Chapter 1: Rational Numbers
 
❖ Natural numbers are a collection of all positive numbers starting from 1.
❖ Whole numbers are a collection of all natural numbers including 0.
❖ Rational numbers are the numbers that can be written in $\frac{p}{q}$ form, where p and q are integers and q ≠ 0
❖ Closure property of numbers
• Whole numbers are closed under addition and multiplication. However, they are not closed under subtraction and division.
This means if a andb are whole numbers, then a + b and a × b are also whole numbers.
Integers are also closed under addition and multiplication. However, they are not closed under subtraction and division.
This means if a and b are integers, then a + b, a – b and a × b are also integers.
• Rational numbers are closed under addition, subtraction and multiplication. They are not closed under division.
This means if a and b are rational numbers, then a + b, a – b and a × b are also rational numbers.
❖ Commutative property of numbers
• Whole numbers, integers and rational numbers are commutative under addition and multiplication. However, they are not commutative under subtraction and division.
$$\begin{gathered} \mathrm{i}.\mathrm{e}., a+b=b+a \\ a\times b=b\times a \\ a-b\ne b-a \\ a÷b\ne b÷a \end{gathered}$$

Where a and b are both whole numbers or integers or rational numbers.
❖ Associative property of numbers
• Whole numbers, integers and rational numbers are associative under addition and multiplication. However, none of the number system is associative under subtraction and division.
$$\begin{gathered} \mathrm{i}.\mathrm{e}., a+\left(b+c\right)=\left(a+b\right)+c \\ a\times \left(b\times c\right)=\left(a\times b\right)\times c \\ a-\left(b-c\right)\ne \left(a-b\right)-c \\ a÷\left(b÷c\right)\ne \left(a÷b\right)÷c \end{gathered}$$

Where a, b and c are all whole numbers or integers or rational numbers
❖ 0 is the additive identity of whole numbers, integers, and rational numbers, since 0 + a = a + 0 = a, where a is a whole number or an integer or a rational number.
❖ 1 is the multiplicative identity of whole numbers, integers, and rational numbers, since a × 1 = 1 × a = a, where a is a whole number or an integer or a rational number.
❖ Additive inverse of a number is the number, which when added to a number, gives 0. It is also called the negative of a number.
• a and -a are the additive inverse of each other as $a+\left(-a\right)=\left(-a\right)+a=0$
❖ Reciprocal or multiplicative inverse of a number is the number, which when multiplied by the number, gives 1.
• a and $\frac{1}{a}$ are the multiplicative inverse of each other as $a\times \frac{1}{a}=1$.
❖ Rational numbers are distributive over addition and subtraction.
• For rational numbers a, b, and c:
$$\begin{gathered} a\left(b+c\right)=ab+c \\ a\left(b+c\right)=ab-ac \end{gathered}$$

❖ Rational numbers can be represented on the number line in the same way as the fractions are represented.

❖ In between two rational numbers, there exists infinite rational numbers.Chapter 2: Linear Equations in One Variable
 
❖ An algebraic equation is an equality involving variables.
• In an equation, the value of expression on the left hand side (LHS) is equal to the value of expression on the right hand side (RHS).
• The equations in which the highest power of variable is 1 are known as linear equations in one variable.
❖ Some linear equations can be solved by transposing terms from one side to the other.
Example: Solve: 2x – 4 = 3x – 2
Solution: Transposing 3x from RHS to LHS, we obtain:
2x – 4 – 3x = –2
Transposing –4 from LHS to RHS, we obtain:
$$\begin{gathered} 2x-3x=-2+4 \\ \Rightarrow -x=2 \\ \therefore x=-2 \end{gathered}$$
  
❖ There are certain linear equations in which the denominators of the expressions on both sides are not 1. In such cases, both the sides of the equation are multiplied with the LCM of the denominators on both the sides.
For example: To solve the equation $\frac{3x-1}{4}+2=\frac{2x+3}{6}+7,$ we are first required to reduce it to its simplest form. This can be done as follows:
$$\begin{gathered} \left(\frac{3x-1}{4}+2\right)\times 12=\left(\frac{2x+3}{6}+7\right)\times 12 \left[\mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 4 \mathrm{and} 6 \mathrm{is} 12\right] \\ \Rightarrow \left(\frac{3x-1}{4}\right)\times 12+24=\left(\frac{2x+3}{6}\right)\times 12+84 \\ \Rightarrow 9x-3+24=4x+6+84 \\ \Rightarrow 9x+21=4x+90 \end{gathered}$$

This is a linear equation and it can be solved further to obtain the value of x.
❖ There are certain equations that can be reduced to linear equations.
For example: To solve the equation $\frac{2-4x}{3x-2}=\frac{3}{2},$ we are first required to reduce it to its simplest form. This can be done by cross-multiplying the terms of both sides as follows:

$$\begin{gathered} \Rightarrow \left(2-4x\right) 2=3\left(3x-2\right) \\ \Rightarrow 4-8x=9x-6 \end{gathered}$$
This is a linear equation in one variable and it can be solved further to obtain the value of x.Chapter 3: Understanding Quadrilaterals
 
❖ Polygons
A simple closed curve made up of line segments only is called a polygon.
• Polygons can be classified according to their number of sides (or vertices).
 

Number of side/vertices	Classification
3	Triangle
4	Quadrilateral
5	Pentagon
6	Hexagon
7	Heptagon
.	.
.	.
n	n - gon

 
• The line segment connecting two non-consecutive vertices of a polygon are called diagonals.


For polygon ABCD, AC and BD are diagonals and for polygon PQRS, QS and PR are diagonals.
• The polygon, none of whose diagonals lie in its exterior, is called a convex polygon. In the given figure, ABCD is a convex polygon.
• The polygon whose diagonals lie in its exterior is called a concave polygon. PQRS is a concave polygon.
• A polygon, which is both equiangular and equilateral, is called a regular polygon. Otherwise, it is an irregular polygon.
• The sum of all the angles of an n-sided polygon is given by, (n – 2) × 180°.
• The sum of measures of all exterior angles of a polygon is 360°.
 
❖ A quadrilateral with a pair of parallel sides is called a trapezium.
• A trapezium whose non-parallel sides are equal is called an isosceles trapezium.
❖ A kite is a quadrilateral with exactly two distinct consecutive pairs of sides of equal lengths.
❖ A parallelogram is a quadrilateral whose opposite sides are parallel and equal.
• Its opposite angles are of equal measure.
• The adjacent angles in a parallelogram are supplementary.
• The diagonals of a parallelogram are not equal. However, they bisect each other.
❖ A quadrilateral whose opposite sides are parallel and all sides are of equal lengths is called a rhombus.
• Its opposite angles are of equal measure.
• Its diagonals are perpendicular bisectors of one another.
❖ A rectangle is a parallelogram with equal angles.
• Its each angle is of measure 90°.
• Its diagonals are of equal length and they bisect each other.
❖ A square is a rectangle with equal sides.
• Its diagonals are equal and are perpendicular bisectors of each other.Chapter 4: Practical Geometry
 
❖ A unique quadrilateral can be constructed, if any five measurements of the quadrilateral are given.
❖ Construction of a quadrilateral when four sides and a diagonal are given
Example: Construct a quadrilateral WXYZ, where WX = 4.5 cm, XY = 5 cm, YZ = 5.5 cm, ZW = 3 cm, and WY = 6 cm.
Solution:
The steps of constructing quadrilateral WXYZ are as follows:
(1) Draw a line WY of length 6 cm. Draw an arc of radius 4.5 cm with W as centre and another arc of length 5 cm with Y as centre. The intersection of the two arcs will be the point, X. Join WX and XY.
(2) The point, Z, will be on the opposite side of point X with respect to WY. Draw an arc of length 3 cm taking W as centre and another arc of length 5.5 cm taking Y as centre. The intersection of these arcs will be the point, Z. Join WZ and YZ.


Thus, WXYZ is the required quadrilateral.
❖ Construction of a quadrilateral when two diagonals and three sides are given
Example: Construct a quadrilateral PQRS, where PR = 7 cm, QS = 8 cm, PQ = 5 cm, QR = 5 cm, and PS = 5.5 cm.
Solution:
The steps of constructing quadrilateral PQRS are as follows:
(1) Draw a line PR of length 7 cm. Draw an arc of radius 5 cm taking P as centre and an arc of radius 5 cm taking R as centre.The point of intersection of these two arcs will be the point, Q.Join PQ and RQ.
(2) With Q as centre, draw an arc of radius 8 cm. The point, S, will lie on this arc.Then, taking P as centre, draw an arc of radius 5.5 cm. The intersection point of the two arcs will be the point, S.Join PS and RS.


Thus, PQRS is the required quadrilateral.
 
❖ Construction of a quadrilateral when two adjacent sides and three angles are given:
Example: Construct a quadrilateral ABCD, where AB = 6 cm, AD = 4 cm, ∠A = 90°, ∠B = 105°, and ∠D = 60°.
Solution:
The steps of constructing quadrilateral ABCD are as follows:
(1) Draw a line segment AB of length 6 cm. Make ∠ABX = 105° at B and ∠BAY = 90° at A.
(2) With A as centre, draw an arc of radius 4 cm to cut the ray AY at point D. At D, draw ∠ADZ = 60°. The point of intersection of the rays, BX and DZ, will be the point, C.


 
Thus, ABCD is the required quadrilateral.
 
❖ Construction of a quadrilateral when three sides and two included angles are given:
Example: Construct a quadrilateral PQRS with SR = 6.5 cm, PS = 5 cm, QR = 3 cm, ∠R = 120°, and ∠S = 70°.
Solution:
The steps of construction are as follows:
(1) Draw SR = 6.5 cm. Draw ∠SRX = 120° at R and ∠RSY = 70° at S.
(2) With S as centre, draw an arc of radius 5 cm intersecting SY at P. With R as centre, draw an arc of radius 3 cm intersecting RX at Q.
Join PQ to obtain the required quadrilateral PQRS.

 
❖ Some special quadrilaterals such as square, rhombus, rectangle, kite, and parallelogram can be constructed by using their properties.
• In a square, all the sides are equal and all the angles are equal to 90°. Therefore, a square can be constructed, if one of its sides is given.
• In a rhombus, all the sides are equal. Also, the diagonals of a rhombus are perpendicular bisectors of one another. Therefore, a rhombus can be constructed, if only the measures of two diagonals are given.
• In a rectangle, all angles are of measure 90° and the opposite sides are equal. Therefore, a rectangle can be constructed even if two of its adjacent sides are given.
• In a kite, the adjacent sides are equal. Therefore, if two adjacent sides and a diagonal are given, then the required kite can be constructed.
• In a parallelogram, the opposite sides are equal and parallel. Therefore, a parallelogram can be constructed even if two of its sides and one angle or a diagonal are given.
For example,a parallelogram ABCD with AB = 5 cm, BC = 4 cm, ∠B = 110° can be constructed as follows:
First, draw AB = 5 cm and then construct ∠ABX = 110°. With B as centre, draw an arc of radius 4 cm cutting the ray, BX, to obtain the point, C. Then, draw a line CY parallel to AB. With C as centre, cut an arc of radius 5 cm at CY to obtain the point, D.Join AD to obtain the parallelogram, ABCD.
Chapter 5: Data Handling
 
❖ The data in an unorganised form is called raw data. In order to draw meaningful inferences from a data, we need to organise the data systematically.
• We can organise a data in the following ways. They are: frequency distribution table, histogram and pie chart
❖ Frequency distribution table
• The number of times a particular entry occurs is called its frequency.
• The difference between the upper and lower class limits is called the width or size of the class interval.
Example: The ages of children (in years) in a group dance competition are given below:
16, 18, 8, 10, 12, 17, 9, 8, 16, 18, 14, 14, 16, 15, 12, 9, 11, 10, 18, 9, 12, 11, 16.
Arrange the following data in a frequency distribution table whose class intervals are 7 – 10, 10 – 13 …
Solution:

Class interval (Age of children)	Tally mark	Frequency (Number of children)
7 – 10		5
10 – 13		6
13 – 16		4
16 – 19		8

 
❖ Histogram
• A histogram is a bar graph that is used to represent grouped data. In a histogram, the class intervals are represented on the horizontal axis and the heights of the bars represent frequency. Also, there is no gap between the bars in a histogram.
The above frequency distribution table can be displayed in a histogram as follows:


In a histogram, a broken line can be used along the horizontal axis to indicate that the numbers between 0 to7 are not included.
❖ Pie chart
A pie chart or a circle graph shows the relationship between a whole and its parts.
 
For example: Consider the given pie chart which shows the favourite colours of the class-VIII students of a school.


 
In this pie chart, the portion of the sector for the colour red is given by,
$\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{whose} \mathrm{favourite} \mathrm{colour} \mathrm{is} \mathrm{red}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{45}{80}=\frac{9}{16}$

Therefore, the sector representing red colour is ${\left(\frac{9}{16}\right)}^{\mathrm{th}}$ part of the circle.
❖ Construction of pie charts
Example: Construct a pie chart for the following data which gives the brands of laptop preferred by the people of a locality.

Brand A	:	100
Brand B	:	120
Brand C	:	180

 
Solution:
The total number of people is 100 + 180 + 120 = 400.
We can form the following table to find the central angle of each sector:
 

Brand of laptop	Number of people	Fraction	Central angle
A	100	$\frac{100}{400}=\frac{1}{4}$	$\frac{1}{4}\times 360°=90°$
B	180	$\frac{120}{400}=\frac{3}{10}$	$\frac{3}{10}\times 360°=108°$
C	120	$\frac{180}{400}=\frac{9}{20}$	$\frac{9}{20}\times 360°=162°$

 
To represent the given information using a circle graph, first of all draw a circle with any convenient radius. Let O be the centre of the circle and OX be its radius. Draw the angle of the sector for brand A, which is 90°. Using protractor, draw ∠XOY = 90°. Now, draw the angle of the sectors for brands B and C.



❖ A pie chart is interpreted by analysing it.
❖ Probability is the chance of the occurrence of an event.
• A random experiment is the experiment whose outcome cannot be predicted exactly in advance.
• The outcomes of an experiment having the same chances of occurrence are known as equally-likely outcomes. For example, if we toss a coin, then the possible outcomes are head or tail, and both of them have an equal chance of occurring. So, these are equally-likely outcomes.
• Each outcome of an experiment or collection of outcomes is known as an event.
• When the outcomes of the experiment are equally-likely,  the probability of an event is given by: $\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{outcomes}}$
For example, if two coins are tossed together, then any of the following outcomes may be obtained.
(i) Head on first coin, head on second coin(H, H)
(ii) Head on first coin, tail on second coin (H, T)
(iii) Tail on first coin, head on second coin (T, H)
(iv) Tail on first coin, tail on second coin (T, T)
 Now, the probability of getting one head and one tail$=\frac{2}{4}=\frac{1}{2}$Chapter 6: Squares and Square Roots
 
❖ If a natural number m can be expressed as n², where n is also a natural number, then m is a square number. The square numbers are also called perfect squares.
• All the square numbers end in 0, 1, 4, 5, 6, or 9.
• If a number has 1 or 9 at its units place, then its square ends in 1.
• If a square number ends in 6, then the number, whose square it is, will either have 4 or 6 in its units place.
• Square numbers can have even number of zeroes at the end.
❖ If we add two consecutive triangular numbers, then we obtain a square number.
For example, 10 + 15 = 25 = 5²
❖ There are 2n non-perfect square numbers between the squares of the numbers, n and (n + 1).
❖ The sum of first n odd natural numbers is n².
For example, the sum of first 7 odd natural numbers is 49 i.e., 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 7²
❖ Square of any odd number can be expressed as the sum of two consecutive positive integers. For example, 9² = 81 = 40 + 41
❖ We can express the product of two consecutive even or odd natural numbers as follows.
(a + 1) × (a – 1) = a² – 1
 
❖ We can find the squares of numbers having more than one digit by making use of the identity,(a + b)² = a × (a + b) + b × (a + b)
For example (45)² = (40 + 5)²= (40) × (40 + 5) + 5 × (40 + 5)= 40² + 40 × 5 + 5 × 40 + 5²= 1600 + 200 + 200 + 25= 2025
❖ A square number with units digit 5, say (a5), can be written as follows:
(a5)² = a(a + 1) hundred  + 25
For example, 625² = 390625 = (62 × 63) hundred + 25= 390600 + 25
❖ 2m, m² – 1, and m²+ 1 forms a Pythagorean Triplet, where m is a natural number greater than 1.
❖ Square root is the inverse operation of squaring. The positive square root of a number is denoted by the symbol $\sqrt{}$. Some methods to find the square root of perfect squares are: repeated subtraction method, prime factorisation method and division method.
 
Example: Find the square root of 36 in repeated subtraction method.
Solution:
36 – 1 = 35, 35 – 3 = 32, 32 – 5 = 27, 27 – 7 = 20, 20 – 9 = 11, 11 – 11= 0
Therefore, from 36, we subtracted successive odd numbers and obtained 0 in the 6^th step. Therefore, $\sqrt{36}=6$
 
Example: Find the smallest number by which 252 can be multiplied to make it a perfect square.
Solution:
We have, 252 = 2 × 2 × 3 × 3 × 7
The number 7 does not occur in pair. Therefore, if we multiply 252 by 7, then it will become a perfect square.
Therefore, 252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
That is, 1764 is a perfect square and $\sqrt{1764}=42$
 
Example: Find the square root of 1369 by division method.
Solution:
Step1: Firstly, place bars over every pair of digits starting from the digit at ones place. We obtain $\overline{13} \overline{69}.$
 
Step2: Find the largest number whose square is less than or equal to the number under the extreme left bar.
Take this number as the divisor and the number under the extreme left bar as the dividend. Divide and obtain the remainder.


Step3: Bring down the number under the next bar to the right of the remainder.
Therefore, the new dividend is 469.
Double the divisor and enter it with the blank on its right.


Step 4: Guess the largest possible digit to fill the blank, which becomes the new digit in the quotient, such that when the new digit is multiplied to the new quotient, the product is less than or equal to the dividend.
In this case, 67 × 7 = 469
Therefore, the quotient is 7.
Also, the remainder becomes 0 and no bar is left.


Therefore, $\sqrt{1369}=37$
 
❖ Square root of a non-perfect square can be estimated to a nearest whole number. Let m be non-perfect square and let n²<m<(n + 1)².
• If m–n²< (n + 1)²–m, then $\sqrt{m}$ is approximately equal to n.
• If m–n²> (n + 1)²–m, then $\sqrt{m}$ is approximately equal to n + 1.
 
Example: Estimate the value of $\sqrt{180}$ to the nearest whole number.
Solution:
It is known that, 13² = 169 and 14² = 196

∴ 169 < 180 < 196
$\Rightarrow {13}^2<\sqrt{180}<{14}^2$
Now, 180  – 13² = 11 and 14² – 180 = 16.
Since, 11 < 16 so 69 is closer to 180 than 196.
Therefore, $\sqrt{180}$ is approximately equal to 13.Chapter 7: Cubes and Cube Roots
 
❖ Numbers obtained when any number is multiplied by itself three times are known as cube numbers. For example, 343 is a cube number, since 343 = 7 × 7 × 7
 
❖ Properties of cubes of numbers
• Cubes of even numbers are even and the cubes of odd numbers are odd.
• If a number has 0 or 1 or 4 or 5 or 6 or 9 in its ones place, then its cube will have the same digit in ones place.
• If a number has 2 in its ones place, then its cube will have 8 in ones place and vice versa.
• If a number has 3 in its ones place, then its cube will have 7 in ones place and vice versa.
Pattern of addition of consecutive odd numbers
1³ = 1 = 1
2³ = 8 = 3 + 5
3³ = 27 = 7 + 9 + 11
4³ = 64 = 13 + 15 + 17 + 19 so on
❖ Each prime factor appears three times in the prime factorization of its cube.
❖ Cube root is the inverse operation of finding a cube. The symbol $\sqrt[3]{3}$ denotes cube-root. If a³ = b, then $\sqrt[3]{b} =a$.
❖ Cube root of a perfect cube can be found by prime factorising it.
 
Example: Find the cube root of 1728
Solution:
$$\begin{gathered} 1728=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{3\times 3\times 3} \\ \therefore \sqrt[3]{1728}=2\times 2\times 3=12 \end{gathered}$$

❖ Cube root of a perfect cube can be found through estimation.
 
Example: Find the cube root of 39304 through estimation.
Solution: The steps involved in finding the cube root of a given number are as follows:
Step I: Make groups of three digits starting from the right-most digit of the number.
$\overline{39} \overline{304}$
Step II: The number 304 ends with 4. We know that 4 comes at unit’s place of a number only when it is cube root ends in 4. So, we get 4 at the unit’s place of the cube root.
Step III: Now, consider the other group, that is, 39.
We know that 3³ = 27 and 4³ = 64.
Also 27 < 39 < 64
Therefore, we take the smallest number amongst 3 and 4, that is, 3 as the ten’s digit of the required cube root.
Thus, we have $\sqrt[3]{39304}=34$Chapter 8: Comparing Quantities
 
❖ Ratio means comparing two quantities.
❖ Percentages are numerators of fractions with denominator 100.
❖ Increase and decrease percent
• $\mathrm{Increase} \mathrm{in} \mathrm{p}\mathrm{ercent}=\frac{\mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{value}}{\mathrm{Original} \mathrm{value}}\times 100$
• $\mathrm{De}\mathrm{crease} \mathrm{in} \mathrm{p}\mathrm{ercent}=\frac{\mathrm{De}\mathrm{crease} \mathrm{in} \mathrm{the} \mathrm{value}}{\mathrm{Original} \mathrm{value}}\times 100$
❖ Discount is the reduction given on the Marked Price (M.P) of an article.
• Discount = Marked Price – Sale price
• Discount = Discount % of Marked Price
• $\mathrm{Discount} \mathrm{p}\mathrm{ercent}=\frac{\mathrm{Discount}}{\mathrm{M}.\mathrm{P}.}\times 100$

❖Additional expenses made after buying an article are included in the cost price and are known as overhead expenses. These may include expenses such as amount spent on repairs, labour charges, transportation, etc. These expenses are added to the cost price (C.P.) to obtain the real C.P. of an article.
∴ C.P. = Buying price + Overhead expenses
$• \mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}.}\times 100$
$• \mathrm{Loss}\%=\frac{\mathrm{Loss}}{\mathrm{C}.\mathrm{P}.}\times 100$
❖ Sales tax is charged on the sale of an item by the government and is added to the bill amount.
Sales Tax (or VAT) = Tax % of bill amount
❖ Interest is the extra money paid by institutions such as banks or post offices on money deposited with them. It is also paid by people when they borrow money from these institutions.
$• \mathrm{Simple} \mathrm{Interest}=\frac{\mathrm{Principal}\times \mathrm{Rate}\times \mathrm{Time}}{100}$
• Amount = Principal + Interest
❖ The interest calculated on the amount of the previous year is known as compound interest.
• Amount (A) when interest is compounded annually is $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^n,$ Where, P = Principal, R = Rate of interest, n = Time period.
• Amount when interest is compounded half yearly is given by, $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{2n}$ Where, $\frac{\mathrm{R}}{2}$ = Half-yearly rate and 2n = Number of half years
Example: A sum of Rs 12,000 was kept in a bank for 2 years and 6 months at 5% per annum. Find the compound interest to be paid by the bank when the interest is compounded annually.
 
Solution:
We have, P = 12000, R = 5 %
$2 \mathrm{years} \mathrm{and} 6 months=2\frac{6}{12} \mathrm{years} 2\frac{1}{2} \mathrm{years}$
For 2 years, the amount is given by,
$$\begin{gathered} \mathrm{A}=\mathrm{Rs} 12000{\left(1+\frac{5}{100}\right)}^2 \\ =\mathrm{Rs} 12000\left(\frac{21}{20}\right)\left(\frac{21}{20}\right) \\ =\mathrm{Rs} 13230 \\ \therefore \mathrm{C}.\mathrm{I}.=\mathrm{Rs}\left(13230-12000\right)=\mathrm{Rs}1230 \end{gathered}$$

For the next $\frac{1}{2}$ year, the interest can be calculated as simple interest.
$\mathrm{S}.\mathrm{I}.=\mathrm{Rs}\frac{13230\times {\displaystyle \frac{1}{2}}\times 5}{100}=\mathrm{Rs}\frac{13230\times 1\times 5}{200}=\mathrm{Rs} 330.75$
∴ Total compound interest = Rs (1230 + 330.75) = Rs 1560.75Chapter 9: Algebraic Expressions and Identities
 
❖ Expressions are formed from variables and constants.
• Terms are added to form expressions.
• Terms themselves can be formed as the product of factors.
❖ An expression that contains only one term is called a monomial. An expression that contains two terms is called a binomial. An expression that contains three terms is called a trinomial. An expression containing one or more terms with non-zero coefficient is called a polynomial.
❖ Like terms are formed by same variables and the powers of these variables are same too.
❖ Addition and subtraction of algebraic expressions
• While performing addition, we write each expression to be added in a separate row. While doing so, we write like terms one below the other and add them.
• Subtraction of a number is same as addition of its additive inverse.
For example, 2x² + 3yz – 4xy is subtracted from 3x² + 8xy + 3y² as:


❖ Multiplication of algebraic expressions
• The multiplication of a monomial by a monomial gives a monomial. While performing multiplication, the coefficients of the two monomials are multiplied and the powers of different variables in the two monomials are multiplied by using the rules of exponents and powers.
For example the expression (–2ab²c) and (3abc²) are multiplied as:
(–2ab²c) × (3abc²) = –6 × (a × a × b² × b × c × c²) = –6a²b³c³
• While multiplying a monomial by a binomial or trinomial or polynomial, we multiply every term in the polynomial by the monomial by making use of distributive law.
For example, 5a × (2b + c) = (5a × 2b) + (5a × c) = 10ab + 5ac
• While multiplying a polynomial by a binomial (or trinomial), we multiply it term by term. That is, every term of the polynomial is multiplied by every term of the binomial (or trinomial).
For example, (5a – c) × (2b + c) = 5a × (2b + c) – c × (2b + c) = (5a × 2b) + (5a × c) – (c × 2b) – (c × c) = 10ab + 5ac – 2bc – c²
 
❖ An identity is an equality which is true for all values of the variables in it. However, an equation is true for only certain values of the variables in it. It is not true for all the values of the variable and hence, it is not an identity.
 
❖ Some standard identities
(i) (a + b)² = a² + 2ab + b²
(ii) (a – b)² = a² – 2ab + b²
(iii) (a + b) (a – b) = a² – b²
(iv) (x + a) (x + b) = x² + (a + b) x + ab
 
Example: Simplify (5x + 2y)² – (3x – y)².
 
Solution:
Using identities (i) and (ii), we obtain:
(5x + 2y)² = (5x)² + 2 (5x) (2y) + (2y)² = 25x² + 20xy + 4y²
(3x – y)² = (3x)² – 2 (3x) (y) + (y)² = 9x² – 6xy + y²
∴ (5x + 2y)² – (3x – y)² = 25x² + 20xy + 4y² – 9x² + 6xy – y² = 16x² + 26xy + 3y²Chapter 10: Visualising Solid Shapes
 
❖ Three-dimensional objects look differently from different positions.
For example, the following figure (made by joining some cubes) is viewed are given as:


❖ A map depicts the location of a particular object/place in relation to other object(s)/place(s).
• Symbols in a map are used to depict the different objects/places.
• There is no reference or perspectives in a map, i.e., objects that are closer to the observer are shown to be of the same size as those that are further away.
• Maps use scale, which is fixed for a particular map. It reduces the real distances proportionately to the distances on the paper.
 
For example, the following map shown is the map of a locality.
 

 
By looking at this map, some questions can be answered. For example, if the question is which is further west: a market or a temple, then the answer is market.
❖ Three-dimensional objects look differently from different positions.
For example, the following figure (made by joining some cubes) is viewed are given as:
❖ Polyhedron is a solid made up of polygonal regions (called its faces). These faces meet at edges (which are line segments) and the edges meet at the vertices (which are points).
Among the solids shown below, the first two solids are polyhedron.


❖ A polyhedron is a convex polyhedron if the line joining any of its two vertices lies within or on the polyhedron. Otherwise it is called a concave polyhedron.
Among the solids shown below, the first two polyhedrons are convex polyhedrons and the last one is a concave polyhedron.



❖ Regular polyhedrons are polyhedrons whose faces are made up of regular polygons and the same numbers of faces meet at each vertex.
Among the solids shown below, the first polyhedron is a regular polyhedron where as the other one is a concave polyhedron.


       
 ❖ A prism is a polyhedron whose base and top are congruent polygons and whose lateral faces are parallelograms in shape. A pyramid is a polyhedron whose base is a polygon (of any number of sides) and whose lateral faces are triangles with a common vertex.
Among the following solids, the first one is a prism and the second one is a pyramid.
❖ For any polyhedron, F + V – E = 2, where F is the number of faces, V is the number of vertices and E is the number of edges. This relationship is called Euler’s formula.Chapter 11: Mensuration
 
❖ Area and perimeter of various plane figures
 

❖ Shape	❖ Area	❖ Perimeter
Rectangle with adjacent sides a and b	a × b	2× (a + b)
Square with side a	a2	4a
Circle with radius r	πr2	2πr
Triangle with base b and its corresponding height h	$\frac{1}{2}\times b\times h$	Sum of the three sides
Parallelogram with base b and its corresponding height h	b × h	Sum of the four sides
Trapezium with parallel sides a and b and height h.	$\frac{1}{2}\times \left(a+b\right)\times h$	Sum of the four sides
Rhombus with base b and its corresponding height h	b × h	Sum of the four sides
Rhombus with diagonals d1 and d2	$\frac{1}{2}\times d_1\times d_2$	Sum of the four sides
Quadrilateral with diagonal d and the altitudes h1 and h2 to this diagonal from the opposite vertices.	$\frac{1}{2}\times d\times \left(h_1+h_2\right)$	Sum of the four sides
Polygon	Break the polygon into any type of triangle (s) or any type of quadrilateral (s). Add the areas of these triangle (s) and quadrilateral (s).	Sum of all the sides of the polygon

 
Example: Find the area of the given polygon, where ABCD is a trapezium.
 
 
Solution:
Area of ABCED = Area of trapezium ABCD + Area of ∆CDE
$=\left[\frac{1}{2}\left(3+5\right)\times 3+\frac{1}{2}\times 5\times 4\right]{\mathrm{cm}}^2$
= (12 + 10) cm²
= 22 cm²
 
❖ Lateral surface area of a solid is the sum of the areas of its lateral faces.
❖ Total surface area of a solid is the sum of the areas of its all faces.
❖ Volume of a solid is the amount of region occupied by the solid.
❖ Surface area and volumes of various solids
 

❖ Solid	❖ Lateral Surface Area	❖ Total Surface Area	❖ Volume
Cuboid with length l, breadth b and height h	2h × (l + b)	2(lb + bh + hl)	l × b × h
Cube with edge l	4l2	6l2	l3
Right circular cylinder with radius of circular base r and height h	2πrh	2πr (r + h)	πr2h

 
❖ Capacity is the quantity that a container can hold.
❖ Relationship between common units
1 mL = 1 cm³
1 L = 1000 cm³
1 m³ = 1000000 cm³ = 1000 LChapter 12: Exponents and Powers
 
❖ Very large numbers and very small numbers are difficult to read, understand, and compare. To make this easier, we use exponents by converting many of the large numbers and small numbers into a shorter form.
For example, 10,000,000,000,000 can be written as (10)¹³.
Here, 10 is called the base and 13 is called the exponent.
❖ For any non-zero integer a, $a^{-m}=\frac{1}{a^m}$, where m is a positive integer, a^–m is called the multiplicative inverse of a^m and vice-versa.
❖ Decimal numbers can be written in expanded form using exponents.
For example, 32845.912
$$\begin{gathered} =3\times {10}^4+2\times {10}^3+8\times {10}^2+4\times {10}^1+5\times 1+9\times \frac{1}{10}+1\times \frac{1}{{10}^2}+2\times \frac{1}{{10}^3} \\ =3\times {10}^4+2\times {10}^3+8\times {10}^2+4\times {10}^1+5\times 1+9\times {10}^{-1}+1\times {10}^{-2}+2\times {10}^{-3} \end{gathered}$$
  
 
❖ Laws of exponents
(i) a^m × aⁿ = a^{m + n}
(ii) $\frac{a^m}{a^n}=a^{m-n} \left(a\ne 0\right)$
(iii) (a^m)ⁿ = a^mn
(iv) a^m × b^m = (ab)^m
(v) $\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m$
(vi) a⁰ = 1  (a ≠ 0)
 
❖ aⁿ = 1 implies that n = 0 for any a ≠ 0, except at a = 1 or a = –1
• (1)ⁿ = 1 for any n
• (–1)^p = 1 for any even integer p
❖ Very smaller numbers can be expressed in a simpler way using exponents.
For example the number 0.00000003812 is written in standard form as:
$0.00000003812=\frac{3812}{100000000000}=\frac{3812}{{10}^{11}}=\frac{3.812\times {10}^3}{{10}^{11}}=\frac{3.812}{{10}^8}=3.812\times {10}^{-8}$
 
Here, the decimal in  is moved 8 places to the right.
❖ When we have to compare or to add or subtract numbers in standard form, we convert them into numbers with the same exponents.
 
Example: The thickness of a cloth is 0.251 cm and the thickness of a piece of a paper is 0.0502 cm. Compare the thickness of the cloth to the thickness of the piece of paper.
Solution:
$$\begin{gathered} \mathrm{Thickness} \mathrm{of} \mathrm{cloth} =\; 0.251 \mathrm{cm}=\frac{251}{1000}\mathrm{cm}=251\times {10}^{-3}\mathrm{cm} \\ \mathrm{Thickness} \mathrm{of} \mathrm{piece} \mathrm{of} \mathrm{paper}=0.0502 \mathrm{cm}\frac{502}{10000} \mathrm{cm}=\frac{502}{10}\times {10}^{-3}\mathrm{cm}=50.2\times {10}^{-3}\mathrm{cm} \\ \therefore \frac{251\times {10}^{-3}}{50.2\times {10}^{-3}}=\frac{251}{50.2}=\frac{251}{502}\times 10=\frac{10}{2}=5 \end{gathered}$$

Thus, the thickness of the cloth is 5 times the thickness of the paper.Chapter 13: Direct and Inverse Proportions
 
❖ Two quantities, x and y, are said to be in direct proportion, if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant.
That is, $\frac{x}{y}=k,$ where k is a positive number.
For example, let r be the radius of a circle.
Now, $\frac{\mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle}}{\mathrm{Radius} \mathrm{of} \mathrm{that} \mathrm{circle}}=\frac{2\pi r}{r}=2\pi ,$ which is a constant
Thus, the circumference of the circle and its radius are in direct proportion.
• If y₁, y₂ are the values of y corresponding to the values x₁, x₂ of x respectively, then $\frac{x_1}{y_1}=\frac{x_2}{y_2}$ is a case of direct proportion.
❖ Two quantities, x and y, are said to be in inverse proportion, if an increase in x causes a proportional decrease in y (and vice-versa) in such a manner that the product of their corresponding values remains constant. That is, xy = k, where k is a positive number.
 
For example, consider the number of employers and the time taken to finish the work in an organisation. More the number of employers less will be the time taken to finish the work. Therefore, this is a case of inverse proportion.
 
• If y₁, y₂ are the values of y corresponding to the values x₁, x₂ of x respectively, then x₁y₁ = x₂y₂ is a case of indirect proportion.
 
Example: If the cost of 5 movie tickets is Rs 875, then find the cost of 2 tickets.
Solution:
Suppose the cost of 2 tickets and the cost of 7 tickets be Rs x and Rs y respectively. We can form a table as shown below.
 

Number of tickets	2	5
Cost of tickets (Rs)	x	875

 
As the number of tickets increases, the cost of tickets will also increase in the same ratio. It is a case of direct proportion.
$$\begin{gathered} \frac{2}{x}=\frac{5}{875} \\ \Rightarrow x=\frac{2\times 875}{5}=2\times 175=350 \end{gathered}$$
Thus, the cost of 2 tickets is Rs 350.Chapter 14: Factorisation
 
❖ Factors of an algebraic term can be numbers or algebraic variables or algebraic expressions.
For example, the factors of 2a²b are 2, a, a, b, since 2a²b = 2 × a × a × b
The factors, 2, a, a, b, are said to be irreducible factors of 2a²b since they cannot be expressed further as a product of factors.
❖ Factorisation of expressions by the method of common factors
This method involves the following steps.
Step 1: Write each term of the expression as a product of irreducible factors.
Step 2: Observe the factors, which are common to the terms and separate them.
Step 3: Combine the remaining factors of each term by making use of distributive law.
 
Example: Factorize $12p^{2}q+8p{q}^2+18pq.$
Solution:
We have, 12p²q = 2 × 2 × 3 × p × p × q
8pq² = 2 × 2 × 2 × p × q × q
18pq = 2 × 3 × 3 × p × q
The common factors are 2, p, and q.
∴12p²q + 8pq² + 18pq = 2 × p × q [(2 × 3 × p) + (2 × 2 × q) + (3 × 3)] = 2pq (6p + 4q + 9)
❖ Factorisation by regrouping terms
Sometimes, all terms in a given expression do not have a common factor. However, the terms can be grouped by trial and error method in such a way that all the terms in each group have a common factor. Then, there happens to occur a common factor amongst each group, which leads to the required factorization.
 
Example: Factorise 2a² – b + 2a – ab.
Solution:
2a² – b + 2a – ab
= 2a² + 2a – b – ab   (The terms, 2a² and 2a, have common factors, 2 and a.)
= 2a (a + 1) – b (1 + a) (The terms, –b and –ab, have common factors, –1 and b.)
= (a + 1) (2a – b)
❖ Factorisation using identities: Some of the expressions can also be factorized by making use of the following identities.
• a² + 2ab + b² = (a + b)²
• a² – 2ab + b² = (a – b)²
• a² – b² = (a + b) (a – b)
• x² + (a +b)x + ab = (x + a) (x + b)
❖ Division of any polynomial by a monomial is carried out either by dividing each term of the polynomial by the monomial or by the common factor method.
$\mathrm{For} example, \left(8x^3+4x^{2}y+6x{y}^2\right)÷2x=\frac{2\times x\left(4x^2+2xy+3y^2\right)}{2 \times x}=4x^2+2xy+3y^2$
❖ Division of a polynomial by a polynomial is carried out by factorizing both the polynomials and then cancelling out the common factors.
 
Example: Divide (35x² – 70x – 105) by (7x + 7).
Solution:
We have, 35x² – 70x – 105 = 35 (x² – 2x – 3) = 35(x² + x – 3x – 3) = 35[x(x + 1) – 3(x + 1)]= 35(x + 1) (x – 3) and 7x + 7 = 7(x + 1)
$\therefore \frac{35x^2-70x-105}{7x+7}=\frac{35\times \left(x+1\right)\times \left(x-3\right)}{7\times \left(x+1\right)}=\frac{5\times 7\times \left(x+1\right)\times \left(x-3\right)}{7\times \left(x+1\right)}=5\left(x-3\right)$
❖ Points to be remembered while carrying out calculations in algebra
• The coefficient, 1, of a term is usually not written. However, while adding or subtracting like terms, we include them also.
• While substituting a negative value, we make use of brackets.
• While multiplying an expression enclosed in a bracket by a constant (or a variable) outside, each term of the expression has to be multiplied by the constant (or the variable).
• While squaring a monomial, the numerical coefficient and each factor have to be squared.
• Before applying an identity, it has to be made sure whether it can be really applied or not.
For example: (x – 5)² ≠ x² – 5². However, (x – 5)² = x² – 10x + 25Chapter 15: Introduction to Graphs
 
❖ A given data can be represented graphically to make it easier to read and understand.

❖ A data that changes continuously over periods of time can be displayed by a line graph.
 
For example, consider that the given table shows the profit earned by a shopkeeper in the last five years of his business.
 

Year	2008	2007	2006	2005	2004
Profit (in Rs)	50,000	60,000	40,000	80,000	90,000

 
On representing the years on the horizontal axis and the profit earned on the vertical axis,
The given information can be represented by a line graph as
 

From the above line graph, a lot of information can be interpreted. For example, it can be observed from the line graph that the profit earned by the shopkeeper was maximum in the year 2004 and minimum in the year 2006.

❖ A line graph with a whole unbroken line is called a linear graph.

❖ For locating a point on a graph sheet, the x and y-coordinates of the point are required.
• The horizontal axis is called the x-axis and the vertical axis is called the y-axis.
• The x-coordinate of a point refers to the number of units to be moved to the right from the vertical axis and the y-coordinate of a point refers to the number of units to be moved up from the horizontal axis.
 
For example, the point (2, 4) can be located by moving 2 units from the left edge and then 4 units from the bottom edge in the graph. This can be shown as
 

 
The point O with coordinates (0, 0) is called the origin.

❖ There are many applications of linear graphs. The relation between two variables, one of which is an independent variable (called control variable) and the other is a dependent variable, can be represented through a linear graph.
 
For example, the relationship between quantity and cost, time and distance can be represented by a linear graph. In this case, quantity and time are independent variable and the cost and the distance are dependent variable as they depend on the quantity and time respectively.

❖ Relationship between principal and simple interest can be represented by a linear graph. In this case, principal is an independent variable and simple interest is a dependent variable.
 
Example: The amount deposited in a bank and the corresponding simple interest for a year given by the bank are given as follows.
 

Principal (in Rs)	1500	2000	3000	4000
Simple interest (in Rs)	75	100	150	200

 
Draw a graph for the given data. From the graph, find the simple interest, if the principal is Rs 1000 and also find the principal whose simple interest is Rs 175.
 
Solution:
We represent principal on the horizontal axis and simple interest on the vertical axis.
Scale: 1 unit = Rs 1000 on the horizontal axis and 1 unit = Rs 50 on the vertical axis
Plot the points, (1500, 75), (2000, 100), (4000, 200), on a graph paper. Join the four points.
Note that the graph will pass through the origin since the simple interest on the principal of Rs 0 is Rs 0.
 

 
From the graph, it can be observed that corresponding to Rs 1000 on the horizontal axis, we obtain the interest as Rs 50 on the vertical axis.
  Therefore, for the principal of Rs 1000, the simple interest is Rs 50.
Also, corresponding to Rs 175 on the vertical axis, we obtain Rs 3500 on the horizontal axis.
  Therefore, the simple interest of Rs 175 can be earned if the principal is Rs 3500.Chapter 16: Playing with Numbers
 
❖ Any two-digit number, ab, made of digits, a and b, can be written in general form as
ab = 10a + b whereas as a three-digit number, abc, made of digits, a, b, and c, can be written as abc = 100a + 10b + c
❖ Some properties of two and three digit numbers
• The sum of a two-digit number and the number formed by reversing the digits is always a multiple of 11.
For example, 28 + 82 = 110 = 11 × 10 is a multiple of 11
• The difference between a two-digit number and the number formed by reversing the digits is always a multiple of 9.
For example, 91 – 19 = 72 = 9 × 8 is a multiple of 9
• The difference between a three-digit number and the number formed by reversing the digits is always a multiple of 99.
For example, 923 – 329 = 594 = 99 × 6 is a multiple of 99
• If abc is a three-digit number, then (abc+ cab + bca) will be a multiple of 37.
For example, 158 + 581 + 815 = 1554 = 37 × 42 is a multiple of 37
❖ We can solve puzzles related to addition and multiplication in which some of the digits are denoted by letters and we have to find the number by which the letters are replaced.
Rules while solving puzzles are as follows.
• Each letter in the puzzle stands for just one digit.
• The first digit of a number cannot be zero.
Example: Find the digits, P and Q, in the given puzzle.
$$\begin{gathered} \begin{array}{c}\mathrm{P} 2\\ \times \mathrm{Q}\end{array} \\ \underline{\overline{ 3 6 0 }} \end{gathered}$$
Solution:
The ones digit of the number should be obtained as 0 on multiplying 2 by Q.
Therefore, the possibility for Q is either 0 or 5.
However, Q cannot be 0 because then, the result obtained after multiplication will be 0. Therefore, Q = 5
In the tens column, there will be a carry of 1, which was obtained as tens digit on multiplication of 2 by 5 in the ones column.
Therefore, we should obtain, 1 + P × 5 = 36
⇒ P × 5 = 35
⇒ P = 7
 Therefore, the only possibility for P is 7.
Thus, the puzzle can now be solved as below.
$$\begin{gathered} \begin{array}{c}7 2\\ \times 5 \end{array} \\ \underline{\overline{ 3 6 0 }} \end{gathered}$$
❖ Divisibility rules
• A number is divisible by 10, if its ones digit is 0.
• A number is divisible by 5, if its ones digit is either 0 or 5.
• A number is divisible by 2, if its ones digit is 0, 2, 4, 6, or 8 i.e., if the ones digit is even.
• A number is divisible by 9, if the sum of its digits is divisible by 9.
• A number is divisible by 3, if the sum of its digits is divisible by 3.
 
Example: If the four-digit number, 835a, is divisible by 9, then what is the value of a?
Solution:
It is known that if a number is divisible by 9, then the sum of its digits is divisible by 9. It is given that the number, 835a, is divisible by 9.
Therefore, 8 + 3 + 5 + a = 16 + a is divisible by 9.
This is possible when 16 + a = 9 or 18 or 27….
However, since a is a digit,
16 + a = 18
∴ a = 2