Countdown Maths Class 4 Exercise 2e Solution Allied School , Smart School , Dare Arqam School , Farhan School. Countdown Maths Class 4 , 5 , 6 , 7 Notes Available 3rd Edition. Full Solution

1.  𝐅𝐚𝐜𝐭𝐨𝐫𝐬 𝐚𝐧𝐝 𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐞𝐬 
Countdown Mathematics Class 4
Third Edition
𝐂𝐡𝐚𝐩𝐭𝐞𝐫 # 𝟐
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐞 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Please note that all previous notes are deleted due to some reason,
So if you want notes then please contact me. Thanks 😊
Countdown Maths Third Edition Notes also available.
Note
My Email: adilaslam5959@gmail.com
My WhatApp: 03466248138

2. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
1 | P a g e
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐞
𝐁𝐨𝐨𝐤 𝐏𝐚𝐠𝐞 # 𝟒𝟐 , 𝟒𝟑 , 𝟒𝟒
𝐍𝐨𝐭𝐞:
𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧 # 𝟏 , 𝟐 , 𝟑 , 𝟗 𝐒𝐨𝐥𝐯𝐞 𝐨𝐧 𝐛𝐨𝐨𝐤.
𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧 # 𝟒 , 𝟓 , 𝟔 , 𝟕, 𝟖 𝐒𝐨𝐥𝐯𝐞 𝐨𝐧 𝐧𝐨𝐭𝐞 𝐛𝐨𝐨𝐤.
𝐃𝐨 𝐚𝐥𝐥 𝐰𝐨𝐫𝐤 𝐧𝐞𝐚𝐭 𝐚𝐧𝐝 𝐜𝐥𝐞𝐚𝐧.
𝟏. 𝐅𝐢𝐥𝐥 𝐢𝐧 𝐭𝐡𝐞 𝐛𝐥𝐚𝐧𝐤𝐬.
𝐚. The common factor of 2 and 3 is ____________________.
𝐛. The common factors of 32 and 4 are 2 and _____________________.
𝐜. The common factors of 8 and 16 are __________________________.
𝐝. The common factor of 8 , 10 and 12 is ____________________.
𝐞. The common factor of 12 and 18 are 2 , 3 and _________________.
𝟐. 𝐒𝐭𝐚𝐭𝐞 𝐰𝐡𝐞𝐭𝐡𝐞𝐫 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐚𝐫𝐞 𝐭𝐫𝐮𝐞 𝐨𝐫 𝐟𝐚𝐥𝐬𝐞.
𝐚. 4 and 6 have no common factors. ( __________ )
𝐛. 1 , 2 , 3 and 4 are the only common factors of 12 and 24. ( __________ )
𝐜. 20 is the common factor of 10 and 20. ( ___________ )
𝐝. Two or more prime numbers have no common factors. ( ____________ )
𝐞. The common factors of 20 , 30 and 40 are 2 , 5 and 10. ( ___________ )
𝟏
𝐅
𝟒
𝟐 , 𝟒 𝐚𝐧𝐝 𝟖
𝟐
𝟔
𝐅
𝐅
𝐓
𝐅

3. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
2 | P a g e
𝟑. 𝐒𝐞𝐥𝐞𝐜𝐭 𝐭𝐡𝐞 𝐜𝐨𝐫𝐫𝐞𝐜𝐭 𝐚𝐧𝐬𝐰𝐞𝐫 𝐟𝐫𝐨𝐦 𝐭𝐡𝐞 𝐠𝐢𝐯𝐞𝐧 𝐨𝐩𝐭𝐢𝐨𝐧𝐬.
𝐚. The common factor of 42 , 56 and 63 is
 7  6  3  21
𝐛. 13 is the common factor of
 39 and 42  26 and 52  52 and 56  45 and 65
𝐜. The product of 3 and common factor of 81 and 45 is
 45  81  5  15
𝐝. The common factors of 26 and 78 are
 13 only  1
 1 , 2 and 13 only  1 , 2 , 13 and 26
𝐞. Which of the following is a common factor of 85 and 95
 5  17  15  19
𝟒. 𝐖𝐫𝐢𝐭𝐞 𝐭𝐡𝐞 𝐟𝐚𝐜𝐭𝐨𝐫𝐬 𝐨𝐟 𝐭𝐡𝐞𝐬𝐞 𝐩𝐚𝐢𝐫 𝐨𝐟 𝐧𝐮𝐦𝐛𝐞𝐫𝐬, 𝐚𝐧𝐝 𝐜𝐢𝐫𝐜𝐥𝐞 𝐭𝐡𝐞
𝐜𝐨𝐦𝐦𝐨𝐧 𝐟𝐚𝐜𝐭𝐨𝐫𝐬.
𝐚. 𝟏𝟎 , 𝟏𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 10 = 10 1 × 18 = 18
2 × 5 = 10 2 × 9 = 18
3 × 6 = 18
Factors of 10 = 1 , 2 , 5 , 10
Factors of 18 = 1 , 2 , 3 , 6 , 9 , 18
✓
✓
✓
✓
✓

4. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
3 | P a g e
𝐛. 𝟏𝟐 , 𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 12 = 12 1 × 1 = 1
2 × 6 = 12
3 × 4 = 12
Factors of 12 = 1 , 2 , 3 , 4 , 6 , 12
Factors of 1 = 1
𝐜. 𝟐𝟓 , 𝟏𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 25 = 25 1 × 15 = 15
5 × 5 = 25 3 × 5 = 15
Factors of 25 = 1 , 5 , 25
Factors of 15 = 1 , 3 , 5 , 15
𝐝. 𝟏𝟔 , 𝟐𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 16 = 16 1 × 20 = 20
2 × 8 = 16 2 × 10 = 20
4 × 4 = 16 4 × 5 = 20
Factors of 16 = 1 , 2 , 4 , 8 , 16
Factors of 20 = 1 , 2 , 4 , 5 , 10 , 20

5. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
4 | P a g e
𝐞. 𝟏𝟎 , 𝟑𝟐
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 10 = 10 1 × 32 = 32
2 × 5 = 10 2 × 16 = 32
4 × 8 = 32
Factors of 10 = 1 , 2 , 5 , 10
Factors of 21 = 1 , 2 , 4 , 8 , 16 , 32
𝐞. 𝟏𝟒 , 𝟐𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 14 = 14 1 × 21 = 21
2 × 7 = 14 3 × 7 = 21
Factors of 14 = 1 , 2 , 7 , 14
Factors of 32 = 1 , 3 , 7 , 21
𝟓. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐇𝐂𝐅 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐩𝐚𝐢𝐫𝐬 𝐨𝐟 𝐧𝐮𝐦𝐛𝐞𝐫𝐬.
𝐚. 𝟑𝟐 , 𝟐𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 32 = 32 1 × 24 = 24
2 × 16 = 32 2 × 12 = 24
4 × 8 = 32 3 × 8 = 24
4 × 6 = 24
Factors of 32 = 1 , 2 , 4 , 8 , 16 , 32
Factors of 24 = 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24

6. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
5 | P a g e
Common factors of 32 and 24 = 1 , 2 , 4 , 8
The HCF of 32 and 24 = 8
𝐛. 𝟓𝟎 , 𝟐𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 50 = 50 1 × 25 = 25
2 × 25 = 50 5 × 5 = 25
5 × 10 = 50
Factors of 50 = 1 , 2 , 5 , 10 , 25 , 50
Factors of 25 = 1 , 5 , 25
Common factors of 50 and 25 = 1 , 5 , 25
The HCF of 50 and 25 = 25
𝐜. 𝟒𝟖 , 𝟑𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 48 = 48 1 × 30 = 30
2 × 24 = 48 2 × 15 = 30
3 × 16 = 48 3 × 10 = 30
4 × 12 = 48 5 × 6 = 30
6 × 8 = 48
Factors of 48 = 1 , 2 , 3 , 4 , 6 , 8 , 12 , 16 , 24 , 48
Factors of 30 = 1 , 2 , 3 , 5 , 6 , 10 , 15 , 30
Common factors of 48 and 30 = 1 , 2 , 3 , 6
The HCF of 48 and 30 = 6

7. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
6 | P a g e
𝐝. 𝟒𝟐 , 𝟕𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 42 = 42 1 × 70 = 70
2 × 21 = 42 2 × 35 = 70
3 × 14 = 42 5 × 14 = 70
6 × 7 = 42 7 × 10 = 70
Factors of 42 = 1 , 2 , 3 , 6 , 7 , 14 , 21 , 42
Factors of 70 = 1 , 2 , 5 , 7 , 10 , 14 , 35 , 70
Common factors of 42 and 70 = 1 , 2, 7 , 14
The HCF of 42 and 70 = 14
𝐞. 𝟔𝟒 , 𝟖𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 64 = 64 1 × 88 = 88
2 × 32 = 64 2 × 44 = 88
4 × 16 = 42 4 × 22 = 88
8 × 8 = 64 8 × 11 = 88
Factors of 64 = 1 , 2 , 4 , 8 , 16 , 32 , 64
Factors of 88 = 1 , 2 , 4 , 8 , 11 , 22 , 44 , 88
Common factors of 64 and 88 = 1 , 2, 4 , 8
The HCF of 64 and 88 = 8

8. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
7 | P a g e
𝐟. 𝟕𝟔 , 𝟐𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 76 = 76 1 × 28 = 28
2 × 38 = 76 2 × 14 = 28
4 × 19 = 76 4 × 7 = 28
Factors of 76 = 1 , 2 , 4 , 19 , 38 , 76
Factors of 28 = 1 , 2 , 4 , 7 , 14 , 28
Common factors of 76 and 28 = 1 , 2, 4
The HCF of 76 and 28 = 4
𝟔. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐜𝐨𝐦𝐦𝐨𝐧 𝐟𝐚𝐜𝐭𝐨𝐫𝐬 𝐚𝐧𝐝 𝐇𝐂𝐅 𝐨𝐟 𝐞𝐚𝐜𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐬𝐞𝐭𝐬.
𝐚. 𝟗 , 𝟏𝟐 𝐚𝐧𝐝 𝟏𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 9 = 9 1 × 12 = 12 1 × 15 = 15
3 × 3 = 9 2 × 6 = 12 3 × 5 = 15
3 × 4 = 12
Factors of 9 = 1 , 3 , 9
Factors of 12 = 1 , 2 , 3 , 4 , 6 , 12
Factors of 15 = 1 , 3 , 5 , 15
Common factors of 9 , 12 and 15 = 1 , 3
The HCF of 9 , 12 and 15 = 3

9. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
8 | P a g e
𝐛. 𝟓 , 𝟐𝟓 𝐚𝐧𝐝 𝟑𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 5 = 5 1 × 25 = 25 1 × 35 = 35
5 × 5 = 25 5 × 7 = 35
Factors of 5 = 1 , 5
Factors of 25 = 1 , 5 , 25
Factors of 35 = 1 , 5 , 7 , 35
Common factors of 5 , 25 and 35 = 1 , 5
The HCF of 5 , 25 and 35 = 5
𝐜. 𝟏𝟖 , 𝟐𝟎 𝐚𝐧𝐝 𝟐𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 18 = 18 1 × 20 = 20 1 × 24 = 24
2 × 9 = 18 2 × 10 = 20 2 × 12 = 24
3 × 6 = 18 4 × 5 = 20 3 × 8 = 24
4 × 6 = 24
Factors of 18 = 1 , 2 , 3 , 6 , 9 , 18
Factors of 20 = 1 , 2 , 4 , 5 , 10 , 20
Factors of 24 = 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24
Common factors of 18 , 20 and 24 = 1 , 2
The HCF of 18 , 20 and 24 = 2

10. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
9 | P a g e
𝐝. 𝟏𝟖 , 𝟐𝟕 𝐚𝐧𝐝 𝟑𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 18 = 18 1 × 27 = 27 1 × 39 = 39
2 × 9 = 18 3 × 9 = 27 3 × 13 = 39
3 × 6 = 18
Factors of 18 = 1 , 2 , 3 , 6 , 9 , 18
Factors of 27 = 1 , 3 , 9 , 27
Factors of 39 = 1 , 3 , 13 , 39
Common factors of 18 , 27 and 39 = 1 , 3
The HCF of 18 , 27 and 39 = 3
𝐞. 𝟏𝟐 , 𝟏𝟔 𝐚𝐧𝐝 𝟐𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 12 = 12 1 × 16 = 16 1 × 20 = 20
2 × 6 = 12 2 × 8 = 16 2 × 10 = 20
3 × 4 = 12 4 × 4 = 16 4 × 5 = 20
Factors of 12 = 1 , 2 , 3 , 4 , 6 , 12
Factors of 16 = 1 , 2 , 4 , 8 , 16
Factors of 20 = 1 , 2 , 4 , 5 , 10 , 20
Common factors of 12 , 16 and 20 = 1 , 2 , 4
The HCF of 12 , 16 and 20 = 4

11. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
10 | P a g e
𝐟. 𝟏𝟒 , 𝟒𝟗 𝐚𝐧𝐝 𝟐𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1 × 14 = 14 1 × 49 = 49 1 × 28 = 28
2 × 7 = 14 7 × 7 = 49 2 × 14 = 28
4 × 7 = 28
Factors of 14 = 1 , 2 , 7 , 14
Factors of 49 = 1 , 7 , 49
Factors of 28 = 1 , 2 , 4 , 7 , 14 , 28
Common factors of 14 , 49 and 28 = 1 , 7
The HCF of 14 , 49 and 28 = 7
𝟕. 𝐔𝐬𝐞 𝐭𝐡𝐞 𝐝𝐢𝐯𝐢𝐬𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝 𝐭𝐨 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐩𝐫𝐢𝐦𝐞 𝐟𝐚𝐜𝐭𝐨𝐫𝐬 𝐨𝐟 𝐭𝐡𝐞𝐬𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐬.
(𝐇𝐢𝐧𝐭: 𝐔𝐬𝐞 𝐫𝐮𝐥𝐞𝐬 𝐨𝐟 𝐝𝐢𝐯𝐢𝐬𝐢𝐛𝐢𝐥𝐢𝐭𝐲)
𝐚. 𝟐𝟑𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus prime factors of 230 are: 2 × 5 × 23
2 230
5 115
23 23
1

12. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
11 | P a g e
𝐛. 𝟑𝟐𝟎𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus prime factors of 3200 are: 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
𝐜. 𝟒𝟓𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus prime factors of 459 are: 3 × 3 × 3 × 17
2 3200
2 1600
2 800
2 400
2 200
2 100
2 50
5 25
5 5
1
3 459
3 153
3 51
17 17
1

13. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
12 | P a g e
𝐝. 𝟒𝟓𝟒𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus prime factors of 230 are: 2 × 5 × 23
𝟖. 𝐔𝐬𝐢𝐧𝐠 𝐭𝐡𝐞 𝐝𝐢𝐯𝐢𝐬𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝, 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐇𝐂𝐅 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐧𝐮𝐦𝐛𝐞𝐫𝐬.
𝐚. 𝟏𝟐 𝐚𝐧𝐝 𝟐𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus the HCF of 12 and 20 = 4
3 4545
3 1515
5 505
101 101
1
12 20
− 12
8 12
− 8
4 8
− 8
1
1
2


14. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
13 | P a g e
𝐛. 𝟑𝟔 𝐚𝐧𝐝 𝟐𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus the HCF of 36 and 24 = 12
𝐜. 𝟏𝟒 𝐚𝐧𝐝 𝟑𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Thus the HCF of 14 and 35 = 7
𝐝. 𝟒𝟖 , 𝟑𝟎 𝐚𝐧𝐝 𝟒𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
24 36
− 24
12 24
− 24
1
2

14 35
− 28
7 14
− 14

2
2
30 48
− 30
18 30
− 18
12 18
− 12
6 12
− 12

1
1
1
2

15. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
14 | P a g e
Now we can find the HCF of 6 and 40
Thus the HCF of 48 , 30 and 40 = 2
𝐞. 𝟏𝟓 , 𝟐𝟏 𝐚𝐧𝐝 𝟑𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now we can find the HCF of 3 and 33
Thus the HCF of 15, 21 and 33 = 3
6 40
− 36
4 6
− 4
2 4
− 4

6
1
2
15 21
− 15
6 16
− 12
3 6
− 6

1
2
2
3 33
− 3
3
− 3
11


16. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 4
15 | P a g e
𝐟. 𝟏𝟐𝟏 , 𝟏𝟑𝟐 𝐚𝐧𝐝 𝟏𝟔𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now we can find the HCF of 11 and 165
Thus the HCF of 121, 132 and 165 = 11
𝟗. 𝐂𝐢𝐫𝐜𝐥𝐞 𝐭𝐡𝐞 𝐩𝐚𝐢𝐫𝐬 𝐭𝐡𝐚𝐭 𝐚𝐫𝐞 𝐜𝐨 − 𝐩𝐫𝐢𝐦𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐬.
𝐚. 𝟒 𝐚𝐧𝐝 𝟖 𝐛. 𝟐 𝐚𝐧𝐝 𝟏𝟎
𝐜. 𝟓 𝐚𝐧𝐝 𝟗 𝐝. 𝟐 𝐚𝐧𝐝 𝟏𝟕
𝐞. 𝟓 𝐚𝐧𝐝 𝟏𝟎 𝐟. 𝟏𝟒 𝐚𝐧𝐝 𝟒
121 132
− 121
11 121
− 121

1
11
11 165
− 11
55
− 55

15