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 am 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.
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.
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.
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.
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.
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
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:
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.
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.
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.
The number which can be expressed as the sum of two cubes in different ways is said to be a Hardy – Ramanujan number.
1729 = 1728 + 1 = 123 + 13
1729 = 1000 + 729 = 103 + 93
As 1729 is the smallest such type of number so it is called the smallest Hardy-Ramanujan number. There is infinite such type of numbers. Like- 4104 (2, 16; 9, 15), 13832 (18, 20; 2, 24), etc.
Cubes
Cube is a 3-dimensional figure with all equal sides. If one cube has all the equal sides of 1 cm then how many such cubes are needed to make a new cube of side 2 cm?
8 such cubes are needed, and what if we need to make a cube of side 3 cm with the cubes of side 1 cm? The numbers 1, 8, 27 …etc can be shown below in the cube.
These are known as perfect cubes or cube numbers. This shows that we got the cube numbers by multiplying the number three times by itself.
Cubes of Some Natural Numbers
Number
Cubes
Numbers
Cubes
1
13 = 1
11
113 = 1331
2
23 = 8
12
123 = 1728
3
33 = 27
13
133 = 2197
4
43 = 64
14
143 = 2744
5
53 = 125
15
153 = 3375
6
63 = 216
16
163 = 4096
7
73 = 343
17
173 = 4913
8
83 = 512
18
183 = 5832
9
93 = 729
19
193 = 6859
10
103 = 1000
20
203 = 8000
This table shows that
There are only 10 perfect cubes between 1-1000.
The cube of an even number is also even.
The cube of an odd number is also an odd number.
One’s digit of the Cubes
One’s digit of the Cubes of a number having a particular number at the end will always remain same. Let’s see in the following table:
Unit’s digit of number
Last digit of its cube number
Example
1
1
113 = 1331, 213 = 9261, etc.
2
8
23 = 8, 123 = 1728, 323 = 32768, etc.
3
7
133 = 2197, 533 = 148877, etc.
4
4
243 = 13824, 743 = 405224, etc.
5
5
153 = 3375, 253 = 15625, etc.
6
6
63 = 216, 263 = 17576,etc.
7
3
173 = 4913, 373 = 50653,etc.
8
2
83 = 512, 183 = 5832, etc.
9
9
193 = 6859, 393 = 59319, etc.
10
20
103 = 1000, 203 = 8000, etc.
Some Interesting Patterns
1. Adding Consecutive Odd Numbers
This shows that if we add the consecutive odd numbers then we get the cube of the next number.
2. Cubes and their Prime Factors
Prime factorization of a number is done by finding the prime factors of the number and then pairing it in the group of three. If all the prime factors are in the pair of three then the number is a perfect cube.
Example
Calculate the cube root of 13824 by using prime factorization method.
Solution
First of all write the prime factors of the given number then pair them in the group of three.
Since all the factors are in the pair of three the number 13824 is a perfect cube.
Smallest Multiple that is a Perfect Cube
As we have seen that the group of three prime factors makes a number perfect cube, so to make a number perfect cube we need to multiply it with the smallest multiple of that number.
Example
Check whether 1188 is a perfect cube or not. If not then which smallest natural number should be multiplied to 1188 to make it a perfect cube?
Solution
1188 = 2 × 2 × 3 × 3 × 3 × 11
This shows that the prime numbers 2 and 11 are not in the groups of three. So, 1188 is not a perfect cube
To make it a perfect cube we need to multiply it with 2 × 11 × 11 = 242, so, it will make the pair of 2, 3 and 11.
Hence the smallest natural number by which 1188 should be multiplied to make it a perfect cube is 242.
And the resulting perfect cube is 1188 × 242 = 287496 ( = 663).
Cube Roots
Finding cube root is the inverse operation of finding the cube.
If 33 =27 then cube root of 27 is 3.
We write it as ∛27 = 3
Symbol of the Cube Root
Some of the cube roots are:
Statement
Inference
Statement
Inference
13 = 1
∛1 = 1
63 = 216
∛216 = ∛63 = 6
23 = 8
∛8 = ∛23 = 2
73 = 343
∛343 = ∛73 = 7
33 = 27
∛27 = ∛33 = 3
83 = 512
∛512 = ∛83 = 8
43 = 64
∛64 = ∛43 = 4
93 = 729
∛729 = ∛93 = 9
53 = 125
∛125 = ∛53 = 5
103 = 1000
∛1000 = ∛103 = 10
Method of finding a Cube Root
There are two methods of finding a cube root
1. Prime Factorization Method
Step 1: Write the prime factors of the given number.
Step 2: Make the pair of three if possible.
Step 3: Then replace them with a single digit.
Step 4: Multiply these single digits to find the cube root.
Example
Find the cube root of 15625 by the prime factorization method.
2. Estimation Method
This method is based on the estimation. Let’s take the above example.
Step 1: If 15625 is the number then make the group of three digits starting from the right.
15 625
Step2: Here 625 is the first group which tells us the unit’s digit of the cube root. As the number is ending with 5 and we know that 5 comes at the unit’s place of a number only when its cube root ends in 5.
So the unit place is 5.
Step 3: Now take the other group, i.e., 15. Cube of 2 is 8 and a cube of 3 is 27. 15 lie between 8 and 27. The number which is smaller among 2 and 3 is 2. The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of 15625. Thus,
Any natural number ‘p’ which can be represented as y2, where y is a natural number, then ‘p’ is called a Square Number.
Example
4 = 22
9 = 32
16 = 42
Where 2, 3, 4 are the natural numbers and 4, 9, 16 are the respective square numbers.
Such types of numbers are also known as Perfect Squares.
Some of the Square Numbers
Properties of Square Numbers
We can see that the square numbers are ending with 0, 1, 4, 5, 6 or 9 only. None of the square number is ending with 2, 3, 7 or 8.
Any number having 1 or 9 in its one’s place will always have a square ending with 1.
Number
Square Number
1
1
9
81
11
121
19
361
21
441
Any number which has 4 or 6 in its unit’s place, its square will always end with 6.
Number
Square Number
4
16
16
256
24
576
36
1296
44
1936
Any number which has 0 in its unit’s place, its square will always have an even number of zeros at the end.
Number
Square number
10
100
50
2500
100
10000
150
22500
400
160000
Some More Interesting Patterns
1. Adding Triangular Numbers
If we could arrange the dotted pattern of the numbers in a triangular form then these numbers are called Triangular Numbers. If we add two consecutive triangular numbers then we can get the square number.
2. Numbers between Square Numbers
If we take two consecutive numbers n and n + 1, then there will be (2n) non-perfect square numbers between their squares numbers.
Example
Let’s take n = 5 and 52 = 25
n + 1 = 5 + 1 = 6 and 62 = 36
2n = 2(5) = 10
There must be 10 numbers between 25 and 36.
The numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.
3. Adding Odd Numbers
Sum of first n natural odd numbers is n2.
Any square number must be the sum of consecutive odd numbers starting from 1.
And if any natural number which is not a sum of successive odd natural numbers starting with 1, then it will not be a perfect square.
4. A Sum of Consecutive Natural Numbers
Every square number is the summation of two consecutive positive natural numbers.
If we are finding the square of n the to find the two consecutive natural numbers we can use the formula
Example
52 = 25
12 + 13 = 25
Likewise, you can check for other numbers like
112 = 121 = 60 + 61
5. The Product of Two Consecutive Even or Odd Natural Numbers
If we have two consecutive odd or even numbers (a + 1) and (a -1) then their product will be (a2– 1)
Example
Let take two consecutive odd numbers 21 and 23.
21 × 23 = (20 – 1) × (20 + 1) = 202 – 1
6. Some More Interesting Patterns about Square Numbers
Finding the Square of a Number
To find the square of any number we needed to divide the number into two parts then we can solve it easily.
If number is ‘x’ then x = (p + q) and x2 = (p + q)2
You can also use the formula (p + q)2 = p2 + 2pq + q2
Example
Find the square of 53.
Solution:
Divide the number in two parts.
53 = 50 + 3
532 = (50 + 3)2
= (50 + 3) (50 + 3)
= 50(50 + 3) +3(50 + 3)
= 2500 + 150 + 150 + 9
= 2809
1. Other pattern for the number ending with 5
For numbers ending with 5 we can use the pattern
(a5)2 = a × (a + 1)100 + 25
Example
252 = 625 = (2 × 3) 100 + 25
452 = 2025 = (4 × 5) 100 + 25
952 = 9025 = (9 × 10) 100 + 25
1252 = 15625 = (12 × 13) 100 + 25
2. Pythagorean Triplets
If the sum of two square numbers is also a square number, then these three numbers form a Pythagorean triplet.
For any natural number p >1, we have (2p) 2 + (p2 -1)2 = (p2 + 1)2. So, 2p, p2-1 and p2+1 forms a Pythagorean triplet.
Example
Write a Pythagorean triplet having 22 as one its member.
Solution:
Let 2p = 6
P = 3
p2 + 1 = 10
p2 – 1 = 8.
Thus, the Pythagorean triplet is 6, 8 and 10.
62 + 82 = 102
36 + 64 = 100
Square Roots
The square root is the inverse operation of squaring. To find the number with the given square is called the Square Root.
22 = 4, so the square root of 4 is 2
102 = 100, therefore square root of 100 is 10
There are two square roots of any number. One is positive and other is negative.
The square root of 100 could be 10 or -10.
Symbol of Positive Square Root
Finding Square Root
1. Through Repeated Subtraction
As we know that every square number is the sum of consecutive odd natural numbers starting from 1, so we can find the square root by doing opposite because root is the inverse of the square.
We need to subtract the odd natural numbers starting from 1 from the given square number until the remainder is zero to get its square root.
The number of steps will be the square root of that square number.
Example
Calculate the square root of 64 by repeated addition.
Solution:
64 – 1 = 63
63 – 3 = 60
60 – 5 = 55
55 – 7 = 48
28 – 13 = 15
48 – 9 = 39
15 – 15 = 0
39 – 11 = 28
2. Prime Factorization
In this method, we need to list the prime factors of the given number and then make the pair of two same numbers.
Then write one number for each pair and multiply to find the square root.
Example
Calculate the square root of 784 using prime factorization method.
Solution:
List the prime factors of 784.
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 7 = 28
3. Division Method
Steps to find the square root by division method
Step 1: First we have to start making the pair of digits starting from the right and if there are odd number of digits then the single digit left over at the left will also have bar .
Step 2: Take the largest possible number whose square is less than or equal to the number which is on the first bar from the left. Write the same number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide to get the remainder.
Step 3: Like a normal division process bring the digits in next bar down and write next to the remainder.
Step 4: In next part the quotient will get double and we will right in next line with a blank on its right.
Step 5: Now we have to take a number to fill the blank so that the if we take it as quotient then the product of the new divisor and the new digit in quotient is less than or equal to the dividend.
Step 6: If there are large number of digits then you can repeat the steps 3, 4, 5 until the remainder does not become 0.
Example
Calculate the square root of √729 using division method.
Solution:
Thus, √729 = 27.
Square Roots of Decimals
To find the square root of a decimal number we have to put bars on the primary part of the number in the same manner as we did above. And for the digits on the right of the decimal we have to put bars starting from the first decimal place. Rest of the method is same as above. We just need to put the decimal in between when the decimal will come in the division.
Example
Find √7.29 using division method.
Solution:
Thus, √7.29 = 2.7
Remark: To put the bar on a number like 174.241, we will put a bar on 74 and a bar on 1 as it is a single digit left. And in the numbers after decimal, we will put a bar on 24 and put zero after 1 to make it double-digit.
174. 24 10
Estimating Square Root
Sometimes we have to estimate the square root of a number if it’s not possible to calculate the exact square root.
Example
Estimate the square root of 300.
Solution:
We know that, 300 comes between 100 and 400 i.e. 100 < 300 < 400. Now, √100 = 10 and √400 = 20.
So, we can say that
10 < √300 < 20.
We can further estimate the numbers as we know that 172 = 289 and 182 = 324. Thus, we can say that the square root of √300 = 17 as 289 is much closer to 300 than 324.
Data handling means to collect and present the data so that it could be used in further studies and to find some results.
Data
Any information collected in the form of numbers, words, measurements, symbols, or in any other form is called data.
Graphical Representation of Data
The grouped data can be represented graphically for its clear picture and it is the easiest way to understand the data.
Types of Graph
1. Pictograph
When we represent the data through pictures or symbols then it is called Pictograph.
Here one tree represents 10 trees. And we can easily read the pictograph.
The graph shows that there are 30 trees of apple and so on.
2. Bar Graphs
In the bar graph, the information represented by the bars of the same width with equal gaps but the height of the bars represent the respective values.
Here, the names of pets are represented on the horizontal line and the values of the respective pets are shown by the height of the bars. There is an equal gap between each bar.
3. Double Bar Graph
To compare some data we can use the double bar graph as it shows the information of two quantities simultaneously.
Here, in the above graph, it represents the marks of the students in two different tests altogether. So we can compare the marks easily.
Organizing Data
Any data which is available in the unorganized form is called Raw Data.
This raw data is arranged or grouped in a systematic manner to make it meaningful which is called the Presentation of Data.
Terms Related to Data Organizing
1. Frequency
Frequency tells us the no. of times a particular quantity repeats itself.
2. Frequency Distribution Table
Frequency can be represented by the frequency distribution table.
The above table shows the no. of times a particular colour repeat in the bag of skittles.
Frequency can also be shown by the tally marks. A cut over four lines represents the number 5.
1. Grouping Data
If we have a large number of quantities then we need to group the observation and then make the table. Such a table is called a Grouped Frequency Distribution Table.
Some Important terms related to grouped Frequency Distribution Table
Class Interval or Class: When all the observations are classified in several groups according to their size then these groups are called Class Interval
Lower-class Limit: The lowest number in every class interval is known as its Lower-class Limit.
Upper-class Limit: The highest number in every class interval is known as its Upper-class Limit.
Width or Size or Magnitude of the Class Interval: The difference between the upper-class limit and the lower class limit is called the Size of the Class Interval.
Example
There is a list of marks of 40 students in a school. Arrange this in grouped frequency distribution table.
Solution
As we can see that the lowest number in the above data is 27 and the highest number is 78, so we can make intervals if 20 – 30, 30 – 40 so on.
Remark: As number 30 comes in two class interval but we cannot count it in both the intervals. So it is to remember that the common observation will always be counted in the higher class. Hence 30 will come in 30-40, not in 20-30.
Histogram
Basically, the bar graph of the grouped frequency distribution or continuous class interval is called Histogram.
The class intervals are shown on the horizontal line and the frequency of the class interval is shown as the height of the bars.
There is no gap between each bar.
Example
Draw a histogram for the wages of 30 workers in a company. The wages are as follows: 830, 840, 868, 890, 806, 840, 835, 890,840, 885, 835, 835, 836, 878, 810, 835, 836, 869, 845, 855, 845, 804, 808, 860, 832, 833, 812, 898, 890, 820.
Solution
Make the grouped frequency distribution of the given data.
Class Interval
Frequency
800 – 810
3
810 – 820
2
820 – 830
1
830 – 840
9
840 – 850
5
850 – 860
1
860 – 870
3
870 – 880
1
880 – 890
1
890 – 900
4
Draw the histogram by taking the class interval on the horizontal line and the frequency on the vertical line.
Remark: As the class interval does not start from zero, so we will put a jagged line which shows that there is no number between 0 – 800.
Circle Graph or Pie Chart
If we represent the data in a circle form then it is said to be a pie chart. This graph shows the relationship between the whole and its part. We have to divide the circle into sectors and each sector is proportional to its respective activity.
We use it when we have information on percentage or fraction.
Drawing of a Pie Chart
If we have the information in percentage then we need to calculate the respective angles to show them in the pie chart.
As we know that a complete circle is of 360°, so we need to calculate the fraction of 360° for every sector.
Example
Draw a pie chart of the following percentage of genres of movies liked by the public.
Genres of Movie
Percentage of the no. of people
Comedy
27%
Action
18%
Romance
14%
Drama
14%
Horror
11%
Foreign
8%
Science fiction
8%
Solution
To draw the pie chart first we need to calculate the angle by taking the fraction of 360°.
Genres of Movie
Percentage of the no. of people
In fractions
Fraction of 360°
Comedy
27%
27/100
27/100 × 360° = 97.2°
Action
18%
18/100
18/100 × 360° = 64.8°
Romance
14%
14/100
14/100 × 360° = 50.4°
Drama
14%
14/100
14/100 × 360° = 50.4°
Horror
11%
11/100
11/100 × 360° = 39.6°
Foreign
8%
8/100
8/100 × 360° = 28.8°
Science fiction
8%
8/100
8/100 × 360° = 28.8°
By using these angles draw a pie chart.
Chance and Probability
Probability tells the degree of uncertainty. It measures the likelihood that an event will occur.
Random Experiment
If the result of the experiment is not known then it is known as a random experiment.
Example
If we throw a dice then the result could be any number from 1 – 6.
Outcomes
When we do an experiment then there could be different results, these possible results of the random experiment are called outcomes.
Example
There are two possible outcomes when we toss a coin i.e. head and tail.
Equally Likely Outcomes
If every outcome has the same possibility of occurring these outcomes are called Equally Likely Outcomes.
Example
If we throw a dice then there is an equal chance of every no. to come while doing the random experiment. i.e. a dice has the same possibility of getting 1, 2, 3, 4, 5 and 6.
Linking Chances to the Probability
Example
What is the chance of getting 3 when we throw a dice?
Solution
There is only one chance to get 3 in one throw and the total possible outcomes are 6.
Hence the probability of getting =1/6.
Outcomes as Events
Each outcome or collection of outcomes of an experiment is known as an event.
Example
If we throw a dice then getting each outcome 1, 2, 3, 4, 5 and 6 are events.
Example
What is the event of getting odd numbers when we throw a dice?
Solution
The probability of getting an odd number is 3(odd numbers are 1, 3, 5)
The total number of outcomes is 6.
The probability of getting an odd number = 3/6.
Example
What is the probability of spinning yellow?
Here number of chance to come yellow while spinning is 3.
A quadrilateral has some measurements like – 4 sides, 4 angles and 2 diagonals.
We can construct a unique quadrilateral if we know the five measurements.
1. If the four sides and a diagonal of the quadrilateral are given.
Example
Construct a quadrilateral ABCD in which AB = 5 cm, BC = 7 cm, CD = 6 cm, DA = 6.5 cm and AC = 8 cm.
Solution
Step 1: ∆ABC can be constructed using SSS criterion of the construction of triangle.
Step 2: Here we can see that AC is diagonal, so D will be somewhere opposite to B with reference to AC.
AD = 6.5 cm so draw an arc from A as the centre with radius 6.5 cm.
Step 3: Now draw an arc with C as the centre and by taking radius 6 cm so that it intersects the above arc.
Step 4: The point of intersection of the two arcs is point D. Now join AD and DC to complete the quadrilateral.
Hence, ABCD is the required quadrilateral.
2. If two diagonals and three sides of the quadrilateral are given
Example
Construct a quadrilateral ABCD if the two diagonals are AC = 6.5 cm and BD = 8 cm. The other sides are BC = 5.5 cm, AD = 6.5 cm and CD = 6 cm.
Solution
First of all, draw a rough sketch of the quadrilateral by using the given measurements. Then start constructing the real one.
Step 1: We can see that AD, AC and DC are given so we can construct a triangle ΔACD by using SSS criterion.
Step 2: Now, we know that BD is given so we can draw the point B keeping D as the centre and draw an arc of radius 8 cm just opposite to the point D with reference to AC.
Step 3: BC is given so we can draw an arc keeping C as centre and radius 5.5 cm so that it intersects the other arc.
Step 4: That point of intersection of the arcs is point B. Join AB and BC to complete the quadrilateral.
ABCD is the required quadrilateral.
3.If three angles and two adjacent sides of the quadrilateral are given.
Example
Construct a quadrilateral ABCD in which the two adjacent sides are AB = 4.5 cm and BC = 7.5 cm. The given three angles are ∠A = 75ᵒ, ∠B = 105ᵒ and ∠C = 120ᵒ.
Solution
Draw a rough sketch so that we can construct easily.
Step 1: Draw AB = 4.5 cm. Then measure ∠B = 105° using protractor and draw BC = 7.5 cm.
Step 2: Draw ∠C = 120°.
Step 3: Measure ∠A = 75° and make a line until it touches the line coming from point C.
ABCD is the required quadrilateral.
4. If the three sides with two included angles of the quadrilateral are given.
Example
Construct a quadrilateral ABCD in which the three sides are AB = 5 cm, BC = 6 cm and CD = 7.5 cm. The two included angles are ∠B = 105° and ∠C = 80°.
Solution
Draw a rough sketch.
Step 1: Draw the line BC = 6 cm. Then draw ∠B = 105° and mark the length of AB = 5 cm.
Step 2: Draw ∠C = 80° using protractor towards point B.
Step 3: Mark the length of CD i.e.7.5 cm from C to make CD = 7.5 cm.
Step 4: Join AD which will complete the quadrilateral ABCD.
Hence ABCD is the required quadrilateral.
Some Special Cases
There are some special cases in which we can construct the quadrilateral with less number of measurements also.
Example
Construct a square READ with RE = 5.1 cm.
Solution
Given Re = 5.1 cm.
As it is a special quadrilateral called square, we can get more details out of it.
a. All sides of square are equal, so RE = EA = AD = RD = 5.1 cm.
b. All the angles of a square are 90°, so ∠R = ∠E = ∠A = ∠D = 90°
Step 1: Draw a rough sketch of the square.
Step 2: To construct a square, draw a line segment RE = 5.1 cm. Then draw the angle of 90° at both ends R and E of the line segment RE.
Step 3: As all the sides of the square READ are equal, draw the arc of 5.1 cm from the vertex R and E to cut the lines RD and EA respectively.
Any expression involving constant, variable and some operations like addition, multiplication etc is called Algebraic Expression.
A variable is an unknown number and generally, it is represented by a letter like x, y, n etc.
Any number without any variable is called Constant.
A number followed by a variable is called Coefficient of that variable.
A term is any number or variable separated by operators.
Equation
A statement which says that the two expressions are equal is called Equation.
Linear Expression
A linear expression is an expression whose highest power of the variable is one only.
Example
2x + 5, 3y etc.
The expressions like x2 + 1, z2 + 2z + 3 are not the linear expressions as their highest power of the variable is greater than 1.
Linear Equations
The equation of a straight line is the linear equation. It could be in one variable or two variables.
Linear Equation in One Variable
If there is only one variable in the equation then it is called a linear equation in one variable.
The general form is
ax + b = c, where a, b and c are real numbers and a ≠ 0.
Example
x + 5 = 10
y – 3 = 19
These are called linear equations in one variable because the highest degree of the variable is one and there is only one variable.
Some Important points related to Linear Equations
There is an equality sign in the linear equation. The expression on the left of the equal sign is called the LHS (left-hand side) and the expression on the right of the equal sign is called the RHS (right-hand side).
In the linear equation, the LHS is equal to RHS but this happens for some values only and these values are the solution of these linear equations.
Graph of the Linear Equation in One Variable
We can mark the point of the linear equation in one variable on the number line.
x = 2 can be marked on the number line as follows-
Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
There are two methods to solve such type of problems-
1. Balancing Method
In this method, we have to add or subtract with the same number on both the sides without disturbing the balance to find the solution.
Example
Find the solution for 3x – 10 = 14
Solution
Step 1: We need to add 10 to both the sides so that the numbers and variables come on the different sides without disturbing the balance.
3x – 10 +10 =10+14
3x = 24
Step 2: Now to balance the equation, we need to divide by 3 into both the sides.
3x/3 = 24/3
x = 8
Hence x = 8 is the solution of the equation.
We can recheck our answer by substituting the value of x in the equation.
3x – 10 = 14
3(8) – 10 = 14
24-10 = 14
14 = 14
Here, LHS = RHS, so our solution is correct.
2. Transposing Method
In this method, we need to transpose or transfer the constants or variables from one side to another until we get the solution. When we transpose the terms the sign will get changed.
Example
Find the solution for 2z +10 = 4.
Solution
Step 1: We transpose 10 from LHS to RHS so that all the constants come in the same side.
2z = 4 -10 (sign will get changed)
2z = -6
Step 2: Now divide both the sides by 2.
2z/2 = – 6/2
z = – 3
Here z = -3 is the solution of the equation.
Some Applications of Linear Equation
We can use the concept of linear equations in our daily routine also. There are some situations where we need to use the variable to find the solution. Like,
What number should be added to 23 to get 75?
If the sum of two numbers is 100 and one of the no. is 63 then what will be the other number?
Example
What is the height of the rectangle whose perimeter is 96 cm2 and the length is 12 cm?
Solution
Let the height of the rectangle be ‘s’.
Area of rectangle = Length × Breadth
96 = S × 12
Now, this is a linear equation with variable s.
We need to divide both sides by 12 to find the solution.
96/12 = 12s/12
s = 8
Hence the height of the rectangle is 8 cm.
Solving Equations having the Variable on both Sides
As the equation can have the variable on both the sides also so we should know how to solve such problems.
In this type of problems, we need to bring all the constants on one side and all the terms having variables on the other side. Then they can be solved easily.
Example
Find the solution of 2x−3 = 6 − x.
Solution
Step 1: Bring all the terms including variable x on LHS and the constants on the RHS.
2x + x = 6 + 3 (sign will change while changing the position of the terms)
Step 2: Solve the equation
3x = 9
Step 3: Divide both the sides by 3 to get the solution.
3x/3 = 9/3
x = 3
Hence the solution of the equation is x = 3.
Some More Applications
Example
Renu’s age is four times that of her younger brother. Five years back her age was 9 times her brother’s age. Find their present ages.
Solution
Let the Renu’s brother age = x
Renu’s age = 4x (as her age is 4 times that of her younger brother)
Five years back her age was = 9(x – 5) which is equal to 4x – 5
9(x – 5) = 4x – 5
9x – 45 = 4x – 5
9x – 4x = – 5 + 45 (by transferring the variable and constants on different sides)
5x = 40
x = 40/5 = 8
Renu’s brother age = x = 8 years
Renu’s age = 4x = 4(8) = 32 years.
Reducing Equations to Simpler Form
When linear equations are in fractions then we can reduce them to a simpler form by-
Taking the LCM of the denominator
Multiply the LCM on both the sides, so that the number will reduce without the denominator and we can solve them by the above methods.
Example Solve the linear equation
Solution As the equation is in complex form, we have to reduce it into a simpler form.
Step 1: Take the L.C.M. of the denominators, 2, 3, 4, and 5, which is 60. Step 2: Multiply both the sides by 60,
30x −12 = 20x + 15 + 60
Step 3. Bring all the variables on the LHS and all the constants on the RHS
30x − 20x = 15 + 12 + 60
10x = 87
Step 4: Dividing both the sides by 10
x = 8.7
Equations Reducible to the Linear Form
Sometimes there are some equations which are not linear equations but can be reduced to the linear form and then can be solved by the above methods.
Example
Solution
This is not a linear equation but can be reduced to linear form
