MCQ Questions for Class 9 Maths Number Systems Quiz 6

The conjugate of the surd

$\sqrt{5}-\sqrt{2}$ ?

0%
$\sqrt{5}+\sqrt2$
0%
$-\sqrt{5}+\sqrt2$
0%
Either $(A)$ or $(B)$
0%
None of these

Explanation

The sum and difference of two quadratic surds is called as conjugate to each other.

In the given case, difference of two surds

$\sqrt{5}$ and

$\sqrt2$ is given. Hence, the conjugate will be sum of these two surds, i.e.

$\sqrt5+\sqrt2$ .

Option A is the correct answer.

$\dfrac {p}{q}$ is a rational number when

0%
$p=0,q\neq0$
0%
$p=0,q=0$
0%
$p\ne 0,q=0$
0%
$p=1,q=0$

Explanation

By the definition of rational number,

$\dfrac{p}{q}$ is rational where

$p$ and

$q$ are integers and

$q$ is not equal to zero.

In all three options except A,

$q=0$ .

Hence,

$p=0, q\ne 0$ is true.

Add :

$6 + 3\sqrt 3$ and

$8 + 4\sqrt 3$

0%
$1+7\sqrt{3}$
0%
$8+7\sqrt{3}$
0%
$14+7\sqrt{3}$
0%
$4+7\sqrt{3}$

Explanation

$6+3\sqrt{3}+8+4\sqrt{3}$

$=6+8+3\sqrt{3}+4\sqrt{3}$

$=14+3\sqrt{3}+4\sqrt{3}$

$=14+\sqrt{3}(3+4)$

$=14+\sqrt{3}\times 7$

$=14+7\sqrt{3}$

$a^{0}=1$ . Is it true or false?

0%
True
0%
False

Explanation

Any thing power zero is equal to

$1$

$\Rightarrow \therefore a^o=1$

Note: All values except

$a=0$ .

$\sqrt{a}$ is an irrational number, what is a?

0%
Rational
0%
Irrational
0%
$0$
0%
Real

Explanation

Consider the given irrational number $\sqrt{a}$ ,

Definition of rational number- which number can be write in the form of $\dfrac{p}{q}$ but $q\ne 0$ is called rational number.

Hence, $a=\dfrac{a}{1}$

That why $a$ is rational number

Hence, this is the answer.

The number

$23+\sqrt{7}$ is

0%
Natural number
0%
Irrational number
0%
Rational number
0%
None of these

Explanation

As we've

$\sqrt{7}$ is an irrational number and

$23$ is a rational number then the sum of an irrational number and a rational number is again an irrational number.

$\surd 4+\surd 83,$ the correct option is

0%
Does not exists as quadratic surd
0%
$2+\surd 83$
0%
$\surd 83-2$
0%
Does not exist as real numbers

Explanation

$\sqrt{4}+\sqrt{83}=\sqrt{2\times 2}+\sqrt{83}=2+\sqrt{83}$

The value of

$1.999...$ in the form

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

$p$ and

$q$ are integers and

$q\neq 0$ , is:

0%
$\dfrac {1}{9}$
0%
$\dfrac {19}{10}$
0%
$\dfrac {1999}{1000}$
0%
$2$

Explanation

Let

$x=1.999...$

Since, one digit is repeating, we multiply

$x$ by

$10$ , we get

$10x = 19.99...$

So,

$10x=18+1.999...=18+x$
Therefore,

$10x-x=18$ , i.e.,

$9x=18$
i.e.,

$x=\dfrac { 18 }{ 9 } =\dfrac { 2 }{ 1 } =2$

Hence, option

$D$ is correct answer.

$\pi$ is a/an _______ .

0%
Rational number
0%
Integer
0%
Irrational number
0%
Whole number

Explanation

The value of

$\pi$ is equal to

$3.14159265358\dots$ which is a non-terminating and non-repeating decimal hence,

$\pi$ is an irrational number.

State whether the given statement is true/false:

An irrational number between two numbers

$\dfrac{1}{7}$ and

$\dfrac{2}{7}$ is

$0.1501500 15000...$

0%
True
0%
False

Explanation

The decimal forms of the given numbers as follows:

$\dfrac { 1 }{ 7 } =0.\overline { 142857 }$ and

$\dfrac { 2 }{ 7 } =0.\overline { 285714 }$ .

Since, they are non-terminating recurring, the given numbers are rational.

We know, an irrational number is a number which is non-terminating non-recurring.

Between two rational numbers, we can find infinitely many irrational numbers.

The given number,

$0.150150015000...$ is irrational as it is non-terminating non-repeating.

Also,

$0.\overline {142857} < 0.150150015000... < 0.\overline {285714}$ .

That is,

$0.150150015000...$ lies between

$\dfrac { 1 }{ 7 }$ and

$\dfrac { 2 }{ 7 }$ .

Hence,

$0.150150015000...$ is an irrational number between two numbers

$\dfrac {1}{7}$ and and

$\dfrac {2}{7}$ and the statement is true.

State true or false.

Five rational numbers between.

$\dfrac{2}{3}$ and

$\dfrac{4}{5}$ are

$\dfrac{41}{60},\dfrac{42}{60},\dfrac{43}{60},\dfrac{44}{60},\dfrac{45}{60}$

0%
True
0%
False

Explanation

To get the rational numbers between

$\displaystyle\frac{2}{3}$ and

$\displaystyle\frac{4}{5}$

Take an LCM of these two numbers:

$\displaystyle\frac{10}{15}$ and

$\displaystyle\frac{12}{15}$

Multiply numerator and denominator by 4:

$\displaystyle\frac{40}{60}$ and

$\displaystyle\frac{48}{60}$

All the numbers between

$\displaystyle\frac{40}{60}$ and

$\displaystyle\frac{48}{60}$ form the answer

Some of these numbers are

$\displaystyle\frac{41}{60}$ ,

$\displaystyle\frac{42}{60}$ ,

$\displaystyle\frac{43}{60}$ ,

$\displaystyle\frac{44}{60}$ ,

$\displaystyle\frac{45}{60}$

Hence the statement is true

State whether the given statement is true or false:

Five rational numbers between

$\dfrac{-3}{2}$ and

$\dfrac{5}{3}$ are

$\dfrac{-8}{6},\dfrac{-7}{6},0,\dfrac{1}{6},\dfrac{2}{6}$ .

0%
True
0%
False

Explanation

Given rational numbers are

$\displaystyle\frac{-3}{2}$ and

$\displaystyle\frac{5}{3}$

LCM of the denominators is

$6$

Hence

$\displaystyle\frac{-3}{2}$ and

$\displaystyle\frac{5}{3}$ can be written as

$\displaystyle\frac{-9}{6}$ and

$\displaystyle\frac{10}{6}$

The numbers between

$\displaystyle\frac{-9}{6}$ and

$\displaystyle\frac{10}{6}$ are

$\displaystyle\frac{-8}{6}$ ,

$\displaystyle\frac{-7}{6}$ ,

$\displaystyle{0}$ ,

$\displaystyle\frac{1}{6}$ ,

$\displaystyle\frac{2}{6},...$

Hence the statement is true.

State true or false

Five rational numbers between

$\dfrac{1}{4}$ and

$\dfrac{1}{2}$ are

$\displaystyle\frac{9}{32},\frac{10}{32},\frac{11}{32},\frac{12}{32},\frac{13}{32}$ .

0%
True
0%
False

Explanation

To get the rational numbers between

$\displaystyle\frac{1}{4}$ and

$\displaystyle\frac{1}{2}$

Take an LCM of these two numbers:

$\displaystyle\frac{1}{4}$ and

$\displaystyle\frac{2}{4}$

Multiply numerator and denominator by 8:

$\displaystyle\frac{8}{32}$ and

$\displaystyle\frac{16}{32}$

All the numbers between

$\displaystyle\frac{8}{32}$ and

$\displaystyle\frac{16}{32}$ form the answer

Some of these numbers are

$\displaystyle\frac{9}{32}$ ,

$\displaystyle\frac{10}{32}$ ,

$\displaystyle\frac{11}{32}$ ,

$\displaystyle\frac{12}{32}$ ,

$\displaystyle\frac{13}{32}$

State true or false.

Three rational numbers between

$\dfrac{2}{5}$ and

$\dfrac{3}{5}$ is

$\dfrac{9}{20},\dfrac{10}{20},\dfrac{11}{20}$

0%
True
0%
False

Explanation

Given numbers can be written as

$\dfrac { 2\times 5 }{ 5\times 5 } =\dfrac { 10 }{ 25 }\\$

$\dfrac { 3\times 5 }{ 5\times 5 } =\dfrac { 15 }{ 25 }$
therefore 3 rational numbers are

$\dfrac { 11 }{ 25 }, \dfrac { 12 }{ 25 }, \dfrac { 13 }{ 25 }$

Which one of the following is an irrational number ?

0%
0.14
0%
0.1416
0%
0.14169452
0%
0.4014001400014543.....

Explanation

${\textbf{Step -1: Observe the options given and find the non-terminating and non-repeating number.}}$

${\text{As }}0.4014001400014.....{\text{is non - terminating and non-repeating so it is a irrational number}}{\text{.}}$

${\textbf{Final Answer: Hence, the correct answer is option D}}{\text{.}}$

State true or false:

Ten rational numbers between

$\dfrac{3}{5}$ and

$\dfrac {3}{4}$ are

$\displaystyle\frac{97}{160},\frac{98}{160},\frac{99}{160},\frac{100}{160},\frac{101}{160},\frac{102}{160},\frac{103}{160},\frac{104}{160},\frac{105}{160},\frac{106}{160}$

0%
True
0%
False

Explanation

To get the rational numbers between

$\displaystyle\frac{3}{5}$ and

$\displaystyle\frac{3}{4}$

Take an LCM of these two numbers:

$\displaystyle\frac{12}{20}$ and

$\displaystyle\frac{15}{20}$

So now to make the denominator

$(160)$ as per the question we need to
multiply numerator and denominator by

$8$ :

$\displaystyle\frac{96}{160}$ and

$\displaystyle\frac{120}{160}$

All the numbers between

$\displaystyle\frac{96}{160}$ and

$\displaystyle\frac{120}{160}$ form the answer

Some of these numbers are

$\displaystyle\frac{97}{160}$ ,

$\displaystyle\frac{98}{160}$ ,

$\displaystyle\frac{99}{160}$ ,

$\displaystyle\frac{100}{160}$ ,

$\displaystyle\frac{101}{160}$ ,

$\displaystyle\frac{102}{160}$ ,

$\displaystyle\frac{103}{160}$ ,

$\displaystyle\frac{104}{160}$ ,

$\displaystyle\frac{105}{160}$ ,

$\displaystyle\frac{106}{160}$

Two rational numbers between

$\dfrac{1}{5}$ and

$\dfrac{4}{5}$ are :

0%
1 and $\dfrac{3}{5}$
0%
$\dfrac{2}{5}$ and $\dfrac{3}{5}$
0%
$\dfrac{1}{2}$ and $\dfrac{2}{1}$
0%
$\dfrac{3}{5}$ and $\dfrac{6}{5}$

Explanation

Since the denominator of both rational numbers are same. So, for getting the rational numbers between the given rational numbers, we only have to consider the numerators of the rational numbers.

Two numbers between 1 & 4 are 2 and 3.

So, two rational numbers between the given rational numbers will be

$\dfrac { 2 }{ 5 }$ and

$\dfrac { 3 }{ 5 }$

So, correct answer is option B.

$\pi$ is an ________ while

$\dfrac{22}{7}$ is rational.

0%
Integer
0%
Whole Number
0%

Rational Number
0%

Irrational Number

Explanation

The value

$\dfrac{22}7$ is a rational number, as it can be expressed in the form

$\dfrac pq$ .

We consider it as an approximate value of

$\pi$ because

$\pi$ is close to

$\dfrac{22}7$ .

But actually its value is

$3.14159....$ , which is neither terminating nor repeating.

Thus,

$\pi$ is irrational, but

$\dfrac{22}7$ is rational.

Check whether following statement is true or false.

$7\sqrt{5}$ is a rational number.

0%
True
0%
False

Explanation

An irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers and b is non-zero.

$7\sqrt5$ is irrational as it can never be expressed in the form a/b

Choose the rational number which does not lie between rational numbers

$\dfrac{3}{5}$ and

$\dfrac{2}{3}$ .

0%
$\dfrac{46}{75}$
0%
$\dfrac{47}{75}$
0%
$\dfrac{49}{75}$
0%
$\dfrac{50}{75}$

Explanation

All the options have a denominator of

$75$ . Hence, let us convert

$\dfrac35$ and

$\dfrac23$ into equivalent fractions having denominator

$75$ .

$\dfrac{3}{5}$

$=\dfrac{3\times 15}{5\times 15}$

$=\dfrac{45}{75}$

$\dfrac{2}{3}$

$=\dfrac{2\times 25}{3\times 25}$

$=\dfrac{50}{75}$

Hence,

$\dfrac{50}{75}$ does not lie between the given rational numbers.

Which of the following is an irrational number ?

0%
$\sqrt{23}$
0%
$\sqrt{225}$
0%
$0.3796$
0%
$7.478$

Explanation

In the given options,

$\sqrt { 225 }=15$ so, it is not an irrational number.

Options

$C$ and

$D$ are terminating decimals. So, they are also rational numbers.

$\sqrt{23}4.7958\dots$ which is a non-terminating non-repeating decimal hence, it is an irrational number.

So, option

$A$ is the correct answer.

Which of the following numbers is different from others?

0%
$\sqrt{6}$
0%
$\sqrt{11}$
0%
$\sqrt{15}$
0%
$\sqrt{16}$

Explanation

$16^{\frac{1}{2}}=4$ , which is a rational number.
The other options are irrational numbers.
So,

$16^{\frac{1}{2}}$ is different from others.

Say true or false.

$\dfrac {1}{0}$ is not rational.

0%
True
0%
False

Explanation

A rational number is a number that can be defined in the form of

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

$q$ is nonzero.
Now, if

$q$ is

$0$ , although an integer, the solution will not be a rational number. It will give an undefined result, so the statement is true.

Which of the following order is correct for rational numbers between

$\displaystyle \frac{4}{11}$ and

$\displaystyle \frac{9}{16}?$

0%
$\dfrac{4}{11}$ , $\dfrac{67}{176}$ , $\dfrac{79}{176}$ , $\dfrac{9}{16}$
0%
$\dfrac{4}{11}$ , $\dfrac{67}{276}$ , $\dfrac{79}{176}$ , $\dfrac{9}{16}$
0%
$\displaystyle \frac{4}{11}, \frac{17}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$
0%
$\displaystyle \frac{4}{11}, \frac{27}{38}, \frac{13}{27}, \frac{22}{43}, \frac{9}{16}$

Explanation

$L.C.M$ of

$11$ and

$16$ is

$11\times 16 =176$

$\dfrac{4}{11}=\dfrac{4\times 16}{11\times 16}=\dfrac{64}{176}$

$\dfrac{9}{16}=\dfrac{9\times 11}{16\times 11}=\dfrac{99}{176}$

rational numbers between

$\dfrac{4}{11}$ and

$\dfrac{9}{16}$ are

$\dfrac{67}{176}$ ,

$\dfrac{79}{176}$

Simplify:

$5\sqrt{2} -\sqrt{98}+\sqrt{162}$

0%
$7\sqrt{2}$
0%
$6\sqrt{2}$
0%
$3\sqrt{2}$
0%
none

Explanation

$5\sqrt{2} -\sqrt{98}+\sqrt{162}$

$=\sqrt{50} -\sqrt{98}+\sqrt{162}$

$=5\sqrt{2} -7\sqrt{2}+9\sqrt{2}$

$=\sqrt{2}(5-7+9)=7\sqrt{2}$

$\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1} = a+b \sqrt{3}$ , find the values of

$a$ and

$b$ .

0%
$a =2, b= -1$
0%
$a =2, b= 1$
0%
$a =-2, b= -1$
0%
$a =-2, b= 1$

Explanation

$\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}$

$=\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$=\displaystyle \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2}$

$=\displaystyle \frac{3+1-2\sqrt{3}}{3-1}$

$=\displaystyle \frac{4-2 \sqrt{3}}{2}$

$=\displaystyle 2-\sqrt{3}$

$= a+b\sqrt{3}$

$a=2,b=-1$

Simplify:

$4\sqrt{8} + 4\sqrt{2} - 3\sqrt{2}$

0%
$6\sqrt{2}$
0%
$4\sqrt{2}$
0%
$9\sqrt{2}$
0%
$5\sqrt{2}$

Explanation

$4\sqrt{8} + 4\sqrt{2} - 3\sqrt{2}$

$=4\sqrt{2^2 \times 2} + \sqrt{2^4 \times 2} - 3\sqrt{2}$

$=((4 \times 2)+4 -3) \sqrt{2}$

$=9 \sqrt{2}$

Simplify:

$7\sqrt{5} - 4\sqrt{5}+\sqrt{125}$

0%
$3\sqrt{5}$
0%
$4\sqrt{5}$
0%
$5\sqrt{5}$
0%
$8\sqrt{5}$

Explanation

$7\sqrt{5} - 4\sqrt{5}+\sqrt{125}$

$=7\sqrt{5} - 4\sqrt{5}+5\sqrt{5}$

$=(7-4+5)\sqrt{5}=8\sqrt{5}$

Find the nine rational numbers between

$0$ and

$1$ .

0%
$0.1,0.2,0.3, ... ,0.9$
0%
$1.1,0.2 ,10.3, ... ,0.9$
0%
$0.1,0.2,0.3, ... ,20.9$
0%
$0.1 , 0.2 , 10.3 , ... , 0.9$

Explanation

$0 < (0+0.1)=0.1 <(0.1+0.1)= 0.2 < (0.2+0.1)=0.3 < ... < (0.8+1)=0.9 <(0.9+0.1)= 1$

$0 < 0.1 < 0.2 < 0.3 < ... < 0.9 < 1$

$\therefore$ The nine rational numbers between

$0$ and

$1$ are

$0.1,0.2,0.3, ... ,0.9$

There can be a pair of irrational numbers whose sum is rational

Such as:

$\displaystyle 3-\sqrt{2}$ and

$\displaystyle 5+\sqrt{2}$

0%
True
0%
False

Explanation

To get the sum as rational, the numbers need to have an irrational part as well which should be same and of the opposite signs such that they cancel each other.

Example,

the pair of numbers

$3 - \sqrt{2}$ and

$5 + \sqrt {2}$ have the

sum

$3 - \sqrt{2} + 5 + \sqrt {2} = 8$

which is a rational number.

CBSE Questions for Class 9 Maths Number Systems Quiz 6 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Number Systems Quiz 6 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.