Oxford Countdown Maths Class 7 Chapter 2 Solution download.

1.  𝐑𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐍𝐮𝐦𝐛𝐞𝐫𝐬 
Countdown Mathematics Class 7
Third Edition
𝐂𝐡𝐚𝐩𝐭𝐞𝐫 # 𝟐
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐀 , 𝟐𝐁 , 𝟐𝐂 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Contact Information
Via Email: adilaslam5959@gmail.com
Via WhatsApp: 0346 6248138

2. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
1 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐀
𝟏. 𝐖𝐡𝐢𝐜𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐚𝐫𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬?
(𝐢) 𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
3 is a rational number because it can be expressed as
3
1
.
(𝐢𝐢)
𝟑
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
3
1
is a rational number as it is in the form of
p
q
, q ≠ 0.
(𝐢𝐢𝐢)
𝟑
𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
3
0
is not a rational number as q = 0 in the form
p
q
.
(𝐢𝐯) −
𝟑
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−
3
1
is a rational number as it is in the form of
p
q
, q ≠ 0.
 Remember 
All integers, whole numbers, natural numbers, and fractions with integers
are rational numbers.
Notes By Adil Aslam
WhatsApp: 0346-6248138

3. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
2 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐯) −
𝟎
𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−
0
3
is a rational number as it is in the form of
p
q
, q ≠ 0.
(𝐯𝐢)
−𝟑
−𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−3
−1
is a rational number as it is in the form of
p
q
, q ≠ 0.
(𝐯𝐢)
𝟎
𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
0
0
is not a rational number as q = 0 in the form
p
q
.
𝟐. 𝐖𝐫𝐢𝐭𝐞 𝐝𝐨𝐰𝐧 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐬 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬.
(𝐢)
𝟕
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
In integer form
7
1
is 7.
(𝐢𝐢)
−𝟓
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
In integer form
−5
1
is − 5.
Notes By Adil Aslam
WhatsApp: 0346-6248138

4. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
3 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐢𝐢)
−𝟗
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
In integer form
−9
1
is − 9.
(𝐢𝐯)
−𝟐𝟏
−𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−21
−1
×
−1
−1
=
−21 × −1
−1 × −1
=
21
1
In integer form
−21
−1
=
21
1
is 21.
𝟑. 𝐖𝐫𝐢𝐭𝐞 𝐝𝐨𝐰𝐧 𝐭𝐡𝐚 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐰𝐡𝐨𝐬𝐞 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫 (−𝟕) × 𝟐 𝐚𝐧𝐝
𝐰𝐡𝐨𝐬𝐞 𝐝𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫 𝐢𝐬 𝟐𝟏 − 𝟐.
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Numerator = −7 × 2 = −14
Denominator = 21 − 2 = 19
Rational number is in the form of
p
q
.
Required rational number =
Numerator
Denominator
Required rational number =
−14
19
𝟒. (𝐢) 𝐖𝐫𝐢𝐭𝐞 𝐝𝐨𝐰𝐧 𝐭𝐡𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐰𝐡𝐨𝐬𝐞 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫 𝟏𝟓 − 𝟒 𝐚𝐧𝐝
𝐰𝐡𝐨𝐬𝐞 𝐝𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫 𝐢𝐬 𝟑𝟕 × (−𝟐).
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138

5. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
4 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Numerator = 15 − 4 = 11
Denominator = 37 × −2 = −74
Rational number is in the form of
p
q
.
Required rational number =
Numerator
Denominator
Required rational number =
11
−74
𝟒. (𝐢𝐢) 𝐄𝐱𝐩𝐫𝐞𝐬𝐬 𝐭𝐡𝐞 𝐚𝐛𝐨𝐯𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐰𝐢𝐭𝐡 𝐚 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐝𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫.
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The rational number with positive denominator is:
11
−74
×
−1
−1
=
11 × −1
−74 × −1
= −
11
74
𝟓. 𝐄𝐱𝐩𝐫𝐞𝐬𝐬 −
𝟑
𝟕
𝐚𝐬 𝐚 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐰𝐡𝐨𝐬𝐞 𝐧𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫 𝐢𝐬:
(𝐢) 𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number = −
3
7
Required rational number = −
3
7
×
−2
−2
=
−3 × −2
7 × −2
=
6
−14
(𝐢𝐢) − 𝟏𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138

6. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
5 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Given rationa number = −
3
7
Required rational number = −
3
7
×
5
5
=
−3 × 5
7 × 5
=
−15
35
(𝐢𝐢𝐢) 𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number = −
3
7
Required rational number = −
3
7
×
−1
−1
=
−3 × −1
7 × −1
=
3
−7
(𝐢𝐯) 𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number = −
3
7
Required rational number = −
3
7
×
−3
−3
=
−3 × −3
7 × −3
=
9
−21
𝟔. 𝐄𝐱𝐩𝐫𝐞𝐬𝐬
−𝟐
−𝟓
𝐚𝐬 𝐚 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐰𝐡𝐨𝐬𝐞 𝐝𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫 𝐢𝐬:
(𝐢) 𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138

7. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
6 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Given rationa number =
−2
−5
Required rational number =
−2
−5
×
−1
−1
=
−2 × −1
−5 × −1
=
2
5
(𝐢𝐢) − 𝟏𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number =
−2
−5
Required rational number =
−2
−5
×
3
3
=
−2 × 3
−5 × 3
=
−6
−15
(𝐢𝐢𝐢) 𝟒𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number =
−2
−5
Required rational number =
−2
−5
×
−8
−8
=
−2 × −8
−5 × −8
=
16
40
(𝐢𝐯) − 𝟓𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rationa number =
−2
−5
Notes By Adil Aslam
WhatsApp: 0346-6248138

8. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
7 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Required rational number =
−2
−5
×
10
10
=
−2 × 10
−5 × 10
=
−20
−50
𝟕. 𝐄𝐱𝐩𝐫𝐞𝐬𝐬 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐰𝐢𝐭𝐡 𝐚 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐝𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫.
(𝐢) −
𝟕
𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number = −
7
1
Required rational number = −
7
1
(𝐢𝐢)
−𝟓
−𝟐
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number =
−5
−2
Required rational number =
−5
−2
×
−1
−1
=
−5 × −1
−2 × −1
=
5
2
(𝐢𝐢𝐢)
−𝟑𝟏
−𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number =
−31
−3
Required rational number =
−31
−3
×
−1
−1
Notes By Adil Aslam
WhatsApp: 0346-6248138

9. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
8 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
−31 × −1
−3 × −1
=
31
3
(𝐢𝐯) −
𝟗
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number = −
9
5
Required rational number = −
9
5
𝟖. 𝐑𝐞𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐨𝐧 𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐥𝐢𝐧𝐞.
(𝐢) −
𝟏𝟏
𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number = −
11
4
Rational number in decimal = −2.75
(𝐢𝐢) − 𝟐
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given integer = −2
−2.75
−2
Notes By Adil Aslam
WhatsApp: 0346-6248138

10. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
9 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐢𝐢)
𝟐
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number =
2
5
Rational number in decimal = 0.4
(𝐢𝐢𝐢) −
𝟑
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given rational number = −
3
5
Rational number in decimal = −0.6
𝟗. 𝐃𝐨 𝐛𝐨𝐭𝐡
−𝟐
𝟑
𝐚𝐧𝐝
𝟐
−𝟑
𝐫𝐞𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐭𝐡𝐞 𝐬𝐚𝐦𝐞 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫?
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Reducing
−2
3
to the simplest form =
−2
3
Reducing
2
−3
to the simplest form =
−2
3
Hence they represent the same negative rational number.
0.4
−0.6
Notes By Adil Aslam
WhatsApp: 0346-6248138

11. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
10 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐁
𝟏. 𝐅𝐢𝐥𝐥 𝐢𝐧 𝐭𝐡𝐞 𝐛𝐥𝐚𝐧𝐤𝐬.
(𝐢)
56
−96
is expressed in standard form as ______________________.
(𝐢𝐢) Which of the the fraction is greater,
3
4
or
11
15
? _______________.
(𝐢𝐢𝐢) The reciprocal of
5
7
is ___________________.
(𝐢𝐯) Is
5
7
is less than
2
5
? __________________.
(𝐯) The standard form of
−9
−8
is __________________.
𝐄𝐱𝐩𝐥𝐚𝐧𝐚𝐭𝐢𝐨𝐧:
(𝐢) The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
56
−96
×
−1
−1
=
−56
96
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 56 = 2 × 2 × 2 × 7
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
HCF of 56 and 96 = 2 × 2 × 2
Now we divide the numerator and denominator of the given rational
Remember
HCF = Common factors
of given numbers.
−
𝟕
𝟏𝟐
𝟑
𝟒
𝟕
𝟓
𝐧𝐨
𝟗
𝟖
Notes By Adil Aslam
WhatsApp: 0346-6248138

12. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
11 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
number by HCF.
−56 ÷ 8
96 ÷ 8
=
−7
12
= −
7
12
(𝐢𝐢) First we write the given rational number so that their denominator is + ve.
In our case both given numbers having + ve denominator:
3
4
and
11
15
.
Now we can use cross multiplication:
3
4
,
11
15
3 × 15 and 4 × 11
45 and 44
45 > 44
Hence
3
4
is greater than
11
15
.
(𝐢𝐢𝐢) By interchanging the numerator and the denominator we get
7
5
.
5
7
×
7
5
= 1
Hence the reciprocal of
5
7
is
7
5
.
(𝐢𝐯) First we write the given rational number so that their denominator is + ve.
In our case both given numbers having + ve denominator:
5
7
and
2
5
Now we can use cross multiplication:
Remember
The reciprocal of fraction obtained by
interchanging Numerator and
Denominator with each other is known
as the reciprocal of the given fraction.
Notes By Adil Aslam
WhatsApp: 0346-6248138

13. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
12 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
5
7
,
2
5
5 × 5 and 7 × 2
25 and 14
25 > 14
Hence
5
7
is greater than
2
5
.
(𝐯) The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
−9
−8
×
−1
−1
=
9
8
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 9 = 3 × 3
Prime factors of 8 = 2 × 2 × 2
HCF of 8 and 9 = 1
Now we divide the numerator and denominator of the given rational
number by HCF.
9 ÷ 1
8 ÷ 1
=
9
8
Required rational number in standard form =
9
8
.
Notes By Adil Aslam
WhatsApp: 0346-6248138

14. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
13 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝟐. 𝐄𝐱𝐩𝐫𝐞𝐬𝐬 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐧 𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐟𝐨𝐫𝐦.
(𝐢)
𝟒
𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The denominator of given rational number is positive.
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 4 = 2 × 2
Prime factors of 6 = 2 × 3
HCF of 4 and 6 = 2
Now we divide the numerator and denominator of the given rational
number by HCF.
4 ÷ 2
6 ÷ 2
=
2
3
Hence the standard form of
4
6
is
2
3
.
(𝐢𝐢)
𝟒
−𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
4
−9
×
−1
−1
=
−4
9
Notes By Adil Aslam
WhatsApp: 0346-6248138

15. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
14 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 4 = 2 × 2
Prime factors of 9 = 3 × 3
HCF of 4 and 9 = 1
Now we divide the numerator and denominator of the given rational
number by HCF.
−4 ÷ 1
9 ÷ 1
=
−4
9
Hence the standard form of
4
−9
is
−4
9
.
(𝐢𝐢𝐢)
−𝟏𝟏
−𝟏𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
−11
−13
×
−1
−1
=
11
13
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 11 = 1 × 11
Prime factors of 9 = 1 × 13
HCF of 11 and 13 = 1
Notes By Adil Aslam
WhatsApp: 0346-6248138

16. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
15 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Now we divide the numerator and denominator of the given rational
number by HCF.
11 ÷ 1
13 ÷ 1
=
11
13
Hence the standard form of
−11
−13
is
11
13
.
(𝐢𝐯)
−𝟐𝟏
−𝟐𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
−21
−28
×
−1
−1
=
21
28
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 21 = 3 × 7
Prime factors of 28 = 2 × 2 × 7
HCF of 21 and 28 = 7
Now we divide the numerator and denominator of the given rational
number by HCF.
21 ÷ 7
28 ÷ 7
=
3
4
Hence the standard form of
−21
−28
is
3
4
.
Notes By Adil Aslam
WhatsApp: 0346-6248138

17. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
16 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐯)
𝟒𝟐
−𝟒𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The denominator of given rational number is negative, So we multiply
numerator and denominator by − 1, So that denominator becomes + ve.
42
−48
×
−1
−1
=
−42
48
Now we can find HCF of the absolute values of the numerator and the
denominator.
Prime factors of 42 = 2 × 3 × 7
Prime factors of 48 = 2 × 2 × 2 × 2 × 3
HCF of 42 and 48 = 2 × 3 = 6
Now we divide the numerator and denominator of the given rational
number by HCF.
−42 ÷ 6
48 ÷ 6
=
−7
8
Hence the standard form of
42
−48
is
−7
8
.
𝟑. 𝐅𝐢𝐥𝐥 𝐢𝐧 𝐭𝐡𝐞
(𝐢)
𝟒
𝟓
=
−𝟏𝟐
=
𝟐𝟎
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138

18. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
17 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
4
5
=
4 × −3
5 × −3
=
−12
−15
and
4
5
=
4 × 5
5 × 5
=
20
25
Hence:
4
5
=
−12
−15
=
20
25
(𝐢𝐢)
−𝟓
𝟕
=
−𝟏𝟓
=
−𝟑𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−5
7
=
−5 × 3
7 × 3
=
−15
21
and
−5
7
=
−5 × 7
7 × 7
=
−35
49
Hence:
−5
7
=
−15
21
=
−35
49
(𝐢𝐢𝐢)
−𝟑
=
𝟔
−𝟏𝟔
=
𝟐𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
6
−16
=
6 ÷ 2
−16 ÷ 2
=
3
−8
and
−3
8
=
−3 × 3
8 × 3
=
−9
24
Hence:
−3
8
=
6
−16
=
−9
24
𝟒. 𝐃𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐞 𝐰𝐡𝐢𝐜𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐭𝐰𝐨 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐬 𝐠𝐫𝐞𝐚𝐭𝐞𝐫 𝐢𝐧 𝐞𝐚𝐜𝐡.
(𝐢) −
𝟑
𝟕
,
𝟐
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−
3
7
,
2
7
By using cross multiplication
⟹ −3 × 7 and 7 × 2
Remember
3
−8
=
−3
8
Notes By Adil Aslam
WhatsApp: 0346-6248138

19. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
18 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
⟹ −21 and 14
As − 21 < 24
Hence
2
7
is greater than −
3
7
.
(𝐢𝐢)
−𝟒
𝟗
,
−𝟓
𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−4
9
,
−5
6
By using cross multiplication
⟹ −4 × 6 and − 5 × 9
⟹ −24 and − 45
As − 24 > −45
Hence
−4
9
is greater than
−5
6
.
(𝐢𝐢𝐢)
𝟏
𝟐
,
𝟒
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
1
2
,
4
7
By using cross multiplication
⟹ 1 × 7 and 2 × 4
⟹ 7 and 8
As 7 < 8
Hence
4
7
is greater than
1
2
.
Notes By Adil Aslam
WhatsApp: 0346-6248138

20. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
19 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐯)
−𝟓
𝟖
,
−𝟑
𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−5
8
,
−3
4
By using cross multiplication
⟹ −5 × 4 and − 3 × 8
⟹ −20 and − 24
As − 20 > −24
Hence
−5
8
is greater than
−3
4
.
(𝐯)
−𝟕
𝟏𝟏
,
𝟓
−𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−7
11
,
5
−8
By using cross multiplication
⟹ −7 × −8 and 5 × 11
⟹ 56 and 55
As 56 > 55
Hence
−7
−11
is greater than
5
−8
.
(𝐯𝐢)
−𝟑
−𝟏𝟑
,
−𝟓
−𝟐𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−3
−13
,
−5
−21
By using cross multiplication
Notes By Adil Aslam
WhatsApp: 0346-6248138

21. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
20 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
⟹ −3 × −21 and − 13 × −5
⟹ 63 and 65
As 63 < 65
Hence
−5
−21
is greater than
−3
−13
.
𝟓. 𝐅𝐢𝐥𝐥 𝐢𝐧 𝐭𝐡𝐞 𝐛𝐨𝐱 𝐛𝐲 𝐮𝐬𝐢𝐧𝐠 𝐜𝐨𝐫𝐫𝐞𝐜𝐭 𝐬𝐲𝐦𝐛𝐨𝐥 𝐨𝐮𝐭 𝐨𝐟 < , = 𝐨𝐫 >.
(𝐢) −
𝟕
𝟏𝟐
𝟓
−𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−
7
12
5
−8
By using cross multiplication
⟹ −7 × −8 and 12 × 5
⟹ 56 and 60
As 56 < 60
So, −
7
12
<
5
−8
.
(𝐢𝐢)
−𝟒
𝟗
−𝟑
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−4
9
−3
7
By using cross multiplication
⟹ −4 × 7 and − 3 × 9
⟹ −28 and − 27
Notes By Adil Aslam
WhatsApp: 0346-6248138

22. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
21 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
As − 28 < −27
So,
−4
9
<
−3
7
.
(𝐢𝐢𝐢)
−𝟕
−𝟖
𝟏𝟒
𝟏𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−7
−8
14
17
By using cross multiplication
⟹ −7 × 17 and − 8 × 14
⟹ −119 and − 112
As − 119 > −112
So,
−7
−8
>
11
17
.
(𝐢𝐯)
−𝟐
𝟗
𝟖
−𝟑𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−2
9
8
−36
By using cross multiplication
⟹ −2 × −36 and 8 × 9
⟹ 72 and 72
As 72 = 72
So,
−2
9
=
8
−36
.
Notes By Adil Aslam
WhatsApp: 0346-6248138
Notes By Adil Aslam
WhatsApp: 0346-6248138

23. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
22 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐯)
𝟓
𝟖
𝟐𝟓
𝟒𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
5
8
25
45
By using cross multiplication
⟹ 5 × 45 and 8 × 25
⟹ 225 and 200
As 225 > 200
So,
5
8
>
25
45
.
(𝐯𝐢)
𝟒
𝟔
𝟏
𝟏𝟐
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
4
6
1
12
By using cross multiplication
⟹ 4 × 12 and 6 × 1
⟹ 48 and 6
As 48 > 6
So,
4
6
>
1
12
.
𝟔. 𝐀𝐫𝐫𝐚𝐧𝐠𝐞 𝐭𝐡𝐞 𝐠𝐢𝐯𝐞𝐧 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐧 𝐚𝐬𝐜𝐞𝐧𝐝𝐢𝐧𝐠 𝐨𝐫𝐝𝐞𝐫.
2
5
,
−1
2
,
8
−15
,
−3
−10
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
First we write given rational numbers so that their denominator are + ve.
Notes By Adil Aslam
WhatsApp: 0346-6248138

24. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
23 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
8
−15
=
8 × −1
−15 × −1
=
−8
15
and
−3
−10
=
−3 × −1
−10 × −1
=
3
10
Thus the rational numbers with positive denominators are:
2
5
,
−1
2
,
−8
15
,
3
10
Now Find the LCM of denominators 5 , 2 , 15 and 10.
LCM of 5 , 2 , 15 and 10 = 2 × 3 × 5 = 30
Now we make all denominators as 30.
2 × 6
5 × 6
=
12
30
and
−1 × 15
2 × 15
=
−15
30
−8 × 2
15 × 2
=
−16
30
and
3 × 3
10 × 3
=
9
30
Now comparing the numerators of these fractions, we get
−16 < −15 < 9 < 12
Hence rational number in ascending order:
8
−15
,
−1
2
,
−3
−10
,
2
5
𝟕. 𝐀𝐫𝐫𝐚𝐧𝐠𝐞 𝐭𝐡𝐞 𝐠𝐢𝐯𝐞𝐧 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐢𝐧 𝐝𝐞𝐬𝐜𝐞𝐧𝐝𝐢𝐧𝐠 𝐨𝐫𝐝𝐞𝐫.
−7
10
,
8
−15
,
19
30
,
−2
−5
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
First we write given rational numbers so that their denominator are + ve.
8
−15
=
8 × −1
−15 × −1
=
−8
15
and
−2
−5
=
−2 × −1
−5 × −1
=
2
5
Thus the rational numbers with positive denominators are:
Notes By Adil Aslam
WhatsApp: 0346-6248138

25. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
24 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
−7
10
,
−8
15
,
19
30
,
2
5
Now Find the LCM of denominators 10 , 15 , 30 and 5.
LCM of 10 , 15 , 30 and 5 = 2 × 3 × 5 = 30
Now we make all denominators as 30.
−7 × 3
10 × 3
=
−21
30
and
−8 × 2
15 × 2
=
−16
30
19 × 1
30 × 1
=
19
30
and
2 × 6
5 × 6
=
12
30
Now comparing the numerators of these fractions, we get
19 > 12 > −16 > −21
Hence rational number in descending order:
19
30
,
−2
−5
,
8
−15
,
−7
10
.
𝟖. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐫𝐞𝐜𝐢𝐩𝐫𝐨𝐜𝐚𝐥𝐬 𝐨𝐟.
(𝐢) − 𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The reciprocal of
−3
1
is
−1
3
.
(𝐢𝐢)
𝟏
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The reciprocal of
1
7
is
7
1
.
Notes By Adil Aslam
WhatsApp: 0346-6248138

26. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
25 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐢𝐢) −
𝟐
𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The reciprocal of −
2
9
is −
9
2
.
(𝐢𝐯)
𝟏𝟏
−𝟏𝟐
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The reciprocal of
11
−12
is
−12
11
.
(𝐯)
−𝟑
−𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
The reciprocal of
−3
−8
is
−8
−3
or
8
3
.
(𝐯𝐢)
𝟐
−𝟑
×
𝟒
−𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
2
−3
×
4
−5
=
2 × 4
−3 × −5
=
8
15
The reciprocal of
8
−15
is
15
8
.
(𝐯𝐢𝐢)
−𝟑
𝟒
×
−𝟓
−𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138

27. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
26 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
−3
4
×
−5
−6
=
−3 × −5
4 × −6
=
15
−24
=
15 ÷ 3
−24 ÷ 3
=
5
−8
The reciprocal of
5
−8
is
−8
5
.
𝟗. 𝐑𝐨𝐮𝐧𝐝 𝐨𝐟𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐭𝐨 𝐭𝐡𝐞 𝐧𝐞𝐚𝐫𝐞𝐬𝐭 𝐭𝐞𝐧𝐭𝐡 ,
𝐡𝐮𝐧𝐝𝐫𝐞𝐝𝐭𝐡 , 𝐚𝐧𝐝 𝐭𝐡𝐨𝐮𝐬𝐚𝐧𝐝𝐭𝐡 𝐚𝐬 𝐩𝐨𝐬𝐬𝐢𝐛𝐥𝐞.
(𝐢)
𝟐𝟓
𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Give rational number =
25
4
Give ratinal number in decimal = 6.25
Round off nearest tenth = 6.3
Round off nearest hundredth = 6.25
Round off nearest thousandth = 6.250
(𝐢𝐢)
𝟏𝟎𝟐
𝟏𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Give rational number =
102
16
Give ratinal number in decimal = 6.375
Round off nearest tenth = 6.4
Round off nearest hundredth = 6.38
Notes By Adil Aslam
WhatsApp: 0346-6248138

28. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
27 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Round off nearest thousandth = 6.375
(𝐢𝐢𝐢)
𝟕𝟕𝟕
𝟏𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Give rational number =
777
16
Give ratinal number in decimal = 48.5625
Round off nearest tenth = 48.6
Round off nearest hundredth = 48.56
Round off nearest thousandth = 48.563
Notes By Adil Aslam
WhatsApp: 0346-6248138
Notes By Adil Aslam
WhatsApp: 0346-6248138

29. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
28 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟐𝐁
𝟏. 𝐒𝐭𝐚𝐭𝐞 𝐰𝐡𝐞𝐭𝐡𝐞𝐫 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐰𝐨𝐢𝐧𝐠 𝐚𝐫𝐞 𝐭𝐮𝐫𝐞 𝐨𝐫 𝐟𝐚𝐥𝐬𝐞:
(𝐢) The product of rational numbers is a rational number.
(𝐢𝐢) Rational nimbers include integers.
(𝐢𝐢𝐢) Zero is a rational number.
(𝐢𝐯) The product of a rational number and its multiplicative inverse is zero.
(𝐯) Addition of rational numbers is commutative.
𝟐. 𝐀𝐝𝐝:
(𝐢)
𝟑
𝟓
𝐚𝐧𝐝 −
𝟐
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
=
3
5
+ (−
2
5
) =
3
5
−
2
5
=
3 − 2
5
=
1
5
(𝐢𝐢)
−𝟓
𝟖
𝐚𝐧𝐝
𝟏
𝟒
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
=
−5
8
+
1
4
Now find the LCM of the denominators 8 and 4
The LCM of 8 and 4 is 8
𝟏. 𝐒𝐭𝐚𝐭𝐞 𝐰𝐡𝐞𝐭𝐡𝐞𝐫 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐰𝐨𝐢𝐧𝐠 𝐚𝐫𝐞 𝐭𝐮𝐫𝐞 𝐨𝐫 𝐟𝐚𝐥𝐬𝐞:
(𝐢) The product of rational numbers is a rational number.
(𝐢𝐢) Rational numbers include integers.
(𝐢𝐢𝐢) Zero is a rational number.
(𝐢𝐯) The product of a rational number and its multiplicative inverse is zero.
(𝐯) Addition of rational numbers is commutative.
True
True
True
False
True
Notes By Adil Aslam
WhatsApp: 0346-6248138

30. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
29 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Find the equivalent fractions with denominator 8
=
−5
8
+
1 × 2
4 × 2
=
−5
8
+
2
8
=
−5 + 2
8
=
−3
8
(𝐢𝐢𝐢)
𝟑
𝟕
𝐚𝐧𝐝 −
𝟐
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
=
3
7
+ (−
2
7
) =
3
7
−
2
7
Above both fractions have same denominators,
=
3 − 2
7
=
1
7
(𝐢𝐯) −
𝟑
𝟏𝟏
𝐚𝐧𝐝 −
𝟓
𝟏𝟏
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −
3
11
+ (−
5
11
) = −
3
11
−
5
11
Above both fractions have same denominators,
=
−3 − 5
11
= −
8
11
𝟑. 𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲:
(𝐢) −
𝟕
𝟏𝟏
+
𝟏
𝟔
Notes By Adil Aslam
WhatsApp: 0346-6248138

31. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
30 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 11 and 6.
The LCM of 11 and 6 is 66.
Find the equivalent fractions with denominator 66.
−
7 × 6
11 × 6
+
1 × 11
6 × 11
= −
42
66
+
11
66
=
−42 + 11
66
= −
31
66
(𝐢𝐢)
−𝟑
−𝟕
+
𝟐
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−3
−7
+
2
5
=
3
7
+
2
5
Now find the LCM of the denominators 5 and 7.
The LCM of 5 and 7 is 35.
Find the equivalent fractions with denominator 35.
=
3 × 5
7 × 5
+
2 × 7
5 × 7
=
15
35
+
14
35
=
15 + 14
35
=
29
35
(𝐢𝐢𝐢) −
𝟕
𝟗
+
𝟐
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 9 and 7.
Notes By Adil Aslam
WhatsApp: 0346-6248138

32. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
31 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
The LCM of 9 and 7 is 63.
Find the equivalent fractions with denominator 63.
= −
7 × 7
9 × 7
+
2 × 9
7 × 9
= −
49
63
+
18
63
=
−49 + 18
63
= −
31
63
(𝐢𝐯)
𝟑
𝟒
−
𝟐
𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 4 and 5.
The LCM of 4 and 5 is 20.
Find the equivalent fractions with denominator 20.
=
3 × 5
4 × 5
−
2 × 4
5 × 4
=
15
20
−
8
20
=
15 − 8
20
=
7
20
𝟒. 𝐒𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 (−
𝟐
𝟓
+
𝟒
𝟗
) + (−
𝟑
𝟒
) = −
𝟐
𝟓
+ (
𝟒
𝟗
+ (−
𝟑
𝟒
))
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
L. H. S = (−
2
5
+
4
9
) + (−
3
4
)
= (−
2 × 9
5 × 9
+
4 × 5
9 × 5
) −
3
4
= (−
18
45
+
20
45
) −
3
4
Notes By Adil Aslam
WhatsApp: 0346-6248138
Notes By Adil Aslam
WhatsApp: 0346-6248138

33. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
32 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= (
−18 + 20
45
) −
3
4
=
2
45
−
3
4
=
2 × 4
45 × 4
−
3 × 45
4 × 45
=
8
180
−
135
180
=
8 − 135
180
= −
127
180
R. H. S = −
2
5
+ (
4
9
+ (−
3
4
))
= −
2
5
+ (
4
9
−
3
4
)
= −
2
5
+ (
4 × 4
9 × 4
−
3 × 9
4 × 9
)
= −
2
5
+ (
16
36
−
27
36
)
= −
2
5
+ (
16 − 27
36
)
= −
2
5
+ (−
11
36
)
= −
2
5
−
11
36
LCM of 5 and 36 = 2 × 2 × 3 × 3 × 5
LCM of 4 and 45 = 2 × 2 × 3 × 3 × 5
Notes By Adil Aslam
WhatsApp: 0346-6248138

34. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
33 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= −
2 × 36
5 × 36
−
11 × 5
36 × 5
= −
72
180
−
55
180
=
−72 − 55
136
= −
127
180
L. H. S = R. H. S Hence Proved.
𝟓. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟:
(𝐢) −
𝟔
𝟕
−
−𝟐
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
−
6
7
−
−2
7
= −
6
7
+
2
7
= −
6 + 2
7
= −
4
7
(𝐢𝐢)
𝟕
𝟐𝟒
−
𝟏𝟏
𝟑𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 24 and 36.
The LCM of 24 and 36 is 72.
Find the equivalent fractions with denominator 72.
LCM of 24 and 36 = 2 × 2 × 2 × 3 × 3
Notes By Adil Aslam
WhatsApp: 0346-6248138

35. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
34 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
7 × 3
24 × 3
−
11 × 2
36 × 2
=
21
72
−
22
72
=
21 − 22
72
= −
1
72
(𝐢𝐢𝐢)
𝟏𝟎
𝟔𝟑
−
−𝟔
𝟕
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 63 and 7.
The LCM of 63 and 7 is 63.
Find the equivalent fractions with denominator 63.
=
10
63
+
6
7
=
10
63
+
6 × 9
7 × 9
=
10
63
+
54
63
=
10 + 54
63
=
64
63
(𝐢𝐢𝐢) −
𝟏𝟏
𝟏𝟑
−
−𝟓
𝟐𝟔
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 13 and 26.
The LCM of 13 and 26 is 26.
Notes By Adil Aslam
WhatsApp: 0346-6248138

36. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
35 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Find the equivalent fractions with denominator 26.
= −
11
13
+
5
26
= −
11 × 2
13 × 2
+
5
26
= −
22
26
+
5
26
=
−22 + 5
26
= −
17
26
𝟔. 𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲:
(𝐢)
𝟑
𝟕
+
𝟓
𝟗
−
−𝟐
𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 7 , 9 and 3.
The LCM of 7 , 9 and 3 is 63.
Find the equivalent fractions with denominator 63.
=
3
7
+
5
9
+
2
3
=
3 × 9
7 × 9
+
5 × 7
9 × 7
+
2 × 21
3 × 21
=
27
63
+
35
63
+
42
63
Notes By Adil Aslam
WhatsApp: 0346-6248138

37. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
36 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
27 + 35 + 42
63
=
62 + 42
63
=
104
63
(𝐢𝐢) −
𝟒
𝟏𝟏
+
−𝟐
𝟑
−
−𝟓
𝟗
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 11 , 3 and 9.
The LCM of 11 , 3 and 9 is 99.
Find the equivalent fractions with denominator 99.
= −
4
11
−
2
3
+
5
9
= −
4 × 9
11 × 9
−
2 × 33
3 × 33
+
5 × 11
9 × 11
= −
36
99
−
66
99
+
55
99
=
−36 − 66 + 55
99
=
−102 + 55
99
=
−47
99
Notes By Adil Aslam
WhatsApp: 0346-6248138

38. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
37 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐢𝐢) −
𝟏
𝟔
+
−𝟐
𝟑
−
𝟏
𝟑
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Now find the LCM of the denominators 6 , 3 and 3.
The LCM of 6 , 3 and 3 is 6.
Find the equivalent fractions with denominator 6.
= −
1
6
−
2
3
−
1
3
= −
1
6
−
2 × 2
3 × 2
−
1 × 2
3 × 2
= −
1
6
−
4
6
−
2
6
=
−1 − 4 − 2
6
= −
7
6
𝟕. 𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐲:
(𝐢)
𝟒
𝟏𝟓
𝐛𝐲
𝟑
𝟖
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
=
14
15
×
3
8
=
4 ÷ 4
15 ÷ 3
×
3 ÷ 3
8 ÷ 4
Notes By Adil Aslam
WhatsApp: 0346-6248138

39. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
38 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
1
5
×
1
2
=
1 × 1
5 × 2
=
1
10
(𝐢𝐢)
𝟑
𝟏𝟏
𝐛𝐲 (−
𝟐
𝟗
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
=
3
11
× −
2
9
=
3 ÷ 3
11
× −
2
9 ÷ 3
=
1
11
−
2
3
=
1 × −2
11 × 3
= −
2
33
(𝐢𝐢𝐢) (−
𝟓
𝟕
) 𝐛𝐲
𝟏𝟒
𝟏𝟓
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −
5
7
×
14
15
= −
5 ÷ 5
7 ÷ 7
×
14 ÷ 7
15 ÷ 5
Notes By Adil Aslam
WhatsApp: 0346-6248138

40. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
39 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= −
1
1
×
2
3
=
−1 × 2
1 × 3
= −
2
3
𝟖. 𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲:
(𝐢) (−
𝟖
𝟓
×
𝟑
𝟒
) + (
𝟕
𝟖
× −
𝟏𝟔
𝟐𝟓
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given: (−
8
5
×
3
4
) + (
7
8
× −
16
25
)
= (−
8
5
×
3
4
) + (
7
8
× −
16
25
)
= (−
2
5
×
3
1
) + (
7
1
× −
2
25
)
= (
−2 × 3
5 × 1
) + (
7 × −2
1 × 25
)
= (−
6
5
) + (−
14
25
)
= −
6
5
−
14
25
LCM of 5 and 25 is 25.
Find the equivalent fractions with denominator 25.
= −
6 × 5
5 × 5
−
14
25
LCM of 5 and 25 = 5 × 5
2
1
2
1
Notes By Adil Aslam
WhatsApp: 0346-6248138

41. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
40 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= −
30
25
−
14
25
=
−30 − 14
25
= −
44
25
(𝐢𝐢) (
𝟕
𝟐𝟓
×
−𝟏𝟓
𝟐𝟖
) − (−
𝟑
𝟓
×
𝟒
𝟗
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given: (
7
25
×
−15
28
) − (−
3
5
×
4
9
)
= (
7
25
×
−15
28
) − (−
3
5
×
4
9
)
= (
1
5
×
−3
4
) − (−
1
5
×
4
3
)
= (
1 × −3
5 × 4
) − (
−1 × 4
5 × 3
)
= (
−3
20
) − (
−4
15
)
= −
3
20
+
4
15
LCM of 20 and 15 is 60.
Find the equivalent fractions with denominator 60.
= −
3 × 3
20 × 3
+
4 × 4
15 × 4
LCM of 15 and 20 = 2 × 2 × 3 × 5
1 3
5 4
1
3
Notes By Adil Aslam
WhatsApp: 0346-6248138

42. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
41 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= −
9
60
+
16
60
=
−9 + 16
60
=
7
60
(𝐢𝐢𝐢) (−
𝟑
𝟒
×
−𝟖
𝟏𝟓
) − (
𝟐
𝟑
×
−𝟑
𝟓
) − (−
𝟒
𝟕
×
−𝟏𝟒
𝟏𝟓
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given: (−
3
4
×
−8
15
) − (
2
3
×
−3
5
) − (−
4
7
×
−14
15
)
= (−
3
4
×
−8
15
) − (
2
3
×
−3
5
) − (−
4
7
×
−14
15
)
= (−
1
1
×
−2
5
) − (
2
1
×
−1
5
) − (−
4
1
×
−2
15
)
= (
−1 × −2
1 × 5
) − (
2 × −1
1 × 5
) − (
−4 × −2
1 × 15
)
= (
2
5
) − (
−2
5
) − (
8
15
)
=
2
5
+
2
5
−
8
15
LCM of 5, 5 and 15 is 15.
Find the equivalent fractions with denominator 15.
=
2 × 3
5 × 3
+
2 × 3
5 × 3
−
8
15
1
5
1
1
2
1
2
LCM of 5, 5 and 15 = 3 × 5
Notes By Adil Aslam
WhatsApp: 0346-6248138

43. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
42 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
6
15
+
6
15
−
8
15
=
6 + 6 − 8
15
=
12 − 8
15
=
4
15
𝟗. 𝐒𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 (−
𝟓
𝟖
×
𝟒
𝟏𝟓
) ×
−𝟑
𝟒
= −
𝟓
𝟖
× (
𝟒
𝟏𝟓
×
−𝟑
𝟒
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given: (−
5
8
×
4
15
) ×
−3
4
= −
5
8
× (
4
15
×
−3
4
)
L. H. S = (−
5
8
×
4
15
) ×
−3
4
= (−
1
2
×
1
3
) ×
−3
4
= (
−1 × 1
2 × 3
) ×
−3
4
= (
−1
6
) ×
−3
4
=
−1
6
×
−3
4
=
−1 × −3
6 × 4
=
3
24
1 1
2 3
1
8
Notes By Adil Aslam
WhatsApp: 0346-6248138

44. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
43 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
1
8
R. H. S = −
5
8
× (
4
15
×
−3
4
)
= −
5
8
× (
1
5
×
−1
1
)
= −
5
8
× (
1 × −1
5 × 1
)
= −
5
8
× (
−1
5
)
= −
5
8
×
−1
5
=
−5 × −1
8 × 5
=
5
40
=
1
8
L. H. S = R. H. S Hence Proved.
𝟏𝟎. 𝐒𝐡𝐨𝐰 𝐭𝐡𝐚𝐭 −
𝟐
𝟑
(
𝟒
𝟓
+
−𝟖
𝟏𝟓
) = (−
𝟐
𝟑
×
𝟒
𝟓
) + (−
𝟐
𝟑
×
−𝟖
𝟏𝟓
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Given: −
2
3
(
4
5
+
−8
15
) = (−
2
3
×
4
5
) + (−
2
3
×
−8
15
)
L. H. S = −
2
3
(
4
5
+
−8
15
)
1 1
1
5
1
8
Notes By Adil Aslam
WhatsApp: 0346-6248138

45. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
44 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
LCM of 5 and 15 is 15.
Find the equivalent fractions with denominator 15.
= −
2
3
(
4 × 3
5 × 3
+
−8
15
)
= −
2
3
(
12
15
+
−8
15
)
= −
2
3
(
12 − 8
15
)
= −
2
3
(
4
15
)
= −
2
3
×
4
15
=
−2 × 4
3 × 15
= −
8
45
R. H. S = (−
2
3
×
4
5
) + (−
2
3
×
−8
15
)
= (
−2 × 4
3 × 5
) + (
−2 × −8
3 × 15
)
= (
−8
15
) + (
16
45
)
= −
8
15
+
16
45
LCM of 15 and 45 is 45.
Find the equivalent fractions with denominator 45.
LCM of 15 and 45 = 3 × 3 × 5
Notes By Adil Aslam
WhatsApp: 0346-6248138

46. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
45 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
= −
8 × 3
15 × 3
+
16
45
= −
24
45
+
16
45
=
−24 + 16
45
= −
8
45
L. H. S = R. H. S Hence Proved.
𝟏𝟏. 𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲:
(𝐢) − 𝟒 ÷ (−
𝟐
𝟓
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −4 ÷ −
2
5
= −4 × −
5
2
=
−4 × −5
2
=
20
2
= 10
(𝐢𝐢)
𝟏𝟓
𝟕
÷ (−
𝟓
𝟏𝟒
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
Notes By Adil Aslam
WhatsApp: 0346-6248138
Notes By Adil Aslam
WhatsApp: 0346-6248138

47. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
46 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
=
15
7
÷ −
5
14
=
15
7
× −
14
5
=
3
1
× −
2
1
=
3 × −2
1 × 1
=
−6
1
= −6
(𝐢𝐢𝐢) −
𝟑
𝟖
÷ (−
𝟑
𝟒
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −
3
8
÷ −
3
4
= −
3
8
× −
4
3
= −
1
2
× −
1
1
=
−1 × −1
2 × 1
=
1
2
(𝐢𝐯) (−
𝟒
𝟏𝟑
) ÷ (−
𝟖
𝟑𝟗
)
1 1
3 2
1 1
1
2
Notes By Adil Aslam
WhatsApp: 0346-6248138

48. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
47 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −
4
13
÷ −
8
39
= −
4
13
× −
39
8
= −
1
1
× −
3
2
=
−1 × −3
1 × 2
=
3
2
(𝐯) (−
𝟔
𝟏𝟓
) ÷ (
𝟒
𝟑𝟓
)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= −
6
15
÷
4
35
= −
6
15
×
35
4
= −
3
3
×
7
2
=
−3 × 7
3 × 2
= −
21
6
= −
7
2
1
1
3
2
3
2
7
3
Notes By Adil Aslam
WhatsApp: 0346-6248138

49. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
48 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
𝟏𝟐. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐚𝐧𝐝 𝐫𝐞𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐭𝐡𝐞𝐦 𝐨𝐧
𝐭𝐡𝐞 𝐧𝐮𝐦𝐛𝐞𝐫 𝐥𝐢𝐧𝐞.
(𝐢) |
𝟕
𝟖
|
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= |
7
8
| =
7
8
Since,
7
8
is a positive rational number.
Therefore,
7
8
lies between 0 and 1 on the number line.
Since in
7
8
, denominator is 8.
We divide line between 0 and 1 into 8 equal parts.
(𝐢𝐢) |
−𝟑
𝟓
|
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= |
−3
5
| =
3
5
Since,
3
5
is a positive rational number.
Notes By Adil Aslam
WhatsApp: 0346-6248138

50. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
49 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Therefore,
3
5
lies between 0 and 1 on the number line.
Since in
3
5
, denominator is 5.
We divide line between 0 and 1 into 5 equal parts.
(𝐢𝐢𝐢) |
𝟒
𝟏𝟏
|
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= |
4
11
| =
4
11
Since,
4
11
is a positive rational number.
Therefore,
4
11
lies between 0 and 1 on the number line.
Since in
4
11
, denominator is 11.
We divide line between 0 and 1 into 11 equal parts.
Notes By Adil Aslam
WhatsApp: 0346-6248138

51. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
50 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
(𝐢𝐯) |
−𝟓
𝟔
|
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= |
−5
6
| =
5
6
Since,
5
6
is a positive rational number.
Therefore,
5
6
lies between 0 and 1 on the number line.
Since in
5
6
, denominator is 6.
We divide line between 0 and 1 into 6 equal parts.
(𝐯) |
−𝟖
𝟗
|
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
= |
−8
9
| =
8
9
Since,
8
9
is a positive rational number.
Therefore,
8
9
lies between 0 and 1 on the number line.
Notes By Adil Aslam
WhatsApp: 0346-6248138
Notes By Adil Aslam
WhatsApp: 0346-6248138

52. 𝐍𝐨𝐭𝐞𝐬 𝐁𝐲: Adil Aslam 𝐒𝐮𝐛𝐣𝐞𝐜𝐭: Mathematics Class 5
51 | P a g e W h a t s A p p : 0 3 4 6 6 2 4 8 1 3 8 E m a i l : a d i l a s l a m 5 9 5 9 @ m a i l . c o m
Since in
8
9
, denominator is 9.
We divide line between 0 and 1 into 9 equal parts.
 Best of Luck 
Notes By Adil Aslam
WhatsApp: 0346-6248138