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CBSE 9th Standard Mathematics Number Systems Model Questions Papers Study Material - QB365 Set B

Published on: 29/10/2025

Question Set B

CBSE 9th Standard Mathematics Number Systems Model Questions Papers Study Material - QB365

Download CBSE Class 9th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 9th Standard CBSE Mathematics

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Std: 9th Standard CBSE - Mathematics

chapter: Number Systems - Model Questions Papers

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Questions:

Generic 2

5 x 2 = 10

Express \(0.12\overline { 3 } \) into the form p/q, where p,q are integers and \(q\neq 0\)

Express \(18.\overline{48} \) in the form of p/q, where p and q are integers, \(q\neq 0\) .

Express \(0.\overline { 001 } \) as a rational number in the form p/q, where p and q are integers and \(q\neq 0\)

Write three rational numbers between -2/5 and -1/5.

Find two rational numbers between 0.1 and 0.2.

Generic 3

4 x 3 = 12

Simplify each of the following expressions: \({ \left( \sqrt { 5 } -\sqrt { 2 } \right) }{ \left( \sqrt { 5 } +\sqrt { 2 } \right) }\)

Classify the following numbers as rational or irrational: \((3+\sqrt { 23 } )-\sqrt { 23 } \)

Evaluate after rationalising the denominator of \(\left( \frac { 25 }{ \sqrt { 40 } -\sqrt { 80 } } \right) \)It is being given that \(\sqrt { 5 } \)

Give an example of two irrational numbers whose:
(a) Sum is rational
(b) Product is rational
(c) Quotient is rational

Multiple Choice Question

9 x 1 = 9

10.

The decimal expansion of \(\sqrt { 2 } \) is

(a)

finite decimal

(b)

1.4121

(c)

non-terminating recurring

(d)

non-terminating non-recurring

11.

The process of visualisation of representation of numbers on the number line through a magnifying glass is called

(a)

successive magnification

(b)

approximation

(c)

imagination

(d)

none of these

12.

A number is an irrational if and only if its decimal representation is:

(a)

non-terminating

(b)

non-terminating and repeating

(c)

non terminating and non repeating

(d)

terminating

13.

If \(\sqrt { x } \) is an irrational number, then x is:

(a)

rational

(b)

irrational

(c)

(d)

real

14.

Two rational numbers between \(\frac { 2 }{ 3 } \) and \(\frac { 5 }{ 3 } \)are:

(a)

1/6 and 2/6

(b)

1/2 and 2/7

(c)

5/6 and 7/6

(d)

2/3 and 4/3

15.

A rational number lying between \(\sqrt { 2 } \) and \(\sqrt { 3 } \) is:

(a)

\(\frac { \sqrt { 2 } +\sqrt { 3 } }{ 2 } \)

(b)

\(\sqrt { 6 } \)

(c)

1.6

(d)

1.9

16.

A rational number lying between -3 and 3 is:

(a)

(b)

-4.3

(c)

-3.4

(d)

1.101 1001 10001...

17.

Every rational number is:

(a)

a natural number

(b)

an integer

(c)

a real number

(d)

a whole number

18.

Which of the following is a rational number?

(a)

\(1+\sqrt { 3 } \)

(b)

\(\pi \)

(c)

\(2\sqrt { 3 } \)

(d)

Generic 5

3 x 5 = 15

19.

Rationalise \(\frac { 1 }{ \sqrt { 7 } +\sqrt { 3 } -\sqrt { 2 } } .\)

20.

Find the value of \(\frac { 4 }{ { (216) }^{ -2/3 } } +\frac { 1 }{ { (256) }^{ -3/4 } } +\frac { 2 }{ { (243) }^{ -1/5 } } .\)

21.

If \(x=(5+2\sqrt { 6 } ),\) then show that \(\sqrt { x } +\frac { 1 }{ \sqrt { x } } =2\sqrt { 3 } .\)

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Answers Key:

Generic 2

5 x 2 = 10

37/300

610/33

1/999

-7/20,-3/10,-1/4

0.125, 0.15

Generic 3

4 x 3 = 12

\({ \left( \sqrt { 5 } -\sqrt { 2 } \right) }{ \left( \sqrt { 5 } +\sqrt { 2 } \right) }={ \left( \sqrt { 5 } \right) }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 }\)
\(\\ =5-2=3\)

\((3+\sqrt { 23 } )-\sqrt { 23 } \)=\((3+\sqrt { 23 } )-\sqrt { 23 } \)= 3 which is a rational number.

- 9.5425

\((a)\ \sqrt { 3 } ,-\sqrt { 3 } \)
\(\\ (b)\ \sqrt { 3 } ,-\sqrt { 3 }\)
\( \\ (c)\ \sqrt { 3 } ,-\sqrt { 3 } \)

Multiple Choice Question

9 x 1 = 9

10.

(d)
non-terminating non-recurring

11.

(a)
successive magnification

12.

13.

(d)
real

14.

15.

16.

(a)
0

17.

18.

(d)
0

Generic 5

3 x 5 = 15

19.

\(\frac { 1 }{ \sqrt { 7 } +\sqrt { 3 } -\sqrt { 2 } } \)
\(=\frac { 1 }{ \sqrt { 7 } +\sqrt { 3 } -\sqrt { 2 } } \times \frac { (\sqrt { 7 } +\sqrt { 3 } )+\sqrt { 2 } }{ (\sqrt { 7 } +\sqrt { 3 } )+\sqrt { 2 } } \)
\(=\frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ [{ (\sqrt { 7 } +\sqrt { 3 } ) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 } } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\quad \frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ { (\sqrt { 7 } ) }^{ 2 }+{ (\sqrt { 3 } ) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 } } \left[ \because \quad { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=\frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ 7+3+2\sqrt { 21 } -2 } =\frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ 8+2\sqrt { 21 } } \)
\(=\frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ 2(4+\sqrt { 21 } ) } \)
\(=\frac { \sqrt { 7 } +\sqrt { 3 } +\sqrt { 2 } }{ 2(4+\sqrt { 21 } ) } \times \frac { 4-\sqrt { 21 } }{ 4-\sqrt { 21 } } \) [by rationalising]
\(=\frac { 4\sqrt { 7 } +4\sqrt { 3 } +4\sqrt { 2 } -7\sqrt { 3 } -3\sqrt { 7 } -\sqrt { 42 } }{ 2(16-21) } \quad \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { \sqrt { 7 } -3\sqrt { 3 } +4\sqrt { 2 } -\sqrt { 42 } }{ -10 } \)
\(=\frac { 3\sqrt { 3 } -4\sqrt { 2 } +\sqrt { 42 } -\sqrt { 7 } }{ 10 } \)

20.

\(\frac { 4 }{ { (216) }^{ -2/3 } } +\frac { 1 }{ { (256) }^{ -3/4 } } +\frac { 2 }{ { (243) }^{ -1/5 } } \)
= 4 x (216)^2/3+(256)^3/4+2 x (243)^1/5 \(\left[ \therefore \quad \frac { 1 }{ { (a) }^{ -m } } ={ a }^{ m } \right] \)
\(=4\times { (6\times 6\times 6) }^{ 2/3 }+{ (4\times 4\times 4\times 4) }^{ 3/4 }+2{ (3\times 3\times 3\times 3\times 3) }^{ 1/5 }\)
\(=4\times { ({ 6 }^{ 3 }) }^{ 2/3 }+{ ({ 4 }^{ 4 }) }^{ 3/4 }+2{ ({ 3 }^{ 5 }) }^{ 1/5 }\)
\(=4\times { (6) }^{ 2 }+{ (4) }^{ 3 }+2\times { (3) }^{ 1 }\quad \quad \left[ \because \quad { ({ a }^{ m }) }^{ n }={ a }^{ mn } \right] \)
= 4 x 36 + 64 + 6
= 144 + 64 + 6 = 150 + 64 = 214

21.

We have, \(x=(5+2\sqrt { 6 } )\)
Now, \(\frac { 1 }{ x } =\frac { 1 }{ 5+2\sqrt { 6 } } \times \frac { 5-2\sqrt { 6 } }{ 5-2\sqrt { 6 } } \) [by rationalising]
\(=\frac { 5-2\sqrt { 6 } }{ { (5) }^{ 2 }-{ (2\sqrt { 6 } ) }^{ 2 } } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { 5-2\sqrt { 6 } }{ 25-24 } =5-2\sqrt { 6 } \)
\(\therefore \ x+\frac { 1 }{ x } =5+2\sqrt { 6 } +5-2\sqrt { 6 } =10\)
\(\Rightarrow \ x+\frac { 1 }{ x } +2=12\) [adding 2 on both sides]
\(\Rightarrow \ { \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) }^{ 2 }={ (2\sqrt { 3 } ) }^{ 2 }\)
On taking positive square root both sides, we get
\(\sqrt { x } +\frac { 1 }{ \sqrt { x } } =2\sqrt { 3 } \)
Hence proved.