Category: Class 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th English Guide Pdf Poem 3 Making Life Worth While Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th English Solutions Poem 3 Making Life Worth While

8th English Guide Making Life Worth While Text Book Back Questions and Answers

Warm Up

Observe the pictures and write the moral values. Share your experience. (Text Book Page No. 79)

Answer:
1. Taking care of the animals.
2. Helping the weaker person.
3. Supporting the differently abled person.
4. Helping the old people.

8th English Guide Making Life Worth While Textual Exercise Questions and Answers

1. Comprehension questions. (Text Book Page. 81)

1. What should we learn from every soul?
Answer:
We should learn the good from every soul.

2. What qualities will help us to brave the thickening ills of life?
Answer:
We must pass on good thoughts, kindness, aspiration, courage, and faith to brave the thickening ills of life.

3. Why should we make this life worthwhile?
Answer:
We should make this life worthwhile to have a glimpse of the brighter skies.

4. What does the poet assure us if we make our life worthwhile?
Answer:
The poet assures the inheritance of heaven for the people who live a purposeful life in this world.

2. Fill in the blanks:

1. We should have a ……………. in life.
Answer:
aspiration

2. A ……………. is needed for the darkening sky.
Answer:
a bit of courage

3. One must have a ……………. of brighter skies to make life worthwhile.
Answer:
glimpse

3. Figure of speech.

Pick out any two lines of repetition from the poem. (Text Book Page No. 81)

On page 81, we have four lines of a poem written by Robert Frost. He repeats the line

And miles to go before. I sleep,
and miles to go before I sleep,

He uses this literary device called repetition to make the idea clearer and more memorable. It is used to emphasize a feeling or idea. Repetition creates rhythm and brings attention to the idea focussed.

8th English Guide Making Life Worth While Additional Appreciation Questions and Answers

1. One bit of courage For the darkening sky;
Answer:
What is needed for the darkening sky?
One bit of courage is needed for the darkening sky.

2. One gleam of faith
Answer:
To brave the thickening ills of life;
What is needed to brave the thickening ills of life?
One gleam of faith is needed to brave the thickening ills of life.

3. One glimpse of brighter skies – To make this life worthwhile
Answer:
What is needed to make the life worthwhile?
One glimpse of a bright sky is needed to make this life worthwhile.

4. One aspiration yet unfelt One bit of courage
Answer:
What is unfelt yet?
One aspiration is yet unfelt.

Poetic Devices

Repetition:
The word ‘one’ is found to be repeated throughout the poem.

Rhyming words:
contact – thought

Alliteration:
And heaven a surer heritage.
The consonant ‘h’ is repeated twice in the above line and henceforth it is an alliterated line of this poem.

Basic idea of the poem:
Love your neighbor as you love yourself.

Making Life Worth While Summary in English

Everyone who comes in contact with, us must be influenced by us. They must get something good from us. That kindly act will make our life purposeful. Even little grace, kindly thought, bit of courage, one gleam of faith, one glimpse of brightness can make our life to fight with the ills of life.

Making Life Worth While Summary in Tamil

நம்மை அணுகி வரும் எவரும் நம்மிடமிருந்து ஏதோ ஒரு நன்மையைப் பெற்றுக் கொள்ள வேண்டும். நம்மிடமிருந்து அவர்கள் பெறும் நன்மையே நம் வாழ்வைக் குறிக்கோள் உள்ளதாக இப்பூமியில் ஆக்குகிறது. சிறிது கருணை, அன்பான எண்ணம், சிறிது தைரியம், ஒரு கீற்று நம்பிக்கை, வெளிச்சத்தின் ஒரு சிறிய பார்வை, நாம் இந்த உலக வாழ்வின் தீமைகளை எதிர்த்துப் போராட உதவுகிறது.

Making Life Worth While About the Author in English

George Eliot – Mary Ann Evans (1819 – 1880), known by her pen name George Eliot, was an English novelist, poet, journalist, translator, and one of the leading writers of the Victorian era. She wrote seven novels.

Making Life Worth While About the Author in Tamil

ஜார்ஜ் எலியட் – மேரி ஆன் ஈவன்ஸ் (1819 – 1880), தன் புனைப்பெயரான ஜார்ஜ் எலியட் என்பதன் மூலம் அனைவராலும் அறியப்பட்டார். அவர் ஓர் ஆங்கில நாவலாசிரியர், கவிஞர், பத்திரிக்கை எழுத்தாளர், மொழிபெயர்ப்பாளர் மற்றும் விக்டோரியன் சகாப்தத்தின் முன்னணி எழுத்தாளர்களுள் ஒருவராவார். இவர் ஏழு நாவல்களை எழுதியுள்ளார்.

Subject Matter Experts at SamacheerKalvi.Guide have created English Nadu State Board Samacheer Kalvi 8th English Book Answers Solutions Guide Pdf Free Download are part of Samacheer Kalvi 8th Books Solutions.

Let us look at these TN State Board New Syllabus Samacheer Kalvi 8th Std English Guide Pdf of Text Book Back Questions and Answers Term 1, 2, 3, Chapter Wise Important Questions, Study Material, Question Bank, Notes, and revise our understanding of the subject.

Samacheer Kalvi 8th English Book Solutions Guide Pdf Free Download

Englishnadu State Board Samacheer Kalvi 8th English Book Back Answers Solutions Guide Term 1, 2, 3.

Samacheer Kalvi 8th English Book Back Answers

Samacheer Kalvi 8th English Prose

• Chapter 1 The Nose-Jewel
• Chapter 2 Hobby – Turns A Successful Career
• Chapter 3 Sir Isaac Newton – The Ingenious Scientist
• Chapter 4 My Reminiscence
• Chapter 5 Being Safe
• Chapter 6 Friendship
• Chapter 7 Cyber Safety

Samacheer Kalvi 8th English Poem

• Chapter 1 Special Hero
• Chapter 2 My Hobby: Reading
• Chapter 3 Making Life Worth While
• Chapter 4 A Thing of Beauty
• Chapter 5 Fire Work Night
• Chapter 6 Lessons in Life
• Chapter 7 My Computer Needs A Break

Samacheer Kalvi 8th English Supplementary

• Chapter 1 The Woman on Platform 8
• Chapter 2 Jim Corbett, A Hunter Turned Naturalist
• Chapter 3 The Three Questions
• Chapter 4 Crossing the River
• Chapter 5 When Instinct Works
• Chapter 6 Homeless Man and His Friends: A True Story
• Chapter 7 The Mystery of the Cyber Friend

Samacheer Kalvi 8th English Play

• Chapter 1 Jack and the Beanstalk

We hope these Englishnadu State Board Samacheer Kalvi Class 8th English Book Solutions Answers Guide Pdf Free Download will help you get through your subjective questions in the exam.

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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.3

Question 1.
In the figure, given that ∠1 = ∠2 and ∠3 ≡ ∠4. Prove that ∆ MUG ≡ ∆TUB.

Answer:

Statements	Reasons
1. In △MUG and △TUG Mu = TU	∠3 = ∠4, opposite sides of equal angles
2. UG = UB	∠1 = ∠2 Side opposite to equal angles are equal
3. ∠GUM = ∠BUT	Vertically opposite angle
4. ∆MUG ≡  ∆TUG	SAS criteria By 1,2 and 3

Question 2.
From the figure, prove that ∆SUN ~ ∆RAY.

Answer:
Proof: from the ∆ SUN and ∆RAY
SU = 10
UN = 12
SN = 14
RA = 5
AY = 6
RY = 7

From (1), (2) and (3) we have

The sides are proportional
∴ ∆SUN ~ ∆RAY

Question 3.
The height of a tower is measured by a mirror on the ground at R by which the top of the tower’s reflection is seen. Find the height of the tower. If ∆PQR ~ ∆STR

Answer:
The image and its reflection make similar shapes
∴ ∆PQR ~ ∆STR

⇒ \(\frac{h}{8}=\frac{60}{10}\)
h = \(\frac{60}{10}\) × 8
= 48 feet
∴ Height of the tower = 48 feet.

Question 4.
Find the length of the support cable required to support the tower with the floor.

Answer:
From the figure, by Pythagoras theorem,
x² = 20² + 15²
= 400 + 225 = 625
x² = 25² ⇒ x = 25ft.
∴ The length of the support cable required to support the tower with the floor is 25ft.

Question 5.
Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason.
Answer:

Take the sides of a right angled triangle ∆ABC as
a = 7 inches
b = 25 inches
c = ?
By Pythagoras theorem,
b² = a² + c²
25² = 7² + c²
⇒ c² = 25² – 7² = 625 – 49 = 576
∴ c² = 24²
⇒ c = 24 inches
∴ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches.
∴ The TV will not fit into the cabinet.

Challenging Problems

Question 6.
In the figure, ∠TMA ≡∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that Δ PIN ~ Δ ATM.

Answer:
proof:

Question 7.
In the figure, if ∠FEG ≡ ∠1 then, prove that DG² = DE.DF.

Answer:
Proof:

Question 8.
The diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
Answer:

Here AO = CO = 8cm
BO = DO = 6cm
(∴the diagonals of rhombus bisect each other at right angles)
∴ In ∆ AOB, AB² = AO² + OB²
= 8² + 6² = 64 + 36
= 100 = 10²
∴ AB = 10
Since it is a rhombus, all the four sides are equal.
AB = BC = CD = DA
∴ Its Perimeter = 10 + 10 + 10 + 10 = 40 cm

Question 9.
In the figure, find AR.

Answer:
∆ AFI, ∆ FRI are right triangles.
By Pythagoras theorem,
AF² = AI² – FI²
= 25² – 15²
= 625 – 225 = 400 = 20²
∴ AF = 20ft.
FR² = RI² – FI²
= 17² – 15² = 289 – 225 = 64 = 8²
FR = 8ft.
∴ AR = AF + FR
= 20 + 8 = 28 ft.

Question 10.
In ∆DEF, DN, EO, FM are medians and point P is the centroid. Find the following.
(i) IF DE = 44, then DM = ?
(ii) IFPD=12, then PN= ?
(iii) IfDO = 8, then PD = ?
(iv) IF 0E = 36 then EP = ?

Answer:
Given DN, EO, FM are medians.
∴ FN = EN
DO = FO
EM = DM

(i) If DE = 44,then
DM = \(\frac{44}{2}\) = 22
DM = 22

(ii) If PD = 12,PN = ?
\(\frac{P D}{P N}=\frac{2}{1}\)
\(\frac{12}{\mathrm{PN}}=\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6.
PN = 6

(iii) If DO = 8, then
FD = DO + OF
= 8 + 8
FD = 16

(iv) If OE = 36

PE = 24

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.2

Question 1.
Fill in the blanks:
(i) If in a ∆ PQR, PR² = PQ² + QR², then the right angle of ∆ PQR is at the vertex _______ .
Answer:
Q
Hint:

(ii) If ‘l‘ and ‘m’ are the legs and ‘n’ is the hypotenuse of a right angled triangle then, l² = _______ .
Answer:
n² – m²
Hint:

(iii) If the sides of a triangle are in the ratio 5:12:13 then, it is _______ .
Answer:
a right angled triangle
Hint:
13² = 169
5² = 25
12² = 144
169 = 25 + 144
∴ 13² = 5² + 12²

(iv) The medians of a triangle cross each other at _______ .
Answer:
Centroid

(v) The centroid of a triangle divides each medians in the ratio _______ .
Answer:
2 : 1

Question 2.
Say True or False.
(i) 8, 15, 17 is a Pythagorean triplet.
Answer:
True
Hint:
17² = 289
15² = 225
8² = 64
64 + 225 = 289 ⇒ 17² = 15² + 8²

(ii) In a right angled triangle, the hypotenuse is the greatest side.
Answer:
True
Hint:

(iii) In any triangle the centroid and the incentre are located inside the triangle.
Answer:
True

(iv) The centroid, orthocentre, and incentre of a triangle are collinear.
Answer:
True

(v) The incentre is equidistant from all the vertices of a triangle.
Answer:
False

Question 3.
Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(i) 8, 15, 17
Answer:
Take a = 8 b = 15 and c = 17
Now a² + b² = 8² + 15² = 64 + 225 = 289
172 = 289 = c²
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
∴ Ans: yes

(ii) 12, 13, 15
Answer:
(ii) 12, 13. 15
Take a = 12,b = 13 and c = 15
Now a² + b² = 12² + 13² = 144 + 169 = 313
15² = 225 ≠ 313
By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
∴ Ans: No.

(iii) 30, 40, 50
Answer:
Take a = 30, b = 40 and c = 50
Now a² + b² = 30² + 40² = 900 + 1600 = 2500
C² = 50² = 2500
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right
angled triangle.
∴ Ans: yes

(iv) 9, 40, 41
Answer:
Take a = 9, b = 40 and c = 41
Now a² + b² = 9² + 40² = 81 + 1600 = 1681
c² = 41² = 1681
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right
angled triangle.
∴ Ans: yes

(v) 24, 45, 51
Answer:
Take a = 24,b = 45 and c = 51
Now a² + b² = 24² + 45² = 576 + 2025 = 2601
c² = 51² = 2601
∴ a² + b² = c²
By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
∴ Ans: yes

Question 4.
Find the unknown side in the following triangles.
(i)

Answer:
From ∆ ABC, by Pythagoras theorem
BC² = AB² + AC²
Take AB² + AC² = 9² + 40² = 81 + 1600 = 1681
BC² = AB² + AC² = 1681 = 41²
BC² = 41² ⇒ BC = 41
∴ x = 41

(ii)

Answer:
From ∆ PQR, by Pythagoras theorem.
PR² = PQ² + QR²
34² = y² + 30²
⇒ y² = 34² – 30²
= 1156 – 900
= 256 = 16²
y² = 16² ⇒ y = 16

(iii)

Answer:
From ∆ XYZ, by Pythagoras theorem,
= YZ² = XY² + XZ²
⇒ XY² = YZ² – XZ²
Z² = 39² – 36²
= 1521 – 1296
= 225 = 15²
z² = 15²
⇒ z = 15

Question 5.
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.
Answer:

In an isosceles triangle the altitude dives its base into two equal parts.
Now in the figure, ∆ABC is an isosceles triangle with AD as its height
In the figure, AD is the altitude and ∆ABD is a right triangle.
By Pythagoras theorem,
AB² = AD² + BD²
⇒ AD² = AB² – BD²
= 13² – 12² = 169 – 144 = 25
AD² = 25 = 5²
Height: AD = 5cm

Question 6.
Find the distance between the helicopter and the ship.

Answer:
From the figure AS is the distance between the helicopter and the ship.
∆ APS is a right angled triangle, by Pythagoras theorem,
AS² = AP² + PS²
= 80² + 150² = 6400 + 22500 = 28900 = 170²
∴ The distance between the helicopter and the ship is 170 m

Question 7.
In triangle ABC, line I, is a perpendicular bisector of BC.
If BC = 12cm, SM = 8cm, find CS.

Answer:
Given l₁, is the perpendicular bisector of BC.
∴ ∠SMC = 90°and BM = MC
BC = 12cm
⇒ BM + MC = 12cm
MC + MC = 12cm
2MC = 12
MC = \(\frac{12}{2}\)
MC = 6cm
Given SM = 8 cm
By Pythagoras theorem SC² = SM² + MC²
SC² = 8² + 6²
SC² = 64 + 36
CS² = 100
CS² = 10²
CS = 10 cm

Question 8.
Identify the centroid of ∆PQR.

Answer:
In ∆PQR, PT = TR ⇒ QT is a median from vertex Q.
QS = SR ⇒ PS is a median from vertex P.
QT and PS meet at W and therefore W is the centroid of ∆PQR.

Question 9.
Name the orthocentre of ∆PQR.

Answer:
This is a right triangle
∴ orthocentre = P [∵ In right triangle orthocentre is the vertex containing 90°]

Question 10.
In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Answer:
Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of XYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2:1
\(\frac{\mathrm{XG}}{\mathrm{GA}}=\frac{2}{1}\)
\(\frac{\mathrm{XG}}{3}=\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA = 6 + 3 ⇒ XA = 9cm

Question 11.
If I is the incentre of ∆XYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Answer:
Since I is the incentre of AXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°

∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Objective Type Questions

Question 12.
If ∆GUT is isosceles and right angled, then ∠TUG is ______ .

(A) 30°
(B) 40°
(C) 45°
(D) 55°
Answer:
(C) 45°
Hint:
∠U ∠T = 45° (∵ GUT is an isosceles given)
∴ ∠TUG = 45°

Question 13.
The hypotenuse of a right angled triangle of sides 12cm and 16cm is ______ .
(A) 28 cm
(B) 20 cm
(C) 24 cm
(D) 21 cm
Answer:
(B) 20 cm
Hint:
Side take a = 12 cm
b = 16cm
The hypotenuse c² = a² + b²
= 12² + 16²
= 144 + 256
c² = 400 ⇒ c = 20 cm

Question 14.
The area of a rectangle of length 21cm and diagonal 29 cm is ______ .
(A) 609 cm²
(B) 580 cm²
(C) 420 cm²
(D) 210 cm²
Answer:
(C) 420 cm²

length = 21 cm
diagonal = 29 cm
By the converse of Pythagoras theorem,
AB² + BC² = AC²
21² + x² = 29²
x² = 841 – 441 400 = 20²
x = 20 cm
Now area of the rectangle = length × breadth.
ie AB × BC = 21 cm × 20 cm = 420 cm²

Question 15.
The sides of a right angled triangle are in the ratio 5:12:13 and its perimeter is 120 units then, the sides are .
(A) 25, 36, 59
(B) 10, 24, 26
(C) 36, 39, 45
(D) 20, 48, 52
Answer:
(D) 20,48,52
Hint:
The sides of a right angled triangle are in the ratio 5 : 12 : 13
Take the three sides as 5a, 12a, 13a
Its perimeter is 5a + 12a + 13a = 30a
It is given that 30a = 120 units
a = 4 units
∴ the sides 5a = 5 × 4 = 20 units
12a = 12 × 4 = 48 units
13a = 13 × 4 = 52 units

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.5

Question 1.
A fruit vendor bought some mangoes of which 10% were rotten. He sold 33\(\frac{1}{3}\) % of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.

Answer:
Let the number of mangoes bought by fruit seller initially be x.
Given that 10% or mangoes were rotten
∴ Number of rotten mangoes = \(\frac{10}{100}\) × x
Number of good mangoes = x – no. of rotten mangoes
= \(x-\frac{10}{100} x=\frac{100 x-10 x}{100}=\frac{90}{100} x\) …….. (1)
Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%
∴ Mangoes sold = \(\frac{100}{3} \times \frac{90}{100} \times \times \frac{1}{100}=\frac{30}{100} x\) ……. (2)
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)

∴ Intially he had 400 mangoes

Question 2.
A student gets 31 % marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.
Answer:
Let the maximum marks in the exam be ‘x’
Pass percentage is given as 35%
∴ Pass mark = \(\frac{35}{100} \times x=\frac{35}{100} x\)
Student gets 31% marks = \(\frac{31}{100} \times x=\frac{31}{100} x\)
But student fails by 12 marks → meaning his mark is 12 less than pass mark.

Maimum mark is 300

Question 3.
Sultana bought the following things from a general store. Calculate the total bill amount paid by her.
(i) Medicines costing ₹ 800 with GST at 5%

Answer:
Formula for bill amount is cost \(\left(1+\frac{\mathrm{GST} \%}{100}\right)\)
Medicine: bill amount is \(800\left(1+\frac{5}{100}\right)=800 \times \frac{105}{100}=840\)

(ii) Cosmetics costing ₹ 650 with GST at 12%

Answer:
Formula for bill amount is cost \(\left(1+\frac{\mathrm{GST} \%}{100}\right)\)
Cosmetics: Bill amount is \(550\left(1+\frac{12}{100}\right)=650 \times \frac{112}{100}=728\)

(iii) Cereals costing ₹ 900 with GST at 0%

Answer:
Formula for bill amount is cost \(\left(1+\frac{\mathrm{GST} \%}{100}\right)\)
Cereals: Bill amount is \(900\left(1+\frac{0}{100}\right)\) = 900

(iv) Sunglass costing ₹ 1750 with GST at 18%

Answer:
Formula for bill amount is cost \(\left(1+\frac{\mathrm{GST} \%}{100}\right)\)
Sunglass: Bill amount is \(1750\left(1+\frac{18}{100}\right)=1750 \times \frac{118}{100}=2065\)

(v) Air Conditioner costing ₹ 28500 with GST at 28 %

Answer:
Formula for bill amount is cost \(\left(1+\frac{\mathrm{GST} \%}{100}\right)\)
Air Conditioner: Bill amount is \(28500\left(1+\frac{28}{100}\right)=28500 \times \frac{128}{100}=36480\)
∴ Total Bill amount = 840 + 728 + 900 + 2065 + 36480
= ₹ 41,013 (total bill amount)

Question 4.
P’s income is 25 % more than that of Q. By what percentage is Q’s income less than P’s?
Answer:
Let Q’s income be 100.
P’s income is 25% more than that of Q.
∴ P’s income = 100 + \(\frac{25}{100}\) × 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}}\) × 100 = \(\frac{125-100}{125}\) × 100 = \(\frac{25}{125}\) × 100 = 20%

Question 5.
Vaidegi sold two sarees for ₹ 2200 each. On one she gains 10% and on the other she loses 12%. Find her total gain or loss percentage in the sale of the sarees.

Answer:
Saree 1 :
The selling price is ₹ 2200, let cost price be CP₁, gain is 10%
Cost price? Using the formula

Saree 2 :
The selling price is 2200, let cost price be CP₂, loss is given as 12%. We need to find CP₂
using the formula as before,

∴ Cost price of both together is CP₁ + CP₂
= 2000 + 2500 = 4500 …….. (1)
Selling price of both together is 2 × 2200 = 4400 …… (2)
Since net selling price is less than net cost price, there is a loss.

Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
100 100 20 2
∴ loss % = \(\frac{100}{4500}\) × 100 = \(\frac{100}{45}=\frac{20}{9}=2 \frac{2}{9}\)%
= 2\(\frac{2}{9}\)%loss

Question 6.
If 32 men working 12 hours a day can do a work in 15 days, then how many men working 10 hours a day can do double that work in 24 days?
Answer:

Days (D)	Hours (H)	Men (P)
15	12	32
24	10	x

Let
P₁ = 32
P₂ = x

H₁ = 12
H₂ = 10

D₁ = 15
D₂ = 24

W₁ = 1
W₂ = 1


x = 24 persons
To complete the same work 24 men needed.
To complete double the work 24 × 2 = 48 men are required.

Question 7.
Amutha can weave a saree in 18 days. Anjali is twice as good a weaver as Amutha. If both of them weave together, then in how many days can they complete weaving the saree?
Answer:
Amutha can weave a saree in 18 days.
Anjali is twice as good as Amutha.
ie. 1f Amutha weave for 2 days, Anjali do the same work in 1 day.
If Anjali weave the saree she will take
\(\frac{18}{2}\) = 9 days.
Hence time taken by them together ab
= \(\frac{a b}{a+b}\) days
= \(\frac{18 \times 9}{18+9}=\frac{18 \times 9}{27}\) = 6 days

In 6 clays they complete weaving the saree.

Question 8.
P and Q can do a piece of work in 12 days and 15 days respectively. P started the work alone and then after 3 days, Q joined him till the work was completed. How long did the work last?
Answer:
P can do a piece of work in 12 days.
∴ P’s 1 day work = \(\frac{1}{12}\)
P’s 3 day’s work = 3 × \(\frac{1}{12}=\frac{3}{12}\)
Q can do a piece of work in 15 days.
∴ Q’s 1 day work = \(\frac{1}{15}\)
Remaining work after 3 days = 1 – \(\frac{3}{12}=\frac{9}{12}\)

∴ Number of days required to finish the remaining work 9

Remaining work lasts for 5 days
Total work lasts for 3 + 5 = 8 days.

Question 9.
If the numerator of a fraction is increased by 50% and the denominator is decreased by 20%, then it becomes \(\frac{3}{5}\). Find the original fraction.
Answer:
Original fraction = \(\frac{x}{y}\)
numerator increased by 50%
∴ Numerator = \(\frac{150}{100} x\)
Denominator decreased by 20%
∴ Denominator = \(\frac{80}{100} y\)
Hence

Question 10.
Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20%. Find the cost price of the laptop.

Answer:
Let the cost price of the laptop be ‘x
Gain = 12%

If the selling price was 1200 more
i.e \(\frac{112}{100} x\) + 1200, the gain is 2o%

Cost price of the laptop is ₹ 15,000/-

Question 11.
A shopkeeper gives two successive discounts on an article whose marked price is ₹ 180 and selling price is ₹ 108. Find the first discount percentage if the second discount is 25%.
Answer:
Marked price is given as ₹ 180
Let 1 discount be d₁ % = ? (to find)
2nd discount be d₂ % = 25%
Selling price is 108 (given)
Price after 1^st discount = 180 \(\left(1-\frac{d_{1}}{100}\right)\) = P …… (1)
Price after 2^nd discount = P₁ \(\left(1-\frac{d_{2}}{100}\right)\) = 108
Substituting for P₁ from (1), we get

1st discount = 20%

Question 12.
Find the rate of compound interest at which a principal becomes 1.69 times itself in 2 years.
Answer:
Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ?(required)
Applying the formula,

∴ rate of compound interest is 30%

Question 13.
A small – scale company undertakes an agreement to make 540 motor pumps in 150 days and employs 40 men for the work. After 75 days, the company could make only 180 motor pumps. How many more men should the company employ so that the work is completed on time as per the agreement?
Answer:
Let the number of men to be appointed more be x.

To produce more pumps more men required
∴ It is direct variation.
∴ The multiplying factor is \(\frac{360}{180}\)
More days means less employees needed.
∴ It is Indirect proportion. 75
∴ The multiplying factor is \(\frac{75}{75}\)
Now 40 + x = 40 × \(\frac{360}{180} \times \frac{75}{75}\)
40 + x = 80
x = 80 – 40
x = 40
40 more man should be employed to complete the work on time as per the agreement.

Question 14.
P alone can do \(\frac{1}{2}\) of a work in 6 days and Q alone can do \(\frac{2}{3}\) of the same work in 4 days. in how many days will they finish \(\frac{3}{4}\) of the work, working together?
Answer:
\(\frac{1}{2}\) of the work is done by P in 6 days

\(\frac{2}{3}\) of work done byQin4days.

(P + Q) will finish the whole work in \(\frac{a b}{a+b}\) days= \(\frac{12 \times 6}{12+6}=\frac{12 \times 6}{18}\)= 4 days
(P + Q) will finish \(\frac{-3}{4}\) of the work in 4 × \(\frac{3}{4}\) = 3 days.

Question 15.
X alone can do a piece of work in 6 days and Y alone in 8 days. X and Y undertook the work for ₹ 48000. With the help of Z, they completed the work in 3 days. How much is Z’s share?
Answer:
X can do the work in 6 days.
X’s I day work = \(\frac{1}{6}\)
X’s share for 1 day = \(\frac{1}{6}\) × 48000 = ₹ 800
X’s share for 3 days = 3 × 800 = ₹ 2400
Y can complete the work in 8 days.
Y’s 1 day work = \(\frac{1}{8}\)
Y’s I day share = \(\frac{1}{8}\) × 4800 = ₹ 600
Y’s 3 days share = ₹ 600 × 3 = ₹ 1800
(X+Y)’s 3days share = ₹ 2400 + ₹ 1800 = ₹ 4200
Remaining money is Z’s share
∴ Z’s share = ₹ 4800 – ₹ 4200 = ₹ 600

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.4

Question 1.
Fill in the blanks
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job working together is __________days.
Answer:
2 days

(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in _________ days.
Answer:
5

(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in ________ days.
Answer:
8

(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in ___________days.
Answer:
25

(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for ₹ 200000. The amount that A will get is .
Answer:
₹ 1,20,000

Question 2.
210 men working 12 hours a day can finish ajob in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Answer:
Let the required number of men be x.

Hours	Day	Men
12	18	210
14	20	x

More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = \(210 \times \frac{12}{14} \times \frac{18}{20}\)

x = 162 men
162 men are required.

Question 3.
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Answer:
Let he required number of cement bags be x.

Days	Machines	Cement bags
12	36	7000
18	24	x

Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = \(7000 \times \frac{18}{12} \times \frac{24}{36}\)

x = 7000 cement bags
7000 cement bags can be made

Question 4.
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 more hours a day?
Answer:
Let the required number of days be x.

Soaps	Hours	Days
9600	15	6
14400	(15 + 3) = 18	x

To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = \(6 \times \frac{14400}{9600} \times \frac{15}{18}\)

x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Question 5.
If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Answer:
Let the number of lorries required more = x.

Container lorries	Goods (tonnes)	Days
6	135	5
6 + x	180	4

As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = \(6 \times \frac{180}{135} \times \frac{5}{4}\)

6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Question 6.
A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Answer:
Time taken by A to complete the work = 12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\) —— (1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\) —— (2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}+\frac{1}{3}=\frac{1+4}{12}=\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A+ B + C)’s 1 hour work – (A + C)’s 1 hr work
\(=\frac{5}{12}-\frac{1}{6}=\frac{5-2}{12}=\frac{3}{12}=\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

Question 7.
A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Answer:
(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\) —— (1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\) —— (2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\) —— (3)
Now (1) + (2) + (3) =
[(A + B)+ (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\)
(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}+\frac{4}{60}+\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)

LCM = 5 × 4 × 3 = 60
(A + B + C)’s 1 day work = \(\frac{12}{60 \times 2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s I day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(=\frac{1}{10}-\frac{1}{15}=\frac{3}{30}-\frac{2}{30}=\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work – (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
\(=\frac{1}{10}-\frac{1}{20}=\frac{6}{60}-\frac{3}{60}\)
\(=\frac{6-3}{60}=\frac{3}{60}=\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work (A + B + C)’s I day work – (A + B)’s I day work
\(=\frac{1}{10}-\frac{1}{12}=\frac{6}{60}-\frac{5}{60}=\frac{6-5}{60}=\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Question 8.
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 minutes more than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Answer:
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 118 minutes
∴ As 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A+B)’s 1 minutes work = \(\frac{1}{15}+\frac{1}{18}\)
\(\frac{12}{180}+\frac{10}{180}=\frac{22}{180}=\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chair

LCM = 3 × 5 × 6 = 180

∴ Time taken by (A + B) to fit a chair
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

Question 9.
A can do a work In 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work, if they work together.
Answer:
A completes the work in 45 days.
∴ A’s 1 day work = \(\frac{1}{45}\)

Remaining work = \(1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}\)
B finishes \(\frac{2}{3}\) rd work in 24 days

Let x days required

Question 10.
A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Answer:
If B does the work in 3 days, A will do it in I day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{a b}{a+b}\) days
= \(\frac{24 \times 8}{24+8}\) days
= \(\frac{24 \times 8}{32}\) days = 6 days

They together complete the work in 6 days.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.2

Question 1.
Fill in the blanks:
(i) Loss or gain percentage is always calculated on the ________ .
Answer:
Cost Price

(ii) A mobile phone is sold for ₹ 8400 at a gain of 20%. The cost price of the mobile phone is ________ .
Answer:
₹ 7000
Hint:
Let cost price of mobile be ₹ x
Given that selling price is ₹ 8400 and gain is 20%
As per formula,

(iii) An article is sold for ₹ 555 at a loss of 7\(\frac { 1 }{ 2 }\)%. The cost price of the article is ________ .
Answer:
₹ 600
Hint:
Given selling price is ₹ 555 & loss 7\(\frac { 1 }{ 2 }\)%
as per formula

(iv) A mixer grinder marked at ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ________ .
Answer:
8 %
Hint:
Marked price is ₹ 4500
Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted priòe
= 4500 – 4140 = 360

(v) The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ________ .
Answer:
Cost of price shirt = ₹ 575 (CP)
GST = 5%

Cost of price shirt = ₹ 325 (CP)
GST = 5%

∴ Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945

Question 2.
If selling an article for ₹ 820 causes 10% loss on the selling price, then find its cost price.
Answer:
Given that selling price (SP) = ₹ 820
Loss % = 10 %

Question 3.
If the profit earned on selling an article for ₹ 810 is the same as loss on selling it for ₹ 530, then find the cost price of the article.
Answer:
Case 1: Profit = Selling price (SP) – Cost price (CP)
Case 2: Loss = Cost price (CP) – Selling price (SP)
Given that profit of case 1 = loss of case 2
∴ P = 810 – CP
L = CP – 530
Since profit (P) = loss (L)
810 – CP = CP – 530
∴ 2CP = 810 + 530 = 1340 ⇒ C.P = \(\frac{1340}{2}\)
∴ CP = 670

Question 4.
If the selling price of 10 rulers is the same as the cost price of 15 rulers, then find the profit percentage.
Answer:
Let cost price of one ruler be x
Given that selling price (SP) of 10 rulers.
i.e., same as cost price (CP) of 15 rulers
∴ SP of 10 rulers = 15 × x = 15x
∴ SP of 1 ruler = \(\frac{15 x}{10}\) = 1.5x
∴ Gain = SP of 1 ruler – CP of 1 ruler = 1.5x – x = 0.5x

Question 5.
Some articles are bought at 2 for ₹ 15 and sold at 3 for ₹ 25. Find the gain percentage.
Answer:
Let cost price of one article be C.P
Given that 2 are bought for ₹ 15
∴ 2 × CP = 15 ⇒ CP = \(\frac{15}{2}\)
Let selling price of one article be SP
Given that 3 are sold for ₹ 25
∴ 3 × SP = 25 ⇒ SP = \(\frac{25}{3}\)
∴ Gain = SP – CP = \(\frac{25}{3}-\frac{15}{2}=\frac{50-45}{6}=\frac{5}{6}\)

= \(11 \frac{1}{9}\)

Question 6.
By selling a speaker for ₹ 768, a man loses 20%. In order to gain 20%, how much should he sell the speaker?

Answer:
Selling price (SP) of speaker = ₹ 768
Loss % = 20 %
as per formula

∴ CP = \(\frac{768 \times 100}{80}\) = 960
For gain of 20%, we should now calculate the selling price

= 96 × 12 = ₹ 1152

Question 7.
Find the unknowns x, y and z.

Answer:
(i) Book marked price = ₹ 225 discount 8%
∴ Selling price (x) = Marked price × \(\left(\frac{(100-d \%)}{100}\right)\)
= 225 × \(\frac{(100-8)}{100}\) = 225 × \(\frac{92}{100}\) = ₹ 207

(ii) LED TV selling price = 11970 discount = 5%, Marked price = y
∴ Selling price Marked price y × \(\left(\frac{(100-d \%)}{100}\right)\)
∴ 11970 = y × \(\frac{(100-5)}{100}\)
∴ y = \(\frac{11970 \times 100}{95}\) = 126 × 100 = ₹ 12,600

(iii) Digital clock marked price (MP) = ₹ 750, MP = ₹ 12.600
Selling price (SP) = ₹ 615, Discount = z

100 – z = 82
∴ z = 100 – 82, Discount = 18%

Question 8.
Find the total bill amount for the data given below:

Answer:
Formula for discounted price LW = Marked price (MP) × \(\frac{(100-d \%)}{100}\)
When d is the discount %

For bill amount, we should apply GST on the discounted value of the items.
Formula: Bill amount = Discounted price × \(\left(\frac{(100+\mathrm{GST} \%)}{100}\right)\)
∴ For (i) School bag.
Bill amount 475 × \(\left(\frac{(100+12)}{100}\right)\) = 475 × 1.12 = %‘532
∴ For (ii) Hair drier,
Bill amount = 1800 × \(\left(\frac{(100+28)}{100}\right)\) = 1800 × 1.28 =
∴ Total bill amount Bill amount of School bag + Stationary + Cosmetics + Hair drier
= 532 + 252 + 1357 + 2304
= ₹ 4.445

Question 9.
A branded Air-Conditioner (AC) has a marked price of ₹ 38000. There are 2 options given for the customer.
(i) Selling Price is the same ₹ 38000 but with attractive gifts worth ₹ 3000
(or)
(ii) Discount of 8% on the marked price but no free gifts. Which offer is better?

Answer:
Marked price of AC = ₹ 38,000
Option 1:
Selling price = ₹ 38000 & gifts worth ₹ 3000
∴ Net gain for customer = ₹ 3000 as there is no discount on AC

Option 2:
Discount of 8%, but no gift
∴ Discounted value = MP × \(\left(\frac{(100-d \%)}{100}\right)\)
38000 × \(\frac{(100-8)}{100}\) = 38000 × 0.92 = 34960
∴ Savings for customer = 38000 – 34960 = 3040
Therefore, the customer gets 3000 gift in option I where as he is able to save only ₹ 3040 in option 2. Therefore, option 2 is better.

Question 10.
If a mattress is marked for ₹ 7500 and is available at two successive discount of 10% and 20%, find the amount to be paid by the customer.
Answer:
Marked price of mattress = ₹ 7500
Discount d₁ = 10%
Discount d₂ = 20%

Objective Type Questions

Question 11.
A fruit vendor sells fruits for ₹ 200 gaining ₹ 40. His gain percentage is
(A) 20%
(B) 22%
(C) 25%
(D) 16
Answer:
(C) 25%
Hint:
Selling price ₹ 200
Gain = 40
∴ CP – Selling price – gain = 200 – 40 = 160

Question 12.
By selling a flower pot for Z528, a woman gains 20%. At what price should she sell it to gain 25%?
(A) ₹ 500
(B) ₹ 550
(C) ₹ 553
(D) ₹ 573
Answer:
(B) ₹ 550
Hint:
If selling price (sp) = ₹ 528
Gain % = 20 %
∴ CP = ?

Question 13.
A man buys an article for ₹ 150 and makes overhead expenses which are 12% of the cost price. At what price must he sell it to gain 5%?
(A) ₹ 180
(B) ₹ 168
(C) ₹ 176.40
(D) ₹ 88.20
Answer:
(C) ₹ 176.40
Hint:
Cost price of article = ₹ 150
Over head expenses = 12% of cost price
= \(\frac{12}{100}\) × 150 = ₹ 18
∴ Effective cost of article = 150 + 18 = ₹ 168
Now, to gain 5%, he has to sell at

Question 14.
What is the marked price of a hat which is bought for Z210 at 16% discount?
(A) ₹ 243
(B) ₹ 176
(C) ₹ 230
(D) ₹ 250
Answer:
(D) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:

Question 15.
The single discount in % which is equivalent to two successive discounts of 20% and 25% is
(A) 40%
(B) 45%
(C) 5%
(D) 22.5%
Answer:
(A) 40%
Hint:
Let marked price be MP, after discount 1 of 20%,

Comparing with formula, we get
∴ This is equivalent to a single discount of 40%

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra InText Questions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra InText Questions

Recap (Text Book Page No. 74 & 75)

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

(ii) m²n²c²
Answer:
1 term

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

(iv) 8x² – 4xy + 7xy²
Answer:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
(i) 2x² – 5xy + 6y² + 7x – 10y + 9
Answer:
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

(ii) \(\frac{x}{3}+\frac{2 y}{5}\) – xy + 7
Answer:
Numerical co efficient in \(\frac{x}{3}\) is \(\frac{1}{3}\)
Numerical co efficient in \(\frac{2 y}{5}\) is \(\frac{2}{5}\)
Numerical co efficient in – xy is – 1
Numerical co efficient in 7 is 7

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

Like Terms
The variables of the terms along with their respective exponents must be same
Examples: x², 4x²
a²b², – 5a²b²
2m, – 7m

Answer:

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

Question 5.
Simplify: (5x³y³ – 3x²y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
Answer:
(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 the perimeter of the triangle.
Answer:
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.
Answer:
6 mn – (-2mn) = 6mn + (+ 2mn)
= (6 + 2)mn
= 8mn

Question 8.
Subtract 6a² – 5ab + 3b² from 4a² – 3ab + b².
Answer:
(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 removed from it. What is the length of the remaining log?

Answer:
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?
Answer:
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) = 3x⁴ + (6 + 1)x + 2
= 3x² + 7x + 2litres
Quantity of oil sold = x² + 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

Think (Text Book Page No. 77)

Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Answer:
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 These (Text Book Page No. 78)

Find the product of
(i) 3ab², – 2a²b³
Answer:
(3ab²) × (- 2a²b³) = (+) × (-) × (3 × 2) × (a × a²) × (b² × b³)
= – 6a³b⁵

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

(iii) 2m, – 5n, – 3p
Answer:
(2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30 mnp
= 30 mnp

Think (Text Book Page No. 79)

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

Try These (Text Book Page No. 79)

Question 1.
MuItipIy
(i) (5x² + 7x – 3) by -4x²
Answer:

= – 20x² – 28x² + 12x²

(ii) (10x – 7y + 5z) by 6xyz
Answer:

= 6xyz (10x) + 6xyz (- 7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × – 7) + (x × y × y × z) + (6 × 5)(x × y × z × z)

(iii) (ab + 3bc – 5ca) by 3a²bc
Answer:
(ab + 3bc – 5ca) × (3a²bc ) = ab(3a²bc) + 3bc (3a²bc) – 5ca (3a²bc)
= 3a³b²c + 9a² b² c² – 15a³bc²

(iv) (4m² – 3m + 7) by – 5m³
Answer:
(4m² – 3m + 7) × (- 5m³) = 4m² (- 5m³) – (3m) (- 5m³) + 7(- 5m³)
= – 20m⁵ + 15m⁴ – 35m³

Try These (Text Book Page No. 81)

MuItipIy
(i) (a – 5) and (a + 4)
Answer:
(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)
Answer:
(a + b)(a – b) = a(a – b) + b(a – b)
= (a × a) + (a × -b) + (b × a) + b(-b)
= a² – ab + ab – b = a² – b²

(iii) (m⁴ + n⁴) and (m – n)
Answer:
(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)
Answer:
(2x + 3)(x + 4) = 2x(x + 4) + 3(x + 4)
= (2x² × x) + (2x × 4) + (3 × x) + (3 × 4)
= 2x² + 8x + 3x + 12
= 2x² + 11x + 12

(v) x – 5)(3x + 7)
Answer:
(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)
Answer:
(x – 2)(6x-3) = x(6x – 3) – 2(6x – 3)
= (x × 6x) + (x × (-3) – (2 × 6x) – (2 × 3)
= 6x² – 3x – 12x + 6
= 6x² – 15x + 6

Think (Text Book Page No. 81)

(i) In 3x²(x⁴ – 7x³ + 2) what is the highest power in the expression?
Answer:
3x² (x⁴ – 7x³ + 2) = (3x²) (x⁴) + 3x² (-7x³) + (3x²)2
= 3x⁶ – 21x⁵ + 6x²

(ii) Is -5y² + 2y – 6 = -(5y² + 2y – 6)? If not, correct the mistake.
Answer:
No, -5y² + 2y – 6 = -(5y² + 2y – 6)

Think (Text Book Page No. 83)

Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}\) = x^{8 – 3} = x⁵
Answer:
\(\frac{x^{3}}{x^{8}}\) = x^{8 – 3} = x⁵ (or) \(\frac{x^{3}}{x^{8}}=\frac{1}{x^{8-3}}=\frac{1}{x^{5}}\)
∴ The given answer is wrong

(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
Answer:
\(\frac{10 m^{4}}{10 m^{4}}=\frac{10}{10} m^{4-4}=1\) m⁰ = 1
[∵ m⁰ = 1]
∴ The given answer is not correct.

(iii) When a monomial is divided by itself, we will get 1. ?
Answer:
When a monomial is divided by itself we get 1.
\(\frac{x}{x}\) = x^1-1 = x⁰ = 1
∴ The given statement is correct.

Try These (Text Book Page No. 83)

Divide
(i) 12x³y² by x²y
Answer:
\(\) = 12x^{3 – 2}y^{2 – 1} = 12x¹y¹ = 12xy

(ii) -20a⁵b² by 2a³b⁷
Answer:

(iii) 28a⁴c² by 21ca³
Answer:

(iv) (3x²y)³ √6x²y³
Answer:

(v) 64m⁴ (n²)³ by 4mn
Answer:

(vi) (8x²y²)³ by (8x²y²)²
Answer:

= 8x^{6 – 4} y^{6 – 4}
= 8x²y²

(vii) 81p²q⁴ by \(\sqrt{81 p^{2} q^{4}}\)
Answer:

= 9p^{2 – 1} q^{4 – 2}
= 9pq²

(vii) (4x²y³)⁰ by \(\frac{\left(x^{3}\right)^{2}}{x^{6}}\)
Answer:

Think (Text Book Page No. 84)

Are the following divisions correct?
(i) \(\frac{4 y+3}{4}\) = y + 3
Answer:
\(\frac{4 y+3}{4}=\frac{4 y}{4}+\frac{3}{4}=y+\frac{3}{4}\) is the correct answer.
∴ The given statement is not correct

(ii) \(\frac{5 m^{2}+9}{9}\) = 5m²
Answer:
\(\frac{5 m^{2}+9}{9}=\frac{5 m^{2}}{9}+\frac{9}{9}=\frac{5}{9} m^{2}+1\) is the correct answer.
∴ The given statement is not correct

(iii) \(\frac{2 x^{2}+8}{4}\) = 2x² + 2. If not , correct it.
Answer:
\(\frac{2 x^{2}+8}{4}=\frac{2 x^{2}}{4}+\frac{8}{4}=\frac{1}{2} x^{2}+2\) is the correct answer.
∴ The given statement is not correct

Try These (Text Book Page No. 84)

(i) (16y⁵ – 8y²) ÷ 4y
Answer:

(ii) (p⁵q² + 24p³q – 128q³) ÷ 6q
Answer:

(iii) (4m²n + 9n²m + 3mn) ÷ 4mn
Answer:

Try These (Text Book Page No. 86)

Expand the following
Question 1.
(p + 2)² = ………………….
Answer:
(p + 2)² = p² + 2(p)(2) + 2²
= p² + 4p + 4

Question 2.
(3 – a)² = ………………….
Answer:
(3 – a)² = 3² – 2(3)(a) + a²
= 9 – 6a + a²

Question 3.
(6² – x²) = ………………….
Answer:
(6² – x²) = (6 + x)(6 – x)

Question 4.
(a + b)² – (a – b)² = ………………….
Answer:
(a + b)² – (a – b)² = a² + 2ab + b² – (a² – 2ab + b²)
= a² + 2ab + b² – a² + 2ab – b²
= (1 – 1)a² + (2 + 2)ab + (+1 – 1)b² = 4ab

Question 5.
(a + b)² = (a + b) × ………………….
Answer:
(a + b)² = (a + b) × (a + b)

Question 6.
(m + n)(…..) = m² – n²
Answer:
(m + n)(m – n) = m² – n²

Question 7.
(m + ……)² = m² + 14m + 49
Answer:
(m + 7)² = m² + 14m + 49

Question 8.
(k² – 49) = (k + …)(k – …)
Answer:
k² – 49 = k² – 7² = (k + 7)(k – 7)

Question 9.
m² – 6m + 9 = ………………….
Answer:
m² – 6m + 9 = (m – 3)²

Question 10.
(m – 10)(m + 5) = ………………….
Answer:
(m – 10)(m + 5) = m² + (-10 + 5)m + (-10)(5) = m² – 5m – 50

Think (Text Book Page No. 87)

Which is correct? (3a)² is equal to
(i) 3a²
(ii) 3²a
(iii) 6a²
(iv) 9a²
Answer:
(iv) 9a²
Hint:
(3a)² = 3² a² = 9a²

Try These (Text Book Page No. 88)

Expand using appropriate identities.
Question 1.
(3p + 2q)²
Answer:
(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
Answer:
(105)² = (100 + 5)²
Comparing (100 + 5)² with (a + b)², we get a = 1oo and b = 5.
(a + b)² = a² + 2ab + b²
(100 + 5)² = (100)² + 2(100)(5) + 5²
= 10000 + 1000 + 25
105² = 11,025

Question 3.
(2x – 5d)²
Answer:
(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
= 2²x² – 20 xd + 5²d²
= 4x² – 20xd + 25d²

Question 4.
(98)²
Answer:
(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)
Answer:
(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
Answer:
(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)
Answer:
(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
Answer:
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) units.
Answer:
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)
= x² – 2x – 2x + 4 .
= x² – 4x + 4 units square

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

Try These (Text Book Page No. 91)

Expand :
(i) (x + 5)³
Answer:
Comparing (x + 5)³ with (a + b)³, we have a = x and b = 4.
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 5)³ = x³ + 3x²(5) + 3(x)(5)² + 5³
= x³ + 15x² + 75x + 125

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

(iii) (x + 1)(x + 4)(x + 6)
Answer:
Comparing (x + 1)(x + 4)(x + 6) with (x + a)(x + b)(x + c) we have
a = 1 b = 4 and c = 6
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
= x³ + (1 + 4 + 6)x² + (1) (4) + (4) (6) + (6) (1)x + (1) (4) (6)
= x³ + 11x² + (4 + 24 + 6)x + 24
= x³ + 11x³ + 34x + 24

Try These (Text Book Page No. 94)

Find the factors

Answer:

Think (Text Book Page No. 94)

x² – 4(x – 2) = (x² – 4)(x – 2) Is this correct? If not correct it.
Answer:
(3a)² = 3²a² = 9a²
x² – 4 (x – 2) = x² – 4x + 8

Try These (Text Book Page No. 95)

Question 1.
3y + 6
Answer:
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 + 15y²
Answer:
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)
Answer:
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
Answer:
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)

Question 5.
x² – 3x + 2
Answer:

x² – 3x + 2 = x² – 2x – x + 2
= x(x – 2) – (x – 2)
= (x – 2)(x – 1)

Question 6.
y² – 4y – 32
Answer:

y² – 4y – 32 = y2 – 8y + 4y – 32
= y(y – 8) + 4(y – 8)
= (y – 8) (y + 4)

Question 7.
p² + 2p – 15
Answer:

p² + 2p – 15 = p² + 5p – 3p – 15
= p(p + 5) -3 (p + 5)
= (p + 5)(p – 3)

Question 8.
m² + 14m + 48
Answer:
m² + 14m + 48 = m² + 8m + 6m +48
= m(m + 8) + 6(m + 8)
= (m + 6)(m + 8)

Question 9.
x² – x – 90
Answer:
x² – x – 90 = x² – 10x + 9x – 90
= x(x – 10) + 9(x – 10)
= (x + 9)(x – 10)

Question 10.
9x² – 6x – 8
Answer:
9x² – 6x – 8 = 9x² – 12x + 6x – 8
= 3x(3x -4) + 2(3x -4)
= (3x + 2)(3x – 4)

Try These (Text Book Page No. 99)

Identify which among the following are linear equations.
(i) 2 + x = 19
Answer:
Linear as degree of the variable x is 1

(ii) 7x² – 5 = 3
Answer:
not linear as highest degree of x is 2

(iii) 4p³ = 12
Answer:
not linear as highest degree of p is 3

(iv) 6m + 2
Answer:
Linear, but not an equation

(v) n = 10
Answer:
Linear equation as degree of n is 1

(vi) 7k – 12 = 0
Answer:
Linear equation as degree of k is 1

(vii) \(\frac{6 x}{8}\) + y = 1
Answer:
Linear equation as degree of x & y is 1

(vii) 5 + y = 3x
Answer:
Linear equation as degree of y & x is 1

(ix) 10p + 2q = 3
Answer:
Linear equation as degree of p & q is 1

(x) x² – 2x – 4
Answer:
not linear equation as highest degree of x is 2

Think (Text Book Page No. 99)

(i) Is t(t – 5) = 10 a linear equation? Why?
Answer:
t(t – 5) = 10
= t × t – 5 × t = 10
= t² – 5t = 10
This is not a linear equations as the highest degree of the variable ‘t’ is 2

(ii) Is x² = 2x, a linear equation? Why?
Answer:
x² = 2x
= x² – 2x = 0
This is not a linear equations as the highest degree of the variable ‘x’ is 2

Try These (Text Book Page No. 100)

Convert the following statements into linear equations:

Question 1.
On subtracting 8 from the product of 5 and a number, I get 32.
Answer:
Convert to linear equations:
Given that on subtracting 8 from product of 5 and a, we get 32
∴ 5 × x – 8 = 32
∴ 5x – 8 = 32

Question 2.
The sum of three consecutive integers is 78.
Answer:
Sum of 3 consecutive integers is 78
Let integer be bx
∴ x + (x + 1) + (x + 2) = 78
∴ x + x + 1 + x + 2 = 78
∴ 3x + 3 = 78

Question 3.
Peter had a Two hundred rupee note. After buying 7 copies of a book he was left with 60.
Answer:
Let cost of one book be ‘x’
∴ Given that 200 – 7 × x = 60
∴ 200 – 7x = 60

Question 4.
The base angles of an isosceles triangle are equal and the vertex angle measures 80°.
Answer:
Let base angles each be equal to x & vertex bottom angle is 80°. Applying triangle property, sum of all angles is 180°
∴ x + x + 80 = 180°
∴ 2x + 80 = 180°

Question 5.
In a triangle ABC, ∠A is 100 more than ∠B. Also ∠C is three times ∠A. Express the equation in terms of angle B.
Answer:
Let ∠B = b
Given ∠A = 10° + ∠B = 10 + b
Also given that ∠C = 3 × ∠A = 3 × (10 + b) = 30 + 3b
Sum of the angles = 180°
∠A + ∠B + ∠C = 180°
10 + b + b + 30 + 3b = 180°
∴ 5b + 40 = 180°

Think (Text Book Page No. 101)

Can you get more than one solution for a linear equation?
Answer:
Yes, we can get. Consider the below line or equation.
x + y = 5
here,when x = 1, y = 4
when x = 2, y = 3
x = 3, y = 2
x = 4, y = 1
Hence, we get multiple solutions for the saine linear equation.

Try These (Text Book Page No. 101)

Identify which among the following are linear equations.

(i) 2 + x = 10
Answer:
2 + x = 10
⇒ x = \(\frac{10}{2}\) = 5

(ii) 3 + x = 5
Answer:
3 + x ⇒ 5
x = 5 – 3 = 2

(iii) x – 6 = 10
Answer:
x – 6 = 10
x = 10 + 6 = 16

(iv) 3x + 5 = 2
Answer:
⇒ 3x + 5 = 2
3x = 2 – 5 = -3

(v) \(\frac{2 x}{7}\) = 3
Answer:
⇒ 2x = 3 × 7 = 21
x = \(\frac{2 1}{2}\)

(vi) – 2 = 4m – 6
Answer:
⇒ -2x = 4m – 6
– 2 + 6 = 4m
4 = 4m
m = \(\frac{4}{4}\) = 1

(vii) 4(3x – 1) = 80
Answer:
⇒ 4(3x – 1) = 80
12x – 4 = 80
12x = 80 + 4 = 84
x = \(\frac{84}{12}\) = 7

(viii) 3x – 8 = 7 – 2x
Answer:
⇒ 3x – 8 = 7 – 2x
3x + 2x = 7 +8 = 15
5x = 15
x = \(\frac{15}{5}\) = 3

(ix) 7 – y = 3(5 – y)
Answer:
⇒ 7 – y = 3(5 – y)
7 – y = 15 – 3y
3y – y = 15 – 7
2y = 8
y = \(\frac{8}{2}\) = 4

(x) 4(1 – 2y) – 2(3 – y) = 0
Answer:
⇒ 4(1 – 2y) – 2(3 – y) = 0
4 – 8y – ó – 2y = 0
– 2 – 6y = 0
6y = -2
y = \(\frac{-2}{6}=\frac{-1}{3}\)

Think (Text Book Page No. 102)

Question 1.
“An equation is multiplied or divided by a non zero number on either side:’ Will there be any change in the solution?
Answer:
Not be any change in the solution

Question 2.
“An equation is multiplied or divided by two different numbers on either side. What will happen to the equation?
Answer:
When an equation is multiplied or divided by 2 different numbers on either side, there will be a change in the equation & accordingly, solution will also change.

Think (Text Book Page No. 104)

Suppose we take the second piece to be x and the first piece to be (200 – x), how will the steps vary ? Will the answer be different?
Answer:
Let 2^nd piece be ‘x’ & 1^st piece is 200 – x
Given that 1st piece is 40 cm smaller than hence the other piece
∴ 200 – x = 2 × x – 40

∴ 200 + 40 = 2x + x
240 = 3x
∴ x = \(\frac{240}{3}\) = 80
∴ 1^st piece = 200 – x = 200 – 80 = 120 cm
2^nd piece = x = 80 cm
The answer will not change

Think (Text Book Page No. 109)

If instead of (4,3), we write (3,4) and tn to mark it, will it represent ‘M’ again?
Answer:
Let 3, 4 be M. when we mark, we find that it is a different point and not ‘M’

Try These (Text Book Page No. 111)

Question 1.
Complete the table given below.

Answer:

Question 2.
Write the coordinates of the points marked in the following figure

Answer:
A – (-3, 2)
B – (5, 2)
C – (5, -3)
D – (-3, 3)
E – (-1, 4)
F – (1, 2)
G – (7, 4)
H – (0, 2)
I – (0, 3)
J – (-3, 0)
K – (5, 0)
L – (-1, 0)
M – (-2, 0)
N – (-2 ,-1)
O – (0, 0)
P – (-1, -1)
Q – (1, -1)
R – (2, -1)
S – (0, -3)
T – (7, 0)
U – (7, -2)

Think (Text Book Page No. 114)

Which of the points (5, -10) (0, 5) (5, 20) lie on the straight line x = 5?
Answer:
All points on the line X = 5 will have X-coordinate as 5. Therefore, any point with X – coordinate as 5 will lie on X = 5 line. Hence the points (5, — 10) & (5, 20) will lie on X = 5

Try These (Text Book Page No. 117)

Identify and correct the errors
(i)

Answer:

(ii)

Answer:

(iii)

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1.
Factorise the following by taking out the common factor
(i) 18xy – 12yz
Answer:
18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y(3x – 2z) = 6y(3x – 2z)

(ii) 9x⁵y³ + 6x³y² – 18x²y
Answer:
9x⁵ + 6x³y² – 18x²y = (3 × 3 × x² × x³ × y × y) + (2 × 3 × x² × x × y × y) – (2 × 3 × 3 × x² × y)
Taking out the common factors 3, x², y, we get
= 3 × x² × y (3x³y² + 2xy – 6)
= 3x²y (3x³y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)
Answer:
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)

(iv)(ax + ay) + (bx + by)
Answer:
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay )+ (bx + by) = a(x + y) + b(x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y) (a + b)

(v) 2x²(4x – 1) – 4x + 1
Answer:
Taking out -1 from last two terms
2x² (4x – 1) – 4x + 1 = 2x² (4x – 1) – 1 (4x – 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1) (2x² – 1)

(vi) 3y(x – 2)² – 2(2 – x)
Answer:
3y(x – 2)² – 2(2 – x) = 3y(x – 2)(x – 2) – 2( -1)(x – 2)
[∵ Taking out – 1 from 2 – x]
= 3y(x – 2)(x – 2) + 2(x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y(x – 2) + 2]

(vii) 6xy – 4y² + 12xy – 2yzx
Answer:
= 6xy + 12xy – 4y² – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 x x x y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1) (2) y [2y + zx]
= 6 × y(3) – 2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy(2y + zx)
Taking out 2y from two terms
= 2y(9x – (2y + zx))
= 2y (9x – 2y – xz)

(viii) a³ – 3a² + a – 3
Answer:
a² – 3a² + a – 3 = a²(a – 3) + 1(a – 3) [:Groupingthetermssuitably]
= (a – 3) (a² + 1)

(ix) 3y³ – 48y
Answer:
3y² – 48y = 3 × y × y² – 3 × l6 × y
Taking out 3 × y
= 3y(y² – 16) = 3y(y² – 4²)
Comparing y² – 4² with a² – b²
a = y, b = 4
a² – b² = (a + b) (a – b)
y² – 4² = (y + 4) (y – 4)
∴ 3y(y² – 16) = 3y(y + 4)(y – 4)

(x) ab² – bc² – ab + c²
Answer:
ab² – bc² – ab + c²
Grouping suitably
ab² – bc² – ab + c² = b (ab – c²) – 1 (ab – c²)
Taking out the binomial factor ab – c² = (ab – c²) (b – 1)

Question 2.
Factorise the following expressions
(i) x² + 14x + 49
Answer:
x² + 14x + 49 = x² + 14x + 72
Comparing with a² + 2ab + b² = (a + b)² we have a = x and b = 7
⇒ x² + 2(x)(7) + 7² = (x + 7)²
∴ x² + 14x + 49 = (x + 7)²

(ii) y² – 10y + 25
Answer:
y² – 10y + 25 = y² – 10y + 5²
Comparing with a² – 2ab + b² = (a – b)² we get a = y; b = 5
⇒ y² – 2(y) (5) + 5² = (y – 5)²
∴ y² – 10y + 25 = (y – 5)²

(iii) c² – 4c – 12
Answer:
This is of the form ax² + bx + c
Where a = 1, b = -4 c = -12, x = c
Now the product ac = 1 × – 12 = -12 and the sum b = -4

Product = – 72	Sum = 1
1 × (-12) = -12	1 + (-12) = -11
2 × (-6) = – 12	2 + (-6)  = – 4

∴ The middle term – 4c can be written as 2c – 6c
∴ c² – 4c – 12 = c² + 2c – 6c – 12
= c(c + 2) -6 (c + 2)

Taking out (c + 2)
⇒ (c + 2)(c – 6)
∴ c² – 4c – 12 = (c + 2)(c – 6)

(iv) m² + m – 72
Answer:
m² + m – 72
This is of the form ax + bx + c
where a = 1, b = 1, c = -72

Product = – 72	Sum = 1
1 × -72 = – 72	1 + (-72) =  -71
2 × – 36 = – 72	2 + (-36) = – 34
3 × (-24) = – 72	3 + (-24) =  – 21
4 ×  (-18) = -72	4 + (-18) =  – 14
6 × (-12) = -72	6 + (-12) = – 6
8 × (-9) = -72	8 + (-9) = – 1
9 × (-8) = – 72	9 + (-8) = 1

Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m² + m – 72 = m² + 9m – 8m – 72
= m(m + 9) – 8(m + 9)

Taking out (m + 9)
= (m + 9)(m – 8)
∴ m² + m – 72 = (m + 9)(m – 8)

(v) 4x² – 8x + 3
Answer:
4x² – 8x + 3
This is of the form ax² + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
sum b = -8

Product = 12	Sum = -8
(-1) × (-12) = 12	(-1) + (-12) = – 13
(-2) × (-6) = 12	(-2) + (-6) = – 8

The middle term can be written as – 8x = – 2x – 6x
4x² – 8x + 3 = 4x² – 2x – 6x + 3
= 2x (2x – 1) – 3 (2x – 1)
= (2x – 1)(2x – 3)
4x² – 8x + 3 = (2x – 1) (2x – 3)

Question 3.
Factorise the following expressions using (a + b)³ = a³ + 3a²b + 3ab² + b³ identity
(i) 64x³ + 144x² + 108x + 27
(ii) 27p³ + 54p³q + 36pq² + 8q³
Answer:
(i) 64x³ + 144x² + 108x + 27
= (4x)³ + 3(4x)² (3) + 3(4) (3)² + 3³
= (4x + 3)³

(ii) 27p³ + 54p³q + 36pq² + 8q³
= (3p)³ + 3(3p)² (2q) + 3(3p) (2q)² + (2q)³
= (3p + 2q)³

Question 4.
Factorise the following expressions using (a – b)³ = a³ – 3a²b + 3ab² – b³ identity
(i) y³ – 18y² + 108y – 216
(ii) 8m³ – 60m²n + 150mn² – 125n³
Answer:
(i) y³ – 18y² + 108y – 216
= y³ – 3y²(6) + 3(6)²y – 6³
= (y – 6)³

(ii) 8m³ – 60m²n + 150mn² – 125n³
= (2m)³ – 3(2m)² (5) + 3(2m)(5n)² – (5n)³
= (2m – 5n)³

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions 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
Answer:
(i) (3m + 5)²
Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5
(a + b)² = a² + 2ab + b²
(3m + 5)² = (3m)² + 2(3m) (5) + 5²
= 3² m² + 30 m + 25
= 9m² + 30 m + 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 – 10 p + 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)
= 2² n² + 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
Answer:
(i) (3 + m)³
Cornparing (3 + m)³ with (a + b)³ we have a = 3 ; b = m
(a + b)³ = a² + 3a²b + 3ab² + b³
(3 + m)³ = 3³ + 3(3)² (m) + 3(3)m² + m³
= 27 + 27m + 9m² + m³
= m³ + 9m² + 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) = a³ + 3a²b + 3ab² + b²
(3p + 4q)³ = (3p)³ + 3(3p)² (4q) + 3(3p)(4q)² + (4q)³
= 3³p³ + 3(9p²)(4q) + 9p(16q²) – 4³q³
= 27p³ + 108p²q + 144pq³ + 64q³

(iv) (52)³
(52)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 = 50³ + 3 (50)² 2 + 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
(104)³ = (100 + 4)³
Comparing (100 + 4)³ with (a + b)³ we have a = 100 and b = 4
(a + b)³ = a³ + 3a²b + 3ab² + b³
(100 + 4)³ = (100)³ + 3 (100)² (4) + 3 (100) (4)² + 4³
= 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
Answer:
(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 – 75x + 15 x² – x³

(ii) (2x – 4y)³
Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y
(a – b)³ = a³ – 3a²b + 3ab² – 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 + 3ab (c)² – c³
= a³b³ – 3(a²b²) c + 3abc² – c³
= a³b³ – 3a²b²c + 3abc² – c³

(iv) (48)³
(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
(97xy)³ = 97³ x³ y³ = (100 – 3) x³y³ … (1)
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 (p – 2)(p + 1)(p – 4)
Answer:
(p – 2)(p + 1)(p – 4) = (p + (-2)) (p + 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) = x² + (a + b + c) x² + (ab + bc+ 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
= p² – 5p² + 2p + 8

Question 5.
Find the volume of the cube whose side is (x + 1) cm
Answer:
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 = (a3³ + 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)
Answer:
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 – 2x³ – 5x + 6 units³

Objective Type Questions

Question 7.
If x² – y² = 16 and (x + y) = 8 then (x – y) is ________
(A) 8
(B) 3
(C) 2
(D) 1
Answer:
(C) 2
Hint:
x² – y² = 16
(x + y) (x – y) = 16
8 (x – y) = 16
(x – y) = \(\frac { 16 }{ 8 }\) = 2

Question 8.
\(\frac{(a+b)\left(a^{3}-b^{3}\right)}{\left(a^{2}-b^{2}\right)}\) = _________
(A) a² – ab + b²
(B) a² + ab + b²
(C) a² + 2ab + b²
(D) a2² – 2ab + b²
Answer:
(B) a² + ab + b²
Hint:

= a² + ab + b²

Question 9.
(p + q)(p² – pq + q²) is equal to __________
(A) p³ + q³
(B) (p + q)³
(C) p³ – q³
(D) (p – q)³
Answer:
(A) p³ + q³
Hint:
a³ + b³ = (a + b) (a² – ab + b²)

Question 10.
(a – b) = 3 and ab = 5 then a³ – b³ = __________
(A) 15
(B) 18
(C) 62
(D) 72
Answer:
(D) 72
Hint:
(a – b) = 3
(a – b)² = 3²
a² + b² – 2ab = 9
a² + b² – 2(5) = 9
a² + b² = 9 + 10
a² + b² = 19
a³ – b³ = (a – b)(a² + ab + b²) = 3(19 + 5)
= 3(24) = 72

Question 11.
a³ + b³ = (a + b)³ _________
(A) 3a(a + b)
(B) 3ab(a – b)
(C) -3ab(a + b)
(D) 3ab(a + b)
Answer:
(D) 3ab(a + b)
Hint:
(a + b)³ = a³ + b³ + 3a²b + 3ab²
(a + b)³ – 3a²b – 3ab³ = a³ + b³
(a + b)³ – 3ab(a + b) = a³ + b³