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

---

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

---

Problem: 

Normal Method:

Add 2 on both sides we get

Shortcut Method:

then

that means from the given problem

---

Problem: 

Normal Method:

Subtract 2 from both sides we get

Shortcut Method:

then

that means from the given problem

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Normal Method:

According to Standard Algebraic Identities:

then by substituting the given values we get

Shortcut Method:

Given that 

Let a = 1 and b = 3

then  and 

Let's pick 

then  and 

then

or

therefore

---

Normal Method:

Given that 

Squaring on both sides

According to Standard Algebraic Identities:

then by substituting the given values we get

Shortcut Method:

Given that 

Let's pick 

then  and

therefore

---

Solving Linear Equations without Quadratic Form:

Problem: Solve the below linear equation for x.

Solution:

The value of x can be solved by rewriting the R.H.S into the form of L.H.S

Then,

and

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Then,

and

---

Solving Linear Equations in One Variable Orally:

Type 1:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 2:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 3:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 4:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

Example 4:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations directly by Cross Multiplying:

Problem: Solve the below linear equations for x and y.

6x + 5y = 8

2x – 3y = 12

Solution:

Method:

Then,

By substituting the value of “x” in any given linear equation we can get the value of “y”

By applying above method for given equations we get

Let the given equations be

Example 1:

Solve the below linear equations for x and y.

Example 2:

Solve the below linear equations for x and y.

&

---

Trick #26

Translating Coin Problems into Specific Types of Equations:

Typical Problem in English:

Adam has 72 coins in his piggy bank. The piggy bank contains only dimes and quarters. If he has $14.70 in his piggy bank, write an equation that can be used to determine q, the number of quarters he has?

Mathematical Translation:

The total value of all coins is 1470 cents.

Let the number of quarters be represented by q and the value of quarters

be represented by 25q.

Let the number of dimes be represented by 72 – q and the value of dimes

be represented by 10(75 – q)

Write:

25q +10(72 − q) =1470

Solve for q.

q = 30

Key Points:

Work with cents as units. Remember that each coin has a specific value in cents

---

Translating Consecutive Integer Problems into Specific Types of Equations:

Typical Problem in English:

The sum of three consecutive odd integers is 18 less than five times the middle number. Find the three integers.

Mathematical Translation:

Let x represent the first integer.

Let x + 2 represent the middle integer.

Let x + 4 represent the 3^rd integer.

Write:

Solve for x, x +2, and x + 4.

7, 9, 11

Key Points:

For consecutive integer problems, define your variables as x, x + 1, and

x + 2

For consecutive even or odd integer problems, define your variables as x,

x +2, and x + 4.

---

Translating Age Problems into Specific Types of Equations:

Typical Problem in English:

Jaqueline has two sisters. One of the sisters is 7 years older than Jaqueline. The other sister is 3 years younger than Jaqueline. The product of Jaqueline’s sisters' ages is 24. How old is Jaqueline?

Mathematical Translation:

Let x represent Jaqueline’s age.

Let x+7 represent the older sister’s age.

Let x - 3 represent the younger sister’s age.

Write:

Solve for x.

x = 5

Key Points:

Define your variables. Check your answers.

Remember than “is” means “=”.

---

Translating Missing Number in the Average Problems into Specific Types of Equations:

Typical Problem in English:

FINE Medicals is a small business with five employees. The mean (average) weekly salary for the five employees is $360. If the weekly salaries of four of the employees are $340, $340, $345, and $425, what is the salary of the fifth employee?

Mathematical Translation:

Let  represent the missing salary Write:

Solve for 

 = $350

Key Points:

Substitute given values into the following formula for finding the average.

then solve for the missing value.

---

Translating Number Problems into Specific Types of Equations:

Typical Problem in English:

Twice the larger of two numbers is ten more than five times the smaller, and the sum of four times the larger and three times the smaller is 39. What are the numbers?

Mathematical Translation:

Let x represent the larger.

Let y represent the smaller.

Write two equations:

2x =10 + 5y

And

4x + 3y = 46

Solve as a system of equations.

x = 10 and y = 2 

Key Points:

Define your variables. Check your answers.

Remember than “is” means “=”.

---

Translating Area, Volume and Perimeter Problems into Specific Types of Equations:

Typical Problem in English:

If the length of a rectangular prism is doubled, its width is tripled, and its height remains the same, what is the volume of the new rectangular prism in relation to the volume of the original rectangular prism?

Mathematical Translation:

Use the formula 

Let the volume of the original rectangular prism be represented by lwh.

Let the volume of the new rectangular prism be represented by 

which simplifies to 6 times lwh.

The new rectangular prism has six times the volume of the original rectangular prism.

Key Points:

Use a geometric formula as a guide.

---

Write equations or expressions out of word  problems: 

Example 1:

Three times the sum of a number and four is equal to five times the number, decreased by two. If x represents the number, write an equation that is a correct translation of the statement?

Equation:

3(x − 4) = 5x − 2

Example 2:

The product of a number and 3, increased by 5, is 7 less than twice the number. Write an equation that can be used to find this number, n?

Equation:

3n + 5 = 2n − 7

Example 3:

The width of a rectangle is 3 less than twice the length, x. If the area of the rectangle is 43 square feet, write an equation that can be used to find the length, in feet?

Equation:

x (2x − 3) = 43

Example 4:

If n is an odd integer, write an equation that can be used to find three consecutive odd integers whose sum is -3?

Equation:

Example 5:

The length of a rectangular window is 5 feet more than its width, w. The area of the window is 36 square feet. Write an equation that could be used to find the dimensions of the window?

Equation:

or

Example 6:

Mary has $1.35 in nickels and dimes in her pocket. If she has six more dimes than nickels, write an equation that can be used to determine x, the number of nickels she has?

Equation:

or

Example 7:

If h represents a number, write an equation that is a correct translation of "Sixty more than 9 times a number is 375"?

Equation:

9h + 60 = 375

Example 8:

The ages of three brothers are consecutive even integers. Three times the age of the youngest brother exceeds the oldest brother's age by 48 years.

Write an equation that could be used to find the age of the youngest brother?

Equation:

3x = 48 + (x + 4)

or

3x − (x + 4) = 48 

Example 9:

The width of a rectangle is 4 less than half the length. If l represents the length, write an equation that could be used to find the width, w?

Equation:

---

Solving a special type of linear equation problem:

Problem: Solve the below linear equations for x and y.

Solution:

This type of equations can be solved by adding and subtracting both equations so that we can get two new equations by solving the resultant equations we can get the values of x and y.

Adding:

Subtracting:

can be written as  and   can be written as 

By solving new equations, we get

then x = 2, therefore y = 1

Example: Solve the below linear equations for x and y.

Solution:

Adding:

Subtracting:

 can be written as  and   can be written as 

By solving new equations, we get

then y = 1, therefore x = 3

---

Solving complex linear equations:

Problem: Solve the below linear equation for x.

Solution:

We can solve this type of equations by taking the LCM of the denominators of both sides and cross multiplying both sides with LCM.

Given equation can be written as

Example: Solve the below linear equation for x.

Solution:

Given equation can be written as

---

Solving a special type of linear equation problem:

Problem: Solve the below linear equation for x.

Solution:

We can solve this type of equations by cross multiplying.

Example: Solve the below linear equation for x.

Solution:

---

Solving a linear equation of the form 

Problem: Solve the below linear equation for x.

Solution:

We can solve this type of equations by cross multiplying.

Example1: Solve the below linear equation for x.

Solution:

Example2: Solve the below linear equation for x.

Solution:

Example3: Solve the below linear equation for x.

Solution:

---

Solving A, B, C from A+B, B+C and C+A:

Problem:

If  and  find A,B,C

Solution:

Method:

If  and 

then

Apply the same for given problem

 and 

Therefore,

Example: If  and  find A,B,C

Solution:

Given that

 and 

Therefore,

---

Solving a special type of linear equation problem:

Problem: Solve the below linear equations for x and y.

Solution:

This type of linear equations can be solved by converting one of the given equations into the form of another.

Multiply 1^st equation with  on both sides we get

Subtracting the 2^nd equation from the new equation we get

Therefore 

Example-1: Solve the below linear equations for x and y.

Solution:

Multiply 1^st equation with "b" and 2^nd equation with “a” on both sides we get

Subtracting the 2^nd equation from the 1^st equation we get

Therefore 

Example-2: Solve the below linear equations for x and y.

Solution:

Multiply 1^st equation with 2 on both sides we get

Subtracting the 2^nd equation from the1^st equation we get

 

Therefore 

---

Linear Equations involving fractions:

Problem: Solve the below linear equation for x.

Solution:

Example-1: Solve the below linear equation for x.

Solution:

Example-2: Solve the below equation.

Solution:

We can solve this type of equations by cross multiplying.

Example-3: Solve the below equation.

Solution:

---

Solving Word Problems by translating into linear equations:

Problem:

Students of a class are made to stand in rows. If 1 student is extra in each row, there would be 2 rows less If 1 student is less in each row, there would be 3 rows more. Find the no of students in the class.

Solution:

let no of rows = x

let no of students in each row = y

.^.. No of students = xy

Subtracting the 2^nd equation with 1^st equation we get

Therefore, number of students 

---

Basic factorization:

Steps:

I. Look for common factor.

II. Look for Identity.

III. Look for Method.

In order to factorize any equation, we have to consider the above three steps by looking for a common factor first if not found then need to look for an algebraic identity like  etc. If both the steps are not suitable then go for method to factorize the equation.

Example-1: Factorize the below equation

Solution:

Where 2 is the common factor.

Example-2: Factorize the below equation

Solution:

Where 10xy is the common factor.

Example-3: Factorize the below equation

Solution:

In this we have used two steps of picking a common factor and choosing an algebraic identity.

Example-4: Factorize the below equation

Solution:

In this we have picked an algebraic identity.

Example-5: Factorize the below equation

Solution:

In this we have picked an algebraic identity.

---

Factorization using algebraic identities:

Following are the examples of factorization of equations using algebraic identity

Example-1: Factorize the below equation

Solution:

Example-2: Factorize the below equation

Solution:

Example-3: Factorize the below equation

Solution:

Example-4: Factorize the below equation

Solution:

Example-5: Factorize the below equation

Solution:

---

Factorization using algebraic identities:

Following are the examples of factorization of equations using below algebraic identities

Example-1: Factorize the below equation

Solution:

Example-2: Factorize the below equation

Solution:

Example-3: Factorize the below equation

Solution:

---

Method of Square completion:

Problem: Factorize the below equation

Solution:

Divide throughout by 2

Then,

and

Therefore

Example: Factorize the below equation

Solution:

---

Solving a quadratic equation without factorization:

Problem:

Solve the below quadratic equation

Solution:

Normal Method:

Therefore,

Short cut Method:

If  then

Step 1: Take out the values of “b” and “ac” from the given equation.

Step 2: Find the relevant numbers m, n which satisfies b = m +n and ac = mn.

Step 3: Change the sign of the numbers.

Step 4: Divide both numbers by coefficient of 

By applying the above steps to the given equation, we get 

Changing the signs, we get 14 and -3

divide both numbers by coefficient of  we get

Therefore,

Example-1: Solve the below quadratic equation.

Solution:

Changing the signs, we get -20 and 2

divide both numbers by coefficient of  we get

Therefore,

Example-2: Solve the below quadratic equation.

Solution:

Changing the signs, we get -10 and 3

divide both numbers by coefficient of  we get

Therefore,

Example-3: Solve the below quadratic equation.

Solution:

Changing the signs, we get 28 and -2

divide both numbers by coefficient or  we get

Therefore,

---

Finding a Quadratic Equation is Factorizable or not:

Method:

If  then

If D is perfect square or D = 0 then the equation is factorizable.

Example-1:

The equation is factorizable.

Example-2:

71 is neither 0 nor a perfect square.

The equation is not factorizable.

Example-3:

The equation is factorizable.

Example-4:

29 is neither 0 nor a perfect square.

The equation is not factorizable.

---

Real and Equal Roots:

Method:

If  then

If D = 0 then the equation has real and equal roots. 

Problem:

Find k so that the given quadratic equation has real, equal roots

Solution:

Here a = 2k, b = -8 and c = k

Since the equation has real and equal roots

Example-1:

Find k so that the given quadratic equation has real, equal roots

Solution:

Here a = 12, b = 4k and c = 3

Since the equation has real and equal roots

Example-2:

Find k so that the given quadratic equation has real, equal roots

Solution:

Here a = 4, b =  and c = k + 4

Since the equation has real and equal roots

Example-3:

Find k so that the given quadratic equation has real, equal roots

Solution:

Since the equation has real and equal roots

---

Solving a Quadratic equation using Splitting middle term method:

Problem:

Solve the below quadratic equation

Solution:

Split the middle term into the values that satisfies

Example-1:

Solve the below quadratic equation

Solution:

Split the middle term into the values that satisfies

Example-2:

Solve the below quadratic equation

Solution:

Split the middle term into the values that satisfies

---

Solving a Quadratic equation using Quadratic Formula:

Problem:

Solve the below quadratic equation

Solution:

Method:

If  then

Given that 

then 

then

Example-1: Solve the below quadratic equation

Solution:

Given that 

then 

Example-2: Solve the below quadratic equation

Solution:

Given that 

then 

---

Solving a Quadratic equation of Large Coefficients:

Problem:

Solve the below quadratic equation

Solution:

Changing the signs, we get 35 and -12

divide both numbers by coefficient of  we get

Therefore,.

Example-1: Solve the below quadratic equation

Solution:

Changing the signs, we get 72 and 5

divide both numbers by coefficient of  we get

Therefore,

Example-2: Solve the below quadratic equation

Solution:

Changing the signs, we get 28 and -6

divide both numbers by coefficient of  we get

Therefore,

---

Sum and Product of a Quadratic equation:

Method:

If  then

 and  are the roots

then

Example-1: Find the sum and product of the roots of below quadratic equation

Solution:

As per the above method

Therefore,

Note: If we know the roots, we can frame a new quadratic equation If roots are p and q, then the quadratic equation becomes 

It means

Example:

If -3 and 5 are the roots then find the relevant quadratic equation

Solution:

We know that

Example-2: If p and q are the roots of  then find the value 

Solution:

Given that 

Example-3: Find the value of k if difference between the roots of equation

Solution:

Given that the difference of the roots = 3

We know that

Let a and b are the roots of the given equation

Example-4: Find the value of k if one root is double of the other root of the equation 

Solution:

Given that 

From the given problem  are the roots of the equation

From (1)

From (2)

---

Common Roots in Quadratic Equation:

Problem:

If one of the roots of equation  is same as one root of other equation

 find the values of p.

Solution:

Given that 

Let’s find the roots of above equation using our shortcut method

Changing the signs, we get 5 and -3 

then the roots are 5 and -3

First Case:

 is common root

Second Case:

 is common root

Therefore, 

Example:

Find the value of p if the equation  and  have a common root.

Solution:

Given that

Let’s find the roots of above equation using our shortcut method

Changing the signs, we get 8 and 9

First Case:

Second Case:

---

Factorizing a Cubic Polynomial:

Problem:

Factorize the below equation 

Solution:

Given that

Step – 1:  Find the first factor using converse of factor theorem

Converse of Factor Theorem:

According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and 'a' is  any real number, then, (x-a) is a factor of f(x), if f(a)=0. Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0.

This proves the converse of the theorem.

Let x = 1, then 

Let x = -1, then

Then x = -2 is one of the roots and  is one of the factors of the equation.

Now we can write

We need find the value of a, b and c to find the roots of the quadratic equation.

By Equating coefficient of  on both sides we get

By Equating constants on both sides, we get

By Equating coefficient of  on both sides we get

Then the above equation becomes

We need to find the roots of the equation using our method

Changing the signs, we get -5 and 3

divide both numbers by coefficient of  we get

Therefore,

Therefore  are the roots.

Example:

Factorize the below equation

Solution:

Given that

Step – 1:  Find the first factor using converse of factor theorem

Now we can write

By Equating coefficient of  on both sides we get

By Equating constants on both sides, we get

By Equating coefficient of  on both sides we get

Then the above equation becomes

Therefore  are the roots.

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Translating Product of Consecutive Integer Problems into Quadratic Equations:

Typical Problem in English:

Find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer.

Mathematical Translation:

Three consecutive even integers 

Then the numbers are 6, 8, 10.

Key Points:

For consecutive integer problems, define your variables as x, x + 1, and x + 2

For consecutive even or odd integer problems, define your variables as x, x +2, and x + 4.

 

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Translating Product of Ages Problems into Quadratic

Equations:

Typical Problem in English:

Henry is 3 years older than Glenn. The product of their ages is 40. How old is Glenn? 

Mathematical Translation:

Let d represent Glenn’s age.

Let d+3 represent Henry’s age.

Let d (d + 3) = 40 represent the product of their ages.

Solve for d.

Reject -8 because age cannot be negative.

Glenn is 5 years old.

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Translating Squared Number Problems into Quadratic Equations:

Typical Problem in English:

When 36 is subtracted from the square of a number, the result is five times the number. What is the positive solution?

Mathematical Translation:

Let the square of a number be represented by 

Let five times the number be represented by 5x

Write:

The problem says to select the positive solution then d = 9

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Operations in Inequalities:

Addition/Subtraction:

If a > b,

then

Where 

If a < b,

then

Example:

Solve the below inequality for x

Subtract 5 from both sides we get

Multiplication/Division:

If 

then

then

Example-1:

Solve the below inequality for x.

Multiply 5 on both sides we get

Example-2:

Solve the below inequality for x.

Divide 5 on both sides we get

Squaring:

 

Cubing:

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Quadratic Inequalities:

Shortcut:

If it could be factorized

We have to consider the shaded region above

Then the solution is 

If it could be factorized

We have to consider the shaded region above

Then the solution is 

Example-1:

Solution:

then as per our short cut method

By considering the shaded region we get then 

We can show the same in the below graph

Example-2:

Solution:

 

then as per our short cut method

By considering the shaded region we get 

We can show the same in the below graph

Example-3:

Solution:

then

then as per our short cut method

By considering the shaded region we get

The solution is 

Example-4:

Solution:

then

then as per our short cut method

By considering the shaded region we get

The solution is 

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Cubic Inequalities:

Solution:

then as per our shortcut method

By considering the shaded region we get

We can show the same in the below graph

then as per our shortcut method

By considering the shaded region we get 

We can show the same in the below graph

Example:

Solution:

Then equation becomes

then as per our shortcut method

By considering the shaded region we get 

We can show the same in the below graph

Example:

Solution:

Given that

 is one of the roots of the equation

on comparing the coefficient of  and constant on both sides we get

equate coefficient of  on both sides

By substituting b in the equation, we get

then as per our shortcut method

By considering the shaded region we get 

We can show the same in the below graph

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Reciprocal of Inequalities:

Example-1:

Solution:

For the reciprocal inequalities we have to consider our short cut method in a reciprocal manner to normal inequalities

By considering the shaded region we get 

We can show the same in the below graph

Example-2:

Solution:

For the reciprocal inequalities we have to consider our short cut method in a reciprocal manner to normal inequalities

By considering the shaded region we get 

We can show the same in the below graph

Example-3:

Solution:

This becomes

 as per the reciprocal inequality table above then as per our shortcut method

By considering the shaded region we get 

We can show the same in the below graph

Absolute Values:

Definition:

or

Note:

i. Any value coming out of modulus will either be positive or equal to zero.

ii. Any value coming out of modulus can be never be negative.

Example:

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Understanding Modulus as Distance:

If 

It means that distance of x from origin ‘O’ is equal to 5 units.

It means 

Example-1:

Solution:

By using Distance method, we get

Origin = 8

Example-2:

Solution:

By using Distance method, we get

Origin = 2

Example-3:

Solution:

By using Distance method, we get

Origin = -2.5

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Absolute Inequalities as Distance:

Example-1:

Solution:

By using Distance method, we get

Origin = 1.5

Example-2:

Solution:

By using Distance method, we get

Origin = -2.5

Key Shortcuts:

If a is a positive number then

then 

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Addition of two or more Absolutes:

Example:

Solution:

Theoretical Approach:

Let’s check each case

2.5 is not greater than 3

Therefore, this is not the expected answer.

In mathematics 7 will not be 6.

Therefore, this is also not the expected answer.

-3.5 is not less than -4

Therefore, this is also not the expected answer.

 

Therefore, there is NO solution for the problem.

Shortcut Approach:

Let’s interpret the problem like below

Sum of distances from ORIGINS (-4) and (3) is equal to 6 units.

We know that using distance formula between two points 

Now let’s consider the distance between two points be 7 units

Now we a point of x which distance from two points will be 6 units

Let’s plot a number line like below

Case 1:

Let’s take a point between two origins

It is clear that any point between the origins will be greater than 6 so we can’t figure out the solution using this case.

Case 2:

Let’s take a point outside the origins

It is clear than any point outside the origin will be greater than 6 so we can’t figure out the solution using this case also.

Case 3:

Let’s take a point outside the origins

It is clear than any point outside the origin will be greater than 6 so we can’t figure out the solution using this case also.

So, there will be NO Solution for this problem.

Note: We have to be noted that the minimum or least distance between two origins must be 7 in this problem case otherwise the problem does not have any Solution to find out.

Let’s change the problem to

Then we can clearly find out the solution using our above Case 1

Then the solution will be 

check for any number between the solution range

let x = 2

then

Example 1:

Solution:

Minimum distance = 

Here we can use above case 2 and 3 to find out the solution

We are assuming the distance from origin to out side point as ‘a’ and also the distance from the point to immediate origin is also ‘a’.

In order to get 11 units, the distance from origin to out side point will be 4

as 7 + 4 = 11

Here we got two cases

then the solution for the problem will be

Example 2:

Solution:

Minimum distance = 

As per our approach there will be NO Solution for this problem.

Example 3:

Solution:

Minimum distance = 

As per our approach the solution will be 

Understanding the Addition of two or more Absolutes concept

through analogy of friends:

Let two friends A and B. A stay at -3 and B stays at 4. Then At which point/s/region should they meet so that minimum distance is covered?

Let’s plot the number line.

So, in all the three cases the minimum distance is 7 only. Then we can find the minimum distance in the range 

Then if there are two points or values the minimum distance will be between the given points.

Let three friends A, B and C. A stay at -3  and B stays at 0 and C stays at 5.

Then At which point/s/region should they meet so that minimum distance is covered?

Let’s plot the number line.

Then the minimum distance will be found out if they meet at point ‘B’ or center point.

Key Points:

 If there are 2 friends A and B then they meet between A and B (Minimum distance)

 If there are 3 friends A, B and C then they meet at B (Minimum distance)

 If there are 4 friends A, B, C and D then they meet between B and C (Minimum distance)

 If there are 5 friends A, B, C, D and E then they meet at C (Minimum distance)

Example 1:

Find the least value of the given expression:

Solution:

Let’s plot the number line

Then as per our friends rule the three friends will meet at B = -2(minimum value)

By substituting the value of B = -2 in place of x in the given equation we get

the least value of the given expression:

Example 2:

Find the least value of the given expression:

Solution:

Let’s plot the number line

Then as per our friends rule the four friends will meet between B and C which means 

By substituting the any value between  in place of x in the given equation we get

 let x = 1

the least value of the given expression:

Example 3:

Find the least value of  Also find at what value of x will f (x) assume

Least value

Solution:

Let’s plot the number line

Then as per our friends rule the nine friends will meet at

Center point = 0 (minimum value)

By substituting the value of 0 in place of x in the given equation we get

the least value of the given expression

Example 4:

Find the least value of f (x)

Solution:

Total number of points = 20(Even)

So as per our friends’ rule least value of  will occur at 

By substituting x = 10 or 11 we can get the value of 

We know that

Example 5:

Find the least value of f (x)

Solution:

Total number of points = 25(odd)

So as per our friends’ rule least value of f (x) will occur at x = 13

By substituting x = 13 we can get the value of f (x)

We know that

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Counting Integral Points or Solutions:

Problem:

How many integral points will satisfy 

OR

How many integral solutions to 

Solution:

Algebraic Way:

We can find the integral solutions by picking the numbers for x

 integral solutions.

 integral solutions

 integral solutions

 integral solutions

 integral solutions

Therefore, total integral solutions

 2 + 4 + 4 + 4 = 16 integral solutions

Using Graphical Approach:

Let’s plot a graph like below

From the figure we have 5 integral points from (4,0) to (0,4) Likewise, we have same number of points in other sides also

So, there are  because each side the already counted points are repeating so we can subtract 4.

Total number of integral points = 

Example 1:

How many integral points will satisfy 

Solution:

Using the Graphical Approach

From the figure we have 7 integral points from (6,0) to (0,6)

Likewise, we have same number of points in other sides also

So, there are  because each side the already counted points are repeating so we can subtract 4.

Total number of integral points = 

Key Shortcuts:

then the equations contain 

Ex:

then the equations contain 

then the equations contain 

Example 2:

How many integral points will satisfy 

Solution:

The equation  will have the same number of integral points same as 

because in its ORIGIN has changed but the distance is still the same.

As per our graphical approach we can find the integral points of 

Therefore,  has 

Example 3:

How many integral points will satisfy 

Solution:

First Approach without graph:

Let’s take the possible values of the equation

We will not consider negative numbers less than 5

Therefore, total number of integral solutions = 

We can also use the graph like below to find out the integral solutions but it is not necessary as above method being easy and better way to find out the integral solutions.

 

Second Approach with graph:

Given equation  so we have to count the number of integral points inside the plotted graph as shown below

From (5,0) to  we have  integral solutions because origin (0,0) is also lies between the two points

Likewise, we can count for others also.

Therefore, total number of integral solutions = 

Example 4:

How many integral points will satisfy 

Solution:

We can use our first approach without graph to find out the integral solutions

Let’s take the possible values of the equation

 

We will not consider negative numbers less than 5

Therefore, total number of integral solutions = 

Example 5:

How many integral points will satisfy 

Solution:

The equation  will have the same number of integral points same as 

because in its ORIGIN has changed but the distance is still the same.

Therefore, total number of integral solutions = 

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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:

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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:

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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 = 

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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

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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,

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Example-1:

Solution:

Example-2:

Solution:

Example-3:

Solution:

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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.

---

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

---

 

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.

---

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

---

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

---

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

---

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,

---

  

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,

---

 

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,

---

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

---

 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

---

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,

---

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,

---

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.

---

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.

---

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,

---

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,

---

If A:B = x:y, B:C = p:q and C:D = m:n then,

Example-1:

If A : B = 3 : 4, B : C = 5 : 7 and C : D = 8 : 9 then A : D is equal to?

Solution:

By Applying Trick 79,

Example-2:

Solution:

By Applying Trick 79,

Thus,

---

Example:

 

Solution:

By Applying Trick 80,

---

If an amount R is to be divided between A and B in the ratio m:n then

where m > n

Example-1:

If an amount $250 is to be divided between P and Q in the ratio 3:2 then

how much will Q get on his part?

Solution:

By Applying Trick 81,

Therefore, Part of Q = 100

Example-2:

If $1000 is divided between A and B in the ratio 3 : 2, then A will receive?

Solution:

By Applying Trick 82,

---

If the ratio of A and B is m:n and the difference in their share is ‘R’ units then,

where m > n

 

Example:

If the ratio of A and B is 5:2 and the difference in their share is $75

then how much will A get?

Solution:

By Applying Trick 82,

Here, m:n = 5:2 and R = 75,

---

If the ratio of A and B is m:n and the part of A is ‘R’, then

where m > n

 

Example:

Marks of two candidates P and Q are in the ratio of 2 : 5. If the marks of P is

120, marks of Q will be?

Solution:

By Applying Trick 83,

---

where m>n

 

Example:

Solution:

By Applying Trick 84,

---

---

If an amount is to be divided among A, B and C in the ratio l : m : n and the difference between A and B is ‘R’, then

---

If there are notes of ‘x’ dollars, ‘y’ dollars and ‘z’ dollars in a box in the ratio m:n:r and the total value of notes is ‘R’, then

---

If adding/subtracting a certain quantity gives new ratio, then multiplier 

---

If the ratio of alligation of milk and water in a glass is m:n and in other glass alligation is p:q, then the ratio of milk and water in third glass which contains alligation of both glasses is

---

If the ratio of milk and water in the alligation of A litre is p:q then water must be added in it so that ratio of milk and water would be r:s is

---

The ratio of income of two persons A and B is p:q. If the ratio of their expenditures are r:s, then the monthly income of A and B, when each one of them saves ‘R’ dollars will be

---

Let ‘x’ be a number which is subtracted from a, b, c and d to make them proportional, then

Let ‘x’ be a number which is added to a, b, c and d to make them proportional, then

Here, a, b, c and d should always be in ascending order.

Example-1:

When a particular number is subtracted from each of 7, 9, 11 and 15, the resulting numbers are in proportion. The number to be subtracted is?

Solution:

By Applying Trick 92,

The number will be x

Example-2:

Which number when added to each of the numbers 6, 7, 15, 17 will make the resulting numbers proportional?

Solution:

By Applying Trick 92,

Required number

---

Example:

 

Solution:

By Applying Trick 93,

Here,

---

Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d.

Then the two numbers will be

Example:

If two numbers are in the ratio 2 : 3 and the ratio becomes 3 : 4 when 8 is

added to both the numbers, then the sum of the two numbers is?

Solution:

By Applying Trick 94,

Here, a = 2, b = 3, x= 8, c = 3, d = 4

Sum of numbers = 16 + 24 = 40

---

Two numbers are in the ratio a:b and if x is subtracted from each number the ratio becomes c:d.

The two numbers will be

Example:

Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is?

Solution:

By Applying Trick 95,

Here, a = 5, b = 7, x = 9, c = 7, d = 11

---

If the ratio of present age and the ratio of age after ‘n’ years is given then present age factor is given by

---

If x is the present age factor, and the difference in cross product of ratio is zero then,

---

If the ratio of ‘some years ago’ and ‘after some years’ is given. And Before  years, the ratio of ages of A and B was a : b.

after  years, the ratio of their ages will be c: d.

When, the difference in ratios is equal, then

---

If the product of present ages is given, then,

---

If sum of present age and ratio of the ages is given then, present age factor,

---

If the ratio of ages and difference in ages is given then,

---

The ratio of ages of A and B was x: y ‘n’ years ago.

---

 

---

---

The cost of cheap object is Rs. C/kg and the cost of costly object is Rs. D/kg.

If the mixture of both object costs Rs. M/kg. then

---

Quantity of x in mixture

---

If from x litre of liquid A, p litre is withdrawn and same quantity of liquid B is added. Again, from mixture q litre mixture is withdrawn and same quantity of liquid B is added. Again, from mixture, r litre is withdrawn and same quantity of liquid B is added, then In final mixture, liquid A is

If only one process is repeated n times, then liquid A in final mixture is

---

If x is initial amount of liquid, p is the amount which is drawn, and this process is repeated n-times such that the resultant mixture is in the ratio a: b then,

---

There are two pots of same volume. Both the pots contain mixture of milk and water in the ratio m:n and p:q respectively. If both the mixtures are mixed together in a big pot, then what will be the final ratio of milk and water?

---

 

If ‘d’ unit water is added to it then

---

There is x% milk in ‘a’ unit mixture of milk and water.

The amount of milk that should be added to increase the percentage of milk from x% to y% is given by

---

There is x% water in ‘a’ unit the mixture of sugar and water.

The quantity of water vapourised such that decrease in the percentage of water is from x% to y% is given by

---

Principal or Sum:

The money borrowed or lent out for a certain period is called the principal or the sum.

Interest:

Extra money paid for using other’s money is called interest.

Simple Interest: 

If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest. 

Formulae:

Let Principal = P, Rate = R% per annum and time = T years.

Example-1:

At some rate of simple interest, A lent $6000 to B for 2 years and $1500 to C for 4 years and received $900 as interest from both of them together. The rate of interest per annum was?

Solution:

If rate of interest be R% p.a. then,

Example-2:

$500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the sum of the interest on both the sum after 4 years is $480, the latter sum of money is?

Solution:

Simple interest gained from $500

Let the other Principal be x.

S.I. gained = $(480 – 240) = $240

Example-3:

A lends $2500 to B and a certain sum to C at the same time at 7% annual simple interest. If after 4 years, A altogether receives $1120 as interest from B and C, the sum lent to C is?

Solution:

Let the sum lent to C be x According to the question,

or 2500 × 28 + 28x = 112000 or

2500 + x = 4000 or

x = 4000 – 2500 = 1500

---

If there are distinct rates of interest for distinct time periods i.e.

---

If a certain sum becomes ‘n’ times of itself in T years on Simple Interest, then the rate per cent per annum is.

Example-1:

A sum of money becomes 7/6 of itself in 3 years at a certain rate of simple interest. The rate per annum is?

Solution:

By Applying Trick 114,

Example-2:

In how many years will a sum of money double itself at  simple interest per annum?

Solution:

By Applying Trick 114,

---

 

---

If Simple Interest (S.I.) becomes ‘n’ times of principal i.e.

S.I. = P × n then.

RT = n × 100

 

Example-1:

The simple interest on a sum after 4 years is 1/5 of the sum. The rate of interest per annum is?

Solution:

By Applying Trick 116,

Example-2:

Simple interest on a certain sum for 6 years is 9/25 of the sum. The rate of interest is?

Solution:

By Applying Trick 116,

---

If an Amount (A) becomes ‘n’ times of certain sum (P) i.e.

---

If the difference between two simple interests is ‘x’ calculated at different annual rates and times, then principal (P) is

Example:

The difference between the simple interest received from two different

banks on $500 for 2 years is $2.50. The difference between their (per

annum) rate of interest is?

Solution:

By Applying Trick 118,

Here, P = $500, x = $2.50, Difference in time = 2 years.

Difference in rate =?

Different in rate = 0.25%

---

---

 

and

---

If a sum is to be deposited in equal instalments, then,

Example-1:

What annual instalment will discharge a debt of $6450 due in 4 years at 5% simple interest?

Solution:

By Applying Trick 121,

Example-2:

What equal instalment of annual payment will discharge a debt which is due as $848 at the end of 4 years at 4% per annum simple interest?

Solution:

By Applying Trick 121,

Here, A = $848, T = 4 years, r = 4%

---

To find the rate of interest under current deposit plan,

where n = no. of months

---

   

Example-1:

The simple interest on a certain sum at 5% per annum for 3 years and 4

years differ by $42. The sum is?

Solution:

By Applying Trick 124,

Example-2:

The difference between the simple interest received from two different sources on $1500 for 3 years is 413.50. The difference between their rates of interest is?

Solution:

By Applying Trick 124

---

Example:

A sum of money lent at simple interest amounts to $880 in 2 years and to $920 in 3 years. The sum of money (in dollars) is?

Solution:

---

Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half yearly or quarterly to settle the previous account. In such case, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that period.

Let Principal = P, Rate = R% per annum, Time = n years.

Example-1:

The compound interest on $10,000 in 2 years at 4% per annum, the interest being compounded half-yearly, is?

Solution:

Example-2:

In what time will 1000 becomes 1331 at 10% per annum compounded annually?

Solution:

Let the required time be n years. Then,

Example-3:

The principal, which will amount to $270.40 in 2 years at the rate of 4% per annum compound interest, is?

Solution:

Let the principal be $P

---

If there are distinct ‘rates of interest’ for distinct time periods i.e.,

C.I. = A – P

Example:

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of $10,000 for 3 years will be?

Solution:

By Applying Trick 125,

A = $11575.2

C.I. = $(11575.2–10000)

= $1575.2

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If the time is in fractional form i.e., t = nF, then

Example:

Find compound interest on 10,000 for  years at 10% per annum, compounded yearly?

Solution:

By Applying Trick 126,

A = $13975.5

CI = $(13975.5 – 10,000)

CI = $3975.5

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A certain sum becomes ‘m’ times of itself in ‘t’ years on compound interest then the time it will take to become mn times of itself is t × n years.

 

Example:

A sum of money placed at compound interest doubles itself in 5 years. In how many years, it would amount to eight times of itself at the same rate of interest?

Solution:

By Applying Trick 127,

---

The difference between C.I. and S.I. on a sum ‘P’ in 2 years at the rate of R% rate of compound interest will be

Example:

If the difference between the compound interest and simple interest on a sum at 5% rate of interest per annum for three years is 36.60, then the sum is?

Solution:

By Applying Trick 128,

---

If on compound interest, a sum becomes A in ‘a’ years and B in ‘b’ years then,

where n is a whole number

 

Example:

A sum of money amounts to $4,840 in 2 years and to $5,324 in 3 years at compound interest compounded annually. The rate of interest per annum is?

Solution:

By Applying Trick 129,

Here, b – a = 3 – 2 = 1 B = $5,324, A = $4,840

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If a sum becomes ‘n’ times of itself in ‘t’ years on compound interest, then

Example:

If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is?

Solution:

By Applying Trick 130,

Here, n = 2.25, t = 2 years

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If a sum ‘P’ is borrowed at r% annual compound interest which is to be paid in ‘n’ equal annual installments including interest, then

Example:

A builder borrows $2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments.

How much will each instalment be?

Solution:

By Applying Trick 131,

Here, P = $2550, n = 2, r = 4%

---

The simple interest for a certain sum for 2 years at an annual rate interest R% is S.I., then

Example:

If the compound interest on a sum for 2 years at  per annum is $510, the simple interest on the same sum at the same rate for the same period of time is?

Solution:

By Applying Trick 132,

Here, C.I. = $510

---

Example-1:

A sum of $12,000, deposited at compound interest becomes double after 5 years. How much will it be after 20 years?

Solution:

By Applying Trick 133,

Example-2:

A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself?

Solution:

Example-3:

A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself?

Solution:

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Work Word Problem:

  It involves a number of people or machines working together to complete a task.

  We are usually given individual rates of completion.

   We are asked to find out how long it would take if they work together.

• The Work Problem Concept:

STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc.)

  If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus, in one hour, A would complete  of the work. But what is the logic behind this?

  Let me explain with the help of an example.

  Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task).

  Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work completed will be (1*1)/5

STEP 2: Add up the amount of work done by each person/machine in that one unit of time.

  This would give us the total amount of work completed by both of them in one hour. For example, if A completes  of the work in one hour and B completes  of the work in one hour, then TOGETHER, they can complete  of the work in one hour.

STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together.

  The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order.

  Suppose  

  This means that in one hour, A and B working together will complete  of the work. Therefore, working together, they will complete the work in Z hours.

 

Example-1:

  Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together?

Solution:

  This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer.

  STEP 1: Calculate how much work each person does in one hour.

  Jack → (1/3) of the work

  John → (1/5) of the work

  STEP 2: Add up the amount of work done by each person in one hour.

  Work done in one hour when both are working together  

  STEP 3: Calculate total amount of time taken when both work together.

  If they complete 8/15 of the work in 1 hour, then they would complete 1 job in 15/8 hours.

 

Example-2:

  Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work?

Solution:

  Now the only reason this is trickier than the first problem is because the sequence of events is slightly more complicated. The concept however is the same. So, if our understanding of the concept is clear, we should have no trouble at all dealing with this.

   ‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish 1/12 of the work.

   ‘He finishes 2/3 of the work’, This tells us that 1/3 of the work still remains.

   ‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete 1/3 of the work.

    ‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be 1/12. Therefore, rate at which Y works is 

    ‘In how much time does Y finish his work?’ If Y completes 1/120 of the work in 1 hour, then he will complete 1/3 of the work in 40 hours.

 

  So, as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of theconcept we learnt earlier to the information in the question in order to answer what was being asked.

 

Example-3:

  Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?

Solution:

  This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’ ‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of 1/24 per minute.

   ‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of 1/60 per minute.

  At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works:

  ‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and

  A complete in one minute corresponds to 5 pages. So, let us calculate that difference.

  ‘How many pages does the task contain?’ If 1/120 of the job consists of 5 pages, then the 1 job will consist of 

 

Example-4:

Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B.

Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution:

We are told that B produces 10% more sprockets per hour than A, thus 

---

If M₁ men can finish W₁ work in D₁ days and M₂ men can finish M₂ work in D₂ days then, Relation is

If M₁ men finish W₁ work in D₁ days, working T₁ time each day and M₂ men finish W₂ work in D₂ days, working T₂ time each day, then

---

If A completes a piece of work in ‘x’ days, and B completes the same work in ‘y’ days, then,

Work done by A in 1 day = 1/x, Work done by B in 1 day = 1/y

---

If A can do a work in ‘x’ days, B can do the same work in ‘y’ days, C can do the same work in ‘z’ days then, total time taken by A, B and C to complete

If workers are more than 3 then total time taken by A, B, C ...... so on to 

Example:

A, B and C individually can do a work in 10 days, 12 days and 15 days respectively. If they start working together, then the number of days required to finish the work is?

Solution:

By Applying Trick 136,

---

If A alone can do a certain work in ‘x’ days and A and B together can do the same work in ‘y’ days, then B alone can do the same work in

Example:

A and B together can dig a trench in 12 days, which A alone can dig in 28 days; B alone can dig it in

Solution:

By Applying Trick 137,

---

If A and B can do a work in ‘x’ days, B and C can do the same work in ‘y’ days, C and A can do the same work in ‘z’ days.

Then total time taken, when A, B and C work together

Example:

A and B can do a given piece of work in 8 days, B and C can do the same work in 12 days and A, B, C complete it in 6 days. Number of days required to finish the work by A and C is?

Solution:

By Applying Trick 138,

Let the time taken by A and C is x days

---

Total work = (work of one day) × (total no. of working days) Remaining

Work = 1 – (work done)

Work done by A = (Work done in 1 day by A) × (total no. of days worked by A, B and C and so on

Where A can complete work in x days, B in y days, C in z days and so on.

---

If A can finish m/n part of the work in D days. Then, Total time taken to finish the work by

---

(i) If A can do a work in ‘x’ days and B can do the same work in ‘y’ days and when they started working together, B left the work ‘m’ days before

(ii) A leaves the work ‘m’ days before its completion then total time taken

Example-1:

A can do a piece of work in 18 days and B in 12 days. They began the work together, but B left the work 3 days before its completion. In how many days, in all, was the work completed?

Solution:

By Applying Trick 141(i),

Here, x = 18, y = 12, m = 3

Example-2:

A and B alone can complete work in 9 days and18 days respectively. They worked together; however, 3 days before the completion of the work A left. In how many days was the work completed?

Solution:

By Applying Trick 141(ii),

Here, x = 9, y = 18, m = 3

Total time taken

---

If A and B together can finish a certain work in ‘a’ days. They worked together for ‘b’ days and then ‘B’ (or A) left the work. A (or B) finished the rest work in ‘d’ days, then Total time taken by A (or B) alone to complete the work

Example:

A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days.

A alone can finish the work in?

Solution:

By Applying Trick 142,

Here, a = 30, b = 20, d = 20

---

Example:

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Solution:

By Applying Trick 144,

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If food is available for ‘a’ days for ‘A’ men at a certain place and after ‘b’ days. ‘B’ men join, then the remaining food will serve total men for

If food is available for ‘a’ days for ‘A’ men at a certain place, and after ‘b’ days ‘B’ men leave then the remaining food will serve remaining men for

Example:

40 men can complete a work in 18 days. Eight days after they started working together, 10 more men joined them. How many days will they now take to complete the remaining work?

Solution:

By Applying Trick 143,

Here, A = 40, a = 18

b = 8, B = 10

---

Example:

If 8 men or 12 boys can do a piece of work in 16 days, the number of days required to complete the work by 20 men and 6 boys is?

Solution:

Here, A = 8, B = 12, a = 16

---

Example:

If 1 man or 2 women or 3 boys can complete a piece of work in 88 days,then 1 man, 1 woman and 1 boy together will complete it in?

Solution:

---

If ‘A’ men can do a certain work in ‘a’ days and ‘B’ women can do the same work in ‘b’ days, then the total time taken when A₁ men and  B₁ women work together is

Example-1:

5 men can do a piece of work in 6 days while 10 women can do it in 5 days.

In how many days can 5 women and 3 men do it?

Solution:

Example-2:

A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days?

Solution:

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The comparison of rate of work done is called efficiency of doing work.

Example:

Mike can do a work in 15 days. John is 50 per cent more efficient than

Mike in doing the work. In how many days will John do that work?

Solution:

Efficiency and time taken are inversely proportional

Hence, John completes the work in 10 days.

---

If the efficiency to work of A is twice the efficiency to work of B, then,

Example:

A man and a boy received $800 as wages for 5 days for the work they did

together. The man’s efficiency in the work was three times that of the boy.

What are the daily wages of the boy?

Solution:

---

If A can do a work in ‘x’ days and B is R% more efficient than A, then ‘B’ alone will do the same work in

Example:

A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do the same work is?

Solution:

Here, x = 70, r = 40%

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A, B and C can do a certain work together within ‘x’ days. While, any two of them can do the same work separately in ‘y’ and ‘z’ days, then in how many days can 3rd do the same work

Example:

A, B and C can complete a work in 8 days. B alone can do it in 18 days and C alone can do it in 24 days. In how many days can A alone do the same work?

Solution:

Here, x = 8, y = 18, z = 24

---

A and B can do a work in ‘x’ days, B and C can do the same work in ‘y’ days. C and A can do the same work in ‘z’ days.

Then, all can do alone the work as following:

Example-1:

A and B can do a piece of work in 10 days. B and C can do it in 12 days. A and C can do it in 15 days. How long will A take to do it alone?

Solution:

Example-2:

If A and B together can complete a work in 18 days, A and C together in 12 days and B and C together in 9 days, then B alone can do the work in?

Solution:

Example-3:

A and B can do a piece of work in 10 days. B and C can do it in 12 days. C and A in 15 days. In how many days will C finish it alone?

Solution:

---

A can do a certain work in ‘m’ days and B can do the same work in ‘n’ days.

They worked together for ‘P’ days and after this A left the work, then in how many days did B alone do the rest of work

when after ‘P’ days B left the work, then in how many days did A alone do the rest of work

Example-1:

A can do a piece of work in 12 days and B in 15 days. They work together for 5 days and then B left. The days taken by A to finish the remaining work is?

Solution:

Here, m = 12, n = 15, p = 5

Example-2:

A can do a piece of work in 12 days and B can do it in 18 days. They work together for 2 days and then A leaves. How long will B take to finish the remaining work?

Solution:

Here, m = 12, n = 18, p = 2

Time taken by B

---

Example:

While working 7 hours a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, working 8 hours a day?

Solution:

---

The efficiency of A to work is ‘n’ times more than that of B, both start to work together and finish it in ‘D’ days. Then, A and B will separately complete, the work in

Example:

A can do in one day three times the work done by B in one day. They together finish 2/5 of the work in 9 days. The number of days by which B can do the work alone is?

Solution:

Here, n = 3 and D

(Time taken to finish whole work)

---

Some people finish a certain work in ‘D’ days. If there were ‘a’ less people, then the work would be completed in ‘d’ days more, what was the number of people initially

Example:

A certain number of persons can complete a piece of work in 55 days. If there were 6 persons more, the work could be finished in 11 days less. How many persons were originally there?

Solution:

Here, D = 55

a = 6, d = 11

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A can do a work in ‘m’ days and B can do the same work in ‘n’ days. If they work together and total wages is R, then

Example:

Adam can do a work in 3 days. Wade can do the same work in 2 days.

Both of them finish the work together and get 150. What is the share of Adam?

Solution:

Here, m = 3, n = 2, R = 150

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If A, B and C finish the work in m, n and p days respectively and they receive the total wages R, then the ratio of their wages is

Example:

‘A’ alone can do a piece of work in 6 days and ‘B’ alone in 8 days. A and B undertook to do it for $3200. With the help of ‘C’, they completed the work in 3 days. How much is to be paid to C?

Solution:

Sum of the ratios = 4 + 3 + 1 = 8

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A and B can do a piece of work in x and y days, respectively. Both begin together but after some days. A leaves the job and B completed the remaining work in a days. After how many days did A leave?

Example:

A and B can do a work in 45 days and 40 days respectively. They began the work together but A left after some time and B completed the remainingwork in 23 days. After how many days of the start of the work did A leave?

Solution:

Here, x = 45, y = 40, a = 23

---

If A men and B boys can complete a work in x days, while A, men and B, boys will complete the same work in y days, then

---

---

Note: Both profit and loss are always calculated on cost price only

Example-1:

A man bought an old typewriter for $1200 and spent $200 on its repair. He sold it for $1680. His profit percent is?

Solution:

Example-2:

A merchant buys an article for $27 and sells it at a profit of 10% of the selling price. The selling price of the article is?

Solution:

---

If C.P > S.P., then there will be Loss

Example-1:

Mike had to sell vegetables worth $5,750 for $4,500 due to heavy rainfall.

What is the loss percentage that he has incurred?

Solution:

Example-2:

If a shop–keeper purchases cashew nut at $250 per kg. and sells it at $10 per 50 grams, then he will have?

Solution:

S.P. of 50 gm of cashew nut = $10

---

If an object is sold on r% Profit.

Similarly, if an object is sold on r% loss, then

Example-1:

A man buys a cycle for $1400 and sells it at a loss of 15%. What is the selling price of the cycle?

Solution:

Example-2:

On selling an article for $651, there is a loss of 7%. The cost price of that article is?

Solution:

---

Successive Profits: If A sells an article to B at a% profit and B sells it to C at b% profit

OR

If a% and b% are two successive profits

If A sells an article to B at a% profit and B sells it to C at b% profit and if C paid $x, then amount paid by

---

If a% and b% are two successive losses then (negative sign shows loss and positive sign shows profit)

Example:

If a certain company undergoes losses of 10% and 5% in the first two months of the year, then total loss percent for the two months is?

Solution:

Negative sign shows decrease

---

If a% profit and b% loss occur, simultaneously then overall loss or profit% 

Example:

If a certain company undergoes profit of 20% for March and loss of 10% for April months of the year, then total loss/profit percent for the two months is?

Solution:

By Applying Trick 166,

Therefore 8% profit

Positive sign shows increase

---

Example:

When the price of cloth was reduced by 25%, the quantity of cloth sold increased by 20%. What was the effect on gross receipt of the shop?

Solution:

Required per cent effect

Negative sign shows decrease

---

If cost price of ‘x’ articles is equal to selling price of ‘y’ articles, then Selling

Price = x, Cost Price = y

Example:

The cost price of 15 articles is same as the selling price of 10 articles. The profit percent is?

Solution:

Here, x = 15, y = 10

---

On selling ‘x’ articles the profit or loss is equal to Selling of ‘y’ articles, then

Example:

A cloth merchant on selling 33 metres of cloth obtains a profit equal to the selling price of 11 metres of cloth. The profit percent is?

Solution:

Here, x = 33, y = 11

---

If a man sells two similar objects, one at a loss of x% and another at a gain of x%, then he always incurs loss in this transaction and loss% is 

Example:

A cloth merchant on selling 33 metres of cloth obtains a profit equal to the selling price of 11 metres of cloth. The profit percent is?

Solution:

Here, selling prices are same, Profit-loss percent are same. In such transactions, there is always loss.

---

A man sells his items at a profit/loss of x%. If he had sold it for $R more, he would have gained/lost y%. Then.

‘+’ = When one is profit and other is loss.

‘–’ = When both are either profit or loss.

Example:

A man sold his watch at a loss of 5%. Had he sold it for $56.25 more, he would have gained 10%. What is the cost price of the watch (in $)?

Solution:

Here, x = 5%, R = 56.25, y = 10%

If a man purchases ‘a’ items for $x and sells ‘b’ items for $y, then his profit or loss per cent is given by

Example:

If a man purchases 5 articles for $120 and sells 10 articles for $80, then his profit or loss percent is?

Solution:

---

If a man purchases ‘a’ items for $x and sells ‘b’ items for $y, then his profit or loss per cent is given by

Example:

If a man purchases 5 articles for $120 and sells 10 articles for $80, then his profit or loss percent is?

Solution:

---

If the total cost of ‘a’ articles having equal cost is $x and the total selling price of ‘b’ articles is $y, then in the transaction gain or loss per cent is given by 

Where positive value signifies ‘profit’ and negative value signifies ‘loss’

Example:

Ten articles were bought for $8, and sold at 8 for $10. The gain percent is?

Solution:

---

A dishonest shopkeeper sells his goods at C.P. but uses false weight, then his

or

Example-1:

A dishonest fruit vendor sells his goods at cost price but he uses a weight of 900 gm for a kg. weight. His gain per cent is?

Solution:

Example-2:

A dishonest dealer professes to sell his goods at the cost price but uses a false weight of 850 g instead of 1 kg. His gain percent is?

Solution:

---

If A sells an article to B at a profit (loss) of  and B sells the same article to C at a profit (loss) of  then the cost price of article for C will be given by C.P of article for C

(Positive and negative sign conventions are used for profit and loss.)

 

Example:

David sold a Car at 20% gain to Jenny. Jenny sold it to John at 10% profit. If John had to pay $33,000 for the Car, find the cost price of the Car for David?

Solution:

C.P. for John = C.P. for David

33000 = C.P. for David

C.P. for David =

= $25,000

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If a vendor used to sell his articles at x% loss on cost price but uses y grams instead of z grams, then his profit or loss% is

(Profit or loss as per positive or negative sign).

---

If the speed of certain swimmer (or boat or ship) in still water is v km/h and the speed of stream is u km/h, then

(i) The speed of swimmer or boat or ship in the direction of

(ii) The speed of swimmer or boat or ship in the opposite direction of

---

If the speed of a swimmer/boat/ship in the direction of stream (downstream) is x km/h and in the opposite direction of stream (upstream) is y km/h, then,

Example-1:

A boatman rows 1 km in 5 minutes, along the stream and 6 km in 1 hour against the stream. The speed of the stream is?

Solution:

Example-2:

A man rows 40 km upstream in 8 hours and a distance of 36 km downstream in 6 hours. Then speed of stream is

Solution:

= 0.5 km/hr

Example-3:

A boat travels 24 km upstream in 6 hours and 20 km downstream in 4 hours. Then the speed of boat in still water and the speed of water current are respectively

Solution:

---

Let the speed of boat is x km/h and speed of stream is y km/h. To travel  km downstream and  km upstream, the time taken is ‘t’ hours, then

Example:

A boat covers 12 km upstream and 18 km downstream in 3 hours, while it covers 36 km upstream and 24 km downstream in  hours.

What is the speed of the current?

Solution:

Let the speed of boat in still water be x kmph and that of current be y kmph, then

By equation (i) × 3 – equation (ii),

From equation (i),

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Let the speed of stream be y km/h and speed of boat be x km/h.

A boat travels equal distance upstream as well as downstream in ‘t’ hours, then

Example:

A boat goes 12 km downstream and comes back to the starting point in 3 hours. If the speed of the current is 3 km/hr, then the speed (in km/hr) of the boat in still water is?

Solution:

Let the speed of boat in still water be x kmph, then

Therefore, Speed can't be negative.

Hence, speed of boat in still water = 9 kmph

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If a boat travels in downstream and upstream, then,

---

A swimmer or a boat travels a certain distance upstream in  hours, while it takes  hours to travel same distance downstream, then,

Example:

A boat goes 6 km an hour in still water, but takes thrice as much time in going the same distance against the current. The speed of the current (in km/hour) is?

Solution:

Here, Speed of boat = 6 km/hr

Speed of current = 3 km/hr

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If a swimmer takes same time to travel  km downstream and  km upstream, then,

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If a man or a boat covers x km distance in  hours along the direction of stream (downstream) and covers the same distance in  hours against the stream i.e., upstream, then

Example-1:

A boat goes 6 km an hour in still water, but takes thrice as much time in going the same distance against the current. The speed of the current (in km/hour) is?

Solution:

= 1.5 km/hr

Example-2:

A man can row 30 km downstream and return in a total of 8 hours. If the speed of the boat in still water is four times the speed of the current, then the speed of the current is?

Solution:

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If the speed of a boat or swimmer in still water is a km/hr and river is flowing with a speed of b km/hr, then average speed in going to a certain place and coming back to starting point is given by

Example-1:

A man can row 30 km downstream and return in a total of 8 hours. If the speed of the boat in still water is four times the speed of the current, then the speed of the current is?

Solution:

Here, a = 5, b = 1

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There are two types of taps:

Tap to fill the water (efficiency +) (inlet)

Tap to release the water (efficiency –) (outlet)

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Two taps ‘A’ and ‘B’ can fill a tank in ‘x’ hours and ‘y’ hours respectively. If both the taps are opened together, then the time it will take to fill the tank is

Example-1:

Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened together, the time taken to fill the tank is?

Solution:

Here, x = 20, y = 30

Required time

Example-2:

Two pipes can fill a cistern separately in 10 hours and 15 hours. They can together fill the cistern in?

Solution:

Part of the cistern filled by both pipes in 1 hour

The cistern will be filled in 6 hours.

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If x, y, z, ........... all taps are opened together then, the time required to fill/empty the tank will be:

where T, is the required time

Example-1:

Two pipes A and B can fill a cistern in 3 hours and 5 hours respectively.

Pipe C can empty in 2 hours. If all the three pipes are open, in how many hours the cistern will be full?

Solution:

Part of cistern filled by three pipes in an hour

Hence, the cistern will be filled in 30 hours.

Example-2:

Two pipes can fill a tank in 15 hours and 20 hours respectively, while the third can empty it in 30 hours. If all the pipes are opened simultaneously, the empty tank will be filled in?

Solution:

Part of tank filled in 1 hour when all three pipes are opened simultaneously

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Two taps can fill a tank in ‘x’ and ‘y’ hours respectively. If both the taps are opened together and 1st tap is closed before ‘m’ hours of filling the tank, then in how much time the tank will be filled

If 2nd tap is closed before ‘m’ hours then

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If a pipe fills a tank in ‘x’ hours but it takes ‘t’ more hours to fill it due to leakage in tank. If tank is filled completely, then in how many hours it will be empty? [due to leakage outlet]

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Amount of water released or filled = Rate × time.

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Two taps ‘A; and ‘B’ can empty a tank in ‘x’ hours and ‘y’ hours respectively.

If both the taps are opened together, then time taken to empty the tank will be

Example-1:

A tap can empty a tank in one hour. A second tap can empty it in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

Solution:

Here, x = 60, y = 30

= 20 minutes

Example-2:

A tap can empty a tank in 30 minutes. A second tap can empty it in 45 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

Solution:

Part of tank emptied by both pipes in 1 minute

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A tap ‘A’ can fill a tank in ‘x’ hours and ‘B’ can empty the tank in ‘y’ hours.

Then (a) time taken to fill the tank

Example-1:

A cistern can be filled with water by a pipe in 5 hours and it can be emptied by a second pipe in 4 hours. If both the pipes are opened when the cistern is full, the time in which it will be emptied is?

Solution:

Here, x = 5, y = 4

Example-2:

A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours.

If both the taps are open, the time (in hours) taken to fill the tank will be?

Solution:

Here, x = 8, y = 16

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Two taps A and B can fill a tank in x hours and y hours respectively. If both the pipes are opened together, then the time after which pipe B should be closed so that the tank is full in t hours

Example-1:

Two pipes X and Y can fill a cistern in 24 minutes and 32 minutes respectively. If both the pipes are opened together, then after how much time (in minutes) should Y be closed so that the tank is full in 18 minutes?

Solution:

x = 24, y = 32, t = 18 Required time

Example-2:

A tap takes 36 hours extra to fill a tank due to a leakage equivalent to half of its inflow. The inflow can fill the tank in how many hours?

Solution:

Here, x = 20, y = 30, t = 18

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If pipes A & B can fill a tank in time x, B & C in time y and C & A in time z, then the time required/taken to fill the tank by

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If Marked Price = (MP) Selling Price = (SP)

Then, Discount = MP – SP and

Note: Any kind of Discount is calculated only on marked price and not on selling price or cost price. 

 

Example-1:

An article, which is marked $650, is sold for $572. The discount given is?

Solution:

Here, M.P. = $650

S.P. = $572

Example-2:

If a dining table with marked price $6,000 was sold to a customer for $5,500, then the rate of discount allowed on the table is?

Solution:

M.P. = $6000

S.P. = $5500

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If article is sold on D% discount, then

Example-1:

A washing machine is sold at a discount of 30%. If a man buys it for $6,580, its list price is?

Solution:

Example-2:

A discount of 14% on the marked price of an article is allowed and then the article is sold for $387. The marked price of the article is?

Solution:

Here, D = 14%, S.P. = $387, M.P. =?

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When successive Discounts  so on, are given then

Example-1:

The marked price of an article is $500. It is sold at successive discounts of 20% and 10%. The selling price of the article (in rupees) is?

Solution:

Example-2:

An item is marked for $240 for sale. If two successive discounts of 10% and 5% are allowed on the sale price, the selling price of the article will be?

Solution:

Here, M.P. = $240,

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If  are successive discounts, then equivalent discount/overall discount is (in percentage)

Example-1:

A single discount equivalent to the successive discounts of 10%, 20% and 25% is?

Solution:

Single equivalent discount

Example-2:

The single discount equal to three consecutive discounts of 10%, 12% and 5% is?

Solution:

Single equivalent discount

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When two successive discounts are given, then overall discount is 

Example-1:

Two successive discounts of 5%, 10% are given for an article costing $850.

Present cost of the article is?

Solution:

Single equivalent discount

 Cost of article after discount

Example-2:

Successive discounts of 10% and 30% are equivalent to a single discount of?

Solution:

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If r% of profit or loss occur after giving D% discount on marked price, then 

(Positive sign for profit and negative for loss)

 

Example-1:

Jenny allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a TV if its marked price is $850?

Solution:

Here r = 20%, D = 4%, M.P. = $850, C.P. =?

Example-2:

The marked price of an article is $500. A shopkeeper gives a discount of 5% and still makes a profit of 25%. The cost price of the article is.

Solution:

Here, R = 25%, D = 5%, M.P. = 500, C.P. =?

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‘y’ articles (quantity/number) are given free on purchasing ‘x’ articles.

Then,

Example-1:

If a Man purchases 8 articles, he gets 2 articles free of cost. Then find the discount percentage the man gets from his purchase?

Solution:

Here x = 8 and y = 2

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A tradesman marks his goods r% above his cost price. If he allows his customers a discount of  on the marked price. Then is profit or loss percent is

(Positive sign signifies profit and negative sign signifies loss).

 

Example-1:

A tradesman marks his goods 10% above his cost price. If he allows his customers 10% discount on the marked price, how much profit or loss does he makes, if any?

Solution:

Example-2:

A tradesman marks his goods at 20% above the cost price. He allows his customers a discount of 8% on marked price. Find out his profit per cent.

Solution:

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The marked price of an article is fixed in such a way that after allowing a discount of r% a profit of R% is obtained. Then the marked price of the article is

Example-1:

What price should a shopkeeper mark on an article costing him $200 to gain 35% after allowing a discount of 25%?

Solution:

Here, r = 25%, R = 35%, C.P. = $200

Example-2:

What price should a shopkeeper mark on an article costing him $200 to gain 35% after allowing a discount of 25%?

Solution:

Here, r = 10% R = 20%

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Percentage refers to “Per hundred” i.e., 8% means 8 out of hundred or 8/100. Percentage is denoted by ‘%’.

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Example:

A reduction in the price of apples enables a person to purchase 3 apples for 1 instead of 1.25. What is the % of reduction in price (approximately)

Solution:

Percentage decrease

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Example:

The price of a commodity rises from $6 per kg to $7.50 per kg. If the expenditure cannot increase, the percentage of reduction in consumption is?

Solution:

Percentage increase

Percentage decrease in consumption

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If an amount is increased by a% and then it is reduced by a% again, then percentage change will be a decrease of

Example:

The salary of a person is decreased by 25% and then the decreased salary is increased by 25%. His new salary in comparison with his original salary?

Solution:

Percentage decrease

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If a number is increased by a% and then it is decreased by b%, then resultant change in percentage will be

(Negative for decrease, Positive for increase)

 

Example:

Water tax is increased by 20% but its consumption is decreased by 20%.

Then the increase or decrease in the expenditure of the money is?

Solution:

Percentage effect

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If a number is decreased by a% and then it is increased by b%, then net increase or decrease per cent is

(Negative for decrease, Positive for increase)

 

Example:

The price of an article is reduced by 25% but the daily sale of the article is increased by 30%. The net effect on the daily sale receipts is?

Solution:

Required change

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If a number is first decreased by a% and then by b%, then net decrease percent is

Example:

The price of an article was decreased by 10% and again reduced by 10%. By what per cent should the price have been reduced once, in order to produce the same effect as these two successive reductions?

Solution:

A single equivalent reduction to reduction series of x%, y%

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If a number is first increased by a% and then again increased by b%, then total increase per cent is

Example:

Two successive price increases of 10% and 10% of an article are equivalent to a single price increase of?

Solution:

Single equivalent percentage increase in price

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If the cost of an article is increased by A%, then how much to decrease the consumption of article, so that expenditure remains same is given by

Example:

If x is 10% more than y, then by what per cent is y less than x?

Solution:

Required per cent decrease

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If the cost of an article is decreased by A%, then the increase inconsumption of article to maintain the expenditure will be?

Example:

If x is less than y by 25% then y exceeds x by?

Solution:

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If the length of a rectangle is increased by a% and breadth is increased by b%, then the area of rectangle will increase by

Note: If a side is increased, take positive sign and if it is decreased, take negative sign. It is applied for two dimensional figures.

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If the side of a square is increased by a% then, its area will increase by

The above formula is also implemented for circle where radius is used as side. This formula is used for two dimensional geometrical figures having both length and breadth equal.

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If the side of a square is decreased by a%, then the area of square will decrease by

This formula is also applicable for circles. where decrease % of radius is given.

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If the length, breadth and height of a cuboid are increased by a%, b% and c% respectively, then,

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If every side of cube is increased by a%, then increase % in volume

This formula will also be used in calculating increase in volume of sphere. where increase in radius is given.

 

Example:

If each side of a cube is increased by 10% the volume of the cube will increase by?

Solution:

Increase % in volume

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If a% of a certain sum is taken by 1st man and b% of remaining sum is taken by 2nd man and finally c% of remaining sum is taken by 3rd man, then if 'x' rupee is the remaining amount then,

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If an amount is increased by a% and then again increased by b% and finally increased by c%, So, that resultant amount is ‘x’ rupees, then,

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If the population/cost of a certain town/ article, is P and annual increment rate is r%, then

Example:

The present population of a city is 180000. If it increases at the rate of 10% per annum, its population after 2 years will be?

Solution:

Required population after two years

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If the population/cost of a town/article is P and it decreases/reduces at the rate of r% annually, then,

Example:

The value of a commodity depreciates 10% annually. If it was purchased 3 years ago, and its present value is $5,832, what was its purchase price?

Solution:

Let the original price of the article be $x.

According to the question,

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On increasing/decreasing the cost of a certain article by x%, a person can buy ‘a’ kg article less/more in ‘y’ rupees, then

[Negative sign when decreasing and positive sign when increasing]

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If a person saves ‘R’ rupees after spending x% on food, y% on cloth and z% on entertainment of his income then,

Example:

A person gave 20% of his income to his elder son, 30% of the remaining to the younger son and 10% of the balance, he donated to a trust. He is left with $10080. His income was?

Solution:

Here, R = $10080

x = 20%,

y = 30%

and z = 10%

Monthly income

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The amount of acid/milk is x% in ‘M’ litre mixture. How much water should be mixed in it so that percentage amount of acid/milk would be y%?

Example:

8% of the voters in an election did not cast their votes. In this election, there were only two candidates. The winner by obtaining 48% of the total votes defeated his contestant by 1100 votes. The total number of voters in the election was?

Solution:

Here, x = 1100, A = 48

Total number of votes

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An examinee scored m% marks in an exam, and failed by p marks. In the same examination another examinee obtained n% marks and passed with q more marks than minimum, then

Example:

A candidate secured 30% marks in an examination and failed by 6 marks.

Another secured 40% marks and got 6 marks more than the bare minimum to pass. The maximum marks are?

Solution:

Here, m = 30%, n = 40%, p = 6, q = 6.

Maximum Marks

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In an examination, a% candidates failed in Maths and b% candidates failed in English. If c% candidate failed in both the subjects, then,

Example:

In an examination 34% failed in Mathematics and 42% failed in English. If 20% failed in both the subjects, the percentage of students who passed in both subjects was?

Solution:

a = 34%, b = 42%, c = 20%

Passed candidates in both the subjects

= 100 – (a + b – c)

= 100 – (34 + 42 – 20)

= 100 – 56 = 44%

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In a certain examination passing marks is a%. If any candidate obtains ‘b’ marks and fails by ‘c’ marks, then,

Example:

A student has to secure minimum 35% marks to pass in an examination. If he gets 200 marks and fails by 10 marks, then the maximum marks are

Solution:

a = 35%, b = 200, c = 10

Maximum Marks

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In a certain examination, ‘B’ boys and ‘G’ girls participated. b% of boys and g% of girls passed the examination, then,

Percentage of passed students of the total students = 

Example:

In an examination, there were 1000 boys and 800 girls. 60% of the boys and 50% of the girls passed. Find the percent of the candidates failed?

Solution:

Percentage of passed students

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If a candidate got A% votes in a poll and he won or defeated by ‘x’ votes, then, what was the total no. of votes which was casted in poll?

Example:

In an election there were only two candidates. One of the candidates secured 40% of votes and is defeated by the other candidate by 298 votes.

The total number of votes polled is?

Solution:

Total number of votes

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If a number ‘a’ is increased or decreased by b%, then the new number will be

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If the present population of a town is P and the population increases or decreases at rate of  in first, second and third year respectively, then

The population of town after 3 years =

‘+’ is used when population increases

‘–’ is used when population decreases.

The above formula may be extended for n number of years.

Example:

The value of a machine is 6,250. It decreases by 10% during the first year, 20% during the second year and 30% during the third year. What will be the value of the machine after 3 years?

Solution:

Required price of the machine

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If two numbers are respectively x% and y% less than the third number, first number as a percentage of second is

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If two numbers are respectively x% and y% more than a third number the first as percentage of second is

Example:

Two numbers are respectively 10% and 25% more than a third number.

What per cent is the first of the second?

Solution:

If two numbers are respectively x% and y% more than a third number, the first as a per cent of second is

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If the price of an article is reduced by a% and buyer gets c kg more for some $ b, the new price per kg of article

Example-1:

The Government reduced the price of sugar by 10 per cent. By this a consumer can buy 6.2 kg more sugar for $837. The reduced price per kg of sugar is

Solution:

Reduced price per kg

Example-2:

A reduction of 10% in the price of sugar enables a housewife to buy 6.2 kg more for 1116. The reduced price per kg is?

Solution:

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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,

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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)

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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)

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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

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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

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Example:

The average of the first nine integral multiples of 3 is

Solution:

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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

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Therefore, least of the numbers are 9.

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