Number Systems(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

NUMBER SYSTEMS MCQs

Total Questions : 197 | Page 3 of 20 pages

Question 21. Find the value of

(3 + \sqrt{5}) (3 - \sqrt{5})

Discuss Question

Answer: Option D. -> 4
:
D
We know the identity:

a^{2} - b^{2} = (a + b) (a - b)

(3 + \sqrt{5}) (3 - \sqrt{5}) = (3^{2} - (\sqrt{5})^{2} = 9 - 5 = 4

Hence,

(3 + \sqrt{5}) (3 - \sqrt{5}) = 4

, which is a rational number.

Question 22. Which of the following fractions is equal to

\frac{1}{\sqrt{5} - \sqrt{7}}

−(√5+√7)4
−(√5+√7)2
−(√5−√7)2
−(√5+√7)8

Discuss Question

Answer: Option B. -> −(√5+√7)2
:
B
Rationalizing the expression we get,

\frac{1}{\sqrt{5} - \sqrt{7}} = \frac{\sqrt{5} + \sqrt{7}}{(\sqrt{5} - \sqrt{7}) (\sqrt{5} + \sqrt{7)}}

\frac{\sqrt{5} + \sqrt{7}}{(\sqrt{5})^{2} - (\sqrt{7})^{2}}

\frac{\sqrt{5} + \sqrt{7}}{5 - 7}

\frac{- (\sqrt{5} + \sqrt{7})}{2}

Question 23.

If {(\sqrt{1024})}^{3} = 8^{x}, then x is

________.

Discuss Question

Answer: Option A. -> 5
:
A
Given:

{(\sqrt{1024})}^{3} = 8^{x}

\Rightarrow (\sqrt{(} 2^{10}))^{3} = 8^{x}

\Rightarrow (\sqrt{(} (2^{5})^{2}))^{3} = 8^{x}

. {as

{(a^{m})}^{n}

a^{m n}

}

{\Rightarrow (2^{5})}^{3} = {(2^{3})}^{x}

{\Rightarrow (2^{3})}^{5} = {(2^{3})}^{x}

\Rightarrow (8)^{5}

(8)^{x}

We know that, if

a^{n}

a^{m}

then,

n = m

.
Hence,

\Rightarrow x = 5

Question 24.

\sqrt[n]{2} \times \sqrt[n]{3} = \sqrt[n]{6}

True
False
−(√5−√7)2
−(√5+√7)8

Discuss Question

Answer: Option A. -> True
:
A
We know that

\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a b}

So, \sqrt[n]{2} \times \sqrt[n]{3} = \sqrt[n]{2 \times 3} = \sqrt[n]{6}

Question 25. Find the hypotenuse of a right angled triangle whose perpendicular sides are

\sqrt{6}

cm and

\sqrt{3}

cm.

4 cm
3 cm
2 cm
1 cm

Discuss Question

Answer: Option B. -> 3 cm
:
B
Given:perpendicular sides of a right angled triangle are

\sqrt{6}

and

\sqrt{3}

.
Using Pythagoras theorem,
Hypotenuse =

\sqrt{{\sqrt{6}}^{2} + {\sqrt{3}}^{2}}

\sqrt{6 + 3}

\sqrt{9}

cm
= 3cm

Question 26.

2 \sqrt{3} \times 3 \sqrt{2}

is ___. (rational/irrational)
.

Discuss Question

2 \sqrt{3} \times 3 \sqrt{2} = 2 \times 3 \times \sqrt{2 \times 3} = 6 \sqrt{6}

The presence of factor

\sqrt{6}

makes the expression irrational.

Question 27.

$\frac{3}{4}$ lies between $\frac{1}{2}$ and 1.

True
False

Discuss Question

Answer: Option A. -> True
:
A

To check if a number lies between any two numbers, we first convert the numbers into the decimal form. $\frac{3}{4} = 0.75$ , $\frac{1}{2} = 0.5$ and we have to check if $0.75$ lies between $0.5$ and $1$ and plot the numbers on the number line as follows :

Therefore, we can say that $\frac{3}{4}$ lies between $\frac{1}{2}$ and 1.

Question 28.

A number $^{'} x^{'}$ was found to be represented in the simplest $\frac{p}{q}$ form. The decimal expansion of this number was found to be 5.3333333333..... Find the value of p - q.

Discuss Question

Answer: Option C. -> 13
:
C
Given decimal is 5.3333333......
Let,

x = 5.3333333......

Therefore,

10 x = 53.33333333....

Subtracting

x

from

10 x

, we get

10 x - x = 53.33333.. - 5.33333.. = 48

\Rightarrow 9 x = 48

\Rightarrow x = \frac{48}{9} = \frac{16}{3} = \frac{p}{q}

So,

p - q = 16 - 3

∴ p - q = 13

Question 29.

Which of the following represents whole numbers?

Positive integers
Negative integers
$π$
Integers

Discuss Question

Answer: Option A. -> Positive integers
:
A

Integers are constituted by natural numbers, their negatives and 0. Removing the negative numbers from integers would leave us with the whole numbers. Therefore, whole numbers include 0 as well as all the positive integers.

Question 30.

The product of two irrational numbers is a/an

Irrational number
Rational number
Real number
Integer

Discuss Question

Answer: Option A. -> Irrational number
:
A, B, C, and D

The product of two irrational numbers will be either a rational or an irrational number.
Consider the following example: $(\sqrt{3} + \sqrt{2}) \times (\sqrt{3} - \sqrt{2}) = 1$ .
Both $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are irrational numbers and their product is 1.
1 is an integer and since, all integers are rational numbers as well, we can infer that product of two irrational numbers may be an integer, a rational number or an irrational number and they are all real numbers.
When we multiply the irrational numbers $\sqrt{2}$ and $\sqrt{3}$ , we get $\sqrt{6}$ , which is also an irrational number.