Prasanna

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.1

Fill in the blanks.

(i) If the cost of 8 apples is 56 then the cost of 12 apples is ____.
(ii) If the weight of one fruit box is \(3 \frac{1}{2}\) kg, then the weight of 6 such boxes is ____.
(iii) A car travels 60 km with 3 liters of petrol. If the car has to cover the distance of
200 km, it requires ___ liters of the petrol.
(iv) If 7 m cloth costs ₹ 294, then the cost of 5m of cloth is ____.
(v) If a machine in a cool drinks factory fills 600 bottles in 5 hrs, then it will fill _____ bottles in 3 hours.
Solutions:
(i) 84
(ii) 21 kg
(iii) 10
(iv) ₹ 210
(v) 360

Question 2.
Say True or False
(i) Distance travelled by a bus and time taken are in direct proportion.
(ii) Expenditure of a family to number of members of the family are in direct proportion.
(iii) Number of students in a hostel and consumption of food are not in direct proportion.
(iv) If Mallika walks 1km in 20 minutes, then she can convert 3km in 1 hour.
(v) If 12 men can dig a pond in 8 days, then 18 men can dig it in 6 days.
Solutions:
(i) True
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
A dozen bananas costs ₹ 20. What is the price of 48 bananas ?
Solution:
Let the required price be ₹ x. As the number of bananas increases price also increases
∴ Number of bananas and cost are in direct proportion.

Question 4.
A group of 21 students paid ₹ 840 as the entry fee for a magic show. How many students entered the magic show if the total amount paid was ₹ 1680?
Solution:
Let the required number of students be x.

As the number of students increases the entry fees also increases.
∴ They are in direct proportion .

∴ The number of students entered magic show = 42

Question 5.
A birthday party is arranged in third floor of a hotel. 120 people take 8 trips in a lift to go to the party hall. If 12 trips were made how many people would have attended the party?
Solution:
Let the number of people attended the party be x.

As the number of trips increases, number of people also increases.
∴ They are in direct proportion.


180 people attend the party in 12 trips

Question 6.
The shadow of a pole with height of 8m is 6m. If the shadow of another pole measured at the same time is 30m, find the height of the pole?
Solution:
Let the required height of the pole be ‘x’ m.

Height of the pole and its shadow are in direct proportion

∴ Height of the pole x = 40m.

Question 7.
A postman can sort out 738 letters in 6 hours. How many letters can be sorted in 9 hours?
Solution:
Let the required number of letters be x.

They are in direct proportion.

In 9 hours 1107 letters can be sorted.

Question 8.
If half a meter of cloth costs ₹ 15. Find the cost of \(8 \frac{1}{3}\) meters of the same cloth.
Solution:
Let the cost of cloth required be x.

Cost and length are in direct proportion.

Question 9.
The weight of 72 books is 9 kg. What is the weight of 40 such books (using unitary method)
Solution:
Weight of 72 books = 9 kg = 9000 g
∴ Weight of 1 book = \(\frac{9000}{72}\) = 125 g
∴ Weight of 40 books = 125 × 40 g = 5000 g = 5 kg.
Weight of 40 books = 5 kg

Question 10.
Thamarai pages ₹ 7500 as rent for 3 months. With the same rate how much does she have to pay for 1 year (using unitary method).
Solution:
Rent paid by Thamarai for 3 months = ₹ 7500
∴ Rent paid for 1 month = \(\frac{7500}{3}\) = 2500
Rent paid for 1 year or 12 moths = 2500 × 12 = ₹ 30,000
For 1 year rent to be paid = ₹ 30,000

Question 11.
If 30 men can reap a field in 15 days, then in how many days can 20 men reap the same field? (using unitary method).
Solution:

∴ 20 men can reap the field in 10 days.

Question 12.
Valli purchase 10 pens for ₹ 180 and Kamala boys 8 pens for ₹ 96. Can you say who bought the pen cheaper (using unitary method).
Solution:

∴ Kamala bought the pen cheaper.

Question 13.
A motorbike requires 2 liters of petrol to cover 100 kilometres. How many liters of petrol will be required to cover 250 kilometers? (using unitary method).
Solution:
To cover 100 km quantity of petrol required = 2 litres

5 litres of petrol required to cover 250 km

Objective Type Questions

Question 14.
If the cost of 3 books is ₹ 90, then find the cost of 12 books.
(i) ₹ 300
(ii) ₹ 320
(iii) ₹ 360
(iv) ₹ 400

Solution:
(iii) ₹ 360

Question 15.
If Mani buys 5 kg of potatoes for ₹ 75 then he can buy ₹ 105.
(i) 6
(ii) 7
(iii) 8
(iv) 5

Solution:
(ii) 7

Question 16.
35 cycles were produced in 5 days by a company then ___ cycles will be produced in 21 days.
(i) 150
(ii) 70
(iii) 100
(iv) 147

Solution:
(iv) 147

Question 17.
An aircraft can accommodate 280 people in 2 trips. It can take ______ trips to take 1400 people.
(i) 8
(ii) 10
(iii) 9
(iv) 12
Solution:
(ii) 10

Question 18.
Suppose 3 kg of sugar is used to prepare sweets for 50 members, then ___ kg of sugar is required for 150 members.
(i) 9
(ii) 10
(iii) 15
(iv) 6

Solution:
(i) 9

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Question 1.
From the given figure, name the parallel lines

Solution:
(i) Parallel Lines:
\(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{EF}}\) ; \(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{IJ}}\) ; \(\overrightarrow{\mathrm{EF}}\) and \(\overrightarrow{\mathrm{IJ}}\) are parallel lines.
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\)
(b) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(e) \(\overrightarrow{\mathrm{GH}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(iii) Points of Intersection:
P, Q and R are the points of intersection.

Question 2.
(a) Name the line segments in the figure.
(b) Is Q, the endpoint of each line segment?

Solution:
(a) \(\overline{\mathrm{QP}} \text { and } \overline{\mathrm{QR}}\) are the line segments
(b) Yes, Q is the end point of each line segment

Question 3.
How many lines can pass through
(a) one given point
(b) two given points.
Solution:

(a) An infinite number of lines can pass through one given point.
(b) Exactly one and only one line can pass through two given points.

Question 4.
A line contains how many points?
(a) minimum?
(b) maximum?
Solution:
(a) A line contains a minimum of two points.
(b) A line contain a maximum of infinitely many points.

Question 5.
Write the (a) maximum and (b) the minimum number of point of intersection of three lines.
Solution:

Maximum – 3 points of intersection
Minimum – No point of intersection

Fill in the blanks.

Question 6.
Complementary angle of 20° is _____
Solution:
70°

Question 7.
The supplementary angle of 90° is _____
Solution:
90°

Question 8.
78°, 12°, ______
Solution:
Complementary angle

Answer the following question.

Question 9.
∠ABD =?

Solution:
On Sum of complementary angles = 90°
∠ABC = 90°
∠CBD = 30°
∠ABD = ∠ABC – ∠DBC = 90° – 30° = 60°
∠ABD = 60°
Complementary angle of 30° = 60°

Question 10.
In the following figure, name the angles.

Solution:
∠AOB, ∠BOZ, ∠AOZ

Question 11.
Write the alternate name of the angle ∠XYZ in the given figure.

Solution:
∠Y or ∠ZYX

Question 12.
Draw the diagram of two angles having only one common point.

Solution:
∠COD and ∠AOB have the point ‘O’ in common

Question 13.
What are the supplementary and complementary angles of 60°?
Solution:
Supplementary angle is 120°
Complementary angle is 30°

Question 14.
How many lines can you draw passing through three collinear points? Draw the figure also.
Solution:
Only one.

Question 15.
Write the maximum number of lines that can pass through a single point.
Solution:
Infinite.

Question 16.
Use a protractor to draw an angle 45°.
Solution:

Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray \(\overrightarrow{\mathrm{PQ}}\) with the 0° line. Then drawn and labelled a pointed (R) at the 45° mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at \(\overrightarrow{\mathrm{PR}}\) to complete the angle
Now ∠P = ∠QPR = ∠RPQ = 45°.

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Exercise 2.1

Parallelogram

(Try These Text book Page No. 33)

Question 1.
Find the missing values for the following:

Solution:
(i) Given length l = 12 m; Breadth b = 8 cm
∴ Area of rectangle = l × b sq. units = 12 × 8 m² = 96 m²
Perimeter of the rectangle = 2 × (1 + b) units = 2 × (12 + 8)m = 2 × 20 = 40m

(ii) Given Length l = 15 cm ; Area of the rectangle = 90 sq. cm
l × b = 90; 15 × 6 = 90; b = \(\frac{90}{15}\) = 6 cm
Perimeter of the rectangle = 2 × (l+ b) units = 2 × (15 + 6) cm = 2 × 21 cm = 42 cm

(iii) Given Breadth of rectangle = 50 mm ; Perimeter of the rectangle = 300 mm
2 × (l + b) = 300
2 × (l + 50) = 300
l + 50 = \(\frac{300}{2}\) = 150
l = 150 – 50
l = 100
Area = l × b sq. untis = 100 × 50 mm² = 5000 mm²

(iv) Length of the rectangle = 12 cm ; Perimeter = 44 cm
2(l + b) = 44
2(12 + b) = 44
12 + b = \(\frac{44}{2}\)
12 + b = 22 ; b = 22 – 12; b = 10 cm
Area = l × b sq. units
= 12 × 10 cm² = 120 cm²

Question 2.

Solution:
(i) Given side a = 60 cm
Area of the square = a × a sq.units = 60 × 60 cm² = 3600 cm²
Perimeter of the square = 4 × a units = 4 × 60 cm = 240 cm

(ii) Given area of a square = 64 sq. m
a × a = 64
a × a = 8 × 8
a = 8m
Perimeter = 4 × a
= 4 × 8
= 32 m

(iii) Given perimeter of the square = 100 mm
4 × a = 100
a = \(\frac{100}{4}\) mm
a = 25 mm
Area = a × a sq. units
= 25 × 25 mm²
= 625 mm²

Question 3.

Solution:
(i) Given base of the right angled triangle = 13 m ; height = 5 m

(iii) Given height h = 6 mm ; Area = 84 sq. mm
\(\frac{1}{2}\) × b × h = 84 ; \(\frac{1}{2}\) × b × 6 = 84
b = \(\frac{{84} \times 2}{6}\); b = 28 mm

(Try This Textbook Page No. 35)

Question 1.
Explain the area of the parallelogram as sum of the areas of the two triangles.
Solution:
ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.

Area of the parallelogram ABCD = base × height
= AB × DE

Question 2.
A rectangle is a parallelogram but a parallelogram is not a rectangle. Why?
Solution:
(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to 90°. But for parallelogram opposite angles are equal.
∴ All rectangles are parallelograms. But all parallelograms are not rectan¬gles as their angles need not be equal to 90°.

(Try These Textbook Page No. 36)

Question 1.
Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).

(a) ______ sq. units
(b) ______ sq. units
(c) ______ sq. units
(d) ______ sq. units
Solution:
Converting the given parallelograms into rectangles we get.

(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Question 2.
Draw the heights for the given parallelograms and mark the measure of their bases and find the area. Analyze your result.

Solution:
(a) Area of the parallelogram = b × h sq. units
= 4 × 2 sq. units = 8 sq. units

By counting the small squares also we get number of full
squares + number of square more than half = 6 + 2 = 8 sq. units.

(b) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

(c) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

Also area = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) Number of half squares = 4 + 4 = 8 sq. units

(d) Area of the parallelogram = (base × height) sq. units = 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + \(\frac{1}{2}\) [Number of half squares] + Number of squares more than half = 4 + 0 + 4 = 8sq. units

(e) Area of parallelogram = (base × height) sq. units
= 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) [Number of half squares] = 2 + 6 = 8sq. units

Question 3.
Find the area o the following parallelograms by measuring their base and height, using formula.

(a) _____ sq. units
(b) _____ sq. units
(c) _____ sq. units
(d) _____ sq. units
(e) _____ sq. units
Solution:
(a) Area of the rectangle = (base × height) sq. units
base = 5 units
height = 5 units
∴ Area = (5 × 5 ) = sq. units = 25 sq. units

(b) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 1 units
∴ Area = (4 × 1 ) = sq. units = 4 sq. units

(c) Area of the rectangle = (base × height) sq. units
base = 2 units
height = 3 units
∴ Area = (2 × 3 ) = sq. units = 6 sq. units

(d) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 4 units
∴ Area = (4 × 4 ) = sq. units = 16 sq. units

(e) Area of the parallelogram = (base × height) sq. units
base = 7 units
height = 5 units
= 7 × 5 = 35 sq. units

Question 4.
Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.
Solution:

Area of parallelogram (a), (b) or (c) = 20 sq. units

Exercise 2.2

Rhombus

(Try These Textbook Page No. 41)

Question 1.
Observe the figure and answer the following questions.
(i) Name two pairs of opposite sides.
(ii) Name two pairs of adjacent sides.
(iii) Name the two diagonals.

Solution:
(i) (a) PQ and RS (b) QR and PS
(ii) (a) PQ and QR (b) PS and RS
(iii) (a) PR and Question are diagonals.

Question 2.
Find the area of the rhombus given in (i) and (ii).

Solution:
(i) Area of the rhombus = \(\frac{1}{2}\) (d₁ + d₂) sq. units = \(\frac{1}{2}\)(11 + 13) sq. units
= \(\frac{1}{2}\) × (24) cm² = 12 cm²

(ii) Base = 10 cm ; Height = 7 cm
Area of the rhombus = b × h sq. units = 10 × 7 cm² = 70 cm²

Question 3.
Can you find the perimeter of the rhombus?
Solution:
If we know the length of one side we can find the perimeter using 4 × side units.

Question 4.
Can diagonals of a rhombus be of the same length?
Solution:
When the diagonals of a rhombus become equal it become a square.

Question 5.
A square is a rhombus but a rhombus is not a square. Why?
Solution:
In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to 90°.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.

Question 6.
Can you draw a rhombus in such a way that the side is equal to the diagonal.
Solution:
Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.

Exercise 2.3

(Try These Textbook Page No. 46)

Question 1.
Can you find the perimeter of the trapezium? Discuss.
Solution:
If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.

Question 2.
In which case a trapezium can be divided into two equal triangles?
Solution:
If two parallel sides are equal in length. Then it can be divided into two equal triangles.

Question 3.
Mention any three life situations where the isosceles trapeziums are used?
Solution:
(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Try this Page No. 35

Question 1.
\(\frac{22}{7}\) and 3.14 are rational numbers. Is ‘π’ a rational number? Why?
Solution:
\(\frac{22}{7}\) and 3.14 are rational numbers n has non-terminating and non -repeating decimal expansion. So it is not a rational number. It is an irrational number.

Try this Page No. 38

Question 1.
The given circular figure is divided into six equal parts. Can we call the parts as sectors? Why?

Solution:
No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.
Here the boundaries are not radii.

Try these Page No. 38

Question 1.
Fill the central angle of the shaded sector (each circle is divided into equal sectors)

Try this Page No. 44

Question 1.
If the radius of a circle is doubled, what will the area of the new circle so formed?
Solution:
If r = 2r₁ ⇒ Area of the circle = πr² = π(2r₁)² = π4r₁² = 4πr₁²
Area = 4 × old area.

Try this Page No. 49

Question 1.
All the sides of a rhombus are equal. Is it a regular polygon?
Solution:
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the
sides are equal. But all the angles are not equal
∴ It is not a regular polygon.

Try this Page No. 53

Question 1.
In the above example split the given mat as into two trapeziums and verify your answer.
Solution:
Area of the mat = Area of I trapezium + Area of II trapezium

∴ Cost per sq.feet = ₹ 20
Cost for 28 sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.

Try these Page No. 54

Question 1.
Show that the area of the unshaded regions in each of the squares of side ‘a’ units are the same in all the cases given below.

Solution:

Question 2.
If π = \(\frac{22}{7}\), show that the area of the unshaded part of a square of side ‘a’ units is approximately \(\frac{3}{7}\) a² sq. units and that of the shaded part is approximately \(\frac{4}{7}\) a² sq. units for the given figure.

Solution:

Try this Page No. 57

Question 1.
List out atleast three objects in each category which are in the shape of cube, cuboid,
cylinder, cone and sphere.
Solution:
(i) Cube – dice, building blocks, jewel box.
(ii) Cuboid – books, bricks, containers.
(iii) Cylinder – candles, electric tube, water pipe.
(iv) Cone – Funnel, cap, ice cream cone
(v) Sphere – ball, beads, lemon.

Try this Page No. 58

Question 1.
Tabulate the number of faces(F), vertices(V) and edges(E) for the following polyhedron. Also find F + V – E
Solution:

From the table F + V – E = 2 for all the solid shapes.

Try this Page No. 58

Question 1.
Find the area of the given nets.

Solution:
Area = 6 × Area of a square of side 6 cm
= 6 × (6 × 6) cm²
= 216 cm²
(ii) Area = Area of 2 rectangles of side (8 × 6) cm² + Area of 2 rectangles of side (8 × 4) cm² + Area of 2 rectangles of side (6 × 4) cm²
= (8 × 6) + (8 × 4) + (6 × 4)cm²
= 48 + 32 + 24 cm²
= 104 cm²

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
In the following figure, PQRS is a parallelogram find x and y.

Solution:
We know that in a parallelogram opposite sides are equal.
∴ 3x = 18
x = \(\frac{18}{3}\)
x = 6 and
3y – 1 = 26
3y = 26 + 1 = 27
y = \(\frac{27}{9}\)
y = 9

Question 2.
Two adjacent sides of a parallelogram are 5 cm and 7 cm respectively. Find its perimeter.
Solution:

Perimeter = AB + BC + CD + AD [∵ AB = DC & AD = BC]
= 7 cm + 5 cm + 7 cm + 5 cm = 24 cm

Question 3.
The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:

Given perimeter = 150 cm
Let one side of the parallelogram be ‘b’ cm
Then the other side = b + 25 cm
b + (b + 25) + b + (b + 25) = 150
b + b + 25 + b + b + 25 = 150
4h + 50 = 150 =4b = 100
b = \(\frac{100}{4}\) = 25
∴ One side b = 25 cm
Other side b + 25 = 50 cm

Exercise 2.2

Question 1.
ABCD is a rhombus. Find x, y, and z.

Solution:
We know that all sides of rhombus are equal and its diagonals bisect each other.
∴ x = 5, y = 12 and z = 13.

Question 2.
Find the altitude of the rhombus whose area is 315 cm² and its perimeter is 180 cm.
Solution:
Given perimeter of the rhombus = 180 cm
∴ One side of the rhombus = \(\frac{180}{4}\) = 45 cm
Given area of the rhombus = 315 cm²
b × h = 315
45 × h = 315 = \(\frac{315}{45}\)
h = 7 cm
Altitude of the rhombus = 7 cm

Question 3.
The floor of a building consists of 2000 titles which are rhombus shaped and each of its diagonals are 40 cm and 25 cm. Find the total cost of polishing the floor, if the cost per m² = ₹ 5.
Solution:
Area of each title = \(\frac{1}{2}\) × d₁ × d₂ sq. units
= \(\frac{1}{2}\) × 40 × 25 cm² = 500 cm²
∴ Area of 2000 titles = 500 × 20,000 = 10,000 cm² = 100 m²
Cost of polishing 1 m² = ₹ 5
∴ Cost of polishing 100 m² = 5 × 100 = ₹ 500

Students can Download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Question 1.
Expand
(i) (3m + 5)²
(ii) (5p – 1)²
(iii) (2n – 1)(2n + 3)
(iv) 4p² – 25q²
Solution:
(i) (3m + 5)²
Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5
(a + b)² = a² + 2 ab + b²
(3m + 5)² = (3m)² + 2 (3m) (5) + 5²
= 3²m² + 30m + 25 = 9m² + 30m +25

(ii) (5p – 1)²
Comparing (5p – 1)² with (a – b)² we have a = 5p and b = 1
(a – b)² = a² – 2ab + b²
(5p – 1)² = (5p)² – 2 (5p) (1) + 1²
= 5²p² – 10p + 1 = 25p² – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x² + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)² + (-1 + 3)2n + (-1) (3)
= 2²n² + 2 (2n) – 3 = 4n² + 4n – 3

(iv) 4p² – 25q² = (2p)² – (5q)²
Comparing (2p)² – (5q)² with a² – b² we have a = 2p and b = 5q
(a² – b²) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Question 2.
Expand
(i) (3 + m)³
(ii) (2a + 5)³
(iii) (3p + 4q)³
(iv) (52)³
(v) (104)³
Solution:
(i) (3 + m)³
Comparing (3 + m)³ with (a + b)³ we have a = 3; b = m
(a + b)³ = a² + 3a²b + 3 ab² + b³
(3 + m)³ = 3³ + 3(3)² (m) + 3 (3) m² + m³
= 27 + 27m + 9m² + m³ = m³ + 9 m² + 27m + 27

(ii) (2a + 5)³
Comparing (2a + 5)³ with (a + b)³ we have a = 2a, b = 5
(a + b)³ = a³ + 3a²b + 3ab² + b³ = (2a)³ + 3(2a)² 5 + 3 (2a) 5² + 5³
= 2³a³ + 3(2²a²) 5 + 6a (25) + 125
= 8a³+ 60a² + 150a + 125

(iii) (3p + 4q)³
Comparing (3p + 4q)³ with (a + b)³ we have a = 3p and b = 4q
(a + b) ³ = a³ + 3a²b + 3ab² + b³
(3p + 4q)³ = (3p)³ + 3(3p)² (4q) + 3(3p)(4q)² + (4q)³
= 3³p³ +3 (9p²) (4q) + 9p (16q²) + 43q³
= 27p³ + 108p²q + 144pq² + 64q³

(iv) (52)³ = (50 + 2)³
Comparing (50 + 2)³ with (a + b)³ we have a = 50 and b = 2
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(50 + 2)³ = 50³ + 3 (50)²2 + 3 (50)(2)² + 2³
52³ = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
52³ = 1,40,608

(v) (104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3 a²b + 3 ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + (4)³
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Question 3.
Expand
(i) (5 – x)³
(ii) (2x – 4y)³
(iii) (ab – c)³
(iv) (48)³
(v) (97xy)³
Solution:
(i) (5 – x)³
Comparing (5 – x)³ with (a – b)³ we have a = 5 and b = x
(a – b)³ = a³ – 3a²b + 3ab² – b³
(5 – x)³ = 5³ – 3 (5)² (x) + 3(5)(x²) – x³
= 125 – 3(25)(x) + 15x² – x³ = 125

(ii) (2x – 4y)³
Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y
(a – b)³ = a³ – 3a²b + 3ab³ – b³
(2x – 4y)³ = (2x)³ – 3(2x)² (4y) + 3(2x) (4y)² – (4y)³
= 2³x³ – 3(2²x²) (4y) + 3(2x) (4²y²) – (4³y³)
= 8x³ – 48x²y + 96xy² – 64y³

(iii) (ab – c)³
Comparing (ab – c)³ with (a – b)³ we have a = ab and b = c
(a – b)³ = a³ – 3a²b + 3ab² – b³
(ab – c)³ = (ab)³ – 3 (ab)² c + 3 ab (c)² – c³
= a³b³ – 3(a²b²) c + 3abc² – c³
= a³b³ – 3a²b² c + 3abc² – c³

(iv) (48)³ = (50 – 2)³
Comparing (50 – 2)³ with (a – b)³ we have a = 50 and b = 2
(a – b)³ = a³ – 3a²b + 3ab² – b³
(50 – 2)³ = (50)³ – 3(50)²(2) + 3 (50)(2)² – 2³
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592

(v) (97xy)³
= 97³ x³ y³ = (100 – 3)³ x³y³
Comparing (100 – 3)³ with (a – b)³ we have a = 100, b = 3
(a – b)³ = a³ – 3a²b + 3ab² – b³
(100 – 3)³ = (100)³ – 3(100)² (3) + 3 (100)(3)² – 3³
97³ = 10,00,000 – 90000 + 2700 – 27
97³ = 910000 + 2673
97³ = 912673
97x³y³ = 912673x³y³

Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² (ab + bc + ca) x + abc
= (5y)³ + (1 + 2 + 3) (5y)² + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 5³y³ + 6(5²y²) + (2 + 6 + 3)5y + 6
= 125³ + 150y² + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + be + ca) x + abc
= p³ + (-2 + 1 + (-4))p² + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p³ + (-5 )p² + (-2 + (-4) + 8)p + 8
= p³ – 5p² + 2p + 8

Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)³ cubic units = (x + 1)³ cm³
We have (a + b)³ = (a³ + 3a²b + 3ab² + b³) cm³
(x + 1)³ = (x³ + 3x² (1) + 3x (1)² + 1³) cm³
Volume = (x³ + 3x² + 3x + 1) cm³

Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units³
= (x + 2) (x – 1) (x – 3) units³
We have (x + a) (x + b) (x + c) = x³ + (a + b + c) x² + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x³ + (2 – 1 – 3)x² + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x³ – 2x² + (-2 + 3 – 6)x + 6
Volume = x³ – 2x² – 5x + 6 units³

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Classify the following polynomials based on number of terms.
(i) x³ – x²
(ii) 5x
(iii) 4x⁴ + 2x³ + 1
(iv) 4.x³
(v) x + 2
(vi) 3x²
(vii) y⁴ + 1
(viii) y²⁰ + y¹⁸ + y²
(ix) 6
(x) 2u³ + u² + 3
(xi) u²³ – u⁴
(xii) y
Solution:
5x, 3x², 4x³, y and 6 are monomials because they have only one term.
x³ – x², x + 2, y⁴ + 1 and u²³ – u⁴ are binomials as they contain only two terms.
4x⁴ + 2x³ + 1 , y²⁰ + y¹⁸ + y² and 2u³ + u² + 3 are trinomials as they contain only three terms.

Question 2.
Classify the following polynomials based on their degree.
Solution:
p(x) = 3, p(x) = -7, p(x) = \(\frac{3}{2}\) are constant polynomials
p(x) = x + 3, p(x) = 4x, p(x) = \(\sqrt{3}\)x + 1 are linear polynomials, since the highest degree of the variable x is one.
p(x) = 5x² – 3x + 2, p(y) = \(\frac{5}{2}\) y² + 1, p(x) = 3x² are quadratic polynomials, since the highest degree of the variable is two.
p(x) = 2x³ – x² + 4x + 1, p(x) = x³ + 1, p(y) = y³ + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.
Find the product of given polynomials p(x) = 3x³+ 2x – x² + 8 and q (x) = 7x + 2.

Solution:
(7x +2) (3x³ + 2x – x² + 8) = 7x(3x³ + 2x – x² + 8) + 2x
(3x³ + 2x – x² + 8) = 21x⁴ + 14x² – 7x³ + 56x + 6x³ + 4x – 2x²+ 16 = 21x⁴ – x³ + 12x⁴ + 60x + 16

Question 4.
Let P(x) = 4x⁴ – 3x + 2x³ + 5 and q(x) = x² + 2x + 4 find p (x) – q(x).

Solution:

Exercise 3.2

Question 1.
If p(x) = 5x³ – 3x² + 7x – 9, find
(i) p(-1)
(ii) p(2).
Solution:
Given that p(x) = 5x³ – 3x² + 7x – 9
(i) p(-1) = 5(-1)³ – 3(-1)² + 7(-1) – 9 = -5 – 3 – 7 – 9
= -24
(ii) p(2) = 5(2)³ – 3(2)³ + 7(2) – 9 = 40 – 12 + 14 – 9
∴ p(2) = 33

Question 2.
Find the zeros of the following polynomials.
(i) p(x) = 2x – 3
(ii) p(x) = x – 2
Solution:

Exercise 3.3

Question 1.
Find the remainder using remainder theorem, when
(i) 4x³ – 5x² + 6x – 2 is divided by x – 1.
(ii) x³ – 7x² – x + 6 is divided by x + 2.
Solution:
(i) Let p(x) = 4x³ – 5x² + 6x – 2. The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p( 1). Now,
p(1) = 4(1)³ – 5(1)² + 6(1) – 2 = 4 – 5 + 6 – 2 = 3
∴ The remainder is 3.

(ii) Let p(x) = x³ – 7x² – x + 6. The zero of x + 2 is -2.
When p(x) is divided by x + 2, the remainder is p(-2). Now,
p(-2) = (-2)³ – 7(-2)² – (-2) +6 = – 8 – 7(4) + 2 + 6
= – 8 – 28 + 2 + 6 = -28.
∴ The remainder is -28.

Question 2.
Find the value of a if 2x³ – 6x² + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.
Solution:
Let p(x) = 2x³ – 6x² + 5ax – 9
When p(x) is divided by (x – 2) the remainder is p(2).
Given that p(2) = 13
⇒ 2(2)³ – 6(2)² + 5a(2) – 9 = 13
⇒ 2(8) – 6(4) + 10a – 9 = 13
⇒ 16 – 24 + 10a – 9 = 13
⇒ 10a – 17 = 13
⇒ 10a = 30
⇒ ∴ a = 3

Question 3.
If the polynomials f(x) = ax³+ 4x² + 3x – 4 and g (x) = x³ – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.

Solution:
Let f(x) = ax³ + 4x² + 3x – 4 and g(x) = x³ – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).
Now f(3) = a(3)³4(3)² + 3(3) – 4 = 27a + 36 + 9 – 4
f (3) = 27a + 41 …(1)
When g(x) is divided by (x – 3), the remainder is g(3).
Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 +a
27a – a = 15 – 41
26a = -26

Sustituting a = -1, in f(3), we get
f(3) = 27(-1) + 41 = -27 + 41
f(3) = 14
∴ The remainder is 14.

Question 4.
Show that x + 4 is a factor of x³ + 6x² – 7x – 60.
Solution:

Question 5.
In (5x + 4) a factor of 5x³ + 14x² – 32x – 32
Solution:

Question 6.
Find the value of k, if (x – 3) is a factor of polynomial x³ – 9x² + 26x + k.
Solution:
Let p (x) = x³ – 9x² + 26x + k
By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0
p(3) = 0
3³ – 9(3)² + 26 (3) + k = 0
27 – 81 + 78 + k = 0
k = -24

Question 7.
Show that (x – 3) is a factor of x³ + 9x² – x – 105.
Solution:
Let p(x) = x³ + 9x² – x – 105
By factor theorem, x – 3 is a factor of p(x), if p(3) = 0
p (3) = 3³ + 9(3)³ – 3 – 105
= 27 + 81 – 3 – 105
= 108 – 108
p(3) = 0

Therefore, x – 3 is a factor of x³ + 9x² – x – 105.

Question 8.
Show that (x + 2) is a factor of x³ – 4x² – 2x + 20.
Solution:
Let p(x) = x³ – 4x² – 2x + 20
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)³ – 4(-2)² – 2(-2) + 20 = -8 – 4(4) + 4 + 20
P(- 2) = 0
Therefore, (x + 2) is a factor of x³ – 4x² – 2x + 20

Exercise 3.4

Question 1.
Expand the following using identities :
(i) (7x + 2y)²
(ii) (4m – 3m)²
(iii) (4a + 3b) (4a – 3b)
(iv) (k + 2)(k – 3)
Solution:
(i) (7x + 2y)² = (7x)² + 2 (7x) (2y) + (2y)² = 49x² + 28xy + 4y²
(ii) (4m – 3m)² = (4m)² – 2(4m) (3m) + (3m)² = 16m² – 24mn + 9n²
(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a² – b² ]
Put [a = 4a, b = 3b]
(4a + 3b) (4a – 3b) = (4a)² – (3b)² = 16a² – 9b²
(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x² + (a – b)x – ab]
Put [x = k, a = 2, b = 3]
(k + 2) (k – 3) = k² + (2 – 3)x – 2 × 3 = k² – x – 6

Question 2.
Expand : (a + b – c)²
Solution:
Replacing ‘c’ by ‘-c’ in the expansion of
(a + b + c)² = a² + b² + c² + 2 ab+ 2 bc + 2ca
(a + b + (-c))² = a² + b² + (-c)² + 2ab + 2b(-c) + 2(-c)a
= a² + b² + c² + 2ab – 2bc – 2ca

Question 3.
Expand : (x + 2y + 3z)²
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3 z)² = x² + (2y)² + (3z)² + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12y² + 6zx

Question 4.
Find the area of square whose side length is m + n – q.
Solution:
Area of square = side × side = (m + n – q)²
We know that,
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
[m + n + (-q)]² = m² + n² + (-q)² + 2mn + 2n(-q) + 2(-q)m
= m² + n² + q² + 2mn – 2nq – 2qm
Therefore, Area of square = m² + n² + q² + 2mn – 2nq – 2qm = [m² + n² + q² + 2mn – 2nq – 2qm] sq. units.

EXERCISE 3.5

Question 1.
Factorise the following
(i) 25m^-2 – 16n²
(ii) x⁴ – 9x²
Solution:
(i) 25m² – 16n² = (5m)² – (4n)²
= (5m – 4n) (5m + 4n) [∵ a² – b² = (a – b)(a + b)]
(ii) x⁴ – 9x² = x²(x² – 9) = x²(x² – 3²) = x²(x – 3)(x + 3)

Question 2.
Factorise the following.
(i) 64m³ + 27n³
(ii) 729x³ – 343y³
Solution:
(i) 64m³ + 27n³ = (4m)³ + (3n)³
= (4m + 3n) ((4m)² – (4m) (3n) + (3n)²) [∵ a² + b² = (a + b) (a² – ab + b²)]
= (4m + 3n) (16m² – 12mn + 9n²)
(ii) 729x³ – 343y³ = (9x)³ – (7y)³ = (9x – 7y) ((9x)² + (9x) (7y) + (7y)²
= (9x – 7y) (81x² + 63xy + 49y²)
[∵ a³ – b³ = (a – b) (a² + ab + b²)]

Question 3.
Factorise 81x² + 90x + 25
Solution:
81x² + 90x + 25 = (9x)² + 2 (9x) (5) + 5²
= (9x + 5)² [ ∵ a² + 2ab + b² = (a + b)²]

Question 4.
Find a³ + b² if a + b = 6, ab = 5.
Solution:
Given, a + b = 6, ab = 5
a² + b² = (a + b)² – 3ab(a + b) = (6)³ – 3(5)(6) = 126

Question 5.
Factorise (a – b)² + 7(a – b) + 10
Solution:
Let a – b = p, we get p² + 7p + 10,

p² + 7p + 10 = p² + 5p + 2p + 10
= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)
Put p = a – b we get,
(a – b)² + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.
Factorise 6x² + 17x + 12
Solution:

Exercise 3.6

Question 1.
Factorise x² + 4x – 96
Solution:

Question 2.
Factorise x² + 7xy – 12y²
Solution:

Question 3.
Find the quotient and remainder when 5x³ – 9x² + 10x + 2 is divided by x + 2 using synthetic division.
Solution:

Hence the quotient is 5x² – 19x + 48 and remainder is -94.

Question 4.
Find the quotient and remainder when -7 + 3x – 2x² + 5x³ is divided by -1 + 4x using synthetic division.
Solution:

Question 5.
If the quotient on dividing 5x⁴ + 4x³ + 2x + 1 by x + 3 is 5x³ + 9x² + bx – 97 then find the values of a, b and also remainder.
Solution:

Quotient 5x³ + 11x² + 33x – 97 is compared with given quotient 5x³ + ax² + bx – 97
co-efficient of x² is – 11 = a and co-efficient of x is 33 = b.
Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.
Find the quotient and the remainder when 10 – 4x + 3x² is divided by x – 2.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending
order). Thus the given problem becomes (3x² – 4x + 10) ÷ (x – 2)

∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x² – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form
Dividend = (Divisor × Quotient) + Remainder

Question 2.
If 8x³ – 14x² – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.
Solution:

Exercise 3.8

Question 1.
Factorise 2x³ – x² – 12x – 9 into linear factors.
Solution:
Let p(x) = 2x³ – x² – 12x – 9
Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = -1 – 9 = -10
Sum of co-efficients of odd powers = 2 – 12 = -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.


Then p (x) = (x + 1)(2x² – 3x – 9)
Now 2x² – 3x – 9 = 2x² – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)
= (x – 3)(2x + 3)
Hence 2x³ – x² – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.
Factorize x³ + 3x² – 13x – 15.
Solution:
Let p(x) = x³ + 3x² – 13x – 15
Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = 3 – 15 = – 12
Sum of coefficients of odd powers = 1 – 13 = – 12
Hence x + 1 is a factor of p (x)
Now use synthetic division to find the other factors.

Hence p(x) = (x + 1) (x² + 2x – 15)
Now x³ + 3x² – 13x – 15 = (x + 1)(x² + 2x – 15)
Now x² + 2x – 15 = x² + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)
Hence x³ + 3x² – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.
Factorize x³ + 13x² + 32x + 20 into linear factors.

Solution:
Let, p(x) = x³ + 13x² + 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x – 1) is not a factor.
Sum of coefficients of even powers with constant = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x + 1) is a factor of p(x)
Now we use synthetic division to find the other factors

Exercise 3.9

Question 1.
Find GCD of 25x³y² z, 45x²y⁴z³b
Solution:

Question 2.
Find the GCD of (y³ – 1) and (y – 1).
Solution:
y³ – 1 = (y – 1)(y³ + y + 1)
y – 1 = y – 1
Therefore, GCD = y – 1

Question 3.
Find the GCD of 3x² – 48 and x² – 7x + 12.
Solution:
3x² – 48 = 3(x² – 16) = 3(x² – 4³) = 3(x + 4)(x – 4)
x² – 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)
Therefore, GCD = x – 4

Question 4.
Find the GCD of a^x , a^{x + y}, a^{x + y + z}.
Solution:

Exercise 3.10

Question 5.
Solve graphically. x – y = 3 ; 2x – y = 11
Solution:

Exercise 3.11

Question 1.
Solve using the method of substitution. 5x – y = 5, 3x + y = 11
Solution:

Exercise 3.12

Question 2.
Solve by the method of elimination. 2x + y = 10; 5x – y = 11
Solution:

Exercise 3.13

Question 3.
Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.
Solution:

Exercise 3.14

Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be x
Let the sum of the age of his sons be y
At present x = 2y ⇒ x – 2y = 0 …. (1)
After 20 years x + 20 = y + 40
x – y = 40 – 20

Exercise 3.15

Multiple Choice Questions :

Question 1.
The polynomial 3x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial

Question 2.
The polynomial 4x² + 2x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of x is 2
Solution:
(2) quadratic polynomial]

Question 3.
The zero of the polynomial 2x – 5 is

Hint:

Solution:
(1) \(\frac{5}{2}\)

Question 4.
The root of the polynomial equation 3x – 1 = 0 is

Hint:

Solution:
(2) x = \(\frac{1}{3}\)

Question 5.
Zero of (7 + 4x) is ___

Solution:
(2) \(\frac{-7}{4}\)

Question 6.
Which of the following has as a factor?
(1) x² + 2x
(2) (x – 1)²
(3) (x + 1)²
(4) (x² – 2²)
Solution:
(1) x² + 2x

Question 7.
If x – 2 is a factor of q (x), then the remainder is _____
(1) q (-2)
(2) x – 2
(3) 0
(4) -2
Solution:
(3) 0

Question 8.
(a – b) (a² + ab + b²) =
Solution:
(1) a³ + b³ + c³ – 3abc
(2) a² – b²
(3) a³ + b³
(4) a³ – b³
Solution:
(4) a³ – b³

Question 9.
The polynomial whose factors are (x + 2) (x + 3) is
(1) x² + 5x + 6
(2) x² – 4
(3) x² – 9
(4) x² + 6x + 5
Solution:
(1) x² + 5x + 6

Question 10.
(-a – b – c)² is equal to
(1) (a – b + c)²
(2) (a + b – c)²
(3) (-a + b + c)²
(4) (a + b + c)²
Solution:
(4) (a + b + c)²

Text Book Activities

Activity – 1
Write the Variable, Coefficient and Constant in the given algebraic expression,
Solution:

Activity – 2
Write the following polynomials in standard form.

Activity – 3

Question 1.
Find the value of k for the given system of linear equations satisfying the condition below:
(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution
(ii) kx + 3y = 3; 12x + ky = 6 has no solution
(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution
Solution:

Question 2.
Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b
Solution:

Activity – 4
For the given linear equations, find another linear equation satisfying each of the given condition
Solution:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Intext Questions

Try These (Textbook Page No. 1)

Question 1.
(i) Observe and complete:
1 + 3 = ?
5 + 11 = ?
21 + 47 = ?
___ + ____ = ?
From this observation, we conclude that “the sum of any two odd numbers is always an _____”
(ii) Observe and complete:
5 × 3 = ?
7 × 9 = ?
11 × 13 = ?
_____ × ____ = ?
From this observation, we conclude that “the product of any two odd numbers is always an _____”
Justify the following statements with appropriate examples:
(iii) The sum of an odd number and an even number is always an odd number.
(iv) The product of an odd and an even number is always an even number.
(v) The product of only three odd numbers is always an odd number.
Solution:
(i) 1 + 3 = 4
5 + 11 = 16
21 + 47 = 68
An odd number + another odd number = An Even number
From this observation, we conclude that the sum of any two odd numbers is always an even number.
(ii) 5 × 3 = 15
7 × 9 = 63
11 × 13 = 143
An odd number × Another odd number = An odd number
From this observation, we conclude that “the product of any two odd numbers is always an odd number.”
(iii) Take the odd number 5 and the even number 10
Their sum = 5 + 10 = 15, which is odd.
∴ Sum of an odd number and an even number is always an odd number.
(iv) Take the odd number 5 and the even number 10.
Their product = 5 × 10 = 50, which is even
Thus the product of an odd and an even number is always an even number.
(v) Consider 7 × 5 × 3
We know that the product of any two odd numbers is an odd number
7 × 5 = 35, odd number.
Also we have 35 × 3 = 105
∴ 7 × 5 × 3 = 105, an odd number.
So the product of three odd numbers is always an odd number.

Try These (Textbook Page No. 3)

Question 1.
(i) Say True or False
(a) The smallest odd natural number is 1.
(b) 2 is the smallest even whole number.
(c) 12345 + 5063 is an odd number.
(d) Every number is a factor of itself.
(e) A number which is a multiple of 6 is also a multiple of 2 and 3.
(ii) Is 7, a factor of 27?
(iii) Is 12, a factor or a multiple of 12?
(iv) Is 30, a factor or a multiple of 10?
(v) Which of the following numbers has 3 as a factor?
(a) 8
(b) 10
(c) 12
(d) 14
(vi) The factors of 24 are 1, 2, 3, __, 6, ___, 12 and 24. Find the missing ones.
(vii) Look at the following numbers carefully and find the missing multiples.

Solution:
(i) (a) True
(b) False
(c) False
(d) True
(e) True
(ii) No, 7 is not a factor of 27. Because 7 does not divide 27 exactly
(iii) 12 is both a factor and a multiple of 12
(iv) 30 is a multiple of 10
(v) (a) Factors of 8 are 1, 2, 4, 8
(b) Factors of 10 are 1, 2, 5, 10
(c) Factors of 2 are 1, 2, 3, 4, 6, 12
(d) Factors of 14 are 1, 2, 7, 14
∴ The number 12 has 3 as a factor
(vi) Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Missing Factors 4, 8.
(vii)

Try These (Textbook Page No. 6)

Question 1.
Express 68 and 128 as the sum of two consecutive primes.
Solution:
68 = 31 + 37
128 = 61 + 67

Question 2.
Express 79 and 104 as the sum of any three odd primes.
Solution:
79 = 37 + 31 + 11
79 = 41 + 31 + 7
79 = 61 + 11 + 7
79 = 59 + 13 + 7
79 = 53 + 19 + 7 and so on.
104 cannot be expressed as the sum of three odd primes.
Because we know that “ the sum of any two odd numbers is an even number”.
Also the sum of an odd and even number is always an odd number.
104 = 61 + 41 + 2
104 = 97 + 5 + 2
104 = 89 + 13 + 2 and so on.

Try These (Textbook Page No. 8)

Question 1.
Are the leap years divisible by 2?
Solution:
Leap years are divisible by 4.
Leap years are divisible by 2.

Question 2.
Is the first 4 digit number divisible by 3?
Solution:
The first four-digit number is 1000.
Sum of the digits is 1 + 0 + 0 + 0 = 1, not divisible by 3.
1000 is not divisible by 3.

Question 3.
Is your date of birth (DDMMYYYY) divisible by 3?
Solution:
Date of birth 25.05.2007
Sum of digits = 2 + 5 + 0 + 5 + 2 + 0 + 0 + 7 = 21
Again 2 + 1 = 3, divisible by 3.
My date of birth is divisible by 3.

Question 4.
Identify the numbers in the sequence 2000, 2006, 2010, 2015, 2019, 2025 that are divisible by both 2 and 5.
Solution:
We know that a number is divisible by both 2 and 5 if it is divisible by 10. 2000 and 2010 are divisible by 10.

Question 5.
Check whether the sum of 5 consecutive numbers is divisible by 5.
Solution:
Take the first five consecutive natural numbers 1, 2, 3, 4 and 5.
Their sum 1 + 2 + 3 + 4 + 5 = 15, divisible by 5.
Also, 2 + 3 + 4 + 5 + 6 = 20, divisible by 5.
3 + 4 + 5 + 6 + 7 = 25, divisible by 5.
Generally, the sum of 5 consecutive natural numbers is divisible by 5.

Try These (Textbook Page No. 19)

A small boy went to a town to sell a basket of wood apples. On the way, some robbers grabbed the fruits from him and ate them! The small boy went to the King and complained. The King asked him, “How many wood apples did you bring?”. The boy replied, “Your Majesty! I didn’t know, but I knew that if you divided my fruits into groups of 2, one fruit would be left in the basket”. He continued saying that if the fruits were divided into groups of 3, 4, 5 and 6, the fruits left in the basket would be 2, 3, 4 and 5 respectively. Also, if the fruits were divided into groups of 7, no fruit would be there in the basket. Can you find the number of fruits, the small boy had initially?
(This problem is taken from the famous Mathematics problems collection book in Tamil called “Kanakkathikaram” under the heading of “Wood Apple Problem”)
Solution:
The total number of fruits, when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4 and 5 respectively.
Here (2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1.
∴ The required no. of fruits will be LCM (2, 3, 4, 5, 6) – 1
L CM (2, 3, 4, 5, 6) = 2 × 3 × 2 × 5 = 60

Now take the multiples of 60 and subtract 1 from it.
Also checking the conditions, multiplies of 60 are 60, 120, 180, …..
The multiple -1
59, 119, 179, ……
The required number = 119
∴ The total number of fruits = 119.

Students can Download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc

Question 2.
Match the following

Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1

Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)

Solution:

Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{45^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 16 cm
l = \(\frac{1}{8}\) × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr²
A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16
A = 100.48 cm²
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm

(ii) Central angle 120°, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the arc


Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm

Question 5.
From the measures given below, find the area of the sectors.

Solution:
(i) Area of the sector
A = \(\frac{l r}{2}\) sq. units
l = 48 m
r = 10 m
= \(\frac{48 \times 10}{2}\) m²
= 24 × 10 m²
= 240 m²
Area of the sector = 240 m²

(ii) length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{12.5 \times 6}{2}\)
= 12.5 × 3 cm² cm²
= 37.5 cm²
Area of the sector = 37.5 cm²

(iii) length of the arc l = 50 cm
Radius r = 13.5 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{50 \times 13.5}{2}\)
= 25 × 13.5 cm² cm²
= 337.5 cm²
Area of the sector = 337.5 cm²

Question 6.
Find the central angle of each of the sectors whose measures are given below (π = \(\frac{22}{7}\))

Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm²
\(\frac{l r}{2}\) = 462
\(\frac{l \times 21}{2}\) = 462

∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm²

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m

Question 7.
Answer the following questions:
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5

Note: We can solve this problem using A = \(\frac{1}{n}\) πr² sq. units also.

Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Radius r = 14 mm
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{50 \times 14}{2}\) mm² = 50 × 7 mm² = 350 mm²
Area of the sector = 350 mm²

Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = \(\frac{20}{2}\) = 10 cm²
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{44 \times 10}{2}\) cm² = 22 × 10 cm² = 220 cm²
Area of the sector = 220 cm²

Question 10.
A sector of radius 4.2 cm has an area 9.24 cm². Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm²
\(\frac{l r}{2}\) = 9.24
\(\frac{l \times 4.2}{2}\) = 9.24
l × 2.1 = 9.24
l = \(\frac{9.24}{2.1}\)
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm

Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)

Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet

Area of the quadrant = 3.14 sq. feet (approximately)

Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).

Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Given design in the design of a circular quadrant.
Area of the quadrant = \(\frac{1}{4}\) πr² sq. units
= \(\frac{1}{4}\) × 3.14 × 30 × 30 cm²
= 3.14 × 15 × 15 cm²
∴ Area of the sector design = 706.5 cm² (approximately)

Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = \(\frac{22}{7}\))

Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = \(\frac{1}{n}\) πr² sq. units
= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56 cm² = 1232 cm²
Area of each sector = 1232 cm² (approximately)

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Recap Page No. 66 and 67

Question 1.
Write the numbers of terms in the following expressions.
(i) x + y + z – xyz
Solution:
4 terms

(ii) m²n²c
Solution:
1 term

(iii) a²b²c – ab²c² + a²bc² + 3abc
Solution:
4 terms

(iv) 8x² – 4xy + 7xy²
Solution:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
Question 1.
2x² – 5xy + 6y² + 7x – 10y + 9
Solution:
Numerical co efficient in 2x² is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y² is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is – 10
Numerical co-efficient in 9 is 9

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:

Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

Question 3.
Pick out the like terms from the following.

Solution:

Question 4.
Add : 2x, 6y, 9x – 2y
Solution:
2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y

Question 5.
Simplify : (5x³ y³ – 3x² y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
Solution:
(5x³y³ – 3x²y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
= 5x³y³ + x³y³ – 3x²y² + 2x²y² + xy + 2xy + 7 – 5
= (5 + 1)x³y³ + (-3 + 2)x²y² +(1 +2)xy + 2
= 6x³y³ – x²y² + 3xy + 2

Question 6.
The sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
Solution:
Perimeter of the triangle = Sum of three sides
= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
= 4x – y + 12
∴ Perimeter of the triangle = 4x – y + 12 units.

Question 7.
Subtract -2mn from 6mn.
Solution:
6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn

Question 8.
Subtract 6a² – 5ab + 3b² from 4a² – 3ab + b².
Solution:
(4a² – 3ab+ b²) – (6a²– 5ab + 3b²)
= (4a² – 6a²) + (- 3ab -(-5 ab)] + (b²– 3b²)
= (4 – 6) a² + [-3ab + (+ 5ab)] + (1 – 3) b²
= [4 + (- 6)] a² + (-3 + 5) ab + [1+ (-3)]b²
= -2a² + 2ab – 2b²

Question 9.
The length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
Solution:
Length of the log = 3a + 4b – 2

Length of the piece removed = 2a – b
Remaining length of the log = (3a + 4b – 2) – (2a – b)
= (3a – 2a) + [4b – (-b)] – 2
= (3 – 2)a + (4 + 1)b – 2
= a + 5b – 2

Question 10.
A tin had ‘x’ litre oil. Another tin had (3x² + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x² + 6) litres of oil from the second tin. How much oil was left In the second tin?
Solution:
Quantity of oil in the second tin = 3x² + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x² + 6x – 5) + (x + 7) litres
= 3x² + (6x + x) + (-5 + 7)
= 3x² + (6 + 1)x + 2
= 3x² + 7x + 2 litres
Quantity of oil sold = x + 6 litres
∴ Quantity of oil left in the second tin = (3x² + 7x + 2) – (x² + 6)(3x² – x² ) + 7x + (2 – 6)
= (3 – 1)x² + 7x + (-4) = 2x² + 7x – 4
Quantity of oil left = 2x² + 7x – 4 litres

Try this Page No. 70

Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Solution:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y² + 5y^-1 – 3 is a an algebraic expression. But not a polynomial.

Try this Page No. 71

Question 2.
-(5y² + 2y – 6) Is this correct? If not, correct the mistake.
Solution:
Taking -(5y² + 2y – 6) = 5y² + [(-)(+) 2y] + [(-) × (-)6]
= -5y² – 2y + 6
≠ -5y² – 2y + 6
∴ Correct answer is -5y² + 2y – 6 = -(5y² + 2y + 6)

Try this Page No 71

(i) 3ab², -2a²b³
(ii) 4xy, 5y²x, (-x²)
(iii) 2m, -5n, -3p
Solution:
(i) (3ab²) × (-2a²b²) = (+) × (-) × (3 × 2) × (a × a²) × (b² × b³) = -6a³ b⁵

(ii) (4xy) × (5y²x) × (-x²)
= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x²) × (y × y²)
= -20x⁴y³

(iii) (2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30mnp = 30 mnp

Try this Page No. 71

Question 1.
Why 3 + (4x – 7y) ≠ 12x – 21y?
Solution:
Addition and multiplication are different
3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.

Try this Page No. 72

Question 1.
Which is corrcet? (3a)² is equal to
(i) 3a²
(ii) 3²a
(iii) 6a²
(iv) 9a²
Solution:
(3a) =3²a² = 9a²
(iv) 9a² is the correct answer

Try These Page No.72

Question 1.
Multiply
(i) (5x² + 7x – 3) by 4x²
Solution:
(5x² + 7x – 3) × 4x²
= 4x²(5x² + 7x – 3) Multiplication is commutative
= 4x² (5x² + 4x² (7x) + 4x² (-3)
= (4 × 5)(x² × x²) + (4 × 7)(x² × x) + (4 × -3)(x²)
= 20x⁴ + 28x³ – 12x²

(ii) (10x – 7y + 5z) by 6xyz
Solution:
(10x – 7y + 5z) by 6xyz
(10x – 7y + 5z) × 6xyz = 6xyz (10x – 7y + 5z) [∵ Multiplication is commutative]
= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × -7) + (x × y × y × z) + (6 × 5)(x × y × z × z)
= 60x²yz + (-42xy²z) + 30xyz²
= 60x²yz – 42x²z + 30xyz²

(iii) (ab + 3bc – 5ca) by – 3abc
Solution:
(ab + 3bc – 5ca) × (- 3abc) = (-3abc) (ab + 3bc – 5ca)
[∵ Multiplication is commutativel
= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
= (-3)(a × a × b × b × c) + (- 3 × 3) + (a × b × b × c × c)
= -3a²b²c – 9ab²c² – 30a²bc²

Try these Page No. 74

Question 1.
Multiply
(i) (a – 5) and (a + 4)
Solution:
(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
= (a × a) + (a × 4) + (-5 × a) + (-5 × 4)
= a² + 4a – 5a – 20 = a² – a – 20

(ii) (a + b) and (a – b)
Solution:
(a + b) (a – b) = a(a – b) + b (a – b)
= (a × a) + (a × -b)+(b × a) + b(-b)
= a² – ab + ab – b² = a² – b²

(iii) (m⁴ + n⁴) and (m – n)
Solution:
(m⁴ + n⁴)(m – n) = m⁴(m – n) + n⁴(m – n)
= (m⁴ × m) + (m⁴ × (-n)) + (n⁴ × m (n⁴ × (-n))
= m⁵ – m⁴n + mn⁴ – n⁵

(iv) (2x + 3)(x – 4)
Solution:
(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
= (2x² × x) – (2x × 4) + (3 × x) – (3 × 4)
= 2x² – 8x + 3x – 12 = 2x² – 5x – 12

(v) (x – 5)(3x + 7)
Solution:
(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
= (x × 3x) + (x × 7) + (-5 × 3x) + (-5 × 7)
= 3x² + 7x – 15x – 35
= 3x² – 8x – 35

(vi) (x – 2)(6x – 3)
Solution:
(x – 2)(6x – 3) × (6x – 3) – 2(6x – 3)
= (x × 6x)+(x × (-3) × (2 × 6x) – (2 × 3)
= 6x² – 3x – 12x + 6
= 6x² – 15x + 6

Try this Page No. 74

Question 2.
3x² (x⁴ – 7x³ + 2), what is the highest power in the expression.
Solution:
3x²(x⁴ – 7x³ + 2) = (3x²) (x⁴) + 3x² (-7x³)+ (3x²)²
= 3x⁶ – 21x⁵ + 6x²
Highest power is 6 in x⁶.

Exercise 3.2

Try this Page No. 77

Question 1.
Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\)
(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
(iii) When a monomial is divided by itself, we will get I?
Solution:

Try this Page No. 77

Question 1.
Divide

Solution:

Try this Page No. 78

Question 1.
Are the following divisions correct ?

Solution:

Try this Page No. 78

Question 1.

Solution:

Exercise 3.3

Try these Page No. 81

Question 1.
1. (p + 2)² = …….
2. (3 – a)² = …….
3. (6² – x²) = ………
4. (a + b)² – (a – b)² = …….
= a² + 2ab + b² – a² – 2ab – b²
= (1 – 1)a² + (2 + 2)ab + (+1 – 1 )b² = 4ab
5. (a + b)² = (a + b) × (a + b)
6. (m + n)( m – n) = m² – n²
7. (m + 7)² = m² + 14m + 49
8. (k² – 36) ≡ k² – 6² = (k + 6)(k – 6)
9. m² – 6m + 9 = (m – 3)²
10. (m – 10)(m + 5) = m² + (-10 + 5)m + (-10)(5) = m² – 5m – 50
Solution:

Try these page No. 83

Question 1.
Expand using appropriate identities
Question 1.
(3p + 2q)²
Solution:
(3p + 2q)²
Comparing (3p + 2q)² with (a + b)², we get a = 3p and b = 2q.
(a + b)² = a² + 2ab + b²
(3p + 2q)² = (3p)²+ 2(3p) (2q) + (2q)²
= 9p² + 12pq + 4q²

Question 2.
(105)²
Solution:
(105)² = (100 + 5)²
Comparing (100 + 5)² with (a + b)², we get a = 100 and b = 5.
(a + b)² = a² + 2ab + b²
(100 + 5)² = (100)² + 2(100)(5) + 5² = 1oooo + 1000 + 25
105² = 11,025

Question 3.
( 2x – 5d)²
Solution:
(2x – 5d)²
Comparing with (a – b)², we get a = 2x b = 5d.
(a – b)² = a² – 2ab + b²
(2x – 5d)² = (2x)² – 2(2x)(5d) + (5d)²
= 2x² – 20 xd + 5²d² = 4x² – 20 xd + 25d²

Question 4.
(98)²
Solution:
(98)² = (100 – 2)²
Comparing (100 – 2)² with (a – b)² we get
a = 100, b = 2
(a – b)² = a² – 2ab + b²
(100 – 2)² = 100² – 2(100)(2) + 2²
= 10000 – 400 + 4 = 9600 + 4 = 9604

Question 5.
(y – 5)(y + 5)
Solution:
(y – 5)(y + 5)
Comparing (y – 5) (y + 5) with (a – b) (a + b) we get
a = y; b = 5
(a – b)(a + b) = a² – b²
(y – 5)(y + 5) = y² – 5² = y² – 25

Question 6.
(3x)² – 5²
Solution:
(3x)² – 5²
Comparing (3x)² – 5² with a² – b² we have
a = 3x; b = 5
(a² – b²) = (a + b)(a – b)
(3x)² – 5² = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
= 9x² – 15x + 15x – 25 = 9x² – 25

Question 7.
(2m + n)(2m +p)
Solution:
(2m + n) (2m + p)
Comparing (2m + n) (2m + p) with (x + a) (x + b) we have
x = 2n; a = n ;b = p
(x – a)(x + b) = x² + (a + b)x + ab
(2m +n) (2m +p) = (2m²) + (n +p)(2m) + (n) (p)
= 2²m² + n(2m) + p(2m) + np
= 4m² + 2mn + 2mp + np

Question 8.
203 × 197
Solution:
203 × 197 = (200 + 3)(200 – 3)
Comparing (a + b) (a – b) we have
a = 200, b = 3
(a + b)(a – b) = a² – b²
(200 + 3)(200 – 3) = 200² – 3²
203 × 197 = 40000 – 9
203 × 197 = 39991

Question 9.
Find the area of the square whose side is (x – 2)
Solution:
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
= x – 2x – 2x + 4x² – 4x + 4

Question 10.
Find the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
Solution:
Length of the rectangle = y+ 4
breadth of the rectangle = y – 3
Area of the rectangle = length × breadth
= (y + 4)(y – 3) = y² + (4 +(-3))y + (4)(-3)
= y² + y – 12

Try these Page No. 88

Question 1.
Expand :
Question 1.
(x + 4)³
Solution:
Comparing (x + 4)³ with (a + b)³, we have a = x and b = 4.
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 4)³ = x³ + 3x²(4) + 3(x)(4)² + 4³
= x³ + 12x² + 48x + 64

Question 2.
( y – 2)²
Solution:
Comparing (y – 2) with (a – b)³ we have a = y b = z
(a – b)³ = a³ – 3a²b + 3ab² – b³
(y – 2)² = y³ – 3y(2) + 3y(2)² + 2³
= y³ – 6y² + 12y + 8

Question 3.
(x + 1)(x + 3)(x + 5)
Solution:
Comparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
a = 1
b = 3
and c = 5

Exercise 3.4

Try These Page No.92

Question 1.
Factorize the following:
Question 1.
3y + 6
Solution:
3y + 6
3y + 6 = 3 × y + 2 × 3
Taking out the common factor 3 from each term we get 3 (y + 2)
∴ 3y + 6 = 3(y + 2)

Question 2.
10x² + 15y
Solution:
10x² + 15y²
10x² + 15y² = (2 × 5 × x × x) + (3 × 5 × y × y)
Taking out the common factor 5 we have
10x² + 15y² = 5(2x² + 3y²)

Question 3.
7m(m – 5) + 1(5 – m)
Solution:
7m(m – 5) + 1(5 – m)
7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
= 7m(m – 5) – 1 (m – 5)
Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

Question 4.
64 – x²
Solution:
64 – x²
64 – x² = 8² – x²
This is of the form a² – b²
Comparing with a² – b² we have a = 8, b = x
a² – b² = (a + b)(a – b)
64 – x² = (8 + x)(8 – x)