All the standard Algebraic Identities are derived from the Binomial

Theorem, which is given as:

Below are some of the Standard Algebraic Identities:

Examples for square of sums:

Examples for square of difference:

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• Divisibility Rule of 2:

If a number is even or a number whose last digit is an even number i.e.  2,4,6,8 including 0, it is always completely divisible by 2.

Example: 508 is an even number and is divisible by 2 but 509 is not an even number, hence it is not divisible by 2. Procedure to check whether 508 is divisible by 2 or not is as follows:

• Consider the number 508
• Just take the last digit 8 and divide it by 2
• If the last digit 8 is divisible by 2 then the number 508 is also divisible by 2.
• Divisibility Rules for 3:

Divisibility rule for 3 states that a number is completely divisible by 3 if the sum of its digits is divisible by 3.

Consider a number, 308. To check whether 308 is divisible by 3 or not, take sum of the digits (i.e. 3+0+8= 11). Now check whether the sum is divisible by 3 or not. If the sum is a multiple of 3, then the original number is also divisible by 3. Here, since 11 is not divisible by 3, 308 is also not divisible by 3.

Similarly, 516 is divisible by 3 completely as the sum of its digits i.e. 5+1+6=12, is a multiple of 3.

• Divisibility Rule of 4:

If the last two digits of a number are divisible by 4, then that number is a multiple of 4 and is divisible by 4 completely.

Example: Take the number 2308. Consider the last two digits i.e.  08. As 08 is divisible by 4, the original number 2308 is also divisible by 4.

• Divisibility Rule of 5:

Numbers, which last with digits, 0 or 5 are always divisible by 5.

Example: 10, 10000, 10000005, 595, 396524850, etc.

• Divisibility Rule of 6:

Numbers which are divisible by both 2 and 3 are divisible by 6. That is, if the last digit of the given number is even and the sum of its digits is a multiple of 3, then the given number is also a multiple of 6.

Example: 630, the number is divisible by 2 as the last digit is 0.
The sum of digits is 6+3+0 = 9, which is also divisible by 3.
Hence, 630 is divisible by 6. 

• Divisibility Rules for 7:

The rule for divisibility by 7 is a bit complicated which can be understood by the steps given below:

Example: Is 1073 divisible by 7?

From the rule stated remove 3 from the number and double it, which becomes 6.

Remaining number becomes 107, so 107-6 = 101.

Repeating the process one more time, we have 1 x 2 = 2.

Remaining number 10 – 2 = 8.

As 8 is not divisible by 7, hence the number 1073 is not divisible by 7.

• Divisibility Rule of 8:

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

Example: Take number 24344. Consider the last two digits i.e.  344. As 344 is divisible by 8, the original number 24344 is also divisible by 8.

• Divisibility Rule of 9:

The rule for divisibility by 9 is similar to divisibility rule for 3. That is, if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.

Example: Consider 78532, as the sum of its digits (7+8+5+3+2) is 25, which is not divisible by 9, hence 78532 is not divisible by 9.

• Divisibility Rule of 10:

Divisibility rule for 10 states that any number whose last digit is 0, is divisible by 10.

Example: 10, 20, 30, 1000, 5000, 60000, etc.

• Divisibility Rules for 11:

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

In order to check whether a number like 2143 is divisible by 11, below is the following procedure.

• Group the alternative digits i.e. digits which are in odd places together and digits in even places together. Here 24 and 13 are two groups.
• Take the sum of the digits of each group i.e. 2+4=6 and 1+3= 4
• Now find the difference of the sums; 6-4=2
• If the difference is divisible by 11, then the original number is also divisible by 11. Here 2 is the difference which is not divisible by 11.
• Therefore, 2143 is not divisible by 11.

A few more conditions are there to test the divisibility of a number by 11. They are explained here with the help of examples:

If the number of digits of a number is even, then add the first digit and subtract the last digit from the rest of the number. 

Example: 3784

Number of digits = 4

Now, 78 + 3 – 4 = 77 = 7 × 11

Thus, 3784 is divisible by 11.

If the number of digits of a number is odd, then subtract the first and the last digits from the rest of the number. 

Example: 82907

Number of digits = 5

Now, 290 – 8 – 7 = 275 × 11

Thus, 82907 is divisible by 11.

Form the groups of two digits from the right end digit to the left end of the number and add the resultant groups. If the sum is a multiple of 11, then the number is divisible by 11.

Example: 3774: = 37 + 74 = 111: = 1 + 11 = 12 

3774 is not divisible by 11.

253: = 2 + 53 = 55 = 5 × 11

253 is divisible by 11.

Subtract the last digit of the number from the rest of the number. If the resultant value is a multiple of 11, then the original number will be divisible by 11.

Example: 9647

9647: = 964 – 7 = 957

957: = 95 – 7 = 88 = 8 × 11

Thus, 9647 is divisible by 11.

• Divisibility Rule of 12:

If the number is divisible by both 3 and 4, then the number is divisible by 12 exactly. 

Example: 5864

Sum of the digits = 5 + 8 + 6 + 4 = 23 (not a multiple of 3)

Last two digits = 64 (divisible by 4)

The given number 5846 is divisible by 4 but not by 3; hence, it is not divisible by 12.

• Divisibility Rules for 13:

For any given number, to check if it is divisible by 13, we have to add four times of the last digit of the number to the remaining number and repeat the process until you get a two-digit number.  Now check if that two-digit number is divisible by 13 or not. If it is divisible, then the given number is divisible by 13.

For example: 2795 → 279 + (5 x 4) 

→ 279 + (20) 

→ 299 

→ 29 + (9 x 4) 

→ 29 + 36 

→65

Number 65 is divisible by 13, 13 x 5 = 65.

• Divisibility Rule for 14:

Check whether the given number is divisible by 2 and 7. If the number is divisible by these numbers, then the original number is also divisible by 14.

Example

Find if the number 224 is divisible by 14

Solution:

112 x 2 = 224 32 x 7 = 224 Now, it is clear that 224 is divisible by 2 and 7. Hence, it is also divisible by 14.

• Divisibility Rule for 15:

If a number is divisible by both 3 and 5, then it is divisible by 15.

Example: Check whether 41295 is divisible by 15.

Solution:

We know that if the given number is divisible by both 3 and 5, then it is divisible by 15.

First, check whether the given number is divisible by 3.

Sum of the digits:

4 + 1 + 2 + 9 + 5 = 21

Sum of the digits (21) is a multiple of 3.

So, the given number is divisible by 3. 

Now, check whether the given number is divisible by 5.

In the given number 41295, the digit in one's place is 5. 

So, the number 41295 is divisible by 5. 

Now, it is clear that the given number 41295 is divisible by both 3 and 5. 

Therefore, the number 41295 is divisible by 15.

• Divisibility Rule for 17:

A number is divisible by 17 if you multiply the last digit by 5 and subtract that from the rest. If that result is divisible by 17, then your number is divisible by 17.

Example: 98 - (6 x 5) = 68. Since, 68 is divisible by 17, then 986 is also divisible by 17.

However, 876 is not divisible by 17 because 87 - (6 x 5) = 57 and 57 is not divisible by 17.

• Divisibility Rule for 19:

To determine if a number is divisible by 19, take the last digit and multiply it by 2. Then add that to the rest of the number. If the result is divisible by 19, then the number is divisible by 19.

Example: 475 is divisible by 19 because 47 + (5 x 2) = 57, and 57 is divisible by 19.

However, 575 is not divisible by 19 because 57 + (5 x 2) = 67, and 67 is not divisible by 19.

• Divisibility Rule for 23:

To determine if a number is divisible by 23, take the last digit and multiply it by 7. Then add that to the rest of the number. If the result is divisible by 23, then the number is divisible by 23.

Example: 575 is divisible by 23 because 57 + (5 x 7) = 92 and 92 is divisible by 23.

However, 576 is not divisible by 23 because 57 + (6 x 7) = 99, and 99 is not divisible by 23.

• Divisibility Rule for 29:

Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.

Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.

• Divisibility Rule for 31:

Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.

Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.

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Let  and  be fractions with b ≠ 0 and d ≠ 0.

• Fractional Equality:

Example: If x/3 = 5/2 then find the value of x.

Solution:

Given: x/3 = 5/2

2x = 15 (.^.. from the above formula)

x = 15/2

Therefore, the value of x is 15/2.

• Fractional Equivalency:

Example: The given fractions 5/16 and x/12 are equivalent fractions, then find the value of x.

Solution:

Given: 5/16 = x/12

x = (5 x 12)/16

x = 60/16

x =15/4

Therefore, the value of x is 15/4.

• Addition (like denominators):

Example: If  then find the value of x+y?

Solution:

Given: 

(.^.. from the above formula)

x + y = 6

Therefore, the value of x + y is 6.

• Addition (unlike denominators):

Note: bd is the common denominator 

Example 1: If  then find the value of 2p + 3q when pq = 1?

Solution:

Given: 

 (.^.. from the above formula)

 (.^.. from the given problem pq = 1)

Therefore, the value of  is 4.

Example 2:

Solution:

First, let’s separate out the denominators and simplify them.

and

We are adding the reciprocals of these two fractions:

• Subtraction (like denominators):

Example: If   then find the value of 

Solution:

Given: 

 (.^.. from the above formula)

Therefore, the value of  is 20.

• Subtraction (unlike denominators):

Example: If  then find the value of  when xy = 2?

Solution:

Given: 

 (.^.. from the above formula)

 (.^.. from the given problem xy = 2)

Therefore, the value of  is 4.

• Multiplication:

Example: If   then find the value of xy?

Solution:

Given: 

 (.^.. from the above formula)

 = 100

Therefore, the value of xy is 100.

• Division:

Example:

Solution:

Let’s think about this is in stages.  First, call the entire denominator D; then (0.2)/D = 4.  From this, we must recognize that D must be 1/4 of 0.2, or D = 0.05.

Now, set that denominator equal to 0.05.

0.3 – x = 0.05

x = 0.3 – 0.05 = 0.25 = 1/4

• Division (missing quantity):

Example:

If then find the value of x?

Solution:

Given that 

then  (.^.. from the above formula)

Therefore x = 100

• Reduction of Complex Fraction:

Example:

If  then find the value of ab?

Solution:

Given that 

then  (.^.. from the above formula)

Therefore ab = 50

• Placement of Sign:

Example:

If  then find the value of 

Solution:

Given that  

then 

Therefore 

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• Addition:

Example: 

Solution:

We have 

So  and 

• Subtraction:

Example:

Solution:

Putting these together,

• Multiplication:

Example:  Find the value of the following expression

Solution:

Use the properties of exponents as follows:

• Distributed over a Simple Product:

Example: If  then find the value of 

Solution:

Given that

Therefore the value of  is 6.

• Distributed over a Complex Product:

Example:  Find the value of the following expression

Solution:

Use the properties of exponents as follows:

• Distributed over a Simple Quotient:

Example:  Find the value of the following expression

Solution:

Use the properties of exponents as follows:

• Distributed over a Complex Quotient:

Example:  Find the value of the following expression

Solution:

Use the properties of exponents as follows:

• Definition of Negative Exponent:

Example:  Find the value of the n if

Solution:

 Therefore,  

• Definition of Radical Expression:

Example:  Convert the below expression into radical form

Solution:

• Definition of Zero Exponent:

Example:  Find the value of the n if

Solution:

Given that

Divide both sides by 

Therefore n = 1

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• Basic Definitions:

Example:  Find the value of the below expression

Solution:

Let's combine the two radicals into one radical and simplify.

Therefore  (.^.. from the above formula)

• Complex Radical:

Example: Find the value of the 

Solution:

• Associative:

Example: Find the value of the 

Solution:

 (.^.. from the above formula)

• Simple Product:

Example: Find the value of the 

Solution:

 (.^.. from the above formula)

• Simple Quotient:

Example: Find the value of the 

Solution:

 (.^.. from the above formula)

• Complex Product:

Example: Find the value of the 

Solution:

 (.^.. from the above formula)

• Complex Quotient:

Example: Find the value of the  

Solution:

 (.^.. from the above formula)

• Nesting:

Example: Find the value of the 

Solution:

 (.^.. from the above formula)

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In equations that involve variables, in general, do not divide each side of an equation by a variable. For example, if we are given xy = y, we cannot divide both sides by y, and conclude that x = 1. What if y = 0? In that case, x could take on any value. In general, bring all the terms to one side, equate to zero, and then factor.

which means either y = 0 or x = 1.

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The degree of an algebraic expression is defined as the highest power of the variables present in the expression.

Degree 1: Linear

Degree 2: Quadratic

Degree 3: Cubic

Degree 4: Bi-quadratic

Examples:

x + y the degree is 1.

 the degree is 2.

 the degree is 3.

 the degree of x is 3, degree of z is 5, degree of the expression is 5.

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A quadratic equation has the form  where a,b, and c are real numbers, and  for example  The values of x that satisfy a given quadratic equation are called roots. The roots of any quadratic equation can be obtained by factoring:

When x is either 3 or -2, the above quadratic equation is satisfied.

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Example:

Solve for 

Split 14 into two parts that add up to 14 and whose product

120 can be written as product of 2 numbers as follows:

The sum is 14 when we split it as 20 and -6

Examples:

 

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Examples:

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The quadratic equations that are tested on the GMAT can be factored by following these steps. We will use the example of 

Multiply the coefficient of  (2 here ) and the constant  which gives a value of -66

Find two numbers that multiply to give -66 but add up to the coefficient of x, which is -5. The two numbers are -11 and 6.

Rewrite the middle term as the sum of these two numbers

Factor the largest common term from the first two terms, and also from the last two terms.

The roots are then obtained by solving  and  which gives  and -3 as the two roots of the quadratic equation

Some quadratic equations have only one real solution, and this results when  Graphically this means that the corresponding parabola  is tangent to the x -axis.

Some quadratic equations have no real solutions, and this results when  Graphically this means that the corresponding parabola  does not intersect the x -axis. It either lies completely above or below the x -axis.

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The solutions to the quadratic equation  can also be obtained by using the formula:

Example:  we have  and  The quadratic formula yields

The two solutions are  and 

Example problems on Quadratic Equations:

• Using Quadratic formula:

Example: Solve for x If 

Solution:

Solve using the quadratic formula:

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What kind of a parabola will be formed by a quadratic equation whose D > 0? If the discriminant of the equation, D > 0, the roots will be real and distinct.

i.e., the parabola will have two points where its y value will become zero or it will cut the x-axis at points as shown below.

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What kind of a parabola will be formed by a quadratic equation whose D = 0 ?

If the discriminant of the equation, D = 0, the roots will be real and equal.

i.e., the parabola will have only one point where its y value will become zero or it will touch the x-axis at one point as shown below.

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What kind of a parabola will be formed by a quadratic equation whose D < 0?

If the discriminant of the equation, D < 0, the roots will be imaginary.

i.e., there exists no real value for which its y-value will become 0.

So, neither will it cut the x-axis nor will it touch the x-axis.

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Discriminant 

D > 0. Roots are real and distinct.

Verifying by finding roots:  So, roots are 2 and 5.

Roots are real and equal as shown by discriminant rule.

             

Discriminant  Roots real and equal Verifying by finding roots:  So, roots are 4 and 4. Roots are real and equal as shown by the discriminant rule.

          

Discriminant  Roots imaginary.

The roots are 

Roots are imaginary as shown by the discriminant rule.

 Using Factorization:

Example:  Solve for x if 

Solution:

First, rewrite the quadratic equation in standard form by FOILING out the product on the left, then collecting all of the terms on the left side:

Now factor the quadratic expression  to two binomial factors  replacing the question marks with two integers whose product is 36 and whose sum is  These numbers are  so:

or

The solution set is {3,12}

Example: 

Solution:

To determine the coefficient of the middle term in a binomial (k, in this case), it's helpful to know the factorization of the binomial. The question gives you part of it, so you can set up the following equation:

 is the product of  and x, so x = 7 :

y is the same as 1y, so the value of k must be 1.

Quadratic equation involving coordinate geometry:

Example:  y = x² + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k are integers, what is the least value of b?

Solution:

The given equation is a quadratic equation. A quadratic equation when plotted on a graph sheet (x - y plane) will result in a parabola.

The roots of the quadratic equation are computed by equating the y = 0

So, the roots of the quadratic equation are the points where the parabola cuts the x-axis.

The question mentions that the curve described by the equation cuts the x-axis at (h, 0) and (k, 0). So, h and k are the roots of the quadratic equation.

For quadratic equations of the form quadratic expression to 0. i.e., the roots are the values that 'x' take when form  the sum of the roots  The sum of the roots of this equation is   Note : Higher the value of 'b', i.e., higher the sum of the roots of this quadratic equation, lower the value of b.

For quadratic equations of the form  the product of roots  Therefore, the product of the roots of this equation  i.e.,  and k are both integers.

So, h and k are both integral factors of 256.

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An inequality is an expression that compares the relative sizes of numbers, expressions, points, lines, or curves. Unlike equations, in which both sides are always equal, inequalities have unequal sides. There are four main inequality signs: Note that x > y and y < x convey the same information.

• Inequality signs:

The most familiar inequality sign is “not equal sign (≠)”. But to compare the values on the inequalities, the following symbols are used. 

• Strict Inequality

The strict inequality symbols are less than symbol (<) and greater than symbol (>). These two symbols are called strict inequalities as it shows the numbers are strictly greater than or less than each other.

For example,

• 5 < 9 (5 is strictly less than 9)
• 10 >7 (10 is strictly greater than 7)

 

• Slack Inequality

The slack inequalities are less than or equal to symbol (≤) and greater than or equal to symbol (≥). The slack inequalities represent the relation between two inequalities that are not strict.

For example,

• x ≥ 15 (x is greater than or equal to 15)
• x ≤ 9 (x is less than or equal to 9) 

Now that you are familiar with the inequality’s signs and expressions, let’s look at a scenario in which you are given and inequality:

x ≤ 2

How can you understand it better?

The best way to visualize and understand an inequality is by representing it on a number line.

Considering the inequality, x ≤ 2. Here’s how you can represent it on a number line.

Let us consider another example: x > 5

Do you find any difference between the two number lines?

Yes! We are sure you have noticed it.

The first number line has a closed or shaded circle, whereas the second number line has an open or unshaded circle.

Here’s what they mean:

1. A closed (shaded) circle at the endpoint of the shaded portion of the number line indicates that the graph is inclusive of that endpoint, as in the case of inequality signs, ≤ or ≥.

Here’s another example for you:

-3 ≤ x ≤ 4

2. An open (unshaded) circle at the endpoint of the shaded portion of the number line indicates that the graph is not inclusive of that endpoint, as in the case of < or >.

Here’s another example for you:

-5 ≤ x < 5

• Properties of Inequalities:

The following are the properties of the inequalities: 

Transitive Property

The relation between the three numbers is defined using the transitive property.

If a, b and c are the three numbers, then

If a ≥ b, and b ≥ c, then a ≥ c

Similarly,

If a ≤ b, and b ≤ c, then a ≤ c

In the above-mentioned example, if one relation is defined by strict inequality, then the result should also be in strict inequality.

For example,

If a ≥ b, and b > c, then a > c.

• Basic rules for inequalities:

Inequality rule 1

Same Number may be added to (or subtracted from) both sides of an

Inequation without changing the sign of inequality.

Example: x > 5 ---------> x + 2 > 5 + 2

Inequality rule 2:

Both sides of an Inequation can be multiplied (or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an Inequation are multiplied or divided by a negative number.

Examples: x > 5 ----> 3x > 15; x > 1 ----> -x < -1

Inequality rule 3:

Any term of an Inequation may be taken to the other side with its sign

changed without affecting the sign of inequality.

Example: 

Inequality rule 4:

In inequalities never multiply for a term unless you know it does not equal 0 AND you know its sign.

• Types of inequalities:

Linear Inequations with one variable

Let a be a non-zero real number and x be a variable.

Then Inequations of the form ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0 are known as linear Inequations in one variable.

Examples: 9x – 15 > 0, 2x – 3 ≤ 0

Linear Inequations with two variables

Let a, b be non-zero real numbers and x, y be a variables.

Then Inequations of the form ax + by < c, ax + by ≤ c, ax + by > c, ax + by ≥ c are known as linear Inequations in two variables x and y

Examples: 2x + 3y ≤ 6, 2x + y ≥ 6

Quadratic Inequation

Let a be a non-zero real number.

Then Inequations of the form  are known as Quadratic Inequations with one variable.

Examples: 

The equation  is a sample quadric Inequation with two

variables. 

System of Inequalities:

A system of inequalities is a set of two or more inequalities in one or more variables. Systems of inequalities are used when a problem requires a range of solutions, and there is more than one constraint on those solutions.

Example:

Solution:

Solution to  inequality

Rule: Between the roots, the expression is negative.

Solution set: 

Solution to  inequality

 

Algorithm to solve Quadratic Inequations:

1) Obtain the Inequation

2) Obtain the factors of Inequation

3) Place them on number line. The number line will get divided into the three regions

4) Mark the leftmost region with + sign and rest two regions with – and + signs respectively.

5) If the Inequation is of the form  the region having minus sign will be the solution of inequality

6) If the Inequation is of the form  the region having plus sign will be the solutions of inequality

Example:

If  which of the following describes all possible values of x ?

A. 3 ≥ x ≥ -3

B. x ≥ 3 or x ≤ -3

C. 3 ≥ x ≥ 0

D. -3 ≥ x

E. 3 ≤ x

Solution:

 Addition in Inequalities

If a > b, and c is any real number, then 

Two different inequalities can always be added. This is the most common operation with inequalities on the GMAT.

If a > b and c > d, then a + c > b + d

In words, the sum of two larger quantities exceeds the sum of the two smaller quantities. For example, 7 > 4 and 6 < 2, and  or 13 > 6.

 Subtraction in Inequalities

If a > b, and c is any real number, then 

If a > b, and c > d, then  Note that this is equivalent to adding these two inequalities, which gives  and then rearranging to yield 

Never subtract two inequalities, in general if a > b and c > d, then  For example, 7 > 4 and 6 > 2, however,  or 1.

Examples for Addition and subtraction of inequalities:

Example 1:

If x + 5 > 2 and  the value of x must be between which of the following pairs of numbers?

(A) -3 and 10
(B) -3 and 4
(C) 2 and 7
(D) 3 and 4
(E) 3 and 10

Solution:

x + 5 > 2

Subtract ‘5’ from both side we get

and

Add ‘3’ on both sides we get

Example 2:  Solve the linear inequality 7x+3 < 5x+9

Solution:

Given inequality is 7x+3 < 5x+9.

Subtract 5x on both the sides of inequality

Thus,

⇒ 7x+3-5x < 5x+9-5x

⇒2x+3 <9

⇒2x < 9-3

⇒ 2x < 6

⇒ x < 3

Hence, the simplified form of the linear inequality 7x+3 < 5x+9 is x < 3.

Example 3:  Given that 2x + 7 > 5 and 5x - 13 < 7, all values of x must be between which of the following pairs of integers?

A. -4 and -1
B. -1 and 4
C. -4 and 1
D. -2 and 5
E. 2 and 5

Solution:

Given that

2x + 7 > 5

Subtract ‘5’ from both side we get

2x > 5-7

2x> -2

x > -1

and

5x - 13 < 7

Add ‘13’ on both sides we get

5x < 7 + 13

5x < 20

x < 4

Thus, 

Hence, option B is the correct answer.

 Multiplication in Inequalities

For any real numbers a,b, and any positive number c.

The converse of the above statement is also true.

For any positive numbers a,b,c and d.

 Division in Inequalities

For any real numbers a, b, and any positive number c.

The converse of the above statement is also true.

For any positive numbers a, b, c and d.

Examples for Multiplication and Division of inequalities:

Example 1:

Given that  which of the following cannot be the value of x?

A. -3/2
B. -1
C. -3/4
D. -3/5
E. -1/2

Solution:

We can see that x has to be negative from the inequality and also options have all negative values. So, we can multiply both sides by x and change the inequality sign. 

Divide both sides by -2, and change the inequality sign

Example 2:

Solution:

x can be positive, or negative so we need to take two cases:

Case 1: If  multiply both sides by 3x without flipping the inequality sign

Multiply both sides by -1 (this will flip the inequality sign)

But x was assumed to be non negative here so we have no solution in this range.

Case 2: If  multiply both sides by 3x by flipping the inequality sign.

Multiply both sides by -1 (this will flip the inequality sign)

So here, 

Answer (D)

 

 Rewriting Inequalities

Do not multiply or divide by a variable if you don't know whether the quantity is positive or negative. For example, if we are given  and if we multiply both sides by x, we incorrectly conclude that the only values of x that satisfy this inequality is  A proper way to obtain the solution is:

Therefore, either x ≥ 1 or x < 0. In general, collect all the terms on one side by subtracting or adding, and then determine the values of the variable which satisfy the given inequality.

Example:

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Solution:

So, x is 5

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The absolute value of the real number x, denoted by |x| is defined to be x if x is positive or zero, and to be -x if x is negative.

In other words, 

Properties of Absolute Values:

When a and b are of opposite signs, magnitude of column B will be the sum of their magnitudes. So, column B will be greater than column A.

When a and b are of the same sign and  column A will be negative.

So, column B will be greater than column A.

 When a and b are of opposite signs, magnitude of column B will be the sum of their magnitudes. So, column B will be greater than column A.

When a and b are of the same sign and  column A will be negative.

So, column B will be greater than column A.

When a and b are of the same sign and  column A will be equal to column B.

Conditions of Absolute Values:

 Absolute Value: Geometric Interpretation

The absolute value of the real number x can also be interpreted as the distance from the origin to the point x on the number line.

For example, | − 3| = 3, means that −3 is 3 units away from the origin (x = 0) on the number line.

Examples: 

The distance between a number, x, and number y on the number line is given by |x − y|.

 Square Root and Absolute Value

 Absolute Value Equalities

For any number a,

For any numbers a and b,

For any number a,

 Absolute Value Inequality rules

 

The solution set of  where 

The solution set of  where 

|x| < 0 -------> No solution (Remember: |x|≤ 0 can have a solution for x = 0)

|x| ≥ 0 -------> -∞ < x < ∞ (all real numbers)

  if and only if  or in other words,  or 

  if and only if  or in other words,  or 

, if and only if  or in other words 

, if and only if  or in other words  or 

Note: The phrase “if and only if” implies that both statements are either true or both are false, meaning if we are given |a| + |b| = |a + b|, then ab ≥ 0, and if we are given ab ≥ 0, then |a| + |b| = |a + b|. These types of inequalities are commonly seen in data sufficiency problems.

Examples of absolute value inequalities:

Example 1:

If |x + 1 | > 2x - 1, which of the following represents the correct range of values of x?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Solution:

Scan the answer choices

Notice that some answer choices say that x = 1 is a solution and some say x = 1 is NOT a solution.

So, let's test x = 1

Plug it into the original inequality to get: |1 + 1 | > 2(1) - 1

Simplify to get: 2 > 1

Perfect!

So, x = 1 IS a solution to the inequality.

Since answer choices A and C do NOT include x = 1 as a solution, we can ELIMINATE them.

Now scan the remaining answer choices (B, D and E)

Some answer choices say that x = -1 is a solution and some say x = -1 is NOT a solution.

So, let's test x = -1

Plug it into the original inequality to get: |(-1) + 1 | > 2(-1) - 1

Simplify to get: 0 > -3

Perfect!

So, x = -1 IS a solution to the inequality.

Since answer choices D and E do NOT include x = -1 as a solution, we can ELIMINATE them.

We're left with B, So answer is B.

Example 2:

If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Solution:

Let’s first solve for when (12x - 5) and (7 - 6x) are both positive.

12x - 5 > 7 - 6x

18x > 12

x > 12/18

x > 2/3

Now let’s solve for when (12x - 5) is negative and (7 - 6x) is positive.

-(12x - 5) > 7 - 6x

-12x + 5 > 7 - 6x

-2 > 6x

-1/3 > x

So we have x < -1/3 or x > 2/3.

If x were -1/3 or 2/3, then the product would be -2/9. However, the inequalities specify that x can be neither -1/3 nor 2/3, so we know the product of two possible values of x cannot be -2/9.

---

When we have two equations of the type

If  Infinite solutions

If  No solutions

If  Unique solution

Type -1:  No Solutions:

No solutions

The lines represented by these two equations are parallel lines. They have the same slope but different intercepts. They never meet.

 

Example 1:

When will two equations in two variables have no solution?

Let us revisit example 2. Solve  and 

Coefficients of x and y in equation (1) are 1 and 4 respectively. Coefficients of x and y in equation (2) are 3 and 12 respectively.

The coefficients of x and y in equation (2) are thrice the first one.

However, the constant term of equation (2) is NOT thrice that of equation (1).

If one of the equations is  and the second one is  such that  we will have NO solution.

The equations will have no solution if 

Solve for x and y: x +4y = 46 and 3x + 12y= 90

Multiply the first equation by 3: 3(x + 4y = 46) = 3x + 12y= 138. 

We have to solve 3x + 12y = 138 (1) and 3x + 12y = 90

Subtracting (2) from (1), we get 0 = 48 which is absurd.

There is no set of values for x and y that will satisfy both these equations.

This system of equations is said to be inconsistent. It has no solution. 

Example 2:

Solve the following system of equations  and 

Multiply equation (1) by 3 :

Compare equations (2) and (3):

The x and y coefficients of equations (2) and (3) are the same. The constant terms are different.

This means there is no solution for this system of equations.

Type -2:  Infinite Solutions:

If  infinite solutions

The two lines represented by these two equations are coincident lines.

When will two equations in two variables have infinite solutions?

Let us take one more look at the last example.

 and 

The second equation was 3 times the first one.

If one of the equations is  and the second one is  such that

  we will have infinite solutions. In other words, if one equation is k times the other, the system of equations will have infinitely many solutions.

The equations will have infinite solution if 

Example 1:

Solve the following system of equations:  and 

Multiply equation (1) by 3:

Compare equations (2) and (3). They are the same.

In other words, 

This means there are infinite solutions for the equations. 

Example 2:

Solve for x and y:  and 

Multiply the first equation by 3: 

The second equation is 3 times the first one.

Subtract one equation from the other we get 0 = 0.

These two equations are identical.

We cannot solve for unique values of x and y with two identical equations.

There are many pairs of values for x and y that satisfy both equations. Here are two examples.

This system of equation has infinitely many solutions.

Type -3:  Unique Solution:

 Unique solution

In a set of linear simultaneous equations, a unique solution exists if and only if, (a) the number of unknowns and the number of equations are equal, (b) all equations are consistent, and (c) there is no linear dependence between any two or more equations, that is, all equations are independent.

Example:

Solve :  and 

Observe both equations have  and  term.

Let  and 

Subtract equation (2) from (1), we get  Substitute  in equation (1).

 and 

---

Problem: 

Short cut Method:

Solution:

---

Problem: 

Short cut Method:

Solution:

---

Problem: 

Short cut Method:

Solution:

---

Problem: 

Short cut Method:

Solution:

---

Problem: 

Short cut Method:

Solution:

---

Problem:  find the value of 

Normal Method:

Shortcut Method:

then

that means from the given problem

---

Problem: 

Normal Method:

Squaring on both sides we get

Squaring on both sides we get

Shortcut Method:

then

that means from the given problem

and

Like Wise

---

Problem: 

Normal Method:

Squaring on both sides we get

Shortcut Method:

then

that means from the given problem

---

Average of two or more numbers/quantities is called the mean of these numbers, which is given by

Or

Or

• Example 1:

The average of 5 numbers is 6. The average of 3 of them is 8. What is the

average of the remaining two numbers?

Solution:

The average of 5 quantities is 6.

We know that 

Therefore, the sum of the 5 numbers is 5 × 6 = 30

The average of three of these 5 numbers is 8

Therefore, the sum of these three numbers = 3 × 8 = 24

Average of these two remaining numbers = 6/2 = 3

 

Example 2:

The arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14, x is

12. What is the value of x?

Solution:

As per our Trick (1)

Given that 

Then

Therefore x = 7.

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If the given observations (x) are occurring with certain frequency (A) then,

 where,  are frequencies

 

Example 1:

The average income of 10 persons is $125 and that of another 5 persons is

$80. The average income of the whole group is?

Solution:

As per our Trick (2)

Average income of whole group

Therefore, average income of the whole group is $110.

Example 2:

A man bought 15 articles at $8 each, 13 at $6 each and 12 at $4 each. The

average price per article is?

Solution:

As per our Trick (2)

Required average price

Therefore, average price per article is $6.15.

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The average of ‘n’ consecutive natural numbers starting from 1 to n i.e.

Example 1:

The average of the first 100 positive integers is?

Solution:

As per our Trick (3)

Average of these numbers

  Required average

Example 2:

What is the average of all the natural numbers from 49 to 125?

Solution:

As we know,

---

The average of squares of ‘n’ consecutive natural numbers starting from 1 i.e.

Example:

The average of the squares of first ten natural numbers is?

Solution:

As per our Trick (4)

---

The average of cubes of first ‘n’ consecutive natural numbers i.e.

Example:

The average of the cubes of first ten natural numbers is?

Solution:

As per our Trick (5)

---

The average of first ‘n’ consecutive even natural numbers i.e.

Example:

The average of the first ten even natural numbers is?

Solution:

Normal Method:

Shortcut Method:

Here, 2n = 10

then n = 10

As per our Trick (6)

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The average of first ‘n’ consecutive odd natural numbers i.e.

Example:

The average of odd numbers up to 100 is?

Solution:

Odd numbers are 1, 3, 5, ............., 99

Total odd numbers are= 50

 Average = 50

---

The average of certain consecutive numbers a, b, c, ......... n is 

 

Example:

The average of odd numbers up to 100 is?

Solution:

Odd numbers are 1, 3, 5, ............., 99

As per our above formula

Here a = 1 and n = 99

---

Example:

The average of the first nine integral multiples of 3 is

Solution:

---

Example-1:

The average of the marks obtained in an examination by 8 students was 51

and by 9 other students was 68. The average marks of all 17 students was?

Solution:

Example-2:

If the average marks of three batches of 55, 60 and 45 students respectively

is 50, 55 and 60, then the average marks of all the students is?

Solution:

The required average marks

Example-3:

Solution:

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If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y. (or vice versa) then the average of remaining numbers will be

(ii) Average of remaining numbers

Example:

The average of 20 numbers is 15 and the average of first five is 12. The

average of the rest is?

Solution:

Here, m = 20, x = 15 n = 5, y = 12

Then m > n

---

 

Therefore, least of the numbers are 9.

---

Example:

Out of four numbers the average of the first three is 16 and that of the last three is 15. If the last number is 20 then the first number is?

Solution:

Normal Method:

Shortcut Method:

---

If in the group of N persons, a new person comes at the place of a person of ‘T’ years, so that average age, increases by ‘t’ years

Then, the age of the new person = T + N.t

If the average age decreases by ‘t’ years after entry of new person, then the age of the new person = T – N.t

 

Example:

3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is?

Solution:

Normal Method:

Shortcut Method:

---

The average age of a group of N students is ‘T’ years. If ‘n’ students join, the average age of the group increases by ‘t’ years, then average age of new students

If the average age of the group decreases by ‘t’ years, then average age of new students

Example-1:

The mean of 9 observations is 16. One more observation is included and

the new mean becomes 17. The 10th observation is?

Solution:

Normal Method:

Sum of Ten observations – Sum of nine observations = Tenth observation

 Tenth observation = 10 × 17 – 16 × 9 = 170 – 144 = 26

Shortcut Method:

By Applying Trick 15,

Here, N = 9, T = 16 n = 1, t = 1

Example-2:

The average of 100 numbers is 44. The average of these 100 numbers and 4

other new numbers are 50. The average of the four new numbers will be?

Solution:

By Applying Trick 15,

Here, N = 100, T = 44, n = 4, t = 

---

 

Example:

 

Solution:

By Applying Trick 16,

Here, n = 10, m = 4, x = 22, y= 28 and z = 23

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If in a group, one member is replaced by a new member, then, 

 

Example-1:

The average weight of 8 persons increases by 2.5 kg when a new person

comes in place of one of them weighing 65 kg. The weight of the new

person is?

Solution:

By Applying 

Here, x = 2.5, n = 8

Example-2:

In a class, there are 40 boys and their average age is 16 years. One boy, aged 17 years, leaving the class and another joining, the average age becomes 15.875 years. The age of the new boy is?

Solution:

Example-3:

In a class there are 30 boys and their average age is 17 years. On one boy aged 18 years leaving the class and another joining, the average age becomes 16.9 years. The age of new boy is?

Solution:

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Example-1:

Average weight of 25 students of a class is 50 kg. If the weight of the class teacher is included, the average is increased by 1 kg. The weight of the teacher is?

Solution:

By Applying Trick 18,

Here, Average = 50,

n = 25,

= 50 + 1 (25 + 1)

= 50 + 26

= 76 kg

Example-2:

The mean weight of 34 students of a school is 42 kg. If the weight of the teacher be included, the mean rises by 400 grams. Find the weight of the teacher (in kg)?

Solution:

By Applying Trick 18,

Here, Average = 42,

n = 34,

= 42 + 0.4 (34 + 1)

= 42+ 14

= 56 kg

The average age of four brothers is 12 years. If the age of their mother is also included, the average is increased by 5 years. The age of the mother (in years) is?

Solution:

By Applying Trick 18,

Here, Average = 12,

x = 5, n = 4

Age of mother = Average + x (n+1)

= 12 + 5 (4 + 1)

= 12 + 25

= 37 years

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If a member leaves the group, then income (or age) of left member =  

where, x = increase (+) or decrease (–) in average income (or age) n = Number of members.

 

Example:

The average of five numbers is 140. If one number is excluded, the average of the remaining four numbers is 130. The excluded number is?

Solution:

By Applying Trick 19,

Here, Average = 140,

x = (140 – 130) = 10

n = 5

Excluded number = Average + x (n – 1)

= 140 + 10 × 4

= 180

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If average of n numbers is m later on it was found that a number ‘a’ was misread as ‘b’.

Example-1:

The average of a collection of 20 measurements was calculated to be 56 cm.

But later it was found that a mistake had occurred in one of the measurements which was recorded as 64 cm., but should have been 61 cm.

The correct average must be

Solution:

By Applying Trick 20,

Here, n = 20, m = 56

a = 61, b = 64

Example-2:

The average marks of 100 students were found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct average corresponding to the correct score.

Solution:

By Applying Trick 20,

Here, n = 100, m = 40

a = 53, b = 83

Example-3:

The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg instead of 87kg. The correct average weight is?

Solution:

By Applying Trick 20,

Here, n = 20, m = 89.4

a = 87, b = 78

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If the average of n numbers is m later on it was found that two numbers a and b misread as p and q.

Example-1:

The mean of 50 numbers is 30. Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean.

Solution:

By Applying Trick 21,

Here, n = 50, m = 30

a = 28, b = 31

p = 82, q = 13

Example-2:

The average of marks of 14 student was calculated as 71. But it was later found that the marks of one student had been wrongly entered as 42 instead of 56 and of another as 74 instead of 32. The correct average is?

Solution:

By Applying Trick 21,

Here, n = 14, m = 71

a = 56, b = 42

p = 32, q = 74

Example-3:

Mean of 10 numbers is 30. Later on, it was observed that numbers 15, 23 are wrongly taken as 51, 32. The correct mean is?

Solution:

By Applying Trick 21,

Here, n = 10, m = 30

a = 15, b = 23

p = 51, q = 32

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Least common Multiple (L.C.M):

The least number which is divisible by two or more given numbers, that least number is called L.C.M. of the numbers.

Ex: L.C.M. of 3,5,6 is 30, because all 3 numbers divide 30, 60, 90, ...... and soon perfectly and 30 is minimum of them.

Highest Common Factor (H.C.F):

It is also called Greatest common Divisor (G.C.D). When a greatest number divides perfectly the two or more given numbers then that number is called the H.C.F. of two or more given numbers.

Ex: The H.C.F of 10, 20, 30 is 10 as they are perfectly divided by 10,5 and 2 and 10 is highest or greatest of them.

Factor and Multiple:

If a number m, divides perfectly second number n, then m is called the factor of n and n is called the multiple of m.

 

Trick #22

Finding LCM by Large Number method:

 

Case - 1:

LCM of 6, 12, 24, 48

 

In this case remaining numbers 6, 12, 24 are divisors of 48 then the LCM of the numbers is 48.

Case - 2:

LCM of 5, 10, 20, 50

In this case only 5, 10 are divisors of 50 then eliminate 5, 10 from the list. 20, 50

Note: If there is no common factor between two numbers, then L.C.M. will be the product of both numbers.

100 is the common multiple. Then the LCM of the numbers is 100.

Case - 3:

LCM of 8, 16, 19, 32

In this case only 8, 16 are divisors of 32 then eliminate 8, 16 from the list.19, 32

But 19 is prime then the LCM becomes

Then the LCM of the numbers is 608.

Example-1:

Find the LCM of the numbers 7, 14, 35, 70

Solution:

By Applying our case – 1 we can directly say the LCM is 70 as all the remaining numbers are divisors of highest number 70.

Example-2:

Find the LCM of the numbers 3, 9, 27, 63

Solution:

By Applying our case – 2 we can directly say the LCM is 189 as 9, 18 are divisors of highest number 72 and 189 is common least multiple of remaining numbers 27 and 63.

Therefore LCM = 189.

Example-3:

Find the LCM of the numbers 8, 11, 32, 64

Solution:

By Applying our case – 3 we can directly say the LCM is 704 as 8, 32 are divisors of highest number 64 and 11 is the prime number.

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Finding HCF By Applying minimum difference method:

Case - 1:

HCF of 6, 12, 18

 If the difference between the numbers are same then we need to consider the common difference.

In this case the common difference is 6.

Note: If the common difference is a prime number then the HCF will be the prime number if that prime number is divisible by all of the given numbers else HCF will be ‘1’.

Else

 we need to prime factorize the difference.

 Now we need to divide each factor with all of the given numbers and need select the factors which are qualified in the below division method.

Divide 6, 12, 18 by 2 we get 3, 6, 9

Now divide 3, 6, 9 by 3 we get 1 2, 3

Therefore, qualified factors are 2, 3

 Now multiply the factors

Therefore HCF = 6

Case - 2:

HCF of 12, 42, 66

 If the difference between the numbers are not same then we need to consider the minimum difference among them.

In this case

Difference from 12 to 42 = 30

Difference from 42 to 66 = 24

Difference from 12 to 66 = 54

Therefore, the minimum(least) difference is 24

Note: If the minimum difference is a prime number then the HCF will be the prime number if that prime number is divisible by all of the given numbers else HCF will be ‘1’.

Else

 Now we need to prime factorize the minimum difference.

 Now we need to divide each factor with all of the given numbers and need select the factors which are qualified in the below division method.

Divide 30, 24, 54 by 2 we get 15, 12, 18

Clearly remaining 2’s will not divide 15.

So now we need to check 3

Now divide 15, 12, 18 by 3 we get 5, 4, 6

Therefore, qualified factors are 2, 3

 Now multiply the factors

Therefore HCF = 6

Example-1:

Find the HCF of the numbers 7, 14, 21, 28

Solution:

By Applying our case – 1

The common difference is 7 and it is the prime number and divisible by all of the given numbers

Therefore HCF = 7

Example-2:

Find the HCF of the numbers 12, 30, 84, 102

Solution:

By Applying our case – 2

The minimum difference is 18 and it is not the prime number

Then prime factorization of 

Divide 12, 30, 84, 102 by 2 we get 6, 15, 42, 51

Now divide 6, 15, 42, 51 by 3 we get 2, 5, 14, 17

But remaining 3 will not divide 2, 5, 14, 17

Then factors which are divisible by all of the given numbers are 2, 3

Example-3:

Find the HCF of the numbers 27, 32, 37, 42

Solution:

By Applying our case – 1

The common difference is 5 and it is the prime number but it is not divisible by all of the given numbers

Therefore HCF = 1

---

If we have LCM and HCF of two numbers then

Example-1:

The LCM of two numbers is 864 and their HCF is 144. If one of the numbers are 288, the other number is?

Solution:

By Applying Trick 23,

Example-2:

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the numbers will be?

Solution:

By Applying Trick 23,

Example-3:

The product of two numbers is 4107 and their L.C.M. is 111. The H.C.F. of the numbers will be?

Solution:

By Applying Trick 23,

---

Example-1:

Solution:

Example-2:

Solution:

Example-3:

Solution:

---

Example-1:

Solution:

Example-2:

Solution:

Example-3:

Solution:

 

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When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.

Example-1:

The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is?

Solution:

By Applying Trick 26,

L.C.M. of 4, 6, 8, 12 and 16 = 48

  Required number = 48 + 2 = 50

Example-2:

Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case?

Solution:

By Applying Trick 26,

Required number = (LCM of 15, 20, 36 and 48) + 3

 LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

 Required number = 720 + 3 = 723

Example-3:

The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is?

Solution:

By Applying Trick 26,

LCM of 15, 12, 20, 54 = 540

 Required number = 540 + 4 = 544

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When a number is divided by a, b or c leaving remainders p, q or r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a, b and c.

Example-1:

The least number, which when divided by 4, 5 and 6 leaves remainder 1, 2 and 3 respectively, is?

Solution:

By Applying Trick 27,

Here 4 – 1 = 3,

5 – 2 = 3,

6 – 3 = 3

 The required number = LCM of (4, 5, 6) – 3 = 60 – 3 = 57

Example-2:

The smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively, is?

Solution:

By Applying Trick 27,

Here, 12 – 5 = 7,

16 – 9 = 7

 Required number = (L.C.M. of 12 and 16) – 7 = 48 – 7 = 41

Example-3:

When a number is divided by 15, 20 or 35, each time the remainder is 8.

Then the smallest number is?

Solution:

LCM of 15, 20 and 35 = 420

 Required least number = 420 + 8 = 428

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The largest number which when divide the numbers a, b and c the remainders are same then that largest number is given by H.C.F. of (a – b), (b – c) and (c – a).

Example:

The greatest number, which when divide 24, 30 and 48 leave same remainder is?

Solution:

= HCF of 6, 18 and 24

 HCF = 2

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The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r).

Example-1:

The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is?

Solution:

Required number = HCF of (989 – 5) and (1327 – 7)

= HCF of 984 and 1320 = 24

 HCF = 24

Example-2:

What is the greatest number that will divide 307 and 330 leaving remainders 3 and 7 respectively?

Solution:

Required number = HCF of (307 – 3) and (330 – 7)

= HCF of 304 and 323 = 19

 HCF = 19

Example-3:

Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?

Solution:

Required number = HCF of (3026 – 11) and (5053 – 13)

= HCF of 3015 and 5040 = 45

 HCF = 45

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Greatest n digit number which when divided by three numbers p, q, r leaves no remainder will be

Required Number = (n – digit greatest number) – R

R is the remainder obtained on dividing greatest n digit number by L.C.M of p, q, r. 

Example-1:

The largest 4-digit number exactly divisible by each of 12, 15, 18 and 27 is?

Solution:

The largest number of 4-digits is 9999.

L.C.M. of divisors 12, 15, 18 and 27

Divide 9999 by 540, now we get 279 as remainder. 9999 – 279 = 9720

Hence, 9720 is the largest 4-digit number exactly divisible by each of 12,

15, 18 and 27.

Example-2:

The greatest 4-digit number exactly divisible by 10, 15, 20 is?

Solution:

LCM of 10, 15 and 20 = 60

Largest 4-digit number = 9999

Divide 9999 by 60, now we get 39 as remainder. 9999 – 39 = 9960

Hence, 9960 is the largest 4-digit number exactly divisible by each of 10, 15 and 20.

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The n digit largest number which when divided by p, q, r leaves remainder ‘a’ will be

Required number = [n – digit largest number – R] + a

where, R is the remainder obtained when n – digit largest number is divided by the L.C.M of p, q, r.

Example:

The largest number of five digits which, when divided by 16, 24, 30, or 36 leaves the same remainder 10 in each case, is?

Solution:

We will find the LCM of 16, 24, 30 and 36.

The largest number of five digits = 99999

On dividing 99999 by 720, the remainder = 639

 The largest five-digit number divisible by 720 = 99999 – 639 = 99360

 Required number = 99360 + 10 = 99370

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Example-1:

A train is travelling at the rate of 45km/hr. How many seconds it will take to cover a distance of 

Solution:

By Applying Trick 32,

Example-2:

An aeroplane covers a certain distance at a speed of 240 km hour in 5 hours. To cover the same distance in   it must travel at a speed of?

Solution:

By Applying Trick 32,

Let the required speed is x km/ hr

Example-3:

A man walking at the rate of 5 km/hr. crosses a bridge in 15 minutes. The length of the bridge (in metres) is?

Solution:

By Applying Trick 32,

Speed of the man = 5km/hr

Time taken to cross the bridge

= 15 minutes Length of the bridge

= speed × time

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If a man travels different distances  ....... and so on in different time  respectively then,

Example-1:

A man travels a distance of 24 km at 6 kmph. Another distance of 24 km at 8 kmph and a third distance of 24 km at 12 kmph. His average speed for the whole journey (in kmph) is?

Solution:

By Applying Trick 33,

Total distance = 24 + 24 + 24 = 72 km.

Example-2:

P travels for 6 hours at the rate of 5 km/hour and for 3 hours at the rate of 6 km/hour. The average speed of the journey in km/hour is?

Solution:

By Applying Trick 33,

Total distance = 5 × 6 + 3 × 6 = 30 + 18 = 48 km

Total time = 9 hours

Example-3:

A train travelled at a speed of 35 km/hr for the first 10 minutes and at a speed of 20 km/hr for the next 5 minutes. The average speed of the train for the total 15 minutes is?

Solution:

By Applying Trick 33,

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If a man travels different distances   and so on with different speeds  respectively then,

Example:

A man travels a distance of 24 km at 6 kmph. Another distance of 24 km at 8 kmph and a third distance of 24 km at 12 kmph. His average speed for the whole journey (in kmph) is?

Solution:

By Applying Trick 34,

Total distance = 24 + 24 + 24 = 72 km.

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If a distance is divided into “n” equal parts each travelled with different speeds, then,

where n = number of equal parts

 are speeds

Example:

Four men travelled a distance which is divided into 4 equal parts with

speeds 5 kmph, 10 kmph, 15 kmph and 20 kmph then average speed?

Solution:

By Applying Trick 35,

Given that 

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If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h, then,

Example-1:

Jackson with his family travelled from Texas to New York by car at a speed of 40 km/hr and returned to Texas at a speed of 50 km/hr. The average speed for the whole journey is?

Solution:

By Applying Trick 36,

Example-2:

A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, the distance between his house and school is?

Solution:

By Applying Trick 36,

Here, x = 3, y = 2

Example-3:

A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, the distance between his house and school is?

Solution:

By Applying Trick 36,

Here, x = 25, y = 4

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If  distance is travelled in  time and  distance is travelled in   time then,

Example:

A man travelled 10 km in 2hrs and x km in 4hrs then find the value of x ?

Solution:

By Applying Trick 37,

---

Example:

A man travelled a distance of 24 km in 2hrs at what speed he needs to travel the same distance in 1.5 hrs?

Solution:

By Applying Trick 38,

Here

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If an object increases/decreases its speed from x km / hr to y km / hr to cover a distance in  hours in place of  hours then

Or

Example:

A Truck increases its speed from 45 kmph to 60 kmph in order to cover the distance in 6 hrs to deliver the goods 2 hrs earlier than expected time. Find the distance of the journey?

Solution:

By Applying Trick 39,

Or

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If an object travels certain distance with the speed of A/B of its original speed and reaches its destination ‘t’ hours before or after, then the taken time by object travelling at original speed is

Example:

Solution:

By Applying Trick 40,

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Example-1:

You arrive at your school 5 minutes late if you walk with a speed of 4 kmph,but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 kmph. The distance of your school from your house (in km) is?

Solution:

By Applying Trick 41,

Here

Example-2:

  

Solution:

By Applying Trick 41,

Here,

Example-3:

Solution:

By Applying Trick 41,

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Example-1:

Two trains X and Y start from City 1 to City 2 and from City 2 to City 1 respectively. After passing each other they take 4 hours 48 minutes and 3 hours 20 minutes to reach City 2 and City 1 respectively. If X is moving at 45 kmph, the speed of Y is?

Solution:

By Applying Trick 42,

Example-2:

Two trains started at the same time, one from A to B and the other from B to A. If they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speed of the two trains was?

Solution:

By Applying Trick 42,

Example-3:

The distance between 2 places R and S is 42 km. Mike starts from R with a uniform speed of 4 kmph towards S and at the same time Jenny starts from S towards R also with some uniform speed. They meet each other after 6 hours. The speed of Jenny is?

Solution:

By Applying Trick 42,

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If both objects run in opposite direction then,

Relative speed = Sum of speeds.

If both objects run in the same direction then,

Relative speed = Difference of Speeds.

Example-1:

  

Solution:

By Applying Trick 43,

Example-2:

Motor-cyclist M started his journey at a speed of 30 kmph. After 30 minutes, motor-cyclist N started from the same place but with a speed of 40 kmph. How much time (in hours) will N take to overtake M?

Solution:

By Applying Trick 43,

Distance covered by motor cyclist M in 30 minutes

Relative speed = 40 – 30 = 10 kmph

Example-3:

A thief is noticed by a policeman from a distance of 200m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 kmph and 11 kmph respectively. What is the distance between them after 6 minutes?

Solution:

By Applying Trick 43,

Relative speed of police = 11 – 10 = 1 kmph

 Distance remained between them = 200–100 = 100 m

---

Let a man take ‘t’ hours to travel ‘x’ km. If he travels some distance on foot with the speed u km/h and remaining distance by cycle with the speed v km/h, then time taken to travel on foot.

Distance travelled on foot = Time × u

Example-1:

A man travelled a distance of 80 km in 7 hrs partly on foot at the rate of 8 km per hour and partly on bicycle at 16km per hour. The distance travelled on the foot is

Solution:

By Applying Trick 44,

Here, x = 80, t = 7 u = 8, v = 16

Distance travelled = 4 × 8 = 32 kms

Example-2:

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at the rate 4 kmph and partly on bicycle at the rate 9 kmph. The distance travelled on foot is?

Solution:

By Applying Trick 44,

Here, t = 9, x= 61, u = 4, v = 9

Distance travelled = 4 × 4 = 16 km

---

If anyone overtakes or follows another, then 

Total travelled distance to catch the thief

Example:

A thief is noticed by a policeman from a distance of 500m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 kmph and 11 kmph respectively. How many minutes will it take to policeman to catch the thief?

Solution:

By Applying Trick 45,

Relative speed = 11 – 10 = 1 kmph

Therefore, time taken to catch the thief = 30 minutes

---

 

Example:

In a certain time, a man travelled 20 km at a speed of 2kmph. At what speed he needs to travel the distance of 25 km in the same time?

Solution:

By Applying Trick 46,

---

Example:

One third of a certain journey is covered at the rate of 25 kmph, one- fourth at the rate of 30 kmph and the rest at 50 kmph. The average speed for the whole journey is?

Solution:

By Applying Trick 47,

Here, x = 3, u = 25

y = 4, v = 30

z = 12/5, w = 50

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 Time taken in crossing ‘b’ metre length (i.e. platform, bridge, tunnel,standing train etc) by ‘a’ metre length train = total time taken in travelling(a + b) metre by the train.

 Let a train is travelling with the speed x kmph and in the same direction, another train is travelling on parallel path with the speed y kmph, then, Relative speed of the faster train = (x – y) kmph.

 Suppose that a train is travelling with the speed ‘x’ kmph and from the opposite direction another train is coming on parallel path with the speed ‘y’ kmph, then Relative speed of the train = (x + y) kmph.

---

If a train crosses an electric pole, a sitting/ standing man, km or mile stone etc. then,

Distance = Length of train

Length of train = Speed × Time

 

The length of a train and that of a platform are equal. If with a speed of 90 kmph the train crosses the platform in one minute, then the length of the train (in metres) is?

Solution:

By Applying Trick 48,

Let the length of train be x metre Speed = 90 kmph

= 25 metre/sec

 Distance covered in 60 sec.

= 25 × 60 = 1500 metres

Now, according to question,

2x = 1500

 x 750 metre

---

Let ‘a’ metre long train is going with the speed ‘x’ m/s and ‘b’ metre long train is also going with the speed ‘y’ m/s in the same direction on parallel path, then total time taken by the faster train to cross the slower train

Here, a = 1800, b = 1600

x = = 15 m/s, y = 10 m/s

---

Let ‘a’ metre long train is going with the speed ‘x’ m/s and ‘b’ metre long train is also going with the speed ‘y’ m/s in the opposite direction on parallel path, then total time taken by the trains to cross each other

Example:

A 1200-metre-long train is going with the speed 22 m/s and 1500 metre long train is also going with the speed 28 m/s in the opposite direction on parallel path, then total time taken by the trains to cross each other?

Solution:

By Applying Trick 50,

Time taken by the trains to cross each other

Here, a = 1200, b = 1500

x = = 22 m/s, y = 28 m/s

---

Example:

If a train crosses a standing pole in 15 sec time and crosses 120 meter long platform in 25 sec. time, then length of the train is?

Solution:

By Applying Trick 51,

---

Let ‘a’ metre long train is running with the speed ‘x’ m/s. A man is running in same direction and with the speed ‘y’ m/s, then

Example:

A train 180 m long moving at the speed of 20 m/sec. over-takes a man moving at a speed of 10m/ sec in the same direction. The train passes the man in?

Solution:

By Applying Trick 52,

Here a = 180, x = 20, y = 10

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Let ‘a’ metre long train is running with the speed ‘x’ m/s. A man is running in opposite direction and with the speed ‘y’ m/s, then

Example:

A train, 240 m long crosses a man walking along the line in opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train is?

Solution:

If the speed of train be x kmph then,

---

Example:

A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively.

The speed of the train is?  

Solution:

By Applying Trick 54,

---

If two trains of (same lengths) are coming from same direction and cross a man in  seconds, then

Example:

If two trains A and B of same lengths are coming from same direction and cross a standing pole in 10 and 15 seconds, then time taken by both the trains to cross each other? 

Solution:

By Applying Trick 55,

---

If two trains of (same lengths) are coming from opposite directions and cross a man in  seconds, then

Example:

If two trains A and B of same lengths are coming from opposite directions and cross a standing pole in 20 and 30 seconds, then time taken by both the trains to cross each other? 

Solution:

By Applying Trick 56,

---

If a train of length x meters crosses a platform/ tunnel/bridge of length y meters with the speed u m/s in t seconds, then,

Example:

A train 300 metres long is running at a speed of 25 metres per second. It will cross a bridge of 200 metres in? 

Solution:

By Applying Trick 57,

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Two trains A and B, run from stations X to Y and from Y to X with the speed 

Example:

Two trains started at the same time, one from A to B and the other from B to A. If they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speed of the two trains was?

Solution:

By Applying Trick 58,

---

If a train of length l meters passes a bridge/ platform of 'x' meters in  sec, then the time taken by the same train to cross another bridge/platform of length ‘y’ meters is,

Example:

If a train of length 800 meters passes a bridge of 200 meters in 8 sec, then the time taken by the same train to cross another bridge of length 250 meters is?

Solution:

By Applying Trick 59,

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From stations A and B, two trains start travelling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as

Example:

Two trains start from station A and B and travel towards each other at speed of 16 miles/ hour and 21 miles/ hour respectively. At the time of their meeting, the second train has travelled 60 miles more than the first.

The distance between A and B (in miles) is?

Solution:

By Applying Trick 60,

Here, a = 21, b = 16, d = 60

Distance between A and B

---

The distance between two places A and B is x km. A train starts from A towards B at a speed of ‘a’ km/ hr and after a gap of ‘t’ hours another train with speed ‘b’ km/hr starts from B towards A, then both the trains will meet at a certain point after time T. Then, we have.

t is taken as positive if second train starts after first train and t is taken as

negative if second train starts before the first train.

 

Example:

A train starts from City 1 towards City 2 at a speed of 40 kmph and after a

gap of 2 hours another train with speed 50 kmph starts from City 2

towards City 1, then after how many hours both the trains will meet if

the distance between two cities is 170 km?

Solution:

By Applying Trick 61,

Here second train starts after first train.

Here, x = 170, a = 40, b = 50, t = 2

= 3 hours.

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A train covers a distance between stations A and B in time  If the speed is changed by S. then the time taken to cover the same distance is  Then the distance (D) between A and B is given by

 Where t' : change in the time taken

Example:

A train covers a certain distance between stations P and Q in 4 hours. If the

speed is changed by 10 kmph, then the time taken to cover the same

distance is 3 hours. Then find the distance between P and Q?

Solution:

By Applying Trick 62,

Here, S = 10 kmph, 

= 120

Therefore, distance between P and Q is 120 kms.

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Ratio:

The comparative relation between two amounts/ quantities of same type is called ratio. The ratio of two amounts is equal to a fraction. It shows how much less or more time an amount is in comparison to another.

Ratio always occurs between same units, as –Dollars: Dollars, kg: kg, Hour: Hour, Second: Second etc.

Proportion:

When two ratios are equal to each other, then they are called proportional as  then, a, b, c and d are in proportion

---

It does not change the ratio, when we multiply or divide antecedent and consequent of the ratio by a same non–zero number as

Where c is the non–zero number.

 

Example:

The ratio of two quantities is 2 to 3.  If each of the two quantities is

tripled, what is the ratio of these two new quantities?

Solution:

By Applying Trick 63,

Tripling both quantities in a ratio (or multiplying each by any term, as a

matter of fact) doesn' t change the ratio. If you triple both terms

---

Mixed ratio:

Let x: y and P: Q be two ratios, then Px : Qy is called mixed ratio. 

Example:

Solution:

By Applying Trick 64,

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Duplicate Ratio:

The mixed ratio of two equal ratios is called the duplicate Ratio as

duplicate ratio of a:b is 

---

Sub-duplicate Ratio:

The square root of a certain ratio is called its sub-duplicate.

The sub-duplicate ratio of 

---

Triplicate Ratio:

The cube of a certain ratio is called triplicate ratio.

The triplicate ratio of 

---

Sub-triplicate Ratio:

The cube root of a certain ratio is called sub-triplicate ratio as

The Sub-triplicate Ratio of 

---

Inverse Ratio:

The Reciprocal of quantities of ratio is called its inverse. Reciprocal or inverse ratio of a:b

---

Invertendo:

The proportion in which antecedent and consequent quantities change their places, is called invertendo, as

Invertendo of 

---

Alternendo:

---

Componendo:

---

---

Componendo and dividendo:

To simplify the proportion any one method of componendo, dividendo, componendo and Dividendo can directly be used.

Example:

Solution:

By Applying Trick 74,

---

Mean Proportion:

Let x be the mean proportion between a and b, then  (Real condition)

Example:

The mean proportional between 

Solution:

By Applying Trick 75,

---

Third proportional:

Let ‘x’ be the third proportional of a and b then,

Example:

The third proportional of 12 and 18 is?

Solution:

By Applying Trick 76,

---

Fourth proportional:

Let ‘x’ be the fourth proportional of a, b and c then, 

Example:

The fourth proportional to 0.12, 0.21, 8 is?

Solution:

By Applying Trick 77,

---

It is done as follows:

Example-1:

If a, b, c are three numbers such that a : b = 3 : 4 and b : c = 8 : 9, then a : c is equal to?

Solution:

By Applying Trick 78,

Example-2:

If A : B = 3 : 4 and B : C = 8 : 9, then A : B : C is?

Solution:

By Applying Trick 78,

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  The set of integers consists of the whole numbers (1, 2, 3, . . .) and their negatives, including zero. The set of integers extends infinitely in both positive and negative directions.

  Positive integer refers to all integers greater than zero. Example:  1, 2, 3,

  Negative integer refers to all integers less than zero. Example:  −1, −2, −3,

  The set of non-negative integers is: 0, 1, 2, 3, . . . .

  Zero is neither a positive nor a negative integer, but is an even integer.

---

Numbers that are divisible by 2 are called even, and all other numbers not divisible by 2 are called odd..

The general form of even numbers is 2k and that of odd numbers is 2k + 1, where k is an integer.

  The following list summarizes the outcome of operations of sum, difference, product, and powers when applied to odd and even integers.

Even × Even = Even

Even × Odd = Even

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  Consecutive integers are those integers that follow each other in a sequence, where the difference between any two successive integers is 1.

  They can be algebraically represented by n, n + 1, n + 2, n + 3, .., where n is an integer.

  Example: −3, −2, −1, 0, 1, 2, 3.

  Consecutive even integers can be represented by 2n, 2n + 2, 2n + 4, ...

  Consecutive odd integers can be represented by 2n + 1, 2n + 3, 2n + 5, ...

---

If n is odd, the sum of consecutive integers is always divisible by n

Example:

Given, {9,10,11}, we have n=3 consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

---

If n is even, the sum of consecutive integers is never divisible by n.

Example:

Given, {9,10,11,12}, we have n=4 consecutive integers.

The sum of 9 + 10 + 11 + 12 = 42, therefore, is divisible by 4.

---

The product of n consecutive integers is always divisible by n!

Example:

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Divisibility Rule of 2:

If a number is even or a number whose last digit is an even number i.e.  2,4,6,8 including 0, it is always completely divisible by 2.

Example: 508 is an even number and is divisible by 2 but 509 is not an even number, hence it is not divisible by 2. Procedure to check whether 508 is divisible by 2 or not is as follows:

Consider the number 508

Just take the last digit 8 and divide it by 2

If the last digit 8 is divisible by 2 then the number 508 is also divisible by 2.

---

Divisibility Rules for 3:

Divisibility rule for 3 states that a number is completely divisible by 3 if the sum of its digits is divisible by 3.

Consider a number, 308. To check whether 308 is divisible by 3 or not, take sum of the digits (i. e. 3+0+8= 11). Now check whether the sum is divisible by 3 or not. If the sum is a multiple of 3, then the original number is also divisible by 3. Here, since 11 is not divisible by 3, 308 is also not divisible by 3.

Similarly, 516 is divisible by 3 completely as the sum of its digits i.e. 5+1+6=12, is a multiple of 3.

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Divisibility Rule of 4:

If the last two digits of a number are divisible by 4, then that number is a multiple of 4 and is divisible by 4 completely.

Example: Take the number 2308. Consider the last two digits i.e.  08. As 08 is divisible by 4, the original number 2308 is also divisible by 4.

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Divisibility Rule of 5:

Numbers, which last with digits, 0 or 5 are always divisible by 5.

Example: 10, 10000, 10000005, 595, 396524850, etc.

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Divisibility Rule of 6:

Numbers which are divisible by both 2 and 3 are divisible by 6. That is, if the last digit of the given number is even and the sum of its digits is a multiple of 3, then the given number is also a multiple of 6.

Example: 630, the number is divisible by 2 as the last digit is 0.

The sum of digits is 6+3+0 = 9, which is also divisible by 3.

Hence, 630 is divisible by 6.

---

Divisibility Rules for 7:

The rule for divisibility by 7 is a bit complicated which can be understood by the steps given below:

Example: Is 1073 divisible by 7?

• From the rule stated remove 3 from the number and double it, which becomes 6.
• Remaining number becomes 107, so 107-6 = 101.
• Repeating the process one more time, we have 1 x 2 = 2.
• Remaining number 10 – 2 = 8.
• As 8 is not divisible by 7, hence the number 1073 is not divisible by 7.

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Divisibility Rule of 8:

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

Example: Take number 24344. Consider the last two digits i.e.  344. As 344 is divisible by 8, the original number 24344 is also divisible by 8.

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Divisibility Rule of 9:

The rule for divisibility by 9 is similar to divisibility rule for 3. That is, if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.

Example: Consider 78532, as the sum of its digits (7+8+5+3+2) is 25, which is not divisible by 9, hence 78532 is not divisible by 9.

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Divisibility Rule of 10:

Divisibility rule for 10 states that any number whose last digit is 0, is divisible by 10.

Example: 10, 20, 30, 1000, 5000, 60000, etc.

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Divisibility Rules for 11:

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

In order to check whether a number like 2143 is divisible by 11, below is the following procedure.

• Group the alternative digits i.e. digits which are in odd places together and digits in even places together. Here 24 and 13 are two groups.
• Take the sum of the digits of each group i.e. 2+4=6 and 1+3= 4
• Now find the difference of the sums; 6-4=2
• If the difference is divisible by 11, then the original number is also divisible by 11. Here 2 is the difference which is not divisible by 11.
• Therefore, 2143 is not divisible by 11.

A few more conditions are there to test the divisibility of a number by 11. They are explained here with the help of examples:

If the number of digits of a number is even, then add the first digit and subtract the last digit from the rest of the number. 

Example-1: 3784

Number of digits = 4

Now, 78 + 3 – 4 = 77 = 7 × 11

Thus, 3784 is divisible by 11.

 

If the number of digits of a number is odd, then subtract the first and the last digits from the rest of the number. 

Example-2: 82907

Number of digits = 5

Now, 290 – 8 – 7 = 275 × 11

Thus, 82907 is divisible by 11.

Form the groups of two digits from the right end digit to the left end of the number and add the resultant groups. If the sum is a multiple of 11, then the number is divisible by 11.

Example-3: 3774: = 37 + 74 = 111: = 1 + 11 = 12 

3774 is not divisible by 11.

253: = 2 + 53 = 55 = 5 × 11

253 is divisible by 11.

Subtract the last digit of the number from the rest of the number. If the resultant value is a multiple of 11, then the original number will be divisible by 11.

Example-4: 9647

9647: = 964 – 7 = 957

957: = 95 – 7 = 88 = 8 × 11

Thus, 9647 is divisible by 11.

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Divisibility Rule of 12:

If the number is divisible by both 3 and 4, then the number is divisible by 12 exactly. 

Example: 5864

Sum of the digits = 5 + 8 + 6 + 4 = 23 (not a multiple of 3)

Last two digits = 64 (divisible by 4)

The given number 5846 is divisible by 4 but not by 3; hence, it is not divisible by 12.

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Divisibility Rules for 13:

For any given number, to check if it is divisible by 13, we have to add four times of the last digit of the number to the remaining number and repeat the process until you get a two-digit number.  Now check if that two-digit number is divisible by 13 or not. If it is divisible, then the given number is divisible by 13.

For example: 2795 → 279 + (5 x 4) 

→ 279 + (20) 

→ 299 

→ 29 + (9 x 4) 

→ 29 + 36 

→65

Number 65 is divisible by 13, 13 x 5 = 65.

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Divisibility Rule for 14:

Check whether the given number is divisible by 2 and 7. If the number is divisible by these numbers, then the original number is also divisible by 14.

Example

Find if the number 224 is divisible by 14

Solution:

112 x 2 = 224 32 x 7 = 224 Now, it is clear that 224 is divisible by 2 and 7. Hence, it is also divisible by 14.

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Divisibility Rule for 15:

If a number is divisible by both 3 and 5, then it is divisible by 15.

Example: Check whether 41295 is divisible by 15.

Solution:

We know that if the given number is divisible by both 3 and 5, then it is divisible by 15.

First, check whether the given number is divisible by 3.

Sum of the digits:

4 + 1 + 2 + 9 + 5 = 21

Sum of the digits (21) is a multiple of 3.

So, the given number is divisible by 3. 

Now, check whether the given number is divisible by 5.

In the given number 41295, the digit in one's place is 5. 

So, the number 41295 is divisible by 5. 

Now, it is clear that the given number 41295 is divisible by both 3 and 5. 

Therefore, the number 41295 is divisible by 15.

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Divisibility Rule for 17:

A number is divisible by 17 if you multiply the last digit by 5 and subtract that from the rest. If that result is divisible by 17, then your number is divisible by 17.

Example: 98 - (6 x 5) = 68. Since, 68 is divisible by 17, then 986 is also divisible by 17.

However, 876 is not divisible by 17 because 87 - (6 x 5) = 57 and 57 is not divisible by 17.

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Divisibility Rule for 19:

To determine if a number is divisible by 19, take the last digit and multiply it by 2. Then add that to the rest of the number. If the result is divisible by 19, then the number is divisible by 19.

Example: 475 is divisible by 19 because 47 + (5 x 2) = 57, and 57 is divisible by 19.

However, 575 is not divisible by 19 because 57 + (5 x 2) = 67, and 67 is not divisible by 19.

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Divisibility Rule for 23:

To determine if a number is divisible by 23, take the last digit and multiply it by 7. Then add that to the rest of the number. If the result is divisible by 23, then the number is divisible by 23.

Example: 575 is divisible by 23 because 57 + (5 x 7) = 92 and 92 is divisible by 23.

However, 576 is not divisible by 23 because 57 + (6 x 7) = 99, and 99 is not divisible by 23.

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Divisibility Rule for 29:

Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.

Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.

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Divisibility Rule for 31:

Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.

Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.

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A prime number is any positive integer greater than 1 that has exactly two whole number factors, itself and the number 1. The number 1 itself is not a prime. The table below lists the prime numbers less than 100. A positive integer that is greater than 1 and is not prime is called composite.

How to Test a number is prime or not:

To test if a given number less than 100 is prime, divide the number by 2, 3, 5, and 7, if the number is not divisible by any of these four prime numbers, then the number is prime.

For example, the number 59 is prime because it is not divisible by 2, 3, 5, or 7.

  2 and 3 are the only pair of consecutive integers that are both prime, because any other pair of consecutive integers will always have one number that is even, which will be divisible by 2, and therefore cannot both be prime.

  3, 5, and 7 are the only three consecutive odd integers that are all prime numbers. Again, when we select a set of three consecutive odd integers, at least one of them is divisible by 3, and therefore all sets of three consecutive odd integers will always have a multiple of 3, with the exception of 3, 5, and 7.

  The possible units digit of all prime number greater than 5 are 1, 3, 7, and 9.

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Fundamental Theorem of Arithmetic:

Every integer greater than or equal to 2 is either a prime number or can be written uniquely as the product of two or more prime numbers. The factorization in to the prime numbers is unique except for the order in which they are written. For example, 120 can be written as 

Number of Divisors of a Composite Number:

Any composite number can be resolved into prime factors in only one way

and in the most general case, N can be written as:

The divisors or factors of N are numbers of the form

Another way to think about this problem is that each term of the following product:

  Prime numbers have two factors, 1 and the prime number itself.

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Numbers that have only three factors are square of a prime number.

Example:

Find all the factors of 529?

Solution:

As per trick 20,

23 is a prime number then the factors are 1, 23, 529.

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The number of factors of  where p is prime, and m is a positive integer is equal to m+ 1.

The factors are 

Example:

Find all the factors of 343?

Solution:

As per trick 21,

7 is a prime number then are 3+1 =4 factors.

The factors are 1, 7, 49, 343.

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Highest power of prime factor that divides n!:

To find the highest power of a prime number p contained in n!, divide the number n repeatedly by  . . . to obtain the set of quotients that are greater than or equal to one. The highest power of p contained in the prime factorization of n! is given by:

Where the floor function or the greatest integer function, ⌊x⌋ is defined as the largest integer less than or equal to x.

Example:

What is the largest value of k for which  is a factor of 100!?

Solution:

As per trick 22,

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Finding the powers of a prime number p, in the n!

The formula is:

Example:

What is the power of 2 in 25!?

Solution:

As per trick 26,

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Finding the Number of Factors of an Integer:

First make prime factorization of an integer

 are prime factors of n and p, q and r are their powers.

The number of factors of n will be expressed by the formula

NOTE: this will include 1and n itself.

 

Example:

Finding the number of all factors of 450?

Solution:

As per trick 23,

Total number of factors of 450 including 1 and 450 itself is

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Finding the Sum of the Factors of an Integer

First make prime factorization of an integer  where a,b and c are prime factors of n and p,q and r are their powers.

The sum of factors of n will be expressed by the formula:

Example:

Finding the sum of all factors of 450?

Solution:

As per trick 24,

The sum of all factors of 450 is

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A perfect square, is an integer that can be written as the square of some other integer. For example,  is an perfect square.

There are some tips about the perfect square:

  The number of distinct factors of a perfect square is ALWAYS ODD.

  The sum of distinct factors of a perfect square is ALWAYS ODD.

  A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.

  Perfect square always has even number of powers of prime factors.

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Basic Definition:

 

Example:  Find the value of the below expression

Solution:

Let's combine the two radicals into one radical and simplify.

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