Bhagya

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:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.2

1. Replace each * by correct digit in each of the following :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

2. Solve the following:

Question 1.
48 – 12 + 18
Solution:

Question 2.
86 – 35 – 12
Solution:

Question 3.
637 – 452 + 315
Solution:

Question 4.
637 + 315 – 452
Solution:

Question 5.
1837 + 3043 – 413
Solution:

Question 6.
937 – 413 + 3043
Solution:

Question 7.
1003 – 417 – 284
Solution:

Question 8.
9419 – 4419 + 2105
Solution:

Question 9.
2419 + 5005 – 4419
Solution:

Question 10.
2294 + 1828 – 1374.
Solution:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.1

Question 1.
Add 4 and 2 on a number line.

Solution:

1. First of all put a mark on 4 on number line.
2. To add 2, go from left to right one by one two steps.
3. Now we will reach on 6 and this is our answer.
   ∴ 4 + 2 = 6.

Question 2.
Add 6 and 4 on a number line.

Solution:

1. First of all put a mark on 6 on number line.
2. To add 4, go from left to right one by one four steps.
3. Now, we will reach on 10 and this is our answer.
   ∴ 6 + 4 = 10.

Question 3.
Subtract 2 from 6 on number line.

Solution:

1. First of all put a mark on 6 on number line.
2. To subtract 2, go back from right to left one by one two steps.
3. Now, we will reach on 4 and this our answer.
   ∴ 6 – 2 = 4.

Question 4.
Subtract 6 from 11 on a number line.

Solution:

1. First of all pur a mark on 11 on the number line.
2. To subtract 6, go back from right to left, one by one, six steps.
3. Now, we will reach on 5 and this is our answer.
   ∴ 11 – 6 = 5.

5. Solve the following :

Question 1.
374 + 202
Solution:

Question 2.
356 + 122
Solution:

Question 3.
4251 + 1244
Solution:

Question 4.
7000 + 1789
Solution:

Question 5.
999 – 234
Solution:

Question 6.
798 – 130
Solution:

Question 7.
9825 – 1214
Solution:

Question 8.
7896 – 1234.
Solution:

6. Solve the following:

Question 1.
769 + 584
Solution:

Question 2.
649 + 161
Solution:

Question 3.
3009 + 5691
Solution:

Question 4.
2347 + 7437
Solution:

Question 5.
769 + 444 + 325
Solution:

Question 6.
688 + 100 + 135
Solution:

Question 7.
2807 + 5938 + 1238
Solution:

Question 8.
7644 + 166 + 1234
Solution:

Question 9.
768 – 119
Solution:

Question 10.
6307 – 4156
Solution:

Question 11.
7503 – 1219
Solution:

Question 12.
7000 – 1234.
Solution:

7. Subtract and verify your answer :

Question 1.
7610 – 1733
Solution:

Question 2.
6113 – 1167
Solution:

Question 3.
6501 – 1212
Solution:

Question 4.
4368 – 1239
Solution:

Question 5.
7001 – 1678.
Solution:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 10 Patterns Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 10 Patterns Ex 10.1

Question 1.
Observe the pattern carefully and complete it.

Solution:

Question 2.
Students of a particular school who have liking for different types of fruits.
(a) 9981
(b) 6039
(c) 243
(d) 6308
(e) 6415
(f) 9108
(g) 1728
(h) 8714
(i) 53694
(j) 40819.
Solution:
(a) 9981
First of all omit 9’s from the given number

Remaining digits of number = 8,1
Now add the remaining• digits e.
= 8, 1
8 + 1 = 9
omit this 9
Now there is no digit left’
Hence, this number is a multiple of 9
Second Method: 9981
Sum of the digits = 9 + 9 + 8 + 1 = 27

Because, sum of digits is divisible by 9
Hence, this number is a multiple of 9

(b) 6039
Sum of the digits = 6 + 0 + 3 + 9 = 18
Because sum of digits is divisible by 9

Hence, this number is multiple of 9.

(c) 243
Sum of the digits = 2 + 4 + 3 = 9

Because, sum of digits is divisible by 9
Hence, this number is a multiple of 9

(d) 6308
Sum of the digits = 6 + 3 + 0 + 8 = 17

Because sum of digits is not divisible by 9
Hence, this number is not a multiple of 9
(e) 6415
Sum of the digits = 6 + 4 + 1 + 5 = 16

Because, sum of digits is not divisible by 9
Hence, this number is not a multiple of 9

(f) 9108
Sum of the digits = 9 + 1 + 0 + 8 = 18

Because sum of digits is divisible by 9
Hence, this number is a multiple of 9

(g) 1728
Sum of the digits = 1 + 7 + 2 + 8 = 18

Because sum of digits is divisible by 9
Hence, this number is a multiple of 9

(h) 8714
Sum of the digits = 8 + 7 + 1 + 4 = 20

Because sum of digits is divisible by 9
Hence, this number is a multiple of 9

(i) 53694
Sum of the digits = 5 + 3 + 6 + 9 + 4 = 27

Because sum of digits is divisible by 9
Hence, this number is a multiple of 9

(j) 40819.
Sum of the digits = 4 + 0 + 8 + 1 + 9 = 22

Because sum of digits is not divisible by 9
Hence, this number is nòt a multiple of 9

Question 3.
Multiply the given numbers using pattern:
(a) 35 × 10 = ………
(b) 9 × 10 = ………
(c) 21 × 10 = ……..
(d) 106 × 10 = ……..
(e) 148 × 10 = ………
(f) 2 × 1oo = ……….
(g) 20 × 1oo = ………
(h) 38 × 100 = ……..
(i) 209 × 100 = ……..
(j) 406 × 100 = ………
Solution:
(a) 350,
(b) 90,
(c) 210,
(d) 1060,
(e) 1480,
(f) 200,
(g) 2000,
(h) 3800,
(j) 20900,
(j) 40600.

Question 4.
Divide the given numbers (Using patterns) :
(a) 60 ÷ 10 = ………..
(b) 700 ÷ 10 = ……..
(c) 96 ÷ 10 = ……….
(d) 600 ÷ 100 = ……….
(e) 1500 ÷ 100 = ……….
(f) 1000 ÷ 100 = …….
Solution:
(a) 6,
(b) 70,
(c) 96,
(d) 6,
(e) 15,
(f) 10.

Question 5.
Fill in the blanks:
(a) …….. × 10 = 500
(b) …… ÷ 10 = 96
(c) ……. × 100 = 900
(d) …….. ÷ 100 = 7
Solution:
(a) 50,
(b) 960,
(c) 9,
(d) 700.

Question 6.
Complete the following towers:



Solution:
(a) 5 + 6 = 11, 7 + 9 = 16
(b) 6 + 8 = 14, 10 + 14 = 24, 16 + 24 = 40
(c) 15 + 20 = 35, 25 + 35 = 60, 40 + 60 = 100.

Question 7.
Multiply the given numbers in one line:

Solution:

2 × 3 = 6
∴ 25 × 25 = 625

5 × 6 = 30
∴ 55 × 55 = 3025

7 × 8 = 56
∴ 75 × 75 = 5625

12 × 13 = 156
∴ 125 × 125 = 15625

10 × 11 = 110
∴ 105 × 105 = 11025

40 × 41 = 1640
∴ 405 × 405 = 164025

Question 8.
Solve:
(a) (13 × 13) – (12 × 12)
(b) (18 × 18) – (17 × 17)
(c) (35 × 35) – (34 × 34)
(d) (120 × 120) – (119 × 119)
(e) (151 × 151) – (150 × 150)
solution:
(a) 13 + 12 = 25,
(b) 18 + 17 = 35,
(c) 35 + 34 = 69,
(d) 120 + 119 = 239,
(e) 151 + 150 = 301.

Question 9.
Complete the series according to pattern:
(a) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
11 + 12 + 13 + 14 + …….. + 19 + 20 = 155
21 + 22 + 23 + 24 + …… + 29 + 30 = 255
31 + 32 + 33 + 34 + …… + 39 + 40 = ………..
41 + 42 + 43 + 44 + …….. + 49 + 50 = ………..
51 + 52 + 53 + 54 + …….. + 59 + 60 = ……….

(b) 1 × 1 = 1
11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
11111 × 11111 = ……….
111111 × 111111 = ……….
111111 × 111111 = ………..
Solution:
(a) 355, 455, 555,
(b) 123454321, 12345654321, 1234567654321.

Question 10.
Find an easy method to complete the series. Write next three numbers using this method:
(a) 7, 12, 17, ……, …….., ………..
(b) 2, 4, 8, ……., ………., ………..
(c) 100, 90, 80, ……., ………., ………..
(d) 66, 55, 44, ……., ………., ………..
(e) 108, 208, 308, ……., ………., ………..
(f) 40, 39, 38, ……., ………., ………..
Solution:
(a) 22, 27, 32,
(b) 16, 32, 64,
(c) 70, 60, 50,
(d) 33, 22, 11,
(e) 408, 508, 608
(f) 37, 36, 35

Question 11.
Draw symmetry lines in the following:


Solution:

Question 12.
Complete the given symmetrical patterns:

Solution: