MCQ Questions for Class 9 Maths Number Systems Quiz 9

Which one of the following has a terminating decimal expansion?

0%
$\displaystyle\frac{5}{32}$
0%
$\displaystyle\frac{7}{9}$
0%
$\displaystyle\frac{8}{15}$
0%
$\displaystyle\frac{1}{12}$

Explanation

We know that a terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. For example

$\dfrac {1}{4}=0.25$ , it has only two decimal digits.

Now consider the fraction

$\dfrac {5}{32}$ whose decimal form will be:

$\dfrac {5}{32}=0.15625$

The resulting decimal number ends with five decimal digits and therefore, it is terminating decimal.

Hence,

$\dfrac {5}{32}$ has a terminating decimal expansion.

While,

$\dfrac{7}{9}=0.7777...$ , non-terminating,

$\dfrac{8}{15}=0.5333...$ , non-terminating

and

$\dfrac{7}{9}=0.08333...$ , non-terminating

Therefore, option

$A$ is correct.

Solve :

$(16)-(14\sqrt{2})$

0%
$(2\sqrt{2})$
0%
$2$
0%
$(10\sqrt{2})$
0%
$(16-14\sqrt{2})$

Explanation

We have

$(16)-(14\sqrt2)$ , where 16 is rational number and

$14\sqrt2$ is irrational number.

We can not add rational and irrational numbers directly. So, we can not solve the given numbers directly.

$(16-14\sqrt2)$ is the answer.

Simplify:

$(\sqrt{3}) + (5\sqrt{3}) - (8\sqrt{3}) + (10\sqrt{3})$

0%
$24\sqrt{3}$
0%
$8\sqrt{3}$
0%
$12\sqrt{3}$
0%
$16\sqrt{3}$

Explanation

In the given expression we can take

$\sqrt{3}$ common

Now we have,

$\sqrt3(1+5-8+10)=8\sqrt3$

Simplify:

$(9\sqrt{3}) + (5\sqrt{3}) - (10\sqrt{2}) - (2\sqrt{2})$

0%
$(\sqrt{6})$
0%
$(14\sqrt{3})-(8\sqrt{2})$
0%
$(14\sqrt{3})-(12\sqrt{2})$
0%
$(2\sqrt{6})$

Explanation

We can also write the given expression, After taking

$\sqrt3$ and

$-\sqrt2$ , as

$=\sqrt3(9+5)-\sqrt2(10+2)\\=14\sqrt3-12\sqrt2$

State if the following statement is true or false.

7.4848.. is an irrational number.

0%
True
0%
False

Explanation

Given number is

$7.4848...$

We can see that the decimal expansion of the given number is non-terminating, repeating.

And we know that, any number having a decimal expansion that is non-terminating, repeating, is a rational number.

So, the given number is a rational number.

Hence, the given statement is false.

Simplify :

$\dfrac{(36\sqrt{2})}{\sqrt{3}}$

0%
$12\sqrt{6}$
0%
$12\sqrt{2}$
0%
$12\sqrt{3}$
0%
$36\sqrt{6}$

Explanation

On multiplying and dividing with

$\sqrt3$ , we get

$\dfrac{36\sqrt2\times\sqrt3}{\sqrt3\times\sqrt3}=\dfrac{36\sqrt6}{3}=12\sqrt6$

Which of the following irrational number lies between

$2.5$ and

$2.6$ ?

0%
$2.512$
0%
$2.77321$
0%
$2.5333...$
0%
$2.5135145...$

Explanation

$2.512$ and

$2.77321$ are rational numbers as the decimal expansion terminates.

Option

$C$ can be rewritten as

$2.5\bar{3}$ , which is also a rational number, as it is non-terminating recurring.

But option

$D$ is non-terminating non-recurring and therefore is an irrational number and lies between

$2.5$ and

$2.6$ .

So, option

$D$ is the correct answer.

Solve:

$(16\sqrt{2})\times(14\sqrt{2})$

0%
$224$
0%
$224\sqrt{2}$
0%
$448$
0%
$448\sqrt{2}$

Explanation

We have

$16\sqrt2\times 14\sqrt2$ , this can also be written as

$=16\times14\times \sqrt2\times\sqrt2=224\times 2$

$(\because \sqrt2\times \sqrt2=2)$

$=448$

What is the conjugate of the binomial surd

$5 + \sqrt {3}$

0%
$5 - \sqrt {3}$
0%
$3 + \sqrt {5}$
0%
$\sqrt{5} + \sqrt {3}$
0%
$\sqrt{5} - \sqrt {3}$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$5+\sqrt 3$ is

$5-\sqrt 3$ .

Option A is correct.

The product of

$\sqrt x + 2\sqrt y$ with its conjugate is?

0%
$x + 4y$
0%
$x - 4y$
0%
$x+2y$
0%
$x-2y$

Explanation

The sum and difference of two simple quadratic surds are said to be conjugate of each other.

The conjuate of

$\sqrt x+2\sqrt y$ is

$\sqrt x-2\sqrt y$

Use the algebric properties,

$(a-b)(a+b)=a^2-b^2$

Product of

$\sqrt x+2\sqrt y$ with

$\sqrt x-2\sqrt y$ is equal to

$(\sqrt x)^2-(\sqrt {2y})^2=x-4y$

The conjugate of

$\sqrt x - \sqrt y$ is?

0%
$\sqrt x +y$
0%
$\sqrt x + \sqrt y$
0%
$x + \sqrt y$
0%
$x - y$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

Thus the conjugate of

$\sqrt x-\sqrt y$ is

$\sqrt x+\sqrt y$ .

Option B is correct.

The conjugate of

$3 + \sqrt 5$ is?

0%
$3 - \sqrt 5$
0%
$\sqrt 3 + 5$
0%
$\sqrt 3 +\sqrt 5$
0%
$\sqrt 3 - 5$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$3+\sqrt 5$ is

$3-\sqrt 5$ .

Option A is correct.

$\dfrac {1}{2} + \sqrt {2}$ is a conjugate of?

0%
$\dfrac {1}{2} + {2}$
0%
${2} + \sqrt {2}$
0%
$\dfrac {1}{2} - \sqrt {2}$
0%
$\dfrac {1}{\sqrt2} + \sqrt {2}$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$\dfrac{1}{2}+\sqrt 2$ is

$\dfrac{1}{2}-\sqrt 2$ .

Option C is correct.

$3\sqrt {7} + 7\sqrt {3}$ is a conjugate of?

0%
$7 + 3\sqrt {7}$
0%
$7\sqrt {3} - 3\sqrt {7}$
0%
$7\sqrt {3} + 3\sqrt {7}$
0%
$3\sqrt {7} - 7\sqrt {3}$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$3\sqrt 7+7\sqrt 3$ is

$3\sqrt 7-7\sqrt 3$ .

The conjugate of the binomial surd

$10\sqrt {2} + 3\sqrt {5}$ is

0%
$10\sqrt {5} -3\sqrt {2}$
0%
$10\sqrt {5} + 3\sqrt {2}$
0%
$10\sqrt {2} - 3\sqrt {5}$
0%
$-10\sqrt {5} + 3\sqrt {2}$

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$10\sqrt 2+3\sqrt 5$ is

$10\sqrt 2-3\sqrt 5$ .

State whether the following statement is True or False

$x + 3\sqrt y$ and

$\sqrt x -3y$ are conjugate pairs

0%
True
0%
False

Explanation

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

The conjugate of

$x+3\sqrt y$ is

$x-3\sqrt y$ .

So, the giving statement is not true.

What is the rationalizing factor of

$\dfrac {1}{\sqrt2 - 1}$ ?

0%
$1-\sqrt2$
0%
$2 + \sqrt2$
0%
$\sqrt2+1$
0%
Both $A$ and $B$

Explanation

When the denominator of an expression is a surd which can be reduced to an expression with the rational denominator, this process is called as rationalizing. When the product of two surds is a rational number, then each surd of the two surds is called as rationalizing factors of each other.

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

So, the rationalizing factor will be

$\dfrac{\sqrt 2+1}{(\sqrt 2-1)(\sqrt 2+1)}=\dfrac{\sqrt 2+1}{1}$

State whether the following statement is True or False.
The product of two binomial quadratic surds is always rational.

0%
True
0%
False

Explanation

The product of two binomial quadratic surds is always rational.

$(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=((\sqrt a)^2-(\sqrt b)^2)=a-b$ which is a rational number.

So, it is a true statement.

Which one of the following is a rational number?

0%
${ \left( \sqrt { 2 } \right) }^{ 2 }$
0%
$2\sqrt { 2 }$
0%
$2+\sqrt { 2 }$
0%
$\cfrac { \sqrt { 2 } }{ 2 }$

Explanation

$\textbf{Step -1: Check each option and find the correct one.}$

$\text{A rational number is a number that can be written in the form }\dfrac{p}{q}\text{ where }q\neq0.$

$(\sqrt2)^2=2$

$\text{and }2\text{ can be written as }\dfrac{2}{1}.$

$\therefore (\sqrt2)^2\text{ is a rational number.}$

$\text{Since, }\sqrt2\text{ is an irrational number so, }$

$2\sqrt2,2+\sqrt2,\dfrac{\sqrt2}{2}\text{ are also irrational numbers.}$

$\textbf{ Therefore, option A is correct.}$

The conjugate of the binomial surd

$\dfrac {1}{2}x + \dfrac {1}{2}\sqrt {y}$ is?

0%
$\dfrac {1}{2}x + \dfrac {1}{3}\sqrt {y}$
0%
$\dfrac {1}{3}x + \dfrac {1}{2}\sqrt {y}$
0%
$\dfrac {1}{2}x - \dfrac {1}{2}\sqrt {y}$
0%
$\dfrac {1}{2}y + \dfrac {1}{2}\sqrt {x}$

Explanation

The sum and difference of two simple quadratic surds are said to be conjugate of each other.

The conjugate of

$\dfrac{1}{2}x+\dfrac{1}{2}\sqrt y$ is

$\dfrac{1}{2}x-\dfrac{1}{2}\sqrt y$ .

Simplify:-

$(2\sqrt 2\times 3\sqrt 5)\div 2$ .

0%
$3\sqrt 5$
0%
$15$
0%
$12\sqrt 10$
0%
$3\sqrt 10$

Explanation

$\dfrac{(2\sqrt2\times3\sqrt5)}{2}=\dfrac{(6\sqrt{2\times5})}{2}=\dfrac{6\sqrt10}{2}\\ =3\sqrt{10}$

After rationalizing the numerator of

$\dfrac {\sqrt {3} - \sqrt {2}}{\sqrt {3}}$ the denominator in simplest form is

0%
$\sqrt {3}(\sqrt {3} + \sqrt {2})$
0%
$\sqrt {3}(\sqrt {3} - \sqrt {2})$
0%
$3 - \sqrt {3}\sqrt {2}$
0%
$\dfrac{1}{3 + \sqrt {6}}$
0%
None of these

Explanation

$\dfrac {\sqrt {3} - \sqrt {2}}{\sqrt {3}} \cdot \dfrac {\sqrt {3} + \sqrt {2}}{\sqrt {3} + \sqrt {2}} = \dfrac {1}{3 + \sqrt {6}}$ .

The conjugate surd of

$2-\sqrt{3}$ is ___________.

0%
$-2+\sqrt{3}$
0%
$2-(-\sqrt{3})$
0%
$3+\sqrt{2}$
0%
$\dfrac{1}{2-\sqrt{3}}$

Explanation

The conjugate surd of

$2-\sqrt{3}$ is

$= 2-(-\sqrt{3})$
i.e The conjugate surd of

$2-\sqrt{3}$ is

$2+\sqrt{3}$

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

$=\dfrac{4-3}{2-\sqrt{3}}$

$=\dfrac{1}{2-\sqrt{3}}$

From the pairs of number given below, whose product is a natural number?

0%
$\sqrt12 , \sqrt3$
0%
$\sqrt4 , \sqrt3$
0%
$\sqrt10 , \sqrt3$
0%
$\sqrt2 , \sqrt3$

Explanation

Option A:

$\sqrt12\times\sqrt3=\sqrt{36}=6\Rightarrow$ natural number

Option B:

$\sqrt4\times\sqrt3=\sqrt{12}\Rightarrow$ irrational number

Option C:

$\sqrt{10}\times\sqrt3=\sqrt{30}\Rightarrow$ irrational number

Option D:

$\sqrt2\times\sqrt3=\sqrt6\Rightarrow$ irrational number

State whether the following statement is True or False
The conjugate of

$x\sqrt {a} + y\sqrt {b}$ is

$x\sqrt {a} - y\sqrt {b}$

0%
True
0%
False

Explanation

The sum and difference of two simple quadratic surds are said to be conjugate of each other.

The conjugate of

$x\sqrt a+y\sqrt b$ is

$x\sqrt a-y\sqrt b$

The giving statement is true.

State whether the following statement is True or False
The conjugate of

$\sqrt {3} + \sqrt {2}$ is

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

0%
True
0%
False

Explanation

The sum and difference of two simple quadratic surds are said to be conjugate of each other.

the conjugate of

$\sqrt 3+\sqrt 2$ is

$\sqrt 3-\sqrt 2$

So, the giving statement is not true.

Which of the following numbers has the terminal decimal representation?

0%
$\dfrac{1}{7}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{17}{3}$

Explanation

We know that a terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. For example

$\dfrac {1}{4}=0.25$ , it has only two decimal digits.

Then,

Option

$A$ :

$\dfrac {1}{7}=0.1428..$ , non-terminating decimal number.

Option

$B$ :

$\dfrac {1}{3}=0.3333....$ , non-terminating decimal number.

Option

$C$ :

$\dfrac{3}{5}=0.6\Rightarrow$ terminating decimal number.

Option

$D$ :

$\dfrac {17}{3}=5.6666...$ , non-terminating decimal number.

Hence, option

$C$ is correct.

Simplify:

$(-1)^{19}+(-1)^{20}+(2)^5$

0%
$34$
0%
$32$
0%
$30$
0%
$0$

Explanation

$(-1)^{19}+(-1)^{20}+(2)^5$

$=(-1)+(1)+2\times2\times2\times2\times2$

$=32$

Find two rational numbers lying between

$\dfrac{-1}{3}$ and

$\dfrac{1}{2}$ .

0%
-1/6,1 /6
0%
-2/3, 2/3
0%
none
0%
both

The rational number that lies between

$\dfrac{3}{7}$ and

$\dfrac{2}{3}$ is

0%
$\dfrac{2}{5}$
0%
$\dfrac{4}{7}$
0%
$\dfrac{3}{7}$
0%
$\dfrac{2}{3}$

Explanation

$\dfrac{3}{7}\times \dfrac{3}{3} = \dfrac{9}{21},$

$\dfrac{2}{3}\times \dfrac{7}{7} = \dfrac{14}{21}$

Rational numbers between

$\dfrac{3}{7}$ &

$\dfrac{2}{3}$ are

$\dfrac{10}{21},\dfrac{11}{21},\dfrac{12}{21},\dfrac{13}{21}$

$\dfrac{12}{21}= \dfrac{4}{7}$

Hence,

$\dfrac{4}{7}$ lies between

$\dfrac37$ and

$\dfrac23$ .

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

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Practice Class 9 Maths Quiz Questions and Answers

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

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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