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:

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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,

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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 

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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:

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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 

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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.

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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:

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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:

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Method of Square completion:

Problem: Factorize the below equation

Solution:

Divide throughout by 2

Then,

and

Therefore

Example: Factorize the below equation

Solution:

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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,

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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.

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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

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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

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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 

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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,

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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)

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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:

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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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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

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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

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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:

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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

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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:

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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

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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

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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

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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

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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

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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

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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.

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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:

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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:

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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:

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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:

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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).

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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

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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

---

If a boat travels in downstream and upstream, then,

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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

---

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

---

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

---

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]

---

Amount of water released or filled = Rate × time.

---

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

---

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

---

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

---

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

---

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%

---

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

---

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

---

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

---

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

---

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%

---

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:

---

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.

---

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.

---

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.

---

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

---

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,

---

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,

---

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

---

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]

---

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%

---

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

---

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

---

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

---

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

---

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

---

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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Example:  Find the value of the below expression

Solution:

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

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Example: Find the value of the 

Solution:

---

Solution:

---

Solution:

  

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Solution:

---

Solution:

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Solution:

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Example: Find the value of the 

Solution:

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  Any number can be expressed in terms of their digits by using the base 10 expression. For example, a two- digit number ab, where a is the tens digit and b is the units digit, can be written as:

Number ab = 10 a + b

  

  The same concept can be expanded to larger numbers, for example a three-digit number 

Number abc = 100a + 10b + c

  If we select any two-digit number and reverse the digits, then the difference between the two numbers thus formed will be nine times the difference between the two digits.

For example, if n = 64, then 64 − 46 = 18 = 9(6 – 4). In other words, if n = tu, where t is the tens digit and u is the units digit, then n = 10t+u, and the number formed by reversing the digits, m = ut = 10u + t, therefore,

  If we select any number and subtract the number formed by reversing its digits, then the resulting difference is always divisible by 9 and 11.

Let n = abc = 100a+10b+c, where a is the hundreds digit, b is the tens digit, and c is the unit’s digit.

If we reverse the digits, the new number formed is m = cba = 100c + 10b + a.

The difference between the two numbers is given by

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Let m be a non-zero integer, and n be an arbitrary integer.

If there is an integer, k, such that n = km, then we say that m divides n.

The following lists statements that are equivalent to m divides n.

  m divides n

  m is a factor or divisor of n

  n is divisible by m

  n is a multiple of m

For example, if n = 18 and m = 6, then 6 divides 18, 6 is a factor of 18, 18 is divisible by 6, and 18 is a multiple of 6.

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If r is divisible by t and s is divisible by t, then the expression r + s is divisible by t. 

Example:

For Example, as per the trick 53,

60 + 72 = 132 is divisible by 12 as 60 and 72 are already divisible by 12.

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If r is divisible by t and s is divisible by t, then the expression ra + sb is divisible by t for all integer values of a and b.

Example:

For Example, as per the trick 54,

8(5) + 16(6) = 136 is divisible by 8 as 8 and 16 are already divisible by 8.

---

If r is divisible by t, then rs is divisible by t for all integers s.

Example:

For Example, as per the trick 55,

15(9) = 135 is divisible by 5 as 15 is already divisible by 5.

---

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

Example:

Find the number of positive integers less than or equal to 20 that are divisible by 3?

Solution:

As per the trick 56,

Therefore, number of positive integers less than or equal to 20 that are divisible by 3 are 6.

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Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example:

5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.

5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.

5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

---

Solution:

---

Example:

Solution:

Putting these together,

---

Example:  Find the value of the following expression

Solution:

Use the properties of exponents as follows:

---

Solution:

Given that

Therefore, the value of  is 6.

---

Example:  Find the value of the following expression 

Solution:

Use the properties of exponents as follows: 

---

Example:  Find the value of the following expression 

Solution:

Use the properties of exponents as follows: 

---

Example:  Find the value of the following expression 

Solution:

Use the properties of exponents as follows: 

---

Example:  Find the value of the n if  

Solution:

---

Example:  Convert the below expression into radical form 

Solution:

---

Example:  Find the value of the n if

Solution:

Given that

Divide both sides by 

Therefore n = 1

---

  Calculate the total numbers after decimal point.

  Remove the decimal point from the number.

  Put 1 under the denominator and annex it with "0" as many as the total in step 1.

  Reduce the fraction to its lowest terms.

Example:

Convert 0.56 to a fraction.

Solution:

Total number after decimal point is 2.

56/100

Reducing it to lowest terms:

---

  Separate the recurring number from the decimal fraction

  Annex denominator with "9" as many times as the length of the recurring number

  Reduce the fraction to its lowest terms

Example:

Convert 0.393939…. to a fraction.

Solution:

The recurring number is 39.

Reducing it to lowest terms:

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  Write down the number consisting with non-repeating digits and repeating digits.

  Subtract non-repeating number from above.

  Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's. 

Example:

Convert 0.2512(12) to a fraction.

Solution:

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If n and m are positive integer, then there exist unique integers q and r, called the quotient and remainder, respectively, such that:

The above relationship can also be written as:

If r = 0, then n = mq, and n is a multiple of m.

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If the remainder when x and y are divided by m are  respectively, then the remainder when x + y is divided by m is equal to the remainder when  is divided by m.

Example:

For example, when 26 and 46 are divided by 7, the remainders are 5 and 4, respectively. If we add the two remainders, 5 + 4 = 9, and divide by 7, the remainder is 2, which is equal to the remainder when 26 + 46 = 72 is divided by 7.

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If the remainder when x and y are divided by m are  respectively, then the remainder when xy is divided by m is equal to the remainder when  is divided by m.

Example:

For example, when 26 and 46 are divided by 7, the remainders are 5 and 4, respectively. If we multiply the two remainders, 5 × 4 = 20, and divide by 7, the remainder is 6, which is equal to the remainder when 26 × 46 = 1196 is divided by 7.

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When a number is divided by 5, the remainder is equal to the remainder

when the last digit of the number is divided by 5. 

Example:

For example, the remainder when 27 is divided by 5, is the same as the remainder when 7 is divided by 5, which is 2. The reason for this rule is that, a number like, 64578 can be expressed as: 64578 = 60000 + 4000 + 500 + 70 + 8

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If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on. 

Example:

For example, 123 divided by 10 has the remainder 3 and 123 divided by 100 has the remainder of 23.

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Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

Example:

125,000 has 3 trailing zeros.

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The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

Example:

How many zeros are in the end (after which no other digits follow) of 32!?

Solution:

equal to 32 also notice that we take into account only the quotient of 

Therefore, 32! has 7 trailing zeros. The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

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