M.C.Q

MCQ 11 Mark

Write the correct answer in the following:
$\sqrt[4]{\sqrt[3]{2^2}}$ equals.

A
$2^{-\frac{1}{6}}$
B
$2^{-6}$
✓
$2^{\frac{1}{6}}$
D
$2^{6}$

Answer

Correct option: C.

$2^{\frac{1}{6}}$

$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{(2^2)^{\frac{1}{3}}}=\Big(2^{\frac{2}{3}}\Big)^{\frac{1}{4}}=2^{\frac{2}{3}\times\frac{1}{4}}=2^{\frac{1}{6}}$
Hence, $(c)$ is the correct answer.

View full question & answer→

MCQ 21 Mark

Write the correct answer in the following: $\sqrt{10}\times\sqrt{15}$ is equal to.

A
$6\sqrt{5}$
✓
$5\sqrt{6}$
C
$\sqrt{25}$
D
$10\sqrt{5}$

Answer

Correct option: B.

$5\sqrt{6}$

$\sqrt{10}, \sqrt{15}=\sqrt{2.5}\sqrt{3.5}=\sqrt{2}\sqrt{5}\sqrt{3}\sqrt{5}=5\sqrt{6}$

View full question & answer→

MCQ 31 Mark

Write the correct answer in the following: The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to

A
$\sqrt{2}$
✓
$2$
C
$4$
D
$8$

Answer

Correct option: B.

$2$

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$
$=\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$
$=\frac{4}{2}=2$
Hence, $(b)$ is correct answer.

View full question & answer→

MCQ 41 Mark

Write the correct answer in the following: Every rational number is

A
A natural number.
B
An integer.
✓
A real number.
D
A whole number.

Answer

Correct option: C.

A real number.

Since, real numbers are the combination of rational and irrational numbers.
Hence, every rational number is a real number.

View full question & answer→

MCQ 51 Mark

Write the correct answer in the following: $2\sqrt{3}+\sqrt{3}$ is equal to

A
$2\sqrt{6}$
B
$6$
✓
$3\sqrt{3}$
D
$4\sqrt{6}$

Answer

Correct option: C.

$3\sqrt{3}$

Given $2\sqrt{3}+\sqrt{3}=(2+1)\sqrt{3}=3\sqrt{3}$
Hence, $(c)$ is the correct answer.

View full question & answer→

MCQ 61 Mark

Write the correct answer in the following: Between two rational numbers.

A
There is no rational number.
B
There is exactly one rational number.
✓
There are infinitely many rational numbers.
D
There are only rational numbers and no irrational numbers.

Answer

Correct option: C.

There are infinitely many rational numbers.

There are infinitely many rational numbers Between two rational numbers there are infinitely many rational number.
For example between $4$ and $5$ there are $4.1, 4.2.4.22, 4.223 ............$
Hence, $(C)$ is the correct answer.

View full question & answer→

MCQ 71 Mark

Write the correct answer in the following: The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ equals.

A
$\sqrt{2}$
✓
$2$
C
$\sqrt[12]{2}$
D
$\sqrt[12]{32}$

Answer

Correct option: B.

$2$

$LCM$ of $3, 4$ and $12 = 12$
$\sqrt[3]{2}=\sqrt[12]{2^4}\ [\because\sqrt[\text{m}]{\text{a}}=\sqrt[\text{mn}]{\text{a}^\text{n}}]$
$\sqrt[4]{2}=\sqrt[12]{2^3}$
and $\sqrt[12]{32}=\sqrt[12]{2^5}$
$\therefore\text{product of }\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{2^3}.\sqrt[12]{2^5}=\sqrt[12]{2^4.2^3.2^5}$
$=12\sqrt{2^{4+3+5}}=\sqrt[12]{2^{12}}=2^{12\times\frac{1}{12}}=2\ [\because\sqrt[\text{m}]{\text{a}^\text{n}}=\text{a}^{\frac{\text{n}}{\text{m}}}]$

View full question & answer→

MCQ 81 Mark

Write the correct answer in the following: Decimal representation of a rational number cannot be.

A
Terminating.
B
Non - terminating.
C
Non - terminating repeating.
✓
Non - terminating non - repeating.

Answer

Correct option: D.

Non - terminating non - repeating.

Decimal representation of a rational number cannot be non - terminating non-repeating because the decimal expansion of rational number is either terminating or non - terminating recurring (repeating).

View full question & answer→

MCQ 91 Mark

Write the correct answer in the following: $\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to

A
$\frac{1}{2}(3-2\sqrt{2})$
B
$\frac{1}{3+2\sqrt{2}}$
C
$3-2\sqrt{2}$
✓
$3+2\sqrt{2}$

Answer

Correct option: D.

$3+2\sqrt{2}$

$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ $[\because\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}]$
$[$multiplying numerator and denominator by $3+2\sqrt{2}]$
$\frac{3+2\sqrt{2}}{9-(2-\sqrt{2})^2}$$[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

View full question & answer→

MCQ 101 Mark

Write the correct answer in the following: The product of any two irrational numbers is.

A
Always an irrational number.
B
Always a rational number.
C
Always an integer.
✓
Sometimes rational, sometimes irrational.

Answer

Correct option: D.

Sometimes rational, sometimes irrational.

Sometimes rational, sometimes irrational The product of any two irrational numbers is sometimes rational and sometimes irrational.
Hence, $(D)$ is the correct answer.
For example:
rational
$(2+\sqrt{3})(2-\sqrt{3})$
$(2)^2-(\sqrt{3})^2$
$4-3=1$
irrational
$(2+\sqrt{3})(1-\sqrt{3})$
$2(1-\sqrt{3})+\sqrt{3}(1-\sqrt{3})$
$2-2\sqrt{3}+\sqrt{3}-3$
$-1-\sqrt{3}$

View full question & answer→

MCQ 111 Mark

Write the correct answer in each of the following:
Which of the following is irrational?

A
$0.14$
B
$0.14\overline{16}$
C
$0.\overline{1416}$
✓
$0.4014001400014...$

Answer

Correct option: D.

$0.4014001400014...$

A number is irrational if and only of its decimal representation is non $-$ terminating and nonrecurring.
$a. 0.14$ is a terminating decimal and therefore cannot be an irrational number.
$0.14\overline{16}$ is a non $-$ terminating and recurring decimal and therefore cannot be irrational.
$b. 0.\overline{1416}$ is a non $-$ terminating and recurring decimal and therefore cannot be irrational.
$c. 0.4014001400014...$ is a non $-$ terminating and non $-$ recurring decimal and therefore is an irrational number.

View full question & answer→

MCQ 121 Mark

Write the correct answer in the following: The value of $1.999...$ in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0,$ is

A
$\frac{19}{10}$
B
$\frac{1999}{1000}$
✓
$2$
D
$\frac{1}{9}$

Answer

Correct option: C.

$2$

Let $x = 1.999...$
Now, $10x = 19.999...$
On subtracting Eq. $(i)$ from Eq. $(ii)$, we get
$10x - x = (19.999...) - (1.9999...)$
$\Rightarrow 9x = 18$
$\therefore\text{x}=\frac{18}{9}=2$

View full question & answer→

MCQ 131 Mark

Write the correct answer in the following: After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as

A
$13$
✓
$19$
C
$5$
D
$35$

Answer

Correct option: B.

$19$

$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
$[$multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2}$ $[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$ $\big[\because\text{fraction}=\frac{\text{numerator}}{\text{denominator}}\big]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$
Hence, after rationalising the denominator of $=\frac{7}{(3\sqrt{3}-2\sqrt{2})}$ we get the denominator as $19$.

View full question & answer→

MCQ 141 Mark

Write the correct answer in the following: The decimal expansion of the number $\sqrt{2}$ is.

A
A finite decimal.
B
$1.41421.$
C
Non - terminating recurring.
✓
Non - terminating non - recurring.

Answer

Correct option: D.

Non - terminating non - recurring.

The decimal expansion of the number $\sqrt{2}$ is non-terminating non - recurring. Because $\sqrt{2}$ is an irrational number.
Also, we know that an irrational number is non - terminating non - recurring.

View full question & answer→

MCQ 151 Mark

Write the correct answer in the following: Value of $\sqrt[4]{(81)^{-2}}$ is.

✓
$\frac{1}{9}$
B
$\frac{1}{3}$
C
$9$
D
$\frac{1}{81}$

Answer

Correct option: A.

$\frac{1}{9}$

$\sqrt[4]{(81)^{-2}}=\sqrt[4]{\Big(\frac{1}{81}\Big)^2}=\sqrt[4]{\Big\{\Big(\frac{1}{9}\Big)^2\Big\}^2}=\sqrt[4]{\Big(\frac{1}{9}\Big)^4}=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}=\frac{1}{9}$
Hence, $(a)$ is the correct answer.

View full question & answer→

MCQ 161 Mark

Write the correct answer in the following: Which of the following is irrational?

A
$\sqrt{\frac{4}{9}}$
B
$\frac{\sqrt{12}}{\sqrt{3}}$
✓
$\sqrt{7}$
D
$\sqrt{81}$

Answer

Correct option: C.

$\sqrt{7}$

$\because\sqrt{\frac{4}{9}}=\frac{2}{3}\text{(rational)},$
$\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2\text{(rational)},$
$\sqrt{81}=9\text{(rational)}$
but $\sqrt{7}$ is an irrational number.
Hence, $\sqrt{7}$ is an irrational number.

View full question & answer→

MCQ 171 Mark

Write the correct answer in the following: The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is

✓
$\frac{\sqrt{7}+2}{3}$
B
$\frac{\sqrt{7}-2}{3}$
C
$\frac{\sqrt{7}+2}{5}$
D
$\frac{\sqrt{7}+2}{45}$

Answer

Correct option: A.

$\frac{\sqrt{7}+2}{3}$

$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$
$\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$
Hence, $(a)$ is the correct answer.

View full question & answer→

MCQ 181 Mark

Write the correct answer in the following: Value of $(256)^{0.16} \times(256)^{0.09}$ is.

✓
$4$
B
$16$
C
$64$
D
$256.25$

Answer

Correct option: A.

$4$

$(256)^{0.16}\times(256)^{0.09}=(256)^{\frac{16}{100}}\times(256)^{\frac{9}{100}}$
$=(256)^{\frac{16}{100}+\frac{9}{100}}\ [\because\text{x}^\text{a}\cdot\text{x}^{\text{b}}=\text{x}^{\text{a+b}}]$
$(256)^{\frac{25}{100}}=(256)^{\frac{1}{4}}$
$=(4^4)^{\frac{1}{4}}=4\ [\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$

View full question & answer→

MCQ 191 Mark

Write the correct answer in the following: A rational number between $\sqrt{2}$ and $\sqrt{3}$ is

A
$\frac{\sqrt{2}+\sqrt{3}}{2}$
B
$\frac{\sqrt{2}\cdot\sqrt{3}}{2}$
✓
$1.5$
D
$1.8$

Answer

Correct option: C.

$1.5$

We know that
$\sqrt{2}.=1.4142135...$ and $\sqrt{3}.=1.732050807...$
We see that 1.5 is a rational number which lies between $1.4142135...$ and $1.732050807...$
Hence, $(c)$ is tthe correct answer.

View full question & answer→

MCQ 201 Mark

Write the correct answer in the following: Which of the following is equal to $x$?

A
$\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}$
B
$\sqrt[12]{(\text{x}^4)^{\frac{1}{3}}}$
✓
$(\sqrt{\text{x}^3})^{\frac{2}{3}}$
D
$\text{x}^{\frac{12}{7}}\times\text{x}^{\frac{7}{12}}$

Answer

Correct option: C.

$(\sqrt{\text{x}^3})^{\frac{2}{3}}$

$a. \text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}\neq\text{x}$
$b. \sqrt[12]{\text{(x}^4})^{\frac{1}{3}}=\sqrt[12]{\text{x}^{4\times\frac{1}{3}}}=\Big(\text{x}^\frac{4}{3}\Big)^\frac{1}{12}=\text{x}^{\frac{4}{3}\times\frac{1}{12}}=\text{x}^\frac{1}{9}\neq\text{x}$
$c. \Big((\text{x}^3)^\frac{1}{2}\Big)^\frac{2}{3}=\text{(x)}^{\frac{3}{2}\times\frac{2}{3}}=\text{x}^1=\text{x}$
$d. \text{x}^{\frac{12}{7}}\times\text{x}^\frac{7}{12}=\text{x}^{\frac{12}{7}+\frac{7}{12}}=\text{x}^\frac{193}{84}\neq\text{x}$
Hence, $(c)$ is the correct answer.

View full question & answer→

MCQ 211 Mark

Write the correct answer in the following: If $\sqrt{2}=1.4142,$ then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to

A
$2.4142$
B
$5.8282$
✓
$0.4142$
D
$0.1718$

Answer

Correct option: C.

$0.4142$

$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{{\sqrt{2}+1}}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}}$
$[$Inside the root, multiplying numerator and denominator by $(\sqrt{2}-1)]$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-(1)^2}}$ $ [\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1=(1.4142...)-1=0.4142...$

View full question & answer→

Maths M.C.Q

Take a timed test