Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.2

1. Using number line write the integer which is:

Question (a)
5 less than -1
Solution:
5 less than -1 = ?
We need to find the integer which is 5 less than -1.
So, we shall start with ‘-1 ’ and proceed 5 steps to the left of -1 as shown below:

Therefore 5 less than -1 is -6

Question (b)
5 more than -5
Solution:
5 more than -5 = ?
We need to find the integer which is 5 more than -5.
So, we shall start with -5 and proceed 5 steps to the right of -5 as shown below:

Therefore 5 more than -5 is zero.

Question (c)
2 less than 5
Solution:
2 less than 5 = ?
We need to find the integer which is 2 less than 5.
So, we shall start with 5 and proceed
2 steps to the left of 5 as shown below:

Therefore 2 less than 5 is 3.

Question (d)
3 less than -2.
Solution:
3 less than – 2 = ?
We need to find the integer which is 3 less than -2.
So, we shall start with -2 and proceed 3 steps to the left of -2 as shown below:

Therefore 3 less than -2 is -5.

2. Using number line, add the following integers:

Question (a)
9 + (-3)
Solution:
On number line we shall start from 0 and move 9 steps to the right of zero. Then we shall move three steps to the left of ‘+9’. We finally reach at +6. Thus, +6 is the answer.

Hence, 9 + (-3) = +6

Question (b)
5 + (-11)
Solution:
On number line we shall start from 0 and move 5 steps to the right of zero. Then we shall move eleven steps to the left of ‘+5’. We finally reach at ‘-6’. Thus, -6 is the answer.

Hence, 5 + (-11) = -6

Question (c)
(-1) + (-4)
Solution:
On number line we first move 1 step to the left of zero. Then we moved ahead 4 steps to the left of ‘-1\ We finally reaches at -5. Thus -5 is the answer.

Hence, (-1) + (-4) = -5

Question (d)
(-5) + 12
Solution:
On number line we shall move 5 steps to the left of zero. Then we shall move 12 steps to the right of ‘-5’ and finally reach at ‘+7’. Thus +7 is the answer.

Hence, (-5) + 12 = +7

Question (e)
(-1) + (-2) + (-4)
Solution:
Step I: On number line we shall move one step to the left of zero suggested by minus sign of ‘-1’.
Step II: Then we shall move 2 steps to the left of -1 suggested by minus sign of -2 and we reach at ‘-3’.
Step III: Then again we shall move 4 steps to the left of ‘-3’ suggested by minus sign of -4 and finally we reach at-7.

Hence, (-1) + (-2) + (-4) = -7

Question (f)
(-2) + 4 + (-5)
Solution:
Step I: On number line we shall move 2 steps to the left of zero suggested by minus sign of ‘-2’.
Step II: Then we shall move 4 steps to the right of ‘-2’ suggested by plus sign of ‘44’ and we reach at +2.
Step III: Then we shall move 5 steps to the left of ‘+2’ as suggested by minus sign of ‘-5’ and finally we reach at -3. Thus answer is ‘-3’.

Hence, (-2) + 4 + (-5) = -3

Question (g)
(-3) + (5) + (-4).
Solution:
Step I: On number line we shall move 3 steps to the left of zero as suggested by minus sign of ‘-3’.
Step II: Then we shall move 5 steps to the right of ‘-3’ as suggested by plus sign of ‘+5’ and we reach at ‘+2’.
Step III: Then we shall move 4 steps to the left of ‘+2’ as suggested by minus sign of ‘4’ and finally we reach at ‘-2’. Thus ‘-2’ is the answer.

Hence, (-3) + (5) + (-4) = -2

3. Add without using number line:

Question (a)
18 + 13
Solution:
18 + 13 = 31

Question (b)
18 + (- 13)
Solution:
18 + (- 13) = + (18 – 13) = +5

Question (c)
(-18) + 13
Solution:
(-18) + 13 = – (18 – 13) = -5

Question (d)
(-18) + (-13)
Solution:
(-18) + (-13) = -(18 + 13) = -31

Question (e)
180 + (-200)
Solution:
180 + (-200) = -(200 – 180) = -20

Question (f)
111 + (-67)
Solution:
111 + (-67) = (777 – 67) = 710

Question (g)
1262 + (-366) + (-962)
Solution:
1262 + (-366) + (-962)
= 1262 – (366 + 962)
= 1262 – 1328
= -(1328 – 1262) = -66

Question (h)
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
Solution:
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
= 30 + 21 + 11 + (-27) + (-19) + (-3) + (-9)
= 62 + (-58) = 62 – 58 = 4

Question (i)
(-7) + (-9) + 4 + 16
Solution:
(-7) + (-9) + 4+16 = (-16) + 20 = 4

Question (j)
37 + (-2) + (-65) + (-8).
Solution:
37 + (-2) + (-65) + (-8)
= 37 – (2 + 65 + 8)
= 37 – 75 = -38

4. Write the successor and predecessor of the following:

Question (a)
-15
Solution:
Successor of -15 = -15 + 1 = -14
Predecessor of -15 = -15 – 1 = -16

Question (b)
27
Solution:
Successor of 27 = 27 + 1 = 28
Predecessor of 27 = 27 – 1 = 26

Question (c)
-79
Solution:
Successor of -79 = -79 + 1 = -78
Predecessor of -79 = -79- 1 = -80

Question (d)
0
Solution:
Successor of 0 = 0+ 1 = 1
Predecessor of 0 = 0 – 1 = -1

Question (e)
29
Solution:
Successor of 29 = 29 + 1 = 30
Predecessor of 29 = 29 – 1 = 28

Question (f)
-18
Solution:
Successor of -18 = -18 + 1 = -17
Predecessor of -18 = -18 – 1 = -19

Question (g)
-21
Solution:
Successor of -21 = -21 + 1 = -200
Predecessor of -21 = -21 – 1 = -22

Question (h)
99
Solution:
Successor of 99 = 99 + 1 = 100
Predecessor of 99 = 99 – 1 = 98

Question (i)
-1
Solution:
Successor of-1 = -1 + 1 = 0
Predecessor of -1 = -1 – 1 = -2

Question (j)
-13.
Solution:
Successor of -13 = -13 + 1 = -12
Predecessor of -13 = -13 – 1 = -14

5. Complete the following addition table:

+	-3	-4	-2	+1	+2	+3
-2
-3
0
+1
+2

Solution:

+	-3	-4	-2	+ 1	+2	+3
-2	-5	-6	-4	-1	0	+ 1
-3	-6	-7	-5	-2	-1	0
0	-3	-4	-2	+ 1	+2	+3
+ 1	-2	-3	-1	+2	+3	+4
+2	-1	-2	0	+3	+4	+5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at D show that,

(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Answer:
∆ ABC and ∆ DBC are isosceles triangles on the same base BC.
∴ In ∆ ABC, AB = AC and in ∆ DBC, DB = DC.
In ∆ ABD and ∆ ACD,
AB = AC
DB = DC
and AD = AD (Common)
∴ ∆ ABD s ∆ ACD (SSS rule) [Result (i)]
∴ ∠ BAD = ∠ CAD (CPCT)
In ∆ ABP and ∆ ACP
AB = AC
∠ BAP = ∠ CAP (∵ ∠ BAD = ∠ CAD)
and AP = AP (Common)
∴ ∆ ABP ≅ ∆ ACP (SAS rule) [Result (ii)]
∴ BP = CP (CPCT)
In ∆ DBP and ∆ DCR
DB = DC
BP = CP
and DP = DP (Common)
∴ ∆ DBP ≅ ∆ DCP (SSS rule)
From ∆ ABP ≅ ∆ ACR ∠ BAP = ∠ CAP (CPCT)
∴ AP bisects ∠A.
From ∆ DBP ≅ ∆ DCR ∠BDP = ∠ CDP (CPCT)
∴ DP bisects ∠D.
Thus, AP bisects ∠A as well as ∠D. [Result (iii)]
∆ A ABP ≅ ∆ ACP
∴ BP = CP and ∠ APB = ∠ APC (CPCT)
But, ∠ APB + ∠ APC = 180° (Linear pair)
∴ ∠ APB = ∠ APC = \(\) = 90°
Thus, BP = CP and AP ⊥ BC.
∴ AP is the perpendicular bisector of BC. [Result (iv)]

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A

Answer:
AD is an altitude of A ABC.
∴ ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
hypotenuse AB = hypotenuse AC (Given)
∠ ADB = ∠ ADC (Right angles)
AD = AD (Common)
∴ ∆ ADB ≅ ∆ ADC (RHS rule)
∴ BD = CD and ∠ BAD = ∠ CAD (CPCT)
Now, BD = CD means D is the midpoint of BC.
Hence, AD bisects BC. [Result (i)]
Moreover, ∠ BAD = ∠ CAD and
∠ BAD + ∠ CAD = ∠ BAC.
Hence, AD bisects ∠A. [Result (ii)]

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see the given figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR

Answer:
In ∆ ABC, AM is a median.
∴ BM = CM = \(\frac{1}{2}\) BC
In ∆ PQR, PN is a median.
∴ QN = RN = \(\frac{1}{2}\) QR
Now, BC = QR (Given)
∴ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ ABM and ∆ PQN,
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved)
∴ ∆ ABM ≅ ∆ PQN (SSS rule) [Result (i)]
∴ ∠ ABM = ∠ PQN (CPCT)
∴ ∠ ABC = ∠ PQR
Now, in ∆ ABC and ∆ PQR,
AB = PQ
∠ ABC = ∠ PQR
BC = QR .
∴ ∆ ABC ≅ ∆ PQR (SAS rule) [Result (ii)]

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:
In ∆ FBC and ∆ ECB,
CF = BE (Given)
∠ CFB = ∠ BEC = 90° (Given)
BC = CB (Common)
∴ A FBC ≅ A ECB (RHS rule)
∴ ∠ FBC = ∠ ECB (CPCT)
∴ ∠ ABC = ∠ ACB
Now, in ∆ ABC, ∠ ABC = ∠ ACB
∴ AC = AB (Theorem 7.3)
Hence, ∆ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
Answer:

In ∆ ABC, AP is an altitude.
∴ ∠ APB = ∠ APC = 90°
In ∆ APB and ∆ APC,
∠ APB = ∠ APC = 90°
AB = AC (Given)
AP = AP (Common)
∴ ∆ APB ≅ ∆ APC (RHS rule)
∴ ∠ ABP = ∠ AGP (CPCT)
∴ ∠ ABC = ∠ ACB
Thus, in ∆ ABC, ∠ B = ∠ C.

Punjab State Board Syllabus PSEB 9th Class English Book Solutions Guide Pdf is part of PSEB Solutions for Class 9.

PSEB 9th Class English Guide | English Guide for Class 9 PSEB

English Guide for Class 9 PSEB | PSEB 9th Class English Book Solutions

PSEB 9th Class English Main Course Book Solutions

English Main Course Book Class 9 Solutions PSEB Prose

• Chapter 1 Grooming of a Boy
• Chapter 2 Plants also Breathe and Feel
• Chapter 3 Budgeting Your Time
• Chapter 4 Journey by Night
• Chapter 5 The Discovery of Moon
• Chapter 6 Three Great Indians
• Chapter 7 The Death of Abhimanyu

Class 9th English Main Course Book Solutions PSEB Poetry

• Poem 1 Open Thy Eyes and see Thy God
• Poem 2 No Men are Foreign
• Poem 3 The Nightingale and the Glow-worm

PSEB 9th Class English Literature Book Solutions

9th Class English Literature Book PSEB Supplementary Reader

• Chapter 1 The Magic Violin
• Chapter 2 Wishes Come True
• Chapter 3 In the Flood
• Chapter 4 A Letter to God
• Chapter 5 The Last Leaf
• Chapter 6 The Bewitched Jacket
• Chapter 7 The King who Limped

PSEB 9th Class English Book Grammar

• Determiners
• Modals
• Prepositions
• Conjunctions
• Simple and Complex Sentences
• Voice
• Narration
• Non-Finites
• Tenses (Concord and Sequence)

PSEB 9th Class English Book Vocabulary

• Nature of Words (Context-Meaning)
• Words as different Parts of Speech
• Synonyms
• Antonyms
• Homonyms
• Formation of Words

PSEB 9th Class English Book Reading Skills

• Reading Comprehension Unseen Passages Prose / Poems
• Reading Comprehension Unseen Picture/Poster Based

PSEB 9th Class English Book Composition & Translation

• Note-Making & Messages
• E-mail Message Writing
• Letter Writing
• Paragraph Writing

PSEB 9th Class English Structure of Question Paper

Class – IX (English)
Time: 3 Hrs.

Theory: 80 marks
INA: 20 marks
(Listening and Speaking Skill-based Practical: 18 marks
Book Bank: 2 marks)
Total: 100 marks

Q.No.	Content	Maximum Marks (80)
1.	Section A : Reading Comprehension
	(a) Unseen Reading Comprehension (5 MCQs)	(5Q × 1M = 5 Marks)
	(b) Unseen Picture/Poster Based Comprehension (5 MCQs)	(5Q × 1M = 5 Marks)
2.	Section B: Objective Type Questions
	It will consist of 8 objective-type questions carrying one mark each. Objective-type questions may include questions with one word to one-sentence answers. Fill in the blanks, True/False, or multiple choice type questions. English Main Course Book (4 Questions) English Supplementary Reader (4 Questions)	(8Q × 1M = 8 Marks)
3.	Section C: English Textbooks
	English Main Course Book: Prose
	(a) Short Answer Type Questions (3 out of 5)	(3Q × 2M = 6 Marks)
	(b) Long Answer Type Questions (1 out of 2)	(1Q × 3M = 3 Marks)
4.	English Main Course Book: Poetry
	(a) Stanza (2 out of 3 questions)	(2Q × 1.5M = 3 Marks)
	(b) Central Idea/Long Answer Type Questions (1 out of 2)	(1Q × 3M = 3 Marks)
5.	English Supplementary Reader (3 out of 5 questions)	(3Q × 3M = 9 Marks)
6.	Section D: Vocabulary
	Vocabulary	(4Q × 1M = 4 Marks)
7.	Section E: Translation, Grammar, and Composition
	(a) Translation from English to Punjabi/Hindi (3 out of 5)	(3Q × 1M = 3 Marks)
	(b) Translation from Punjabi/Hindi to English (3 out of 5)	(3Q × 1M = 3 Marks)
8.	Grammar (Do as directed)	(10Q × 1M = 10 Marks)
9.	(a) Note Making/E-mail messages (with internal choice)	(3 Marks)
	(b) Paragraph Writing (1 out of 2)	(4 Marks)
	(c) Letter Writing (1 out of 2)	(6 Marks)
10.	Marks for Good Handwriting	(5 Marks)

Note: abbreviations used: Q – question, M – marks

In Main Coursebook (including poetry) and Supplementary Reader Section, questions will come from the given back exercises of lessons in the prescribed textbooks.

There will be four choices in multiple-choice questions.

For letter writing any two kinds of letters (Personal, Official and Business) with internal choice will be given.

In Question 3(b), one question will be asked from the first-four chapters (‘The Grooming of a Boy’, ‘Plants also Breathe and Feel’, ‘Budgeting your Time’, ‘Journey by Night’); The other question will be asked from the next three chapters (‘The Discovery of Moon’, ‘Three Great Indians’, ‘The Death of Abhimanyu’).

Listening and Speaking skill-based practical: 18 Marks

• Listening – 10 Marks
• Speaking – 08 Marks

For the listening test, students will be given a practice sheet containing 10 questions. They will answer all the 10 questions on the basis of an audio clip. Each question will carry 1 mark.

For the speaking test, students will speak 8 correct sentences on the basis of ‘picture and cue words’ provided in the practice sheet. Each correct sentence carries 1 mark.

PSEB 9th Class English Syllabus

Unit-I: Reading Skills

• Reading Comprehension of unseen passages with five multiple-choice questions.
• Unseen Picture/Poster-based Comprehension with five multiple-choice questions.

Unit-II: Text Books

1. English Main Coursebook
2. Grooming of a Boy
3. Plants also Breathe and Feel
4. Budgeting Your Time
5. Journey by Night
6. The Discovery of Moon
7. Three Great Indians
8. The Death of Abhimanyu

Poetry

1. Open Thy Eyes and See Thy God (Rabindranath Tagore)
2. No Men Are Foreign (James Kirkup)
3. The Nightingale and the Glow-worm (William Cowper)

English Literature book (Supplementary Reader)

1. The Magic Violin
2. Wishes Come True
3. In the Flood
4. A Letter to God
5. The Last Leaf
6. The Bewitched Jacket
7. The King Who Limped

Unit-III: Vocabulary

• Nature of Words (Context Meaning)
• Words as different Parts of Speech
• Synonyms
• Antonyms
• Homonyms
• Formation of Words

Unit-IV: Grammar

• Determiners
• Modals
• Prepositions
• Conjunctions
• Simple and Complex Sentences
• Voice
• Narration
• Non-Finites
• Tenses (Concord and Sequence)

Unit-V: Guided Creative Writing

• Note-making
• E-mail messages
• Letter Writing
• Paragraph Writing (not more than 100 words)

Translation

• Translation from English to Punjabi/Hindi.
• Translation from Punjabi/Hindi to English.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Answer:

In ∆ ABC, AB = AC
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ OBC = ∠ OCB (BO bisects ∠ ABC and CO bisects ∠ ACB)
Now, in ∆ OBC, ∠ OBC = ∠ OCB
∴ OB = OC (Theorem 7.3)
Similarly, ∠ ABC = ∠ ACB gives
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ ABO = ∠ ACO
Now, in ∆ ABO and ∆ ACO,
AB = AC (Given)
∠ ABO = ∠ ACO
and OB = OC
∴ ∆ ABO ≅ ∆ ACO (SAS rule)
∴ ∠ BAO = ∠ CAO (CPCT)
But, ∠ BAO + ∠ CAO = ∠ BAC (Adjacent angles)
∴ ∠ BAO = ∠ CAO = \(\frac{1}{2}\) ∠ BAC
Thus, AO bisects ∠ A.

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Answer:
In ∆ ABC, AD is the perpendicular bisector of BC.
∴ BD = CD and ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
AD = AD (Common)
∠ ADB = ∠ ADC (Right angles)
and BD = CD
∴ ∆ ADB ≅ ∆ ADC (SAS rule)
∴ AB = AC (CPCT)
Now, in ∆ ABC, AB = AC.
Hence, ∆ ABC is an isosceles triangle in which AB = AC.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Answer:
In ∆ ABC, AC = AB
∴ ∠ ABC = ∠ ACB
∴ ∠ FBC = ∠ ECB
Now, in ∆ FBC and ∆ ECB,
∠ FBC = ∠ ECB
∠ BFC = ∠ CEB (Right angles)
BC = CB (Common)
∴ ∆ FBC ≅ ∆ ECB (AAS rule)
∴ CF = BE (CPCT)
Thus, the altitudes CF and BE on equal sides AB and AC respectively of ∆ ABC are equal.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:
In ∆ ABE and ∆ ACF,
∠ AEB = ∠ AFC (Right angles)
∠ A = ∠ A (Common)
BE = CF (Given)
∴ ∆ ABE ≅ ∆ ACF (AAS rule)
∴ AB = AC (CPCT)
Thus, ∆ ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ ABD = ∠ ACD.

Answer:
∠ ABC and ∠ DBC are adjacent angles.
∴ ∠ ABC + ∠ DBC = ∠ ABD ………… (1)
∠ ACB and ∠ DCB are adjacent angles.
∴ ∠ ACB + ∠ DCB = ∠ ACD ………….. (2)
In ∆ ABC, AB = AC.
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
In ∆ DBC, DB = DC.
∴ ∠ DBC = ∠ DCB (Theorem 7.2)
∴ ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∴ ∠ ABD = ∠ ACD [From (1) and (2))

Question 6.
∆ ABC is an isosceles triangle in which AB = AC. side BA is produced to D such that AD = AB (see the given figure). Show that ∠ BCD is a right angle.

Answer:
AB = AC and AD = AB
∴ AC = AD
In ∆ ABC, AB = AC
∴ ∠ ACB = ∠ ABC (Theorem 7.2) ……………… (1)
In A ADC, AC = AD
∴ ∠ ACD = ∠ ADC (Theorem 7.2) ……………… (2)
Adding (1) and (2),
∠ ACB + ∠ ACD = ∠ ABC + ∠ ADC
∴ ∠ BCD = ∠ DBC + ∠ BDC
(Adjacent angles and A lies on BD)
In ∆ BCD,
∠ DBC + ∠ BDC + ∠ BCD = 180°
∴ ∠ BCD + ∠ BCD = 180° (from (3)]
∴ 2 ∠ BCD = 180°
∴ ∠ BCD = 90°
Thus, ∠ BCD is a right angle.

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.
Answer:

In ∆ ABC, AB = AC
∴ Z C = Z B (Theorem 7.2)
In ∆ ABC,
∠ A + ∠ B + ∠ C = 180°
∴ 90° + ∠ B + ∠ B = 180° (Given and ∠ C = ∠ B)
∴ 2 ∠ B = 90°
∴ ∠ B = 45°
∴ ∠ C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:

∆ ABC is an equilateral triangle.
∴ AB = BC = AC
In ∆ ABC, AB = BC
∴ ∠ C = ∠ A (Theorem 7.2)
In ∆ ABC, AB = AC
∴ ∠ C = ∠ B (Theorem 7.2)
Hence, ∠ A = ∠B = ∠ C.
Now, in ∆ ABC, ∠ A + ∠ B + Z C = 180°
∴ ∠ A = ∠ B = ∠ C = \(\frac{180^{\circ}}{3}\) = 60°
Thus, the angles of ah equilateral triangle are 60° each.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

1. Construct ΔABC, given AB = 6 cm, ∠A = 30° and ∠B = 75°.
Solution:
Given. One side of ΔABC as AB = 6 cm, m∠A = 30° and m∠B = 75°.
To construct: A triangle with one side and these two angles.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔABC and indicate the measures of side and two angles.

Step 2. Draw a ray AB of length 6 cm.

Step 3. At A; draw a ray AX making an angle 30° with AB.

Step 4. With the help of compass. At B; draw a ray BY making an angle of 75° with AB.

Step 5. Both rays AX and BY intersect, at a point. So the point of intersection of the two rays is C Then. ΔABC is now obtained.

2. Construct an isosceles ΔABC such that AB = 5.3 cm and each base angle = 45°.
Solution:
Given : Isosceles ΔABC with AB = 5.3 cm each base angle = 45°.
To Construct: A triangle with one side and two base angles.
Steps of construction :
Step 1. Draw a rough sketch of ΔABC with given measures

Step 2. Draw a line segment AB = 5.3 cm.

Step 3. Taking A as centre with the help of compass. Draw a ray AX making an angle 45° with AB.

Step 4. With the help of compass and taking B as a centre. Draw a ray BY making an angle 45° with the line segment AB.

Step 5. Rays AX and BY intersect, at a point say C, then ABC is the required triangle.

3. Construct ΔXYZ if XY = 4 cm, ∠X = 45° and ∠Z = 60°.
[Hint : ∠Y = 180° – 45° – 60° = 75°]
Solution:
One side of ΔXYZ as
XY = 4 cm,
∠X = 45°
and ∠Z = 60°.
As we know that by angle-sum property of a triangle; sum of all three angles of a triangle is equal to 180°.
∴ ∠X + ∠Y + ∠Z = 180°
⇒ 45° + ∠Y + 60° = 180°
⇒ 105° + ∠Y = 180°
⇒ ∠Y = 75°.
Now it will be easy to construct triangle with side
XY = 4 cm,
∠X = 45°
and ∠Y = 75°.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔXYZ and indicate the measure of side and two angles.

Step 2. Draw a ray XY of length 4 cm.

Step 3. At X draw a ray XA making an angle of 45° with XY.

Step 4. At Y; draw a ray YB making an angle of 75° with XY.

Step 5. Z has to lie on both rays XA and YB. So, the point of intersection of two rays is Z.
ΔXYZ is now obtained.

4. Examine whether you can construct ΔPQR such that ∠P = 100°, ∠Q = 90° and PQ = 4.3 cm If not possible given reason.
Solution:
No, we cannot construct given ΔPQR.
Reason :
As we know that by angle sum property of a triangle; sum of all three angles a triangle is equal to 180°. But in given question sum of two angles;
m∠P + m∠Q
= 100° + 90°
= 190°
The sum of these two angles should be less than 180°. So triangle with given measures cannot be constructed as it violates the angle sum property of a triangle.

5. Question (i).
In which of the following cases a unique triangle can be drawn ?
(a) BC = 5 cm, ∠B = 90° and ∠C = 100°
(b) AB = 4 cm, BC = 7 cm and CA = 2 cm
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°
(d) An isosceles triangle with length of each equal side equal to 5 cm.
Answer:
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°

Question (ii).
A triangle can be constructed by taking two of its angles as.
(a) 110°, 40°
(b) 70°, 115°
(c) 135°, 45°
(d) 90°, 90°
Answer:
(a) 110°, 40°

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.1

1. Write two examples from day-to-day life in which we can use positive and negative integers.
Solution:
1. If positive represents above sea level, then negative represents below sea level.
2. If positive represents a deposit, negative represents a withdrawal.

2. Write the opposite of the following:

Question (a)
A profit of ₹ 500
Solution:
A loss of ₹ 500

Question (b)
A withdrawal of ₹ 70 from bank account
Solution:
Deposit of ₹ 70 in bank account

Question (c)
A deposit of ₹ 1000
Solution:
Withdrawal of ₹ 1000

Question (d)
326 B.C
Solution:
326 AD

Question (e)
500 m below sea level
Solution:
500 m above sea level

Question (f)
25° above 0°C.
Solution:
25° below 0°C.

3. Represent the situations mentioned in integers.
Solution:
(a) + 500
(b) – 70
(c) + 1000
(d) – 326
(e) – 500 m
(f) + 25.

4. Represent the following situations in integers.

Question (a)
A deposit of ₹ 500.
Solution:
+ 500

Question (b)
An Aeroplane is flying at a height two thousand metre above the sea level.
Solution:
+ 2000

Question (c)
A withdrawal of ₹ 700 from Bank Account.
Solution:
– 700

Question (d)
A diver dives to a depth of 6 feet below ground level.
Solution:
– 6.

5. Represent the following numbers on number line.

Question (i)
(a) – 5
(b) + 6
(c) o
(d) + 1
(e) – 9
(f) – 4
(g) + 8
(h) + 3.
Solution:

6. Integers are represented on a horizontal number line as shown where A represents – 2. With reference to the number line, answer the following questions:

(a) Which point represent – 3?
(b) Locate the point which represents the opposite of B and name it P.
(c) Write integers for the points C and E.
(d) Which point marked on the number line has the least value?
Solution:

(a) Point B represents – 3.
(b) Point P represents + 3.
(c) Point C represents -7 and Point E represents + 4.
(d) Point C has the least value – 7.

7. In each of the following pairs, which number is to the right of other on the number line?

Question (i)
(a) 2 9
(b) -3, -8
(c) 0, -5
(d) -11, 10
(e) -9, 9
(f) 2, – 200.
Solution:
(a) 9
(b) – 3
(c) 0
(d) 10
(e) 9
(f) 2

8. Write all the integers between the given pairs (write them in increasing order)

Question (a)
0 and -6
Solution:
-5, -4, -3, -2, -1

Question (b)
-6 and +6
Solution:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Question (c)
-9 and -17
Solution:
-16, -15, -14, -13, -12, -11, -10

Question (d)
-19 and -5.
Solution:
-18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6.

9.

Question (a)
Write five negative integers greater than ‘-15’.
Solution:
Five negative integers greater than ‘-15’ are:
-14, -13, -12, -11, -10

Question (b)
Write five integers smaller than ‘-20’.
Solution:
Five integers smaller than ‘-20’ are:
-21, -22, -23, -24, -25

Question (c)
Write five integers greater than 0.
Solution:
Five integers greater than 0 are:
1,2, 3, 4, 5

Question (d)
Write five integers smaller than 0.
Solution:
Five integers smaller than 0 are:
-1, -2, -3, -4, -5.

10. Encircle the greater integer in each given pair.

(a) -5, -7
(b) 0,-3
(e) 5, 7
(d) -9, 0
(e) -9, -11
(f) -4, 4
(g) -10, -100
(h) 10, 100.
Solution:
(a) -5
(b) 0
(c) 7
(d) 0
(e) -9
(f) 4
(g) -10
(h) 100.

11. Arrange the following integers in ascending order:

Question (a)
0, -7, -9, 5, -3, 2, -4
Solution:
Ascending order of given integers is:
-9, -7, -4, -3, 0, 2, 5

Question (b)
8, -3, 7, 0, -9, -6.
Solution:
Ascending order of given integers is:
-9, -6, -3, 0, 7, 8.

12. Arrange the following integers in descending order:

Question (a)
-9, 3, 4, -6, 8, -3
Solution:
8, 4, 3, -3 -6, -9

Question (b)
4, 8,-3,-2, 5, 0.
Solution:
8, 5, 4, 0, -2, -3.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ACBD. AC = AD and AB bisects ∠ A (see the given figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Answer:
In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠ BAC = ∠ BAD (AB bisects ∠ A)
AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (SAS rule)
∴ BC = BD (CPCT)
Thus, BC and BD are equal.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC

Answer:
In ∆ ABD and ∆ BAC,
AD = BC (Given)
∠ DAB = ∠ CBA (Given)
AB = BA (Common)
∴ ∆ ABD ≅ ∆ BAC (SAS rule)
∴ BD = AC (CPCT)
∴ ∠ ABD = ∠ BAC (CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.

Answer:
AD and BC are equal perpendiculars to line segment AB.
∴ AD = BC and ∠ OAD = ∠ OBC = 90°.
Now, in ∆ ADO and ∆ BCO,
AD = BC
∠ OAD = ∠ OBC
∠ AOD = ∠ BOC (Vertically opposite angles)
∴ ∆ ADO ≅ ∆ BCO (AAS rule)
∴ OA = OB (CPCT)
CD intersects AB at O and OA = OB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:
∆ ABC ≅ ∆ CDA.

Answer:
l || m and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
p l| q and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
Now, in ∆ ABC and ∆ CDA,
∠ BCA = ∠ DAC
∠ BAC = ∠ DCA
AC = CA (Common)
∴ ∆ ABC ≅ ∆ CDA (ASA rule)

Question 5.
Ray l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see the given figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Answer:
l is the bisector of ∠ PAQ and B is any point on l.
∴ ∠ PAB = ∠ QAB
BP and BQ are perpendiculars from B to AP and AQ.
∴ ∠ BPA = ∠ BQA = 90°.
Now, in ∆ APB and ∆ AQB,
∠ PAB = ∠ QAB
∠ BPA = ∠ BQA
AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (AAS rule)
∴ BP = BQ (CPCT)
BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.
∴ BP is the distance of B from AP and BQ is the distance of B from AQ.
Thus, B is equidistant from the arms of ∠ A.

Question 6.
In the given figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Answer:
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ DAE (Adjacent angles)
Now, in ∆ BAC and ∆ DAE,
AC = AE (Given)
AB = AD (Given)
∠ BAC = ∠ DAE
∴ ∆ BAC ≅ ∆ DAE (SAS rule)
∴ BC = DE (CPCT)

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

Answer:
∠ BAD = ∠ ABE
∴ ∠ PAD = ∠PBE (∵ P lies on AB.)
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE (Adjacent angles)
P is the midpoint of AB.
∴ AP = BP
Now, in ∆ DAP and ∆ EBP
∠ PAD = ∠ PBE
∠ APD = ∠ BPE
AP = BP
∴ ∆ DAP ≅ ∆ EBP (ASA rule)
∴ AD = BE (CPCT)

Question 8.
In right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = \(\frac{1}{2}\) AB

Answer:
In ∆ AMC and ∆ BMD,
AM = BM (∵ M is the midpoint of AB.)
CM = DM (Given)
∠ AMC = ∠ BMD (Vertically opposite angles)
∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]
∴ ∠ MCA = ∠ MDB (CPCT)
∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.
∴ AC || BD
Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.
∴ ∠ DBC + ∠ ACB = 180°
∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)
∴ ∠ DBC = 90°
Thus, ∠ DBC is a right angle. [Result (ii)]
Now, ∆ AMC ≅ ∆ BMD
∴ AC = BD
In ∆ DBC and ∆ ACB,
BD = CA
∠ DBC = ∠ ACB (Right angles)
BC = CB (Common)
∴ ∆ DBC ≅ ∆ ACB [Result (iii)]
∴ DC = AB (CPCT)
DM = CM and M lies on line’ segment CD.
∴ DC = 2 CM
∴ AB = 2CM
∴ \(\frac{1}{2}\)AB = CM
∴ CM = \(\frac{1}{2}\)AB

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

1. Construct ΔABC such that AB = 4 cm, ∠B = 30°, BC = 4 cm. Also name the type of triangle on the basis of sides.
Solution:
Given : Two sides of ΔABC as AB = 4 cm, BC = 4 cm and ∠B = 30°.
To construct: A triangle with these two sides and included angle.
Step of Construction :
Step 1. We first draw a rough sketch of the ΔABC and indicate the measure of these two sides and included angle.

Step 2. Draw a line segment BC of length 4 cm.

Step 3. At B draw BX making an angle of 30° with BC (The point A must be somewhere on this ray of the angle).

Step 4. (To fix A, the distance AB has been given) With B as centre, draw an arc of radius 3 cm. It cuts BX at the point A.

Step 5. Join AC.
ΔDEF is now obtained.

Since two sides of triangle are equal.
Therefore ΔABC is an isosceles triangle.

2. Construct ΔABC with AB = 7.5 cm, BC = 5 cm and ∠B = 30°.
Solution:
Given. Two sides of ΔABC as AB = 7.5 cm,
BC = 5 cm
and ∠B = 30°
To construct A triangle with these two sides and included angle.
Steps of Construction.
Step 1. We first draw a rough sketch of the ΔABC and indicate the measures of these two sides and included angle.

Step 2. Draw a line segment BC of length 5 cm.

Step 3. At B draw BX making an angle of 30° with BC. (The point A must be somewhere on this ray of the angle)

Step 4. (To fix A; the distance BC has been given) With B as centre draw an arc of radius 7.5 cm. It cuts CX at the point A.

Step 5 : Join AC.
ΔABC is now obtained.

3. Construct a triangle XYZ, such that XY = 6 cm, YZ = 6 cm and ∠Y = 60°. Also name the type of this triangle.
Solution:
Step 1. Draw a rough sketch of XYZ with given measures.

Step 2. Draw a line segment XY of length 6 cm.

Step 3. With the help of compass, at Y, draw a ray YA making an angle 60°

Step 4. With Y as centre and radius 6 cm. draw an arc intersecting the ray YX at point Z.

Step 5. Join XZ.ΔXYZ is required triangle, Measure the third side. We see that ZX = 6 cm
∴ In Δ XYZ
XY = YZ = ZX = 6 cm
Therefore XYZ is an equilateral triangle.

4. Which of the following triangle can be constructed using SAS criterion.
(a) AB = 5 cm, BC = 5 cm, CA = 6 cm
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°
(c) ∠A = 60°, ∠B = 60°, ∠C = 60°
(d) BC = 5 cm, ∠B = ∠C = 45°
Answer:
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

1. Construct ΔABC in which AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
Solution:
Given : Three sides of a triangle as AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
To construct : A triangle with these three sides.
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment BC = 5 cm.

Step 3 : From B; point A is at a distance 3.5 cm. So, with B as centre, draw an arc of radius
3.5 cm (Now point A will be some where on this arc. Our job is to find where exactly A is.)

Step 4. From C, point A is at a distance of 7 cm. So, with C as centre; draw an arc of radius 7 cm. (A will be some where on this arc. We have to fix it).

Step 5. A has to be on both the arcs drawn. So it is the point of intersection of arcs.
Mark the point of intersection of arcs as A. Join AB and AC.
Thus we obtain ΔABC.

2. Construct a triangle ABC in which AB = BC = 6.5 cm and CA = 4 cm. Also name the kind of triangle drawn.
Solution:
Given : Three sides of triangle as AB = BC = 6.5 cm. and CA = 4 cm.
To construct : A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment AC = 4 cm.

Step 3. From A; point B is at a distance of 6.5 cm. So, with A as centre, draw an arc of radius 6.5 cm. (Now point B will be somewhere on this arc. Our job is to find where exactly A is.)

Step 4. From C; point B is at a distance of 6.5 cm. So; with C as centre; draw an arc of radius 6.5 cm. (B will be some where on this arc. We have to fix it.)

Step 5. B has to be on both the arcs drawn. So it is the point of intersection of arcs. Mark the point of intersection of arcs as B. Join AB and BC.

We observe that AB = BC = 6.5 cm.
Since two sides are of equal length. Thus we obtain an isosceles ΔABC.

3. Construct a triangle XYZ such that length of each side is 5 cm. Also name the kind of triangle drawn.
Solution:
Given : A triangle XYZ in which XY = YZ = ZX = 5 cm..
To Construct. A triangle XYZ with each side 5 cm.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle XYZ and indicate the lengths of the three sides.

Step 2. Draw a line segment YZ = 5 cm.

Step 3. From Y; point X is at a distance of 5 cm. So, with Y as centre, draw an arc of radius 5 cm. (Now point X will be somewhere on this arc. Our job is to find where exactly X is.)

Step 4. From Z, point X is at a distance of 5 cm. So, with Z as centre, draw an arc of radius 5 cm. (X will be somewhere oh this arc. We have to fix it.)

Step 5. Point X has to be on both the arcs drawn. So, it is the point of intersection of arcs.
Mark the point of intersection of arcs as X. Join XY and XZ.

Thus we obtain an equilateral ΔXYZ each of whose side is 5 cm.

4. Construct a triangle PQR such that PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm. Measure ∠PQR and also name the kind of triangle drawn.
Solution:
Given. Three sides of triangle as PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm.
To construct. A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the lengths of the three sides.

Step 2. Draw a line segment QR of length 6 cm.

Step 3. From Q; point P is at a distance of 2.5 cm. So, with Q as centre, draw an arc of radius 2.5 cm. (Now point P will be some where on this arc. Our job is to find where exactly P is.)

Step 4. From R; point P is at a distance of 6.5 cm. So; with R as centre; draw an arc of radius 6.5 cm. (P will be somewhere on this arc. We have to fix it.)

Step 5. Point P has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark a point of intersection of arcs as P. Join PQ and PR.

Thus we obtain ΔPQR; on measuring ∠PQR, we observe that ∠PQR = 90° so it is a right angled triangle.

5. Construct a triangle ABC, in which AB = 6 cm, BC = 2 cm, CA = 3 cm. (If possible). If not possible give the reason.
Solution:
Since AB = 6 cm, BC = 2 cm, CA = 3 cm
Here BC + CA = 2 cm + 3 cm
= 5 cm < 6 cm < AB
Which is not possible because the sum of two sides of a triangle is always greater than third side of the triangle.

6. Question (i).
Which of the following can be used to construct a triangle ?
(a) The lengths of the three sides
(b) The perimeter of the triangle
(c) The measures of three angles
(d) The name of three vertices
Answer:
(a) The lengths of the three sides

Question (ii).
A triangle can be constructed by taking its sides as :
(a) 1.8 cm, 2.6 cm, 4.4 cm
(b) 3 cm, 4 cm, 8 cm
(c) 4 cm, 7 cm, 2 cm
(d) 5 cm, 4 cm, 4 cm.
Answer:
(d) 5 cm, 4 cm, 4 cm.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 3 Playing with Numbers MCQ Questions

Multiple Choice Questions

Question 1.
Which number is a factor of every, number?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 2.
How many even numbers are prime?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 3.
The smallest composite number is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 4.
Which of the following number is a perfect number?
(a) 8
(b) 6
(c) 12
(d) 18.
Answer:
(b) 6

Question 5.
Which of the following is not a multiple of 7?
(a) 35
(b) 48
(c) 56
(d) 91.
Answer:
(b) 48

Question 6.
Which of the following is not a factor of 36?
(a) 12
(b) 6
(c) 9
(d) 8.
Answer:
(d) 8.

Question 7.
The number of prime numbers upto 25 are:
(a) 9
(b) 10
(c) 8
(d) 12.
Answer:
(a) 9

Question 8.
Which mathematician gave the method to find prime and composite numbers?
(a) Aryabhatta
(b) Ramayan
(c) Eratosthenes
(d) Goldbach.
Answer:
(c) Eratosthenes

Question 9.
The statement “Every even number greater than 4 can be expressed as the sum of two odd prime numbers” is given by:
(a) Goldbach
(b) Eratosthenes
(c) Aryabhatta
(d) Ramanujan.
Answer:
(a) Goldbach

Question 10.
Which of the following is a prime number?
(a) 221
(b) 195
(c) 97
(d) 111.
Answer:
(c) 97

Question 11.
Which of the following number is divisible by 4?
(a) 52369
(b) 25746
(c) 21564
(d) 83426.
Answer:
(c) 21564

Question 12.
Which of the following is not true?
(a) If a number is factor of two numbers then it is also factor of their sum.
(b) If a number is factor of two numbers then it is also factor of their difference.
(c) 15 and 24 are co-prime to each other.
(d) 1 is neither prime nor composite.
Answer:
(c) 15 and 24 are co-prime to each other.

Question 13.
Which of the following pair is co-prime?
(a) (12, 25)
(b) (18, 27)
(c) (25, 35)
(d) (21, 56).
Answer:
(a) (12, 25)

Question 14.
Which of the following number is divisible by 8?
(a) 123568
(b) 412580
(c) 258124
(d) 453230.
Answer:
(a) 123568

Question 15.
Prime factorisation of 84:
(a) 2 × 2 × 3 × 2 × 7
(b) 7 × 2 × 3 × 3
(c) 2 × 3 × 7 × 2
(d) 3 × 2 × 3 × 2 × 7.
Answer:
(c) 2 × 3 × 7 × 2

Question 16.
H.C.F. of 25 and 45 is:
(a) 15
(b) 5
(c) 225
(d) 135.
Answer:
(b) 5

Question 17.
If L.C.M. of two numbers is 36 then which of the following can not be their H.C.F.?
(a) 9
(b) 12
(c) 8
(d) 18.
Answer:
(c) 8

Question 18.
The L.C.M. of two co-prime numbers is 143. If one number is 11 then find other number.
(a) 132
(b) 154
(c) 18
(d) 13.
Answer:
(d) 13.

Question 19.
Find the greatest number which divides 145 and 235 leaving the remainder 1 in each case.
(a) 24
(b) 18
(c) 19
(d) 17.
Answer:
(b) 18

Question 20.
The greatest 4 digit number which is divisible by 12,15 and 20.
(a) 9990
(b) 9000
(c) 9960
(d) 9999.
Answer:
(c) 9960

Question 21.
Which of the following is a prime number?
(a) 23
(b) 51
(c) 39
(d) 26.
Answer:
(a) 23

Question 22.
Which of die following is a prime number?
(a) 32
(b) 30
(c) 31.
(d) 33.
Answer:
(c) 31.

Question 23.
Which of the following is a composite number?
(a) 12
(b) 19
(c) 29
(d) 31.
Answer:
(a) 12

Question 24.
Which of the following is an even number?
(a) 13
(b) 15
(c) 16
(d) 19.
Answer:
(c) 16

Question 25.
Which of the following is an odd number?
(a) 12
(b) 13
(c) 14
(d) 20.
Answer:
(b) 13

Fill in the blanks:

Question 1.
…………… is an even prime number?
Answer:
2

Question 2.
…………… is the greatest prime number between 1 and 10.
Answer:
7

Question 3.
……………. is neither prime nor composite number.
Answer:
1

Question 4.
A number which has only two factors is called a …………….. number.
Answer:
prime number

Question 5.
A number which has more than two factors is called a ……………… number.
Answer:
composite number

Write True/False:

Question 1.
The sum of three odd number is even. (True/False)
Answer:
False

Question 2.
All prime numbers are odd. (True/False)
Answer:
False

Question 3.
All even numbers are composite numbers. (True/False)
Answer:
False

Question 4.
1 neither prime nor composite. (True/False)
Answer:
True

Question 5.
If a number is factor of two numbers then it is also factor of their sum. (True/False)
Answer:
True