Monthly Archives: October 2022

Introduction

→ Human requires food for their survival as the energy obtained from the food is used for the various metabolic activities as Respiration, Excretion etc. We get our food from plants by Agricultural practices and animals too by the process of Animal husbandry (the branch of agriculture which deals with the rearing of animals-livestock). In order to provide food for a large population, regular production, proper management and distribution of food is necessary.
→ When plants of the same kind are grown and cultivated at one place on a large scale, it is called a crop. For example- wheat, rice, cereals, vegetables, fruits. These are classified on the basis of season in which they grow.
→ India is the India is a vast country. The climatic conditions like temperature, humidity and rainfall vary from one region to another. Accordingly, there is a rich variety of crops grown in different parts of the country.

Types of crops

→ Kharif Crops: The crops which are sown in the rainy season are called kharif crop. The rainy season in India is from June to September. For Example- Paddy, maize, soybean, groundnut, cotton  etc.
→ Rabi Crops: The crops grown in the winter season are Rabi crops. Their time Period is generally from October to March. For Example- Wheat, gram, pea, mustard and linseed.
(Paddy is not cultivated in summer season because it requires lot of water. Therefore, it is cultivated in rainy season)

Cultivation of crops include various Agricultural practice 

• Preparation of soil: The preparation of soil is the first step before growing a crop. One of the most important tasks in agricultural is to turn the soil and loosen it. This allows the roots to penetrate deep into the soil. The loose soil allows the roots to breathe easily even when they go deep into the soil. The loosened soil helps in the growth of earthworm and microbes present in the soil. These organisms are friends of the farmer since they further turn and loose the soil and add humus to it. only a few centimetres of the top layer of soil supports plant growth, turning and loosening of soil brings the nutrient-rich soil to the top so that plants can use these nutrients. Thus, turning and loosening of soil is very important for cultivation of crops.
→ The process of loosening and turning of the soil is called tilling or ploughing. This is done by using a plough. Ploughs are made of wood or iron. The ploughed field may have big pieces of soil called crumbs. Field is levelled for Sowing.

Agricultural Implements

(i) Plough: This is being used since ancient times for tilling the soil, adding fertilisers to the crop, removing the weeds, scraping of soil, etc. It contains a strong triangular iron strip called ploughshare. The main part of the plough is a long log of wood which is called a plough shaft.

The indigenous wooden plough is increasingly being replaced by iron ploughs nowadays.

• Hoe: It is a simple tool which is used for removing weeds and for loosening the soil. It has a long rod of wood or iron. A strong, broad and bent plate of iron is fixed to one of its ends and works like a blade. It is pulled by animals.

• Cultivator: Nowadays ploughing is done by tractor driven cultivator. The use of cultivator saves labour and time.

(ii) Sowing: Sowing is the most important part of crop production. Before sowing, good quality seeds are selected. Farmers prefer to use seeds which give a high yield.

• Traditional tool: The tool used traditionally for sowing seeds is shaped like a funnel. The seeds are filled into the funnel, passed down through two or three pipes having sharp ends. These ends pierce into the soil and place seeds there.

•  Seed drill: the seed drill is used for sowing with the help of tractors. This tool sows the seeds uniformly at proper distances and depths. It ensures that seeds get covered by the soil after sowing. This prevents damage caused by birds. Sowing by using a seed drill saves time and labour.
It is important to maintain the distance to avoid overcrowding of plants to ensure proper nutrients, sufficient sunlight and water.

(iii) Adding Manures and Fertilizers: The substances which are added to the soil in the form of nutrients for the healthy growth of plants are called manure and fertilisers. Continuous growing of crops makes the soil poorer in certain nutrients. Therefore, farmers have to add manure to the fields to replenish the soil with nutrients. This process is called Manuring. Improper or insufficient Manuring results in weak plants.

•  Farmers dump plant and animal waste in pits at open places and allow it to decompose. The decomposition is caused by some microorganisms. The decomposed matter is used as organic manure.

•  Fertilisers are chemical substances which are rich in a particular nutrient. Fertilisers are produced in factories. Some examples of fertilisers are urea, ammonium sulphate, super phosphate, potash, NPK(Nitrogen, Phosphorus, Potassium). But excessive use of fertilisers has made the soil less fertile. Fertilisers have also become a source of water pollution.

•  Another method of replenishing the soil with nutrients is through crop rotation.

Advantages of Manures
→ The organic manure is considered better than fertilisers.
→ It enhances the water holding capacity of the soil.
→ It makes the soil porous due to which exchange of gases becomes easy.
→ It increases the number of friendly microbes.
→ It improves the texture of the soil.

(iv) Irrigation: The supply of water to crops at different intervals is called irrigation. The time and frequency of irrigation varies from crop to crop, soil to soil and season to season
→ To maintain the moisture of the soil for healthy crop growth, fields have to be watered regularly. Water also protects the crop from both frost and hot air currents.
→ Sources of irrigation: The sources of irrigation are wells, tube wells, ponds, lakes, rivers, dams and canals.
• Traditional system: These methods are cheaper, but less efficient. The various traditional ways are:
→ Moat (pulley-system)
→ Chain pump
→ Dhekli, and
→ Rahat (Lever system)

• Modern method of Irrigation: Modern methods of irrigation help us to use water economically. The main methods used are as follows
→ Sprinkler System
→ Drip system

(v) Protection from weed
→ In a field many other undesirable plants may grow naturally along with the crop. These undesirable plants are called weeds.
→ The removal of weeds is called weeding. Weeding is necessary since weeds compete with the crop plants for water, nutrients, space and light. The best time for the removal of weeds is before they produce flowers and seeds. Weeds are also controlled by using certain chemicals, called weedicides, like 2,4-D. These are sprayed in the fields to kill the weeds. They do not damage the crops. The weedicides are diluted with water to the extent required and sprayed in the fields with a sprayer.

(vi) Harvesting
→ The cutting of crop after it is mature is called harvesting. In harvesting, crops are pulled out or cut close to the ground. It usually takes 3 to 4 months for a cereal crop to mature. Harvesting in our country is either done manually by sickle or by a machine called harvester.

→ In the harvested crop, the grain seeds need to be separated from the chaff. This process is called threshing. This is carried out with the help of a machine called ‘combine’ which is in fact a combined harvester and thresher. Farmers with small holdings of land do the separation of grain and chaff by winnowing.

(vii) Storage: Storage of produce is an important task. Before storing them, the grains are properly dried in the sun to reduce the moisture in them. This prevents the attack by insect pests, bacteria and fungi. Farmers store grains in jute bags or metallic bins. However, large scale storage of grains is done in silos and granaries to protect them from pests like rats and insects dried neem leaves are used for storing food grains at home. For storing large quantities of grains in big godowns, specific chemical treatments are required to protect them from pests and microorganisms.

Animal Husbandry

→ Animals reared at home or in farms, have to be provided with proper food, shelter and care. When this is done on a large scale, it is called animal husbandry.
→ Fish is good for health. We get cod liver oil from fish which is rich in vitamin D.

Numbers in General Form

The general form of a number is obtained by adding the product of the digits with there place values.

1. The General Form of a Two Digit Number

ab = a × 10 + b = 10a + b

Example

93 = 10 × 9 + 3

= 90 + 3

2. The General Form of a Three Digit Number

abc = 100 × a + 10 × b + c = 100a + 10b + c

Example

256 = 100 × 2 + 10 × 5 + 6

= 200 + 50 + 6

Remark: ab doesn’t mean a × b as we generally use it like this.

Games with Numbers

1. Reversing the Digits – Two Digit Number

a. The trick to divide by 11: If we have any two digit number then we will always get the remainder zero if we add the original number with its reverse number and divide it by 11. The trick behind it:

Step 1: Let us choose any two digit number ab, which can be written as (10a + b).

Step 2: Its reverse number will be ba, which can be written as (10b + a).

Step 3: Sum of both the numbers is ab + ba, we can also write as (10a + b) + (10b + a) = 11a+ 11b = 11 (a + b)

Step 4: By dividing the number by 11 we get 11 (a + b)/11 = (a + b) .

This shows that the remainder will always be zero in case of any number. The quotient is a + b, which is the sum of the digits of the given number ab.

Example

Try for the number 78.

Solution:

• The given number is 78.
• The reverse number is 87.
• Sum of both the numbers is 78 + 87 = 165, 165 ÷ 11 = 15 and the remainder are zero.

b. The trick to divide by 9: If we have any two digit number then we will always get the remainder zero if we subtract the original number with its reverse number and divide it by 9. The trick behind it:

Step 1: Let’s choose any two digit number ab, which can be written as (10a + b).

Step 2: Its reverse number will be ba, which can be written as (10b + a).

Step 3: By subtracting both the numbers i.e. ab – ba we get (10a + b) – (10b + a) = 9a – 9b = 9 (a – b)

Step 4: By dividing the number by 9 we get  9 (a – b)/9 = (a – b).

This shows that the remainder will always be zero in case of any number.

Example

Try for the number 78.

Solution:

• The given number is 78.
• The reverse number is 87.
• Subtraction of both the numbers is 87 – 78 = 9, 9 ÷ 9 = 1 and the remainder is zero.

2. Reversing the Digits – Three Digit Number

The trick to divide by 99: If we have any three digit number then we will always get the remainder zero if we subtract the original number with its reverse number and divide it by 99. The trick behind it:

Step 1: Let’s choose any three digit number abc, which can be written as (100a + 10b + c).

Step 2: Its reverse number will be cba, which can be written as (100c + 10b + a).

Step 3: By subtracting both the numbers i.e. abc – cba we get (100a + 10b + c) – (100c + 10b + a) = 99a – 99c = 99 (a – c).

Step 4: By dividing the number by 99 we get 99 (a – c)/9 = (a – c).

This shows that the remainder will always be zero in case of any number. 

Example

Try with the number 456

Solution:

• The given number is 456.
• The reverse number is 654.
• Subtraction of both numbers is 654-456=198
• By dividing it by 99 we get 2 i.e. 198 ÷ 99 = 2 and the remainder is zero.

3. Forming Three-digit Numbers with given three-digits.

If we have a three digit number then we will always get the remainder zero if we rearrange the number in such a way that all the three numbers are different and if we add them all and then divide it by 37. The trick behind it:

Step 1: Let us take any three digit number abc, which can be written as (100a +10b + c).

Step 2: Rearrange the number in such a way that it forms two different numbers. One number can be bca which can be written as (100b + 10c + a) and other can be cab which can be written as (100c + 10a + b).

Step 3: By adding all the three numbers i.e. abc +bca + cab we get (100a +10b + c) + (100b +10c + a) + (100c + 10a + b) = 111 (a + b + c)

Step 4: By dividing the number by 37 we will always get the remainder zero .

Example

Try for the number 397.

Solution:

• Given number is 397.
• By rearranging we will have two other numbers i.e. 973 and 739.
• Sum of all the three numbers is 397+ 973 + 739 = 2109.
• By dividing 2109 by 37 we get 57 i.e. 2109 ÷ 37 = 57 and remainder will be zero.

Letters for Digits

In this type of puzzles, the letter represents the digits and we have to crack the code. Generally, it has puzzles related to addition and multiplication.

Some important rules related to such types of puzzles:

• Every letter can represent only one digit and the vice-versa.
• The first digit cannot be zero Like 25 cannot be written as 025 and 0025.
• Every puzzle must have only one solution.

Example

Find the value of P & Q.

Solution:

Here, at the ones place the addition of P & Q is given 9. This shows that it is the sum of two single digits as the sum of two single digit numbers cannot be 19. Hence there is not any carry in this step. For the next step, 2 + P = 0 and as there was not any carry from the last step it is possible only if P = 8.
Now, 2 + 8 = 10 and 1 will be carry for the next step.
So, 1 + 1 + 6 = P. Hence, clearly, P is 8 and Q can only be 1 to satisfy the given addition.

Hence, the value of P is 8 and Q is 1.

1. Divisibility by 2: If the number ends with an even number i.e. 0, 2, 4, 6 and 8 then it will always be divisible by 2.

Example 

Check the number 23 and 630 are divisible by 2 or not.

Solution:

• 23 is not divisible by 2 as its one’s digit is an odd number.
• 630 is divisible by 2 as its one’s digit is 0 which is even number.

2. Divisibility by 3: If the sum of all the digits of the given number is divisible by 3, then that number will also be divisible by 3.

Example

Check the number 232 and 6300 are divisible by 3 or not.

Solution: 

• 232 is not divisible by 3 as its sum of the digits i.e. 2 + 3 + 2 = 7 is not divisible by 3.
• 630 is divisible by 3 as its sum of the digits i.e 6 + 3 + 0 = 9 is divisible by 3.

3. Divisibility by 4: If the last two digits of the number are divisible by 4 then that number will also be divisible by 4.​

Example

Check the number 1748 and 258 are divisible by 4 or not.

Solution:

• 1748 is divisible by 4 as the last two digits 48 is divisible by 4.
• 258 is not divisible by 4 as the last two digits 58 is not divisible by 4.

4. Divisibility by 5: If the number ends with ‘0’ or ‘5’ then it will always be divisible by 5.

Example

Check the number 23 and 630 are divisible by 5 or not.

Solution: 

• 23 is not divisible by 5 as its one’s digit is 3.
• 630 is divisible by 5 as its one’s digit is 0.

5. Divisibility by 6: To be divisible by 3 a number must be divisible by 2 and 3 both. i.e. you need to check for the divisibility test of 2 and 3 with that number. Or you can say that a number must end with even number and the sum of its digit should be divisible by 3 then that number will be divisible by 6.

Example

Check the number 2341 and 6300 are divisible by 5 or not.

Solution:

• 2341 is not divisible by 6 as its one’s digit is odd and its sum of digit 2+3+4+1 = 10 is not divisible by 3.
• 6300 is divisible by 6 as its one’s digit is even and the sum of its digits 6+3+0+0=9 is divisible by 3 i.e it is divisible by both 2 and 3 hence the number is divisible by 6.

6. Divisibility by 7: To check the number is divisible with 7 or not, we need to double the last digit of the number and then subtract the result from the rest of the digits and check whether the remainder is divisible by 7 or not.If the number of digits are very large then you need to repeat the process until you get the number which could be checked for the divisibility of 7.

Example

Check the number 203 is divisible by 7 or not.

Solution:

Given number is 203

• Double the last digit, 3 × 2 = 6
• Subtract 6 from the remaining number 20 i.e. 20 – 6 = 14
• The remainder 14 is divisible by 7 hence the number 203 is divisible by 7.

7. Divisibility by 8: If the last three digits form a number which is divisible by 8 then the whole number will be divisible by 8.

Example

Check the number 19640 is divisible by 8 or not.

Solution:

The last three digit of the number 19640 is 640.

640/8 = 80

As the number 640 is divisible by 8 hence the number 19640 is also divisible by 8.

8. Divisibility by 9: If the sum of all the digits of the given number is divisible by 9, then that number will also be divisible by 9.

Example

Check the number 232 and 6300 are divisible by 9 or not.

Solution:

• 232 is not divisible by 9 as its sum of the digits i.e. 2 + 3 + 2 = 7 is not divisible by 9.
• 630 is divisible by 9 as its sum of the digits i.e 6 + 3 + 0 = 9 is divisible by 9.

9. Divisibility by 10: If the number ends with zero then it will always be divisible by 10.

Example 1

Check the number 23 and 630 are divisible by 10 or not.

Solution:

• 23 is not divisible by 10 as its one’s digit is 3.
• 630 is divisible by 10 as its one’s digit is 0.

Example 2

If 31a5 is a multiple of 3, where a is a digit, what could be the values of a?

Solution:

As we know that the number will be divisible by 3 only if its digits sum is divisible by 3.

So let’s check the sum of its digits i.e. 3 + 1 + a + 5 = 9 + a which should be multiple of 3.

Here, a is a single digit number and also being added with 9, so any multiple of 3 can take place i.e. a can be 3 or 6 or 9 and as it is added by 9 so it could be zero also.

Hence, the value of a could be 0, 3, 6 or 9.

Graph

The graphs are the visual representation of the data collected in numerical form so that it could be easily understood. Graphs are generally used for comparisons between two or more data.

Types of Graph

1. Bar Graph

A bar graph is a type of graph which is used to show comparisons between different categories. In this type of graph, bars of uniform width are used to represent the different quantities and their height is proportional to their respective values.

This is the bar graph which shows the profit of the company over different years.

2. Double Bar Graph

This is the same as the bar graph but two types of data are shown here simultaneously. Generally, this type of graph is used to show a comparison between two data.

This Double bar graph shows the average temperature in different seasons in two different years. Green bars show the temperature in 1950 and pink bars show the temperature in 2000. Hence we can easily compare the up and downs in average temperature from 1950 to 2000.

3. A Pie Chart (Circle Graph)

When we represent the given data in the circular form then it is said to be a pie chart. It shows the part of a whole.

This pie chart shows the percentage of people watching different kind of movies. From this chart, we can easily get that the maximum number of people like romantic movies and the least number of people like drama movies.

4. Histogram

When we have grouped data in the form of class intervals then we can draw a histogram. It is also a bar graph but there is no gap between the bars as it is a graph of continuous data.

This histogram shows the distribution of people in a society according to their age.

5. Line Graph

When we need to see the changes continuously over a period of time then we use a line graph.

Here the horizontal line (x-axis) shows the number of days and vertical line (y-axis) shows the rainfall in mm on every successive day.

By this line graph, we can easily understand the changes in rainfall during these 9 days of a month.

Linear Graphs

A line graph which is a whole unbroken line is called a Linear Graph.

1. Location of a Point

For making a linear graph we use the Cartesian plane. It is that system on which we mark the points with the help of Vertical and Horizontal Lines.

This is a Cartesian plane which is like a square grid sheet. We make a horizontal line (x-axis) and a vertical line (y-axis) which divides it into four quadrants.

2. Coordinates of a Point

To write the coordinates of a point we need an x – coordinate and a y-coordinate of a point.

• x-coordinate tells how many units to move right or left. It is also called the Abscissa.
• y-coordinate tells how many units to move up or down. It is also called the Ordinate.
• While writing the coordinates of a point in the coordinate plane, the x – coordinate comes first, and then the y – coordinate. We place the coordinates in brackets.
• Coordinates of the point of intersection of x-axis and y-axis is (0, 0).this is called Origin.

In the above figure, OB = CA = x coordinate (Abscissa), and CO = AB = y coordinate (Ordinate).

We write the coordinate as (x, y).

Example

Plot the given points on the graph sheet.

• (2, 3)
• (-3, 1)
• (-1.5, – 2.5)

Solution:

Remark: If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y.

Some Applications of Graph

In our day to day life, there are so many situations where we can use the graph for comparisons and analysis.

1. Independent Variable

Anything which is completely independent and its movement do not depend on any other factor then it is called Independent Variable.

2. Dependent Variable

Anything which increases or decreases with the movement of any other factor or it is dependent on any other factor then it is called Dependent Variable.

The Relationship between the Independent and Dependent Variable

The above graph shows that the revenue will increase with the increase in the number of cars washed. So the revenue is the dependent variable and the number of cars washed is an independent variable.

Factors of Natural Numbers

Factors are the pair of natural numbers which give the resultant number.

Example

24 = 12 × 2 = 6 × 4 = 8 × 3 = 2 × 2 × 2 × 3 = 24 × 1

Hence, 1, 2, 3, 4, 6, 8, 12 and 24 are the factors of 24.

Prime Factor Form

If we write the factors of a number in such a way that all the factors are prime numbers then it is said to be a prime factor form.

Example

The prime factor form of 24 is

24 = 2 × 2 × 2 × 3

Factors of Algebraic Expressions

Like any natural number, an algebraic expression is also the product of its factors. In the case of algebraic expression, it is said to be an irreducible form instead of prime factor form.

Example

7pq = 7 × p × q =7p × q = 7q × p = 7 × pq

These are the factors of 7pq but the irreducible form of it is

7pq = 7 × p × q

Example

2x (5 + x)

Here the irreducible factors are

2x (5 + x) = 2 × x × (5 + x)

Factorisation

The factors of an algebraic expression could be anything like numbers, variables and expressions.

As we have seen above that the factors of algebraic expression can be seen easily but in some case like 2y + 4, x² + 5x etc. the factors are not visible, so we need to decompose the expression to find its factors.

Methods of Factorisation

1. Method of Common Factors

• In this method, we have to write the irreducible factors of all the terms
• Then find the common factors amongst all the irreducible factors.
• The required factor form is the product of the common term we had chosen and the leftover terms.

Example

2. Factorisation by Regrouping Terms

Sometimes it happens that there is no common term in the expressions then

• We have to make the groups of the terms.
• Then choose the common factor among these groups.
• Find the common binomial factor and it will give the required factors.

Example

Factorise 3x² + 2x + 12x + 8 by regrouping the terms.

Solution:

First, we have to make the groups then find the common factor from both the groups.

Now the common binomial factor i.e. (3x + 2) has to be taken out to get the two factors of the expression.

3. Factorisation Using Identities

Remember some identities to factorise the expression

• (a + b)²  = a² + 2ab + b²
• (a – b)²  = a² – 2ab + b²
• (a + b) (a – b)  = a² – b²

We can see the different identities from the same expression.

(2x + 3)² = (2x)²+ 2(2x) (3) + (3)²

= 4x² + 12x + 9

(2x – 3)² = (2x)² – 2(2x)(3) + (3)²

= 4x² -12x + 9

(2x + 3) (2x – 3) = (2x)² – (3)²

= 4x² – 9

Example 1

Factorise x– (2x – 1)² using identity.

Solution:

This is using the identity (a + b) (a – b) = a² – b²

x²– (2x – 1)² = [(x + (2x – 1))] [x – (2x-1))]

= (x + 2x – 1) (x – 2x + 1)

= (3x – 1) (- x + 1)

Example 2

Factorize 9x² – 24xy + 16y² using identity.

Solution:

1. First, write the first and last terms as squares.

9x² – 24xy + 16y² 

= (3x)² – 24xy + (4y)²

2. Now split the middle term.

= (3x)² – 2(3x) (4y) + (4y)²

3. Now check it with the identities

4. This is (3x – 4y)²

5. Hence the factors are (3x – 4y) (3x – 4y).

Example 3

Factorise x² + 10x + 25 using identity.

Solution:

x² + 10x + 25

= (x)² + 2(5) (x) + (5)²

We will use the identity (a + b) ² = a² + 2ab + b² here.
Therefore,

x² + 10x + 25 = (x + 5)²

4. Factors of  the form ( x + a) ( x + b)

 (x + a) (x + b) = x² + (a + b) x + ab.

Example:

Factorise x² + 3x + 2.

Solution:

If we compare it with the identity (x + a) (x + b) = x² + (a + b) x +ab

We get to know that (a + b) = 3 and ab = 2.

This is possible when a = 1 and b = 2.

Substitute these values into the identity,

x² + (1 + 2) x + 1 × 2

(x + 1) (x + 2)

Division of Algebraic Expressions

Division is the inverse operation of multiplication.

1. Process to divide a monomial by another monomial

• Write the irreducible factors of both the monomials
• Cancel out the common factors.
• The balance is the answer to the division.

Example

Solve 54y³ ÷ 9y

Solution:

Write the irreducible factors of the monomials

54y³ = 3 × 3 × 3 × 2 × y × y × y

9y = 3 × 3 × y

2. Process to divide a polynomial by a monomial

• Write the irreducible form of the polynomial and monomial both.
• Take out the common factor from the polynomial.
• Cancel out the common factor if possible.
• The balance will be the required answer.

Example

Solve 4x³ + 2x² + 2x ÷ 2x.

Solution:

Write the irreducible form of all the terms of polynomial

4x³ + 2x² + 2x

= 4(x) (x) (x) + 2(x) (x) + 2x

Take out the common factor i.e.2x

= 2x (2x² + x + 1)

3. Process to divide a polynomial by a polynomial

In the case of polynomials we need to reduce them and find their factors by using identities or by finding common terms or any other form of factorization. Then cancel out the common factors and the remainder will be the required answer.

Example

Solve z (5z² – 80) ÷ 5z (z + 4)

Solution:

Find the factors of the polynomial

 = z (5z² – 80)

= z [(5 × z²) – (5 × 16)]

= z × 5 × (z² – 16)

= 5z × (z + 4) (z – 4)  [using the identity a² – b² = (a + b) (a – b)]

Some Common Errors

• While adding the terms with same variable students left the term with no coefficient but the variable with no coefficient means 1.

2x + x + 3 = 3x + 3 not 2x +3

We will consider x as 1x while adding the like terms.

• If we multiply the expressions enclosed in the bracket then remember to multiply all the terms.

2(3y + 9) = 6y + 18 not 6y + 9

We have to multiply both the terms with the constant.

• If we are substituting any negative value for the variables then remember to use the brackets otherwise it will change the operation and the answer too.

If x = – 5

Then 2x = 2(-5) = -10

Not, 2 – 5 = – 3

• While squaring of the monomial we have to square both the number and the variable.

(4x)² = 16x² not 4x²

We have to square both the numerical coefficient and the variable.

• While squaring a binomial always use the correct formulas.

(2x + 3)² ≠  4x²+ 9 But (2x + 3)² = 4x² + 12x + 9

• While dividing a polynomial by a monomial remember to divide each term of the polynomial in the numerator by the monomial in the denominator.

Introduction to Direct and Inverse Proportions

There are so many situations in our life where we see some direct or indirect relationship between two things. Like

• If the number of things purchased is increasing then the amount to pay will also increase.
• If the speeds of the car will increase then the time to reach the destination will decreases.

Variations

If the value of two objects depends upon each other in such a way that the increase or decrease in the value of one object affects the value of another object then these two objects are said to be in variation.

Direct Proportion

Two quantities a and b are said to be in direct proportion if

• Increase in a increases the b
• Decrease in a decreases the b

But the ratio of their respective values must be the same.

• a and b will be in direct proportion if a/b = k(k is constant) or a = kb.
• In such a case if b₁, b₂ are the values of b corresponding to the values a₁, a₂ of a respectively then, a₁/b₂ = a₂/b₁​

This shows that as the no. of hours worked will increase the amount of salary will also increase with the constant ratio.

Symbol of Proportion

When two quantities a and b are in proportion then they are written as a ∝ b where ∝ represents “is proportion to”.

Methods to solve Direct Proportion Problems

There are two methods to solve the problems related to direct proportion-

1. Tabular Method

As we know that,

so, if one ratio is given then we can find the other values also. (The ratio remains the constant in the direct proportion).

Example

The cost of 4-litre milk is 200 Rs. Tabulate the cost of 2, 3, 5, 8 litres of milk of same quality.

Solution:

Let X litre of milk is of cost Y Rs.

X(Liter)	2	3	4	5	8
Y(Rupees)	Y2	Y3	200	Y4	Y5

We know that as the litre will increase the cost will also increase and if the litre will decrease then the cost will also decrease.

Given,

So the cost of 2 ltr milk is 100 Rs.

So the cost of 3 ltr milk is 150 Rs.

So the cost of 5 ltr milk is 250 Rs.

So the cost of 8 ltr milk is 400 Rs.

2. Unitary Method

If two quantities a and b are in direct proportion then the relation will be

k = a/b or a = kb

We can use this relation in solving the problem.

Example

If a worker gets 2000 Rs. to work for 4 hours then how much time will they work to get 60000 Rs.?

Solution:

Here, 

By using this relation a = kb we can find

Hence, they have to work for 12 hours to get 60000 Rs. 

Inverse Proportion

Two quantities a and b are said to be in Inverse proportion if

1. Increase in a decreases the b
2. Decrease in a increases the b

But the ratio of their respective values must be the same.

• a and b will be in inverse proportion if k= ab
• In such a case if b₁, b₂ are the values of b corresponding to the values a₁, a₂ of a respectively then, a₁b₁ = a₂b₂ = k, 
• When two quantities a and b are in inverse proportion then they are written as a ∝ 1/b

This shows that as the distance of the figure from light increases then the brightness to the figure decreases.

Example

If 15 artists can make a statue in 48 hours then how many artists will be required to do the same work in 30 hours?

Solution:

Let the number of artists required to make a statue in 30 hours be y.

Number of hours	48	30
Number of artists	15	y

We know that as the no. of artists will increase the time to complete the work will reduce. So, the number of hours and number of artists are in inverse proportion.

So 48 × 15 = 30 × y (a₁b₁ = a₂b₂)

Therefore,

So, 24 artists will be required to make the statue in 30 hours.

Introduction to Exponents and Powers

The exponents tells us that how many times the number should be multiplied. It is called the Exponential form. This is written like this:

Here 10 is the base and 9 is the exponent and this complete number is the power. We pronounce it as 10 raised to the power 9. The exponent could be positive or negative.

This tells us that the number 10 will be multiplied 9 times, like, 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10

Powers with Negative Exponents

The exponents could be negative also and we can convert them in positive by the following method.

This shows that for any non-zero negative integers a,

where m is the positive integer and a^m is the multiplicative inverse of a^-m.

Laws of Exponents

If we have a and b as the base and m and n as the exponents, then

Laws of Exponents	Example
am × an = am+n	73 × 74 = 73+4 = 77
(am)n	(73)4 = 73×4 = 712
ambm = (ab)n	7343 (7× 4)3 = 283
a0 = 1	70 = 1
a1 = a	71 = 7

Some More Examples

Use of Exponents to Express Small Numbers in Standard Form

Sometimes we need to write the numbers in very small or large form and we can use the exponents to represent the numbers in small numbers.

1. Standard form to write the natural numbers like xyz000000……

Step 1: First of all count the number of digits from left leaving only the first digit.
Step 2: To write it in exponent or standard form, write down the first digit.
Step 3: If there are more digits in the number then put a decimal after the first digit and then write down the other digits until the zero comes. And if there are no digits after the first digit then skip this step.

Step 4: Now place a multiplication sign and then write down the counted digits in the first step as the exponent to the base number 10.

Example:

Express 1730000000000 in exponent form.

Solution: 

In standard form, the number 1730000000000 will be written as 1.73 x 10¹².

2. Standard form to write decimal numbers like 0.00000…..xyz.

Step 1: First of all count the number of digits from the decimal point to the last digit.
Step 2: If there is only one digit after the zeros then simply write down that digit. Place a multiplication sign and write down the counted digits in step-1 with a negative sign as the exponent to base number 10.
Step 3: If there are two or more non-zero digits at the end of the number. Then, write down the digits followed by a decimal point after the first digit and the other non-zero digits.
Step 4: Now calculate the number of digits in the first step and minus the number of digits appearing after the decimal point.
Step 5: Place a multiplication sign and write down the counted digits in step-4 with a negative sign as an exponent to base number 10.

Example:

Express 0.000000000000073 in exponent form.

Solution:

In standard form, the number will be written as 7.3 x 10^-14.

Comparing Very Large and Very Small Numbers

To compare the very large or very small numbers we need to make their exponents same. When their exponents are the same then we can compare the numbers and check which number is large or small.

Example

Compare the two numbers 4.56 × 10⁸ and 392 × 10⁷.

Solution

To compare these numbers we need to make their exponents same.

4.56 × 10⁸

392 × 10⁷ = 39.2 × 10⁸

As the exponents are the same, we can easily see that which number is larger.

392 × 10⁷ > 4.56 × 10⁸

Remark: To add and subtract also we need to make their exponents same and then they can be easily added or subtracted.

Mensuration

It is all about the measurement of area, perimeter and volume of the plane and solid figures.

Area

The surface covered by the border line of the figure is the area of the plain shape.

Unit of the area is square if the length unit.

Perimeter

The perimeter is the length of the boundary of the plane shape.

The unit of the perimeter is same as the length unit.

The green part is the area of the square and the distance all the way around the outside is the perimeter.

Area and Perimeter of Some 2D Shapes

Shape	Image	Area	Perimeter
Square		(Side)2	4 × Side
Rectangle		Length × Breadth	2(Length + Breadth)
Triangle		(1/2) × Base × Height where, a, b and c are the three sides of the triangle)	a + b + c
Parallelogram		Base × Height	2(sum of adjacent sides)
Circle		πr2	2πr Where, r = radius of the circle

Area of Trapezium

A trapezium is a quadrilateral whose two sides are parallel. And if its non-parallel sides are equal then it is said to be an isosceles trapezium.

Area of Trapezium can be found,

1. By Splitting the figure

One way to find the Area of trapezium is to divide it into two or three plane figures and then find the area.

In the trapezium ABCD,

It can be divided into two parts i.e. a rectangle and a triangle.

Area of ABCD = Area of ABED + Area of DEC

2. By using formula

Another way is to calculate the area by using formula.

Area of trapezium is half of the product of the summation of the parallel sides and the perpendicular distance between them.

Example

Find the area of the trapezium whose parallel sides are 6 cm and 16 cm, with a height of 5 cm. Calculate the area using both the methods.

Solution:

Splitting the trapezium we get –

Area of the trapezium = Area of rectangle + Area of a triangle

= (6 x 5) + (1/2) x 5 x 10

= 30 + 25

= 55 cm²

Remark: We should use the formula most of the time if possible as it is the quick and easy method.

Area of a General Quadrilateral

To find the area of any quadrilateral we can divide it into two triangles and then the area can be easily calculated by calculating the area of both the triangles separately.

Area of ABCD = Area of ∆ABC + Area of ∆ACD

= (1/2) × AC × h₁ + (1/2) × AC× h₂

The formula for the Area of a General Quadrilateral

Where h₁ and h₂ are the height of both the triangles and d is the length of common diagonal i.e.AC.

Example

Find the area of quadrilateral ABCD.

Solution:

In the quadrilateral ABCD,

BD is the common diagonal so d = 5 cm.

Height of the two triangles are h₁ = 2 cm and h₂ = 1 cm.

Area of Special Quadrilaterals (Rhombus)

A rhombus is a quadrilateral with all the sides are equal and parallel but not the right angle. Its two diagonals are the perpendicular bisector to each other.

In this also we can split the rhombus into two triangles and can find the area of rhombus easily.

Formula of Area of Rhombus

Area of rhombus is half of the product of its two diagonals.

Area of a Polygon

There is no particular formula for the area of the polygon so we need to divide it in a possible number of figures like a triangle, rectangle, trapezium and so on. By adding the area of all the split figures we will get the area of the required polygon.

Example

Find the area of the given octagon.

Solution:

We can divide the given octagon into three parts.

Two trapezium A and B and one rectangle shown by part B.

Two trapezium A and B and one rectangle shown by part B.
Area of A = Area of B = (1/2) × (a + b) × h

= (1/2) x (10 + 3) × 2

= 13 cm².

Area of B = Length x Breadth

= 10 x 3

= 30 cm².

So, the area of Octagon = 2A + B

= 2 × 13 + 30

= 56 cm².

Solid Shapes

The 3-dimensional shapes which occupy some space are called solid shapes. Example- Cube, Cylinder, Sphere etc.

Surface Area

If we draw the net of the solid shape then we can see it’s all the faces clearly and if we add the areas of all the faces then we get the total surface area of that solid shape. The unit of surface area is a square unit.

Lateral or Curved Surface Area

If we leave the top and bottom faces of the solid shape then the area of the rest of the figure is the lateral surface of the shape. The unit of lateral surface area is a square unit.

Surface Area of Cube, Cuboid and Cylinder

Name	Figure	Lateral or Curved Surface Area	Total Surface Area	Nomenclature
Cube		4l2	6l2	l = Edge of the cube
Cuboid		2h(l + b)	2(lb + bh + lh)	l = Length, b = Breadth, h = Height
Cylinder		2πrh	2πr2+ 2πrh = 2πr(r + h)	r = Radius, h = Height

Volume

Volume is the space occupied by any solid figure i.e. the amount of capacity to carry something is the volume of that solid shape. The unit of volume is a cubic unit.

Volume of Cube, Cuboid and Cylinder

Name	Volume	Nomenclature
Cube	l3	l = Edge of the cube
Cuboid	lbh	l = Length, b = Breadth, h = Height
Cylinder	πr2h	r = Radius, h = Height

Example 1

There is a shoe box whose length, breadth and height is 9 cm, 3 cm and 4 cm respectively. Find the surface area and volume of the shoe box.

Solution:

Given,

length = 9 cm

Breadth = 3 cm

Height = 4 cm

Area of cuboid = 2(lb + bh + lh)

= 2(9 × 3 + 3×4 + 9 × 4)

= 2(27 + 12 + 36)

= 2(75)

= 150 cm²

Volume of cuboid = lbh

= 9 × 3 × 4

= 108 cm³

Example 2

If there is a cold drink can whose height is 7 cm and the radius of its round top is 3 cm then what will be the lateral surface area and volume of that cylinder? (π = 3.14)

Solution:

Given,

radius = 3 cm

Height = 7 cm

Lateral surface area of cylinder = 2πrh

= 2 × 3.14 × 3 × 7

= 131.88 cm²

Volume of cylinder = πr²h

= 3.14 × 3 × 3 × 7

= 197.82 cm³

Example 3

If there is a box of cube shape with the length of 4 cm then what will be the capacity of this box. Also, find the surface area of the box if it is open from the top.

Solution:

Given, side = 4 cm

Capacity or volume of the box = s³

= 4³ = 64 cm³

The total surface area of the box = 6s²

But, if the box is open from the top then the surface area will be total surface area minus the area of one face of the cube.

Surface Area = Total Surface Area – Area of one face

= 6s² – s²

= 5s² = 5 × 4²

= 80 cm²

Volume and Capacity

Volume and capacity are one and the same thing.

Volume is the amount of space occupied by a shape.

Capacity is the quantity that a container can hold.

 Capacity can be measured in form of liters.

We can see the relation between liter and cm³ as,

1 L = 1000 mL

1 mL = 1 cm³,

1 L = 1000 cm³.

Thus, 1 m³ = 1000000 cm³ = 1000 L.

Two-dimensional shapes

Plane figures with only two measurements –length and width are called 2-D shapes.

Three-dimensional shapes

Solid figures with three measurements –length, width and height are called 3D shapes.

Views of 3D-Shapes

As the 3-D shapes are solid in nature so they may have a different view from different sides.

When we draw the top view, front view and side view on paper then it will look like this.

Example

Draw the front view, side view and the top view of the given figure.

Solution

Mapping Space around Us

A map shows the location of a particular thing with respect to others.

Some important points related to map:

• To represent different objects or place different symbols are used.
• A map represents everything proportional to their actual size not on the basis of perspective. It means that the size of the object will remain the same irrespective of the observer’s viewpoint.
•  A particular scale is used to draw a map so that the lengths drawn are proportional with respect to the size of the original figures.

This is the map which shows the different routes from Nehru road.

Faces, Edges and Vertices

• Faces – All the flat surfaces of the three 3-D shapes are the faces. Solid shapes are made up of these plane figures called faces.
• Edges – The line segments which make the structure of the solid shapes are called edges. The two faces meet at the edges of the 3D shapes.
• Vertex – The corner of the solid shapes is called vertex. The two edges meet at the vertex. The plural of the vertex is vertices.

Polyhedrons

Polygons are the flat surface made up of line segments. The 3-D shapes made up of polygons are called polyhedron.

• These solid shapes have faces, edges and vertices.
• The polygons are the faces of the solid shape.
• Three or more edges meet at a point to form a vertex.
• The plural of word polyhedron is polyhedral.

Non-polyhedron

The solid shape who’s all the faces are not polygon are called non-polyhedron. i.e. it has one of the curved faces.

Convex Polyhedrons

If the line segment formed by joining any two vertices of the polyhedron lies inside the figure then it is said to be a convex polyhedron.

Non-convex or Concave Polyhedron

If anyone or more line segments formed by joining any two vertices of the polyhedron lie outside the figure then it is said to be a non-convex polyhedron.

Regular Polyhedron

If all the faces of a polyhedron are regular polygons and its same number of faces meets at each vertex then it is called regular polyhedron.

Non-regular Polyhedron

The polyhedron which is not regular is called non-regular polyhedron. Its vertices are not made by the same number of faces.

In this figure, 4 faces meet at the top point and 3 faces meet at all the bottom points.

Prism

If the top and bottom of a polyhedron are a congruent polygon and its lateral faces are parallelogram in shape, then it is said to be a prism.

Pyramid

If the base of a polyhedron is the polygon and its lateral faces are triangular in shape with a common vertex, then it is said to be a pyramid.

Number of faces, vertices and edges of some polyhedrons

Solid	Number of Faces	Number of Edges	Number of Vertices
Cube	6	12	8
Rectangular Prism	6	12	8
Triangular Prism	5	9	6
Pentagonal Prism	7	15	10
Hexagonal Prism	8	18	12
Square Pyramid	5	8	5
Triangular Pyramid	4	6	6
Pentagonal Pyramid	6	10	6
Hexagonal Pyramid	7	12	7

Euler’s formula

Euler’s formula shows the relationship between edges, faces and vertices of a polyhedron.

Every polyhedron will satisfy the criterion F + V – E = 2,

Where F is the number of faces of the polyhedron, V is the vertices of the polyhedron and E is the number of edges of the polyhedron.

Example

Using Euler’s formula, find the number of faces if the number of vertices is 6 and the number of edges is 12.

Solution

Given, V = 6 and E = 12.

We know Euler’s formula, F + V – E = 2

So, F + 6 – 12 = 2.

Hence, F = 8.

Algebraic Expression

Any mathematical expression which consists of numbers, variables and operations are called Algebraic Expression.

1. Terms

Every expression is separated by an operation which is called Terms. Like 7n and 2 are the two terms in the above figure.

2. Factors

Every term is formed by the product of the factors.7n is the product of 7 and n which are the factors of 7n.

3. Coefficient

The number placed before the variable or the numerical factor of the term is called Coefficient of that variable.7 is the numerical factor of 7n so 7 is coefficient here.

4. Variable

Any letter like x, y etc. are called Variables. The variable in the above figure is n.

5. Operations

Addition, subtraction etc. are the operations which separate each term.

6. Constant

The number without any variable is constant. 2 is constant here.

Number Line and an Expression

An expression can be represented on the number line.

Example

How to represent x + 5 and x – 5 on the number line?

Solution:

First, mark the distance x and then x + 5 will be 5 unit to the right of x.

In the case of x – 5 we will start from the right and move towards the negative side. x – 5 will be 5 units to the left of x.

Monomials, Binomials and Polynomials

Expressions	Meaning	Example
Monomial	The algebraic expression having only one term.	5x2
Binomial	The algebraic expression having two terms.	5x2 + 2y
Trinomial	The algebraic expression having three terms.	5x2 + 2y + 9xy
Polynomial	The algebraic expression having one or more terms with the variable having non-negative integers as an exponent.	5x2 + 2y + 9xy + 4 and all the above expressions are also polynomial.

Like and Unlike Terms

Terms having the same variable are called Like Terms.

Examples of Like Terms

• 2x and -9x
• 24xy and 5yx
• 6x² and 12x²

The terms having different variable are called, Unlike Terms.

Examples of Unlike Terms

• 2x and – 9y
• 24xy and 5pq
• 6x² and 12y²

Addition and Subtraction of Algebraic Expressions

Steps to add or Subtract Algebraic Expression

• First of all, we have to write the algebraic expressions in different rows in such a way that the like terms come in the same column.
• Add them as we add other numbers.
• If any term of the same variable is not there in another expression then write is as it is in the solution.

Example

Add 15p² – 4p + 5 and 9p – 11

Solution:

Write down the expressions in separate rows with like terms in the same column and add. 

Example

Subtract 5a² – 4b² + 6b – 3 from 7a² – 4ab + 8b² + 5a – 3b.

Solution:

For subtraction also write the expressions in different rows. But to subtract we have to change their signs from negative to positive and vice versa.

Multiplication of Algebraic Expressions

While multiplying we need to take care of some points about the multiplication of like and unlike terms.

1. Multiplication of Like Terms

• The coefficients will get multiplied.
• The power will not get multiplied but the resultant variable will be the addition of the individual powers.

Example

• The product of 4x and 3x will be 12x².
• The product of 5x, 3x and 4x will be 60x³.

2. Multiplication of Unlike Terms

• The coefficients will get multiplied.
• The power will remain the same if the variable is different.
• If some of the variables are the same then their powers will be added.

Example

• The product of 2p and 3q will be 6pq
• The product of 2x²y, 3x and 9 will  be 54x³y

Multiplying a Monomial by a Monomial

1. Multiplying Two Monomials

While multiplying two polynomials the resultant variable will come by

• The coefficient of product = Coefficient of the first monomial × Coefficient of the second monomial
• The algebraic factor of product = Algebraic factor of the first monomial × Algebraic factor of the second monomial.

Example

25y × 3xy = 125xy²

2. Multiplying Three or More Monomials

While multiplying three or more monomial the criterion will remain the same.

Example

4xy × 5x²y² × 6x³y³ = (4xy × 5x²y²) × 6x³y³

= 20x³y³ × 6x³y³

= 120x³y³ × x³y³

= 120 (x³ × x³) × (y³ × y³)

= 120x⁶ × y⁶

= 120x⁶y⁶

We can do it in other way also

4xy × 5x²y² × 6x³ y³

= (4 × 5 × 6) × (x × x² × x³) × (y × y² × y³)

= 120 x⁶y⁶

Multiplying a Monomial by a Polynomial

1. Multiplying a Monomial by a Binomial

To multiply a monomial with a binomial we have to multiply the monomial with each term of the binomial.

Example

• Multiplication of 8 and (x + y) will be (8x + 8y).
• Multiplication of 3x and (4y + 7) will be (12xy + 21x).
• Multiplication of 7x³ and (2x⁴ + y⁴) will be (14x⁷+ 7x³y⁴).

2. Multiplication of Monomial by a trinomial

This is also the same as above.

Example

• Multiplication of 8 and (x + y + z) will be (8x + 8y + 8z).
• Multiplication of 4x and (2x + y + z) will be (8x² + 4xy + 4xz).
• Multiplication of 7x³ and (2x⁴+ y⁴+ 2) will be (14x⁷ + 7x³y⁴ + 14x³).

Multiplying a Polynomial by a Polynomial

1. Multiplying a Binomial by a Binomial

We use the distributive law of multiplication in this case. Multiply each term of a binomial with every term of another binomial. After multiplying the polynomials we have to look for the like terms and combine them.

Example

Simplify (3a + 4b) × (2a + 3b)

Solution:

(3a + 4b) × (2a + 3b)

= 3a × (2a + 3b) + 4b × (2a + 3b)    [distributive law]

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6 a² + 9ab + 8ba + 12b²

= 6 a² + 17ab + 12b²     [Since ba = ab]

2. Multiplying a Binomial by a Trinomial

In this also we have to multiply each term of the binomial with every term of trinomial.

Example

Simplify (p + q) (2p – 3q + r) – (2p – 3q) r.

Solution:

 We have a binomial (p + q) and one trinomial (2p – 3q + r)

(p + q) (2p – 3q + r)

= p(2p – 3q + r) + q (2p – 3q + r)

= 2p² – 3pq + pr + 2pq – 3q² + qr

= 2p² – pq – 3q² + qr + pr    (–3pq and 2pq are like terms)

(2p – 3q) r = 2pr – 3qr

Therefore,

(p + q) (2p – 3q + r) – (2p – 3q) r

= 2p² – pq – 3q² + qr + pr – (2pr – 3qr)

= 2p² – pq – 3q² + qr + pr – 2pr + 3qr

= 2p² – pq – 3q² + (qr + 3qr) + (pr – 2pr)

= 2p² – 3q² – pq + 4qr – pr

Identities

An identity is an equality which is true for every value of the variable but an equation is true for only some of the values of the variables.

So an equation is not an identity.

Like, x² = 1, is valid if x is 1 but is not true if x is 2.so it is an equation but not an identity.

Some of the Standard Identities

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

a² – b² = (a + b) (a – b)

(x + a) (x + b) = x² + (a + b)x + ab

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

These identities are useful in carrying out squares and products of algebraic expressions. They give alternative methods to calculate products of numbers and so on.

Applying Identities

Example

(4x – 3y)²

= (4x)² – 2(4x) (3y) + (3y)² 

= 16x² – 24xy + 9y²

Example

Use the Identity (x + a) (x + b) = x² + (a + b) x + ab to find the value of 501 × 502

Solution:

 501 × 502

= (500 + 1) × (500 + 2)

 = 500² + (1 + 2) × 500 + 1 × 2

= 250000 + 1500 + 2

= 251502  

Discount is a reduction given on marked price.
Discount = Marked Price – Selling price.

Discount can be calculated when the discount percentage is given.
Discount = Discount % of marked Price.

Additional expenses made after buying an article are included in the cost price and are known as overhead expenses.
C.P = Buying Price + Overhead expenses

Sales tax is charged on the sale of an item by the government and is added to the bill amount.
Sales tax = Tax % of bill amount

VAT (value added tax) is charged on the selling price of an article.

Percent: The word percent is an abbreviation of the Latin phrase ‘per centum’ which means per hundred or hundredths.

M.P. = Marked Price
S.P. = Selling Price
M.P = S.E + Discount
Discount = M.P – S.P

When profit % is given, then S.P > C.P and

When loss % is given S.P < C.P and

Increase and Decrease Percent

Simple Interest(SI): When the interest is paid to the lender regularly every year or half year on the same interest, we call it a simple interest. In other words, interest is said to simple, if it is calculated on the original principle throughout the loan period.
S.I.=P×R×T100
Where, P = Principal, R = Rate of Interest, T = Time.

Compound Interest (CI): If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half-year or a quarter of a year, etc.), so that the amount at the end of an interval becomes the principal for the next-interval, then the total interest over all the interval calculated in this way is called the compound interest.
Also, CI = Amount – Principal

(a) When interest is compounded annually, then

where P is Principal, R is the rate of interest and n is time period.

(b) When Interest is Compounded Half-Yearly, then

When R₁, R₂ and R₃ are different rates for the first, second and third year, then

Recalling Ratios and Percentages
We usually compare two quantities by division, i.e., by using fractions.
Comparison by division is called ratio.
Note that two quantities can be compared only when they have the same units. Consequently, the ratio has no unit. However, if the two quantities are not in the same unit, then we convert them into the same unit before comparison.

Two quantities can also be compared using percentages. By percentage, we mean a fraction where the denominator is 100. The numerator of the fraction is called rate per cent.
For example: 5100 means 5%. The symbol % is often used for the expression ‘per cent’ (p.c.).

To convert ratio into a percentage, we convert it into a fraction whose denominator is 100. [or we multiply by 100 and employ % sign.]

To convert percentage into a fraction, we divide the numerator by 100 and express it in the lowest form.
For example: 5% = 5100 = 120

In unitary method, we find the value of one unit from the given value of some units and then we find the value of required number of units.

Finding the Increase or Decrease Percent

New Price = Original (Old) price + Increase
New Price = Original (Old) price – Reduction

Finding Discounts
Discount = Marked Price – Sale Price

Estimation in Percentages

• Round off the bill to the nearest tens.
• Find the amount of discount.
• Reduce the bill amount by discount amount.

Prices Related to Buying and Selling (Profit and Loss)
Overhead Charges
Sometimes when an article is bought, some additional expenses are made while buying or before selling it.
These expenses are sometimes referred to as overhead charges. These may include expenses like the amount spent on repairs, labour charges, transportation etc. These expenses have to be included in the cost price.

Finding Cost Price/Selling Price, Profit% / Loss%
Cost Price: The buying price of an item is known as its cost price. It is written in short as CP.

Selling Price: The price at which an item is sold is called its selling price. It is written in short as SP.

Profit: If SP > CP, then we make a profit.
Profit = SP – CP

Loss: If CP > SP, then we make a loss.
Loss = CP – SP
Note: If SP = CP, then we are in a no profit no loss situation.

Profit/Loss Percent:
Profit/Loss is always calculated on the CP.

Overall Gain:
Overall gain = Combined SP – Combined CP

Overall Loss:
Overall loss = Combined CP – Combined SP.
Note: We need to find the combined CP and SP to say whether there was an overall profit or loss.

Sales Tax/Value Added Tax
Sales tax is charged at a specified rate on the sale price of an item by the state government and is added to the bill amount. It is different for different items and also for different states.
Amount of Sales Tax = Tax% of the bill amount
These days, the prices include the tax known as Value Added Tax (VAT).

Compound Interest
Interest: Interest is the extra money paid by institutions like banks or post offices on money which is deposited (kept) with them. Interest is also paid by people when they borrow money. The money deposited or borrowed is called the principal. Interest is generally given in per cent for a period of one year.

Simple interest (SI): The interest is called simple when the principal does not change.

The formula for Simple Interest: Simple interest on a principal of ₹ P at R% rate of interest per year for T years is given by

Amount: Amount(A) = Principal (P) + Simple Interest (SI)

Deducing a Formula for Compound Interest
A=P(1+R100)n
where
P = Principal
R = Rate of interest per annum compounded annually
n = Number of years
A = Amount
CI = A – P

Rate Compounded Annually or Half Yearly (Semi-Annually)
The word annually mentioned after the rate means that the interest is charged at the end of every year, whereas the rate is given for one year.

We could also have interest rates compounded half-yearly or quarterly. This means that the rate for the one-half year (i.e., 6 months) is half of the rate given for one year and the time period is of two half years because interest is charged twice a year.
So if a sum of ₹ 50,000 is taken for 1 year at 10% p.a. compounded semi-annually, it means time period = 2 half years (i.e., 1 × 2) and rate = 12 × 10% = 5%.

Note:

1. The time period for which the interest is calculated and added each time to form a new principle is called the conversion period or time period.
2. If interest is compounded half-yearly, then there are two conversion periods in a year each after 6 months. In such situations, we compute the interest two times. So, the time period becomes twice and the rate becomes half of the annual rate.
3. If interest is compounded quarterly, then there are four conversion periods in a year each after 3 months. In such situations, we compute the interest four times. So, the time period becomes four times and the rate becomes one-fourth of the annual rate.

Applications of Compound Interest Formula
We use the compound interest formula to find

• Increase (or decrease) in population.
• The growth of bacteria if the rate of growth is known.
• The value of an item, if its price increases or decreases in the intermediate years.

Note: For increase, R is positive and for decrease, R is negative.