Class 8 : maths : Rational and Irrational numbers :All Practice set

Class 8 : maths : Rational and Irrational numbers : Practice set 1.1 

QUE : (1). Show the following numbers on a number line. Draw a separate number line

for each example.

i. 32,52,−32
ii. 75,−25,−45
iii. −58,118
iv. 1310,−1710

ANSWER :

(i.)  32,52,−32

Here, the denominator of each fraction is 2.
∴ Divide each unit into 2 equal parts.

ii. 75,−25,−45

Here, the denominator of each fraction is 5.
∴ Divide each unit into 5 equal parts.

iii. −58,118

Here, the denominator of each fraction is 8.
∴ Divide each unit into 8 equal parts.

iv. 1310,−1710

Here, the denominator of each fraction is 10.
∴ Divide each unit into 10 equal parts.

QUE : 2. Observe the number line and answer the questions.

Observe the number line and answer the questions.

i. Which number is indicated by point B?
ii. Which point indicates the number 134 ?
iii. State whether the statement, ‘the point D denotes the number 52 is true or false.
Solution:
Here, each emit is divided into 4 equal parts.
i. Point B is marked on the 10th equal part on the left side of O.
∴ The number indicated by point B is −104.

ii.

Point C is marked on the 7th equal part on the right side of O.
∴ The number 134 is indicated by point C.

iii. True
Point D is marked on the 10th equal part on the right side of O.
∴ D denotes the number 104=5×22×2=52

 Class 8 : maths : Rational and Irrational numbers : Practice set 1.2

QUE : 1. Compare the following numbers.

i. 7, -2
ii. 0, −95
iii. 87, 0
iv. −54,14
v. 4029,14129
vi. −1720,−1320
vii. 1512,716
viii. −258,−94
ix. 1215,35
x. −711,−34
Solution:
1]. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

2]. 0, −95
On a number line, −95 is to the left of zero.
∴ 0 > −95

3] . 87, 0
On a number line, zero is to the left of 87 .
∴ 87 > 0

4]. −54,14
We know that, a negative number is always less than a positive number.
∴ −54<14

5]. 4029,14129
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ 4029<14129

6]. −1720,−1320
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ −1720<−1320

7]. 1512,716
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

8]. −258,−94
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

9]. 1215,35
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

10]. −711,−34
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

 Class 8 : maths : Rational and Irrational numbers : Practice set 1.3

QUE :1. Write the following rational numbers in decimal form.

1]. 937
2]. 1842
3]. 914
4]. −1035
5]. −1113
ANSWER :
1]. 937

2]. 1842

3]. 914

4]. −1035

5]. −1113

 Class 8 : maths : Rational and Irrational numbers : Practice set 1.4

QUE :1]. The number √2 is shown on a number line. Steps are given to show 3 on the number line using √2. Fill in the boxes properly and complete the activity.

The point Q on the number line shows the number ……….
A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
Right angled ∆OQR is obtained by drawing seg OR.
l(OQ) = √2, l(QR) = 1
∴By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3
Solution:
The point Q on the number line shows the number √2
A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
Right angled ∆OQR is obtained by drawing seg OR.
l(OQ) = √2, l(QR) = 1
∴By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²

.. .[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number √3.

QUE : 2]  Show the number 5 on the number line.

ANSWER :

Draw a number line and take a point Q at 2
such that l(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.
Draw seg OR.
∆OQR formed is a right angled triangle.
By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²
= 2² + 1²
= 4 + 1
= 5
∴l(OR) = √5 units
…[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C. The point C shows the number √5.

QUE :3]. Show the number 7 on the number line.

ANSWER :

Draw a number line and take a point Q at 2 such that l(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that l(QR) = 1 unit.
Draw seg OR.
∆OQR formed is a right angled triangle.
By Pythagoras theorem,
[l(OR)]² = [l(OQ)]² + [l(QR)]²
= 2² + 1²
= 4 + 1
= 5
∴ l(OR) = √5 units
… [Taking square root of both sides]
Draw an arc with centre O and radius OR.
Mark the point of intersection of the number line and arc as C. The point C shows the number √5.
Similarly, draw a line CD perpendicular to the number line through the point C such that l(CD) = 1 unit.
By Pythagoras theorem,
l(OD) = √6 units
The point E shows the number √6 .
Similarly, draw a line EP perpendicular to the number line through the point E such that l(EP) = 1 unit.
By Pythagoras theorem,
l(OP) = √7 units
The point F shows the number √7.

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