Class 4

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.1

Question 1.
Write the time shown in each clock given below:
(a)

Solution:
1:55

(b)

Solution:
7:10

(c)

Solution:
9:05

(d)

Solution:
2:40

(e)

Solution:
10:40

(f)

Solution:
11:45

Question 2.
Draw clocks in your note book and show the time as given below :
(a) 4:20
(b) 7:35
(c) 4:45
(d) 3:15
(e) 11:40
(f) 9:15.
Solution:

Question 3.
How many minutes a minute hand will take to reach the time shown between the first clock and the second clock ?
(a)

Solution:
15 minutes

(b)

Solution:
25 minutes

Question 4.
Tell the time shown in the given clock and write.

Solution:
4:18

Question 5.
Tell the time shown in the given clock and write.

Solution:
5:58

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.9

Question 1.
In morning assembly 161 students stand in 7 rows equally. How many students stand in each row ?
Solution:
Total number of student = 161
Number of rows = 7
Number of students in each row = 161 ÷ 7 = 23

Question 2.
I have 72 apples I have to put these in 3 buckets equally. How many apples each basket will contain ?
Solution:
The total number of apples = 72
Number of baskets = 3
Number of apples in each basket = 72 ÷ 3 = 24

Question 3.
A farmer has produced 4250 kg wheat in his Held. One bag is required to fill for 50 kg of wheat. How many such bags are required to fill 4250 kg wheat ?
Solution:
Total quantity of wheat produced = 4250 kg
Quantity of wheat in one bag = 50 kg
Number of bags required = 4250 ÷ 50 = 85

Question 4.
By which number to multiply 25 so that product becomes 625 ?
Solution:
Product of the two numbers = 625
One of the number = 25 Second number = 625 ÷ 25 = 25

Question 5.
A gardener has 120 flowers. He has to make garland of 24 flowers. How many such garlands are made from 120 flowers ?
Solution:
The total number of flowers = 120
Number of flowers in 1 garland = 24
Number of garlands that can be made = 120 ÷ 24 = 5

Question 6.
How many ₹ 50 notes are required to make ₹ 2000 ?
Solution:
Total amount = ₹ 2000
Value of 1 note = ₹ 50
Number of notes required = ₹ 2000 ÷ ₹ 50

Question 7.
I want to exchange my ₹ 500 note. How many notes of following denomination will I get ?
(a) If all are ₹ 100 notes …………………….
(b) If all are ₹ 50 notes ……………………
(c) If all are ₹ 10 notes ……………………..
Solution:
(a) If all are ₹ 100 notes then ₹ 500 ÷ ₹ 100

= 5 notes are required to exchange ₹ 500 note.

(b) If all are ₹ 50 notes then ₹ 500 ÷ ₹ 50

= 10 notes are required to exchange
= ₹ 500 note

(c) If all are ₹ 10 notes then ₹ 500 ÷ ₹ 10

= 50 notes are required to exchange a ₹ 500

Question 8.
A labourer picks up 20 bricks in 1 round. How many number of rounds are required to pick up 1000 bricks ?
Solution:
Total number of bricks = 1000
Number of bricks picked in 1 round = 20
Number of rounds = 1000 ÷ 20 = 50

Question 9.
A railway ticket costs ₹ 24. palak gave ₹ 576 to the Station Master. How many mumber of tickets did she get ?
Solution:
The cost of 1 ticket = ₹ 24
Palak gave the money = ₹ 576
No of tickets = ₹ 576 ÷ 24 = 24

Question 10.
Kashvi brought a toffees packet on her birthday. There are 175 toffees in this packet and there are 35 students in her class. How many toffees did each of them get.
Solution:
Total ho. of toffees = 175
No. of children = 35
No. of toffees that each child will get = 175 ÷ 35 = 5

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.8

1. Fill in the boxes of Q. No. 1 and Q.no. 2 as directed :

Question 1.

Question 2.
9 × 4 = 36 __________ ____________
Solution:
36 ÷ 9 = 4, 36 ÷ 4 = 6

Question 3.
6 × 8 = 48 __________ ____________
Solution:
48 ÷ 6 = 8, 48 ÷ 8 = 6

Question 4.
10 × 4 = 40 __________ ____________
Solution:
40 ÷ 10 = 4, 40 ÷ 4 = 10

2.

Question 1.

Question 2.
35 ÷ 7 = 5 _____________ _____________
Solution:
5 × 7 = 35, 7 × 5 = 35

Question 3.
56 ÷ 8 = 7 _____________ _____________
Solution:
7 × 8 = 56, 8 × 7 = 56

Question 4.
150 ÷ 10 = 15 _____________ _____________
Solution:
10 × 15 = 150, 15 × 10 = 150

Question 5.
120 ÷ 10 = 10 _____________ _____________
Solution:
10 × 12 = 120, 12 × 10 = 120

3. Divide and verify the solution :

Question 1.
66 ÷ 6
Solution:

Quotient = 11
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
66 = 11 × 6 + 0
66 = 66

Question 2.
431 ÷ 7
Solution:

Quotient = 61
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
431 = 61 × 7 + 4
431 = 427 + 4
431 = 431

Question 3.
728 ÷ 8
Solution:

Quotient = 91
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
728 = 91 × 8
728 = 728

Question 4.
648 ÷ 9
Solution:

Quotient = 72
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
648 = 72 × 9 + 0
648 = 648

Question 5.
960 ÷ 5
Solution:

Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
960 = 192 × 5 + 0
960 = 960

4. Solve the following :

Question 1.
666 ÷ 6
Solution:

Quotient = 111

Question 2.
655 ÷ 5
Solution:

Quotient = 131

Question 3.
787 ÷ 7
Solution:

Quotient = 112
Remainder = 3

Question 4.
877 ÷ 7
Solution:

Quotient = 125
Remainder = 2

Question 5.
598 ÷ 6
Solution:

Quotient = 99
Remainder = 4

Question 6.
566 ÷ 8
Solution:

Quotient = 70
Remainder = 6

Question 7.
707 ÷ 7
Solution:

Quotient = 101
Remainder = 0

5. Solve the following :

Question 1.
2150 ÷ 2
Solution:

Quotient = 1075

Question 2.
4050 ÷ 3
Solution:

Quotient = 1350

Question 3.
8048 ÷ 8
Solution:

Quotient = 1006

Question 4.
5106 ÷ 6
Solution:

Quotient = 851

Question 5.
3043 ÷ 3
Solution:

Quotient = 1014
Remainder = 1

Question 6.
7890 ÷ 7
Solution:

Quotient = 1127
Remainder = 1

Question 7.
4050 ÷ 5
Solution:

Quotient = 810

6. Divide and verify the following:

Question 1.
96 ÷ 12
Solution:

Quotient = 8
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
96 = 8 × 12 + 0
96 = 96

Question 2.
98 ÷ 14
Solution:

Quotient = 7
Verification : Dividend = Quotient × Divisor + Remainder
98 = 14 × 7 + 0
98 = 98

Question 3.
78 ÷ 16
Solution:

Quotient = 4
Remainder = 14
Verification : Dividend = Quotient × Divisor + Remainder
78 = 4 × 16 + 14
78 = 64 + 14
78 = 78

Question 4.
760 ÷ 19
Solution:

Quotient = 40
Verification : Dividend = Quotient × Divisor + Remainder
760 = 40 × 19 + 0
760 = 760

Question 5.
550 ÷ 13
Solution:

Quotient = 42
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
550 = 42 × 13 + 4
550 = 446 + 4
550 = 550

Question 6.
894 ÷ 24
Solution:

Quotient = 37
Remainder = 6
Verification : Dividend = Quotient × Divisor + Remainder
894 = 37 × 24 + 6
894 = 888 + 6
894 = 894

Question 7.
913 ÷ 66
Solution:

Quotient = 13
Remainder = 55
Verification : Dividend = Quotient × Divisor + Remainder
913 = 13 × 66 + 55
913 = 858 + 55
913 = 913

Question 8.
826 ÷ 34
Solution:

Quotient = 24
Remainder = 10
Verification : Dividend = Quotient × Divisor + Remainder
826 = 24 × 34 + 10
826 = 816 + 10
826 = 826

Question 9.
7645 ÷ 12
Solution:

Quotient = 637
Remainder = 1
Verification : Dividend = Quotient × Divisor + Remainder
7645 = 637 × 12 + 1
7645 = 7644 + 1
7645 = 7645

Question 10.
7813 ÷ 13
Solution:

Quotient = 601
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
7813 = 601 × 13 + 0
7813 = 7813

Question 11.
5375 ÷ 25
Solution:

Quotient = 215
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
5375 = 215 × 25 + 0
5375 = 5375

Question 12.
6767 ÷ 33
Solution:

Quotient = 205
Remainder = 2
Verification : Dividend = Quotient × Divisor + Remainder
6767 = 205 × 33 + 2
6767 = 6765 + 2
6767 = 6767

Question 13.
9600 ÷ 50
Solution:

Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9600 = 192 × 50 + 0
9600 = 9600

Question 14.
9999 ÷ 33
Solution:

Quotient = 303
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9999 = 303 × 33 + 0
9999 = 9999

Question 15.
9660 ÷ 60
Solution:

Quotient = 161
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9660 = 161 × 60 + 0
9660 = 9660

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.7

1. Fill in the blank:

Question 1.
18 ÷ 9 = ____________
Solution:
2

Question 2.
77 ÷ 7 = ___________
Solution:
11

Question 3.
48 ÷ 8 = __________
Solution:
6

Question 4.
78 ÷ ________ = 6
Solution:
13

Question 5.
42 ÷ 7 = __________
Solution:
6

Question 6.
84 ÷ 14 = ____________
Solution:
6

Question 7.
28 ÷ __________ = 7
Solution:
4

Question 8.
0 ÷ 8 = __________
Solution:
0

Question 9.
50 ÷ 5 = _________
Solution:
10

Question 10.
12 ÷ 1 = __________
Solution:
12

Question 11.
54 ÷ __________ = 9
Solution:
6

Question 12.
________ ÷ 15 = 1
Solution:
15

Question 13.
70 ÷ 5 = ___________
Solution:
14

Question 14.
100 ÷ 10 = _________
Solution:
10

Question 15.
81 ÷ 9 = __________
Solution:
9

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.6

Question 1.
Cost of 1 notebook is ₹ 15. Find the cost of 9 such notebooks ?
Solution:
Cost of 1 notebook = ₹ 15
Cost of 9 notebooks = ₹ 15 × 9 = ₹ 135.

Question 2.
There are 75 pencils in a box. How many pencils are there in 19 such boxes ?
Solution:
Number of pencils in a packet = 75
Number of pencils in 19 packets = 75 × 19
= 1425.

Question 3.
There are 79 beads in a chain. How many beads are there in 68 such chains ?
Solution:
Number of beads in one chain = 79
Number of beads in 68 chains = 79 × 68
= 5372.

Question 4.
The cost of a toycycle is ? 1560. Find total cost of 6 such toycycles ?
Solution:
The cost of 1 toycycle = ₹ 1560
The cost of 6 toycycles = ₹ 1560 × 6
= ₹ 9360.

Question 5.
There are 11 players in a cricket team. How many players are there in 12 such teams ?
Solution:
Number of players in 1 cricket team = 11
Number of players in 12 cricket teams = 11 × 12 = 132.

Question 6.
A box contains 1440 soaps. Find the numbers of soaps in 6 such boxes ?
Solution:
Number of soaps in 1 box = 1440
Number of soaps in 6 boxes = 1440 × 6
= 8640

Question 7.

Your mom went to market—
(a) She bought 2 kg apples, 2 kg. guava. How much amount she will pay to the seller ?
Solution:
Cost price of 2 kg apples
= ₹ 120 × 2 = ₹ 240
Cost price of 2 kg guava
= ₹ 35 × 2 = ₹ 70
Amount she will pay to the seller = ₹ 310.

(b) If she bought 3 kg oranges and 2 kg pomegranate. What amount she will pay to seller ?
Solution:
Cost price of 3 kg oranges
= ₹ 45 × 3 = ₹ 135
Cost price 2 kg pomegranate
= ₹ 40 × 2 = ₹ 80
Amount she will pay to seller = ₹ 215

Question 8.
These all notes and coins Karan received on his birthday. How much total amount did Karan receive ?

Solution:
= ₹ 500 × 5 + ₹ 50 × 3 + ₹ 10 × 7 + ₹ 2 × 3
= ₹ 2500 + ₹ 150 + ₹ 70 + ₹ 6
= ₹ 2726.

Question 9.
A car covers a distance of 16 km in a litre. How much distance it will cover in 28 litres ?
Solution:
Distance covered in 1 litre = 16 km
Distance covered in 28 litres = 16 km × 28

= 448 km.

Question 10.
A factory produces 125 soap bars in an hour. How many such soap bars will be produced in 8 hours ?
Solution:
Number of soaps produced in 1 hour = 125
Number of soaps produced in 8 hours = 8 × 125
= 1000

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 7 Shapes Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 7 Shapes Ex 7.2

Question 1.
Which shapes can we get from the following net ?

Solution:

Question 2.
How does a brick look from the top view ?

Solution:

Question 3.
Complete the pattern by filling colours :

Solution:
Filling colours:

Question 4.
Which tile would complete the following desings?

Solution:
I.

II.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.5

1. Fill in the blanks :

Question 1.
4 × 1 =
Solution:
4

Question 2.
5 × 10 =
Solution:
20

Question 3.
6 × 100 =
Solution:
600

Question 4.
190 × 0 =
Solution:
0

Question 5.
19 × = 1900
Solution:
100

Question 6.
× 100 = 1600
Solution:
16

Question 7.
× 791 = 0
Solution:
0

Question 8.
× 9 = 9 × 8
Solution:
8

Question 9.
4 × 10 =
Solution:
40

Question 10.
7 × 100 =
Solution:
700

Question 11.
9 × 1000 =
Solution:
9000

Question 12.
10 × 1000 =
Solution:
10000

Question 13.
15 × = 150
Solution:
10

Question 14.
× 10 = 760
Solution:
76

Question 15.
798 × = 798.
Solution:
1

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.4

1. Find the product of the following :

Question 1.
41 × 4
Solution:

Question 2.
25 × 36
Solution:

Question 3.
445 × 22
Solution:

Question 4.
269 × 36
Solution:

Question 5.
368 × 19
Solution:

Question 6.
145 × 68
Solution:

Question 7.
150 × 59
Solution:

Question 8.
4639 × 2
Solution:

Question 9.
1569 × 6
Solution:

Question 10.
1179 × 8
Solution:

Question 11.
1988 × 5
Solution:

Question 12.
5000 × 2
Solution:

Question 13.
303 × 31
Solution:

Question 14.
425 × 17
Solution:

Question 15.
706 × 12
Solution:

Question 16.
308 × 28
Solution:

2. Find product using expanded notation method:

Question 1.
52 × 7
Solution:
52 × 7 = (50 + 2) × 7
= 50 × 7 + 2 × 7
= 350 + 14
= 364.

Question 2.
63 × 4
Solution:
63 × 4 = (60 + 3) × 4
= 60 × 4 + 3 × 4
= 240 + 12
= 252

Question 3.
81 × 9
Solution:
81 × 9 = (80 + 1) × 9
= 80 × 9 + 1 × 9
= 720 + 9
= 729

Question 4.
123 × 5
Solution:
123 × 5 = (100 + 20 + 3) × 5
= 100 × 5 + 20 × 5 + 3 × 5
= 500 + 100 + 15
= 615

Question 5.
205 × 6
Solution:
205 × 6 = (200 + 5) × 6
= 200 × 6 + 5 × 6
= 1200 + 30
= 1230

3. Find product using Lattice algorithm.

Question 1.
43 × 15

Solution:

Step 1: 43 × 1 (Shown in picture).
Step 2 : 43 × 5 (Shown in picture).
Step 3 : Add diagonally.

Question 2.
426 × 35

Solution:

Step 1: 426 × 3 (Shown in picture).
Step 2 : 426 × 5 (Shown in picture).
Step 3 : By adding diagonal elements
we got 1, 4, 9, 1, 0. which gives us no.
14910.
₹ 426 × 35 = 14910.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.3

1.

Question 1.
Find the sum of 1198,1296 and 796.
Solution:

Question 2.
Find the difference of 7693 and 4566.
Solution:

Question 2.
The price of a fan is ₹ 11467 and the price of a cooler is ₹ 2215. How much total amount is required to buy both the things ?
Solution:
Cost price of the fan = ₹ 1467
Cost price of the cooler = ₹ 2275

Question 3.
Karan had ₹ 19080. He bought clothes worth ₹ 3705. How much amount was left with him ?
Solution:
The amount that Karan had with him = ₹ 9080
Amount spent on clothes = ₹ 3705
Amount left with him = ₹ 5375

Question 4.
In a library there are 3115 Punjabi books, 2876 Maths books and 976 English books. What is the total number of books in library ?
Solution:
Number of Punjabi books in the library = 3115
Number of Maths books in the library = 2876
Number of English books in the library = 976
The total number of books in library = 6967

Question 5.
The sum of two numbers is 9030. One number is 2141. Find the other number.
Solution:
Sum of two numbers = 9030
One number = 2141
Other number = 6889

Question 6.
What should be added in 7569 to get 9000 ?
Solution:
Sum = 9000
Number = – 7569

= 1431

7. Find the number which is :

Question 1.
778 more than 3792
Solution:
Required number = 3792 + 778
= 4570

Question 2.
515 less than 3777.
Solution:
Required number = 3777 – 515
= 3262

Question 8.
The price of an almirah is ₹ 1595 and price of refrigerator is ₹ 6055 more than Almirah.
(a) Find the price of refrigerator
(b) Find the total price of almirah and refrigerator.
Solution:
Cost price of Almirah = ₹ 1595
(a) Cost price of refrigerator
= ₹ 1595 + ₹ 6055
= ₹ 7650.

(b) Total cost price of almirah and refrigerator
= ₹ 1595 + ₹ 7650
= ₹ 9245.

Question 9.
Find the greatest and smallest 4 digit numbers by using digits 1, 4, 6 and 7 and also find the sum and difference of these numbers.
Solution:
The greatest 4 digit numbers = 7641
The smallest 4 digit numbers = 1467
Sum of these numbers = 9108

Difference of these numbers

Question 10.
Find the sum of the smallest 4 digit number and greatest 3 digit number.
Solution:
The smallest 4 digit number = 1000
The smallest 3 digit number = 999
Sum of these numbers = 1999

Question 11.
Find the difference between place value of 8 and place value of 7 in the number 9874.
Solution:
Place value of 8 in the number
9874 = 8 × 100 = 800
Place value of 7 in the number
9874 = 7 × 10 = -70
Difference = 730

Question 12.
Subtract 248 from smallest 4 digit number.
Solution:
The smallest 4 digit number

Question 13.
Satnam had ₹ 765, his uncle gave him ₹ 250. Satnam gave ₹ 370 to his sister from the total amount. How much amount is left with him ?
Solution:
The amount that Satnam had with him = ₹ 765
The amount that his uncle has given to him = ₹ 250
The total amount that Satnam has now = ₹ 1015
The amount given by him to his sister = ₹ 370
The amount left with him = ₹ 645

Question 14.
Rozi had ₹ 1000. She bought a pair of shoes worth ₹ 150 and a suit worth ₹ 360. How much money was left with her ?
Solution:
The cost price of a pair of shoes = ₹ 150
The cost price of a pair of a suit = ₹ 360
The total cost price of both = ₹ 510
The total amount Rozi had with her = ₹ 1000
The amount spent by her = ₹ 510
The amount left with her = ₹ 490

Question 15.
Sandeep has ₹ 785 in his bank account. How much money should he deposit so that his total balance becomes ₹ 1000 ?
Solution:
The amount that should be in account = ₹ 1000
The amount he has in his account = ₹ 785
The amount that should be deposited = ₹ 215

Question 16.
The distance between Ferozepur to Chandigarh is 220 km. However distance between Ferozepur to Bathinda is 98 km. How much distance from Ferozepur to Chandigarh is more than that of Ferozepur to Bathinda ?
Solution:
The distance between Ferozepur and Chandigarh = 220 km
The distance between Ferozepur and Bathinda = 98 km
The distance from Ferozepur to Chandigarh is more than that of Ferozepur to Bathinda = 122 km.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 7 Shapes Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 7 Shapes Ex 7.1

Question 1.
From the given figure write the names of the following :
(a) radius
(b) diameter
(c) chords

Solution:
(a) Radius = OC, OB, OG, OD, OE, OA
(b) Diameter = AB,EG .
(c) Chords = AF, AB, EG.

Question 2.
Find the radius of a circle whose diameter is :
(a) 6 cm
(b) 8.2 cm
(c) 8.6 cm
Solution:
(a) Diameter of the circle = 6 cm

= \(\frac{6}{2}\) cm = 3 cm

(b) Diameter of the circle = 8.2 cm

= \(\frac{8.2}{2}\) cm = 4.1 cm

(c) Diameter of the circle = 8.6 cm

= \(\frac{8.6}{2}\) cm = 4.3 cm

Question 3.
Find the diameter of a circle whose radius is :
(a) 13 cm
(b) 21 cm
(c) 17 cm
(d) 8 cm
Solution:
(a) Radius of the circle = 13 cm
Diameter of the circle = 2 × Radius
= 2 × 13 cm = 26 cm
(b) Radius of the circle = 21 cm
Diameter of the circle = 2 × Radius
= 2 × 21 cm
= 42 cm
(c) Radius of thte circle = 17 cm
Diameter of the circle = 2 × Radius
= 2 × 17 cm
= 34 cm
(d) Radius of the circle = 8 cm
Diameter of the circle = 2 × Radius
= 2 × 8 cm
= 16 cm

Question 4.
With the help of a compass draw a circle whose radius is :
(a) 5 cm
(b) 3 cm
(c) 2 cm
(d) 3.5 cm
(e) 4.6 cm
(f) 2.5 cm
Solution:

Question 5.
Which is the longest chord of a circle ?
Solution:
Diameter.

Question 6.
Fill in the blanks :
(a) A line segment which joins centre of a circle with any point on circumference is called …………
(b) Diameter of a circle = ………. × radius.
(c) The longest chord of a circle is called …….. of circle.
(d) All the radii of circle are …… in length.
Solution:
(a) Radius
(b) 2
(c) Diameter
(d) Equal

Question 7.
Fill the blank

Solution: