation; and combination if otherwise.

If the question is to arrange some objects, it is usually permutation. On the other hand, if it is to select some objects from a given set of objects, then it is combination.

“For detailed theory, refer the book “CSIR-NET General Aptitude – A New Outlook”

1.1 Natural Numbers

The counting numbers are called natural numbers.

Hence 1,2,3,4… are all natural numbers.

1.2 Integers

All counting numbers including zero and the negatives of the counting numbers form the set of integers.

Hence …, -3, -2, -1, 0, 1, 2, 3,… are all integers.

Set of positive integers = {1, 2, 3, 4, 5,…}

Set of negative integers = {-1, -2, -3, -4, -5,…}

Set of all non-negative integers = {0, 1, 2, 3, 4, 5,…}

1.3 Even Numbers

An integer which is exactly divisible by two (2) is called an even number.

Thus -6, -4, -2, 0, 2, 4, 6, 8, 10,… are all even numbers.

1.4 Odd Numbers

An integer which is not exactly divisible by two (2) is called an odd number.

Thus -5, -3, -1, 1, 3, 5, 7, 9, 11,… are all odd numbers.

1.5 Prime Numbers

A counting number is called a prime number if it has exactly two positive factors (divisors), namely itself and 1.

Example: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97 are the primes less than 100.

• Test for finding out whether the given number is prime or not:

Let p be a given number and let n be the smallest counting number such that n2 ≥ p.

Now, test whether p is divisible by any of the prime numbers less than or equal to n.

If yes, then p is not prime. Otherwise, p is prime.

Example: Check whether 67 is a prime number.

We know that, 92 ˃ 67.

Prime numbers less than 9 are 2, 3, 5, and 7.

Clearly, none of them divides 67.

Thus, 67 is a prime number.

• Prime Factorization

The prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is the process of listing of the integer’s prime factors, together with their multiplicities.

Example 1: Prime factorization of 240 = 2 × 2 × 2 × 2 × 3 × 5 = 243151.

in which the factors 2, 3, 5 have multiplicities 4, 1, 1 respectively.

Example 2: Prime factorization of 3150 = 2 × 3 × 3 × 5 × 5 × 7 = 21325271.

in which the factors 2, 3 , 5, 7 have multiplicities 1, 2, 2, 1 respectively.

1.6 Composite Number

The natural numbers, which are not prime, are called composite numbers.

4, 6, 8, 9, 10, 12… are all composite numbers.

Note: 1 is neither prime nor composite.

1.7 Perfect Square

A perfect square is a number that can be written as the product of two equal factors.

Example: 36, 49, 64,… are all perfect squares.

Important properties of perfect squares:

• No perfect square ends with 2, 3, 7, 8.
• No perfect square ends with an odd number of zeros.
• Perfect squares can be recognized by the fact that all of their prime factors have even multiplicities.

Example: Prime factorization of 900 = 223252 in which multiplicities of the factors 2, 3, 5 are 2, 2, 2 respectively, which are all even. Thus 900 is a perfect square.

2   Tests of Divisibility

Divisibility by 2

A number is divisible by 2 if the last digit is even. i.e., 0, 2, 4, 6 or 8.

Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3.

Divisibility by 4

A number is divisible by 4 if the number formed by the last 2 digits is divisible by 4.

Divisibility by 5

A number is divisible by 5 if its last digit is 0 or 5.

Divisibility by 6

A number is divisible by 6 if the number is divisible by both 2 and 3.

Divisibility by 7

To check whether a given number is divisible by 7, subtract twice the unit digit from the number represented by the remaining digits of the number to obtain an integer.

The original number is divisible by 7 if the obtained integer is divisible by 7.

Example: 343 is divisible by 7, we take the unit digit 3, double it to 6, and then subtract it from 34, then 34 – 6 = 28, which is divisible by 7.

Divisibility by 8

A number is divisible by 8 if the number formed by the last 3 digits is divisible by 8.

Divisibility by 9

A number is divisible by 9 if the sum of the digits is divisible by 9.

Divisibility by 10

A number is divisible by 10 if it ends in a 0.

Divisibility by 11

A number is divisible by 11 if the difference of the sum of the digits in the even and odd positions in the number is either 0 or divisible by 11.

Example: 979 is divisible by 11,

Sum of digits in the odd positions = 9 + 9 = 18

Sum of digits in the even positions = 7

Difference = 18 – 7 = 11, which is divisible by 11.

3  GCD and LCM

Factors and Multiples:

If a number, say, ‘a’ exactly divides another number ‘b’, we can say that ‘a’ is a factor of ‘b’. Further, ‘b’ is called a multiple of ‘a’.

Greatest Common Divisor (GCD or HCF):

The GCD or HCF of two or more than two numbers is the greatest number that divides each of them exactly.

Method for finding the GCD of two or more numbers:

Factorization Method: Express each one of the given numbers as the product of prime factors (Prime factorize). The product of least powers of common prime factors gives GCD.

Example: Find the GCD of 504 and 264

Prime factors of 504 = 2 × 2 × 2 × 3 × 3 × 7 = 233271

Prime factors of 264 = 2 × 2 × 2 × 3 × 11 = 2331111

Prime factors common to 504 and 264 = 2331 = 2 × 2 × 2 × 3 = 24

Thus, the GCD of 504 and 264 is 24.

Least Common Multiple (LCM)

The LCM of two or more numbers is the smallest number which is exactly divisible by each one of the given numbers.

Method for finding the LCM of two or more numbers:

Factorization Method: Express each one of the given numbers as the product of prime factors (Prime factorize). The product of highest powers of all the prime factors gives their LCM.

Example: Find LCM of 504 and 264

Prime factors of 504 = 2 × 2 × 2 × 3 × 3 × 7 = 233271.

Prime factors of 264 = 2 × 2 × 2 × 3 × 11 = 2331111.

Product of highest powers of all prime factors of 504 and 264 = 23327111= 2 × 2 × 2 × 3 × 3 × 7 × 11= 5544.

Thus, the LCM of 504 and 264 is 5544.

Important Properties of GCD and LCM

• Product of two numbers = Product of their GCD and LCM
• GCD of given numbers always divides their LCM

4  BODMAS rule

BODMAS (Bracket – of – Division – Multiplication – Addition – Subtraction) rule depicts the correct sequence in which the operations are to be executed, to find out the value of a given mathematical expression.

Thus, in simplifying an expression, first of all the brackets must be removed (by solving the expressions in the brackets) strictly in the order ( ), { }, and [ ].

After removing the brackets, we must use the following operations strictly in the order:

(i) Division     (ii) Multiplication       (iii) Addition    (iv) Subtraction.

5   Modulus of a real number:

Modulus of a real number x is defined as

|x| = x if x > 0

|x| = -x if x < 0

In simple words, modulus function takes the positive value of any real number.

Example 1: |-10| = 10.
Example 2: |15|  = 15.

A quadratic equation is an equation of the form f(x): ax2 + bx + c = 0, where a,b,c are all real and a ≠0, and x is an unknown variable.  The values of x satisfying f(x) = 0 are called its roots.

• In ax2 + bx +c = 0, the expression D = b2 – 4ac is called its discriminant.
• Let α, β be the roots of the equation, ax2 + bx +c = 0.

Sum of roots = (α + β) = – b/a   Product of roots = αβ = c/a

• Whenever the roots of a quadratic equation are given, the quadratic equation is given by x2 – (Sum of roots) x + Product of roots = 0.

Example: If the roots are given as 3 and 5, then the quadratic equation will be

• x2 – (3 + 5)x + (3 × 5) = 0
• x2 – 8x + 15 = 0.

7  Fractions

A fraction represents a part of a whole or, more generally, any number of equal parts.  A simple fraction (examples:   and 5/7) consists of an integer numerator, displayed above a line (or before a slash), and a non-zero integer denominator, displayed below (or after) that line.

Proper fraction: A fraction whose numerator is less than the denominator.

Examples: , 3/8, 7/11

Improper fraction: A fraction whose numerator is equal to or greater than the denominator.

Examples: , 8/3 , 21/13.

Mixed Fraction: An improper fraction can be expressed as a whole number and a proper fraction. This expression is called mixed fraction.

Examples: 32/5 .

Method to convert improper fraction into mixed fraction

Step 1: Divide the numerator by the denominator.

Step 2: Write down the whole number answer (Quotient).

Step 3: Then write down the remainder above the denominator.

8  Laws of Exponents

Properties of exponents

• am × an = am+n
• am / an= am-n
• (am)n= amn
• (ab)m = am.bm
• For any natural numbers a, b, c, d,

ab> ac> ad, if and only if b > c > d.

• For three positive real numbers ‘a’,‘b’,‘c’ with the same positive valued exponent x,

ax> bx> cx if and only if a > b > c.

Some important expansions

• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 – 2ab + b2
• a2 – b2 = (a + b)(a – b)
• (a + b)3 = a3 + 3a2b + 3ab2 + b3
• (a – b)3 = a3 – 3a2b + 3ab2 – b3

9  Last digit of a power

Note: While multiplying two multiple digit numbers, the last digit of the product is the last digit of the product of the last digit of the two numbers. For example, 332 × 332 =110224, the last digit of this product can be simply obtained by multiplying 2 × 2 = 4.
The last digit of a number ab forms a particular sequence or order depending on the unit digit of the number (a) and the power the number is raised to (b) This sequence can be called the power cycle of a number and thus it depends on its unit digit.

Consider the power cycle of 2:

 21 = 2, 25 = 32, 22 = 4, 26 = 64, 23 = 8, 27 = 128, 24 = 16 28 = 256

Clearly, we can observe that the unit digit gets repeated after every 4th power of 2.

Hence, we can say that 2 has a power cycle of 2, 4, 6, 8, with frequency 4.

Thus,

24k+1have last digit as 2.

24k+2 have last digit as 4.

24k+3 have last digit as 8.

24k+4 have last digit as 6.         where, k = 0, 1, 2, 3, …

Note that

4k + 1 means the set of natural numbers, which gives remainder 1 when divided by 4.

4k + 2 means the set of natural numbers, which gives remainder 2 when divided by 4.

4k + 3 means the set of natural numbers, which gives remainder 3 when divided by 4.

4k + 4 means the set of natural numbers, which gives remainder 0 when divided by 4.

Note: For all the numbers ending in 2, the power cycle and its frequency is the same as that of 2.

Thus to find the last digit of a number raised to any power, we need to know the power cycle of digits from 0 to 9 which are given below.

 Last digit Power Cycle Frequency 0 0 1 1 1 1 2 2, 4, 8, 6 4 3 3, 9, 7, 1 4 4 4, 6 2 5 5 1 6 6 1 7 7, 9, 3, 1 4 8 8, 4, 2, 6 4 9 9, 1 2

Note that if you can learn this table, you can find the last digit of any number of the form ab in few seconds.

“For detailed theory, refer the book “CSIR-NET General Aptitude – A New Outlook”