MCQ Questions for Class 9 Maths Number Systems Quiz 4

The rational number between

$\cfrac{1}{2}$ and

$\cfrac{6}{10}$ is

0%
$\cfrac{1}{4}$
0%
$\cfrac{3}{4}$
0%
$\cfrac{21}{40}$
0%
$\cfrac{33}{100}$

Explanation

$\dfrac{1}{2}$ =

$\dfrac{5}{10}$

$\dfrac{5}{10}$ and

$\dfrac{6}{10}$

Multiplying the denominator and numerator by

$4$ we get,

$\dfrac{5}{10}\times 4=\dfrac{20}{40}$ and

$\dfrac{6}{10}\times 4=\dfrac{24}{40}$

Number between

$\dfrac{20}{40}$ and

$\dfrac{24}{40}$ is

$\dfrac{21}{40}$

Write five rational numbers which are smaller than

$2$ .

0%
$1,\,\displaystyle\frac{1}{2},\,0,\,-1,\,-\displaystyle\frac{1}{2}$
0%
$0, 1 , 1.414, \sqrt3, -1$
0%
$0, 1 , \sqrt2, \sqrt3, -1$
0%
$0, 1 , 1.732, \sqrt2, -1$

Explanation

Five rational numbers less than

$2$ may be taken

$1,\,\displaystyle\frac{1}{2},\,0,\,-1,\,-\displaystyle\frac{1}{2}$
(There can be many more such rational numbers).

Simplification of

$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$ is

$\dfrac{729}{8}$

0%
True
0%
False

Explanation

$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$

$=\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{3\times 5}{2} \right )^{3}$

$=\displaystyle \left ( \frac{3^3}{5^3} \right ) \times \left ( \frac{3^3\times 5^3}{2^3} \right )$

$=\displaystyle \left ( \frac{3^3}1 \right ) \times \left ( \frac{3^3}{2^3} \right )$

$=\dfrac{3^6}{2^3}$

$=\dfrac{729}{8}$

The rationalising factor of

$\displaystyle \left ( a+\sqrt{b} \right )$ is

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$\displaystyle a-\sqrt{b}$
0%
$\displaystyle \sqrt{a}-b$
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$\displaystyle \sqrt{a}-\sqrt{b}$
0%
None of this

Explanation

Rationalizing factor of

$a+\sqrt{b}$ is

$a-\sqrt{b}$

$(a+\sqrt{b})\times (a-\sqrt{b})=a^2-b^2$

Which of the following expresses zero law of exponents?

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$0^{x} = 0$
0%
$x^{0} = 1$
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$x^{1} = x$
0%
None of the above

Explanation

According to the zero law of exponents,

any non - zero number raised to the power of zero is equal to

$1$ .

$\therefore x^{0} = 1$ , where

$x\neq 0$

So, option

$B$ is correct.

$4\times 4^{10}$ is represented as:

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$4^{40}$
0%
$4^{10}$
0%
$4^{11}$
0%
$16^{10}$

Explanation

$4\times 4^{10} = 4^{1 + 10}$

$= 4^{11}$

So, option

$C$ is correct.

In simplified form,

$((3)^{0} + (5)^{0})^{0}$ is equal to:

0%
$2$
0%
$1$
0%
$0$
0%
$8$

Explanation

$(3^{0} + 5^{0})^{0} = (1 + 1)^{0}$

$= (2)^{0}$

$= 1$

So, option

$B$ is correct.

$(2+\sqrt 3)$ is

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rational
0%
irrational
0%
an integer
0%
not real

Explanation

$\sqrt{3}$ is an irrational number. So any irrational number added to a rational number gives an irrational number.
So,

$2+\sqrt{3}$ is an irrational number
Answer is Option B

$a^0=1$ is true for all

$a$ except

0%
$-1$
0%
negative integers
0%
0
0%
1

Explanation

The correct answer is C

${ (-1 )}^{ 0 }=1$

${ (-2) }^{ 0 }=1$

${ 0 }^{ 0 }$ is not defined.

${ 1 }^{ 0 }=1$

$\displaystyle \frac{-3}{0}$ is a-

0%
Negative rational number
0%
Positive rational number
0%
Either positive or negative rational number
0%
None of these

The value of

$21^{0}$ is _____.

0%
$0$
0%
$21$
0%
$1$
0%
$-21$

Explanation

$21^{0} = 1$

$\because a^{0} = 1$

This is the law for zero exponent.

So, option

$C$ is correct.

$\displaystyle \frac{-2}{-19}$ is a -

0%
Negative rational number
0%
positive rational number
0%
neither positive nor negative rational number
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None of these

Which of the following expresses product law of exponents?

0%
$(a^{m})^{n} = a^{mn}$
0%
$a^{m}\times a^{n} = a^{m+n}$
0%
$a^{m} \div a^{n} = a^{m-n}$
0%
$a^{m}\div b^{m} = \left (\dfrac {a}{b}\right )^{m}$

Explanation

The product law of exponents states that,

when multiplying two powers that have the same base,

we can add their exponents.

$\therefore a^{m}\times a^{n} = a^{m+n}$

So, option

$B$ is correct.

Which one of the following is the value of

$(101)^{0}$ ?

0%
$0$
0%
$101$
0%
$1010$
0%
$1$

Explanation

$101^{0} = 1$

So, option

$D$ is correct.

The value of

$(10)^{0} \times (20)^{0}\times (30)^{0}$ is:

0%
$1$
0%
$3$
0%
$60$
0%
$0$

Explanation

$(10)^{0} \times (20)^{0} \times (30)^{0} = 1\times 1\times 1 = 1$

So, option

$A$ is correct.

Find the value of

$(6)^{0} - (10)^{0}$

0%
$-4$
0%
$2$
0%
$1$
0%
$0$

Explanation

$(6)^0 - (10)^0 = 1-1 = 0$

So, option

$D$ is correct.

The expression

$((2)^{0} + (3)^{0} + (5)^{0})^{0}$ is equal to _____

0%
$3$
0%
$1$
0%
$10$
0%
$0$

Explanation

$((2)^{0} + (3)^{0} + (5)^{0})^{0} = (1 + 1 + 1)^{0}$

$= 3^{0}$

$= 1$

So, option

$B$ is the correct answer.

Which of the following is equal to

$1$ ?

0%
$(3)^{0} + (3)^{0}$
0%
$(3)^{0} - (3)^{0}$
0%
$(3)^{0}$
0%
$(3^{3})^{0} + (3^{0})^{3}$

Explanation

$\textbf{Step 1: Simplify the following expressions.}$

$(a)\ 3^0+3^0=1+1=2\neq 1$

$(b)\ 3^0-3^0=1-1=0\neq 1$

$(c)\ 3^0=1$

$(d)\ (3^3)^0+(3^0)^3=9^0+1^3=1+1=2\neq 1$

$\textbf{Hence, Option C is correct.}$

Simplify:

$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$

0%
$1$
0%
$3$
0%
$0$
0%
$2$

Explanation

$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$

$= (1 + 1 + 1) \div 1$

$= 3\div 1$

$= 3$

So, option

$B$ is correct.

The value of

$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$ is _____

0%
$2$
0%
$4$
0%
$1$
0%
$0$

Explanation

$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$

$= (1 + 1)\div (1 + 1)$

$= 2\div 2$

$= 1$

So, option

$C$ is correct.

The value of

$(100)^{0}$ is _____.

0%
$1$
0%
$0$
0%
$100$
0%
$1000$

Explanation

According to the law of power of

$0$ ,

$(100)^{0} = 1$ .

So, option

$A$ is correct.

Evaluate:

$({(10)^0} + (12)^{0})\times (18)^{0}$

0%
$1$
0%
$0$
0%
$2$
0%
$18$

Explanation

$((10)^{0} + (12)^{0})\times (18)^{0} = (1 + 1) \times 1$

$= (2)\times 1$

$= 2$

So. option

$C$ is correct.

What are rational numbers?

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A number of the form $\dfrac {p}{q}$ where $p$ and $q$ are any positive integers and $q\neq 0$ is called a rational number.
0%
A number of the form $\dfrac {p}{q}$ where $p$ and $q$ are any negative integers and $q\neq 0$ is called a rational number.
0%
A number of the form $\dfrac {p}{q}$ where $p$ and $q$ are integers is called a rational number.
0%
A number of the form $\dfrac {p}{q}$ where $p$ and $q$ are any integers and $q\neq 0$ is called a rational number.

Explanation

A number of the form

$\dfrac {p}{q}$ where

$p$ and

$q$ are any integers and

$q\neq 0$ is called a rational number.

Ex:- Consider

$\dfrac{2}{3}$ here

$2,3$ are co-primes so the number

$\dfrac{2}{3}$ is a rational number

$\dfrac {7}{9}$ is a/an _______ number.

0%
rational
0%
composite
0%
irrational
0%
prime

Explanation

$\dfrac{7}{9}$ is of the form

$\dfrac {p}{q}$ form , hence it is rational no.

Which of the following is not a rational number?

0%
$\dfrac {0}{1}$
0%
$\dfrac {1}{0}$
0%
$\dfrac {0}{3}$
0%
$\dfrac {0}{2}$

Explanation

A number of the form

$\dfrac {p}{q}$ , where

$'p'$ and

$'q'$ are any integers and

$q\neq 0$ is called a rational number.

$\therefore \dfrac {1}{0}$ is not a rational number because it is not defined.

$\sqrt {23}$ is not a ...... number.

0%
irrational
0%
co-prime
0%
composite
0%
rational

Explanation

As per the theorem, the square root of any prime number is irrational.

$\sqrt {23}$ is a prime number, so is not a rational number. It is irrational.
Therefore,

$D$ is the correct answer.

Irrational number is defined as

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a real number that cannot be made by dividing two integers.
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a real number that can be made by dividing two integer.
0%
a number that can be made derived after multiplying two integers.
0%
a real number that can be written as whole number.

Explanation

An irrational is any real number that cannot be expressed as a ratio of integers.

Therefore,

$A$ is the correct answer.

There exists ....... number of rational numbers between

$\dfrac {2}{5}$ and

$\dfrac {4}{5}.$

0%
$0$
0%
$1$
0%
$5$
0%
infinite

Explanation

There exists infinite number of rational numbers between any two rational numbers. i.e. in this case between

$\dfrac {2}{5}$ and

$\dfrac {4}{5}$ .

If the quotient is terminating decimal, the division is complete only when ...............

0%
we get the remainder $1$
0%
we get the remainder zero
0%
we get the remainder as the repeated numbers
0%
All of the above

Explanation

Any division is complete and terminates only when we get the remainder zero.

e.g.

$\dfrac 2 {10}=0.2$

we can see the remainder becomes zero in this division.

Therefore,

$B$ is the correct answer.

Which of the following rational numbers lies between

$0$ and

$-1$ ?

0%
$0$
0%
$-1$
0%
$\dfrac {-1}{4}$
0%
$\dfrac {1}{4}$

Explanation

Clearly,

$0$ and

$-1$ cannot lie between

$0$ and

$-1$ .

Also,

$0=\dfrac{0}{4}$ and

$-1=\dfrac{-4}{4}$

We can clearly see that

$\dfrac {-1}{4}$ lies between

$0$ and

$-1$ .

CBSE Questions for Class 9 Maths Number Systems Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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