M.C.Q

Question 11 Mark

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

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

Answer

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

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

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Question 21 Mark

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

$6\sqrt{5}$
$5\sqrt{6}$
$\sqrt{25}$
$10\sqrt{5}$

Answer

$5\sqrt{6}$

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

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Question 31 Mark

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

A natural number.
An integer.
A real number.
A whole number.

Answer

A real number.

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

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Question 41 Mark

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

$2\sqrt{6}$
$6$
$3\sqrt{3}$
$4\sqrt{6}$

Answer

$3\sqrt{3}$

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

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Question 51 Mark

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

There is no rational number.
There is exactly one rational number.
There are infinitely many rational numbers.
There are only rational numbers and no irrational numbers.

Answer

There are infinitely many rational numbers.

Solution:
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.

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Question 61 Mark

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

$\sqrt{2}$
$2$
$\sqrt[12]{2}$
$\sqrt[12]{32}$

Answer

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

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Question 71 Mark

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

Terminating.
Non - terminating.
Non - terminating repeating.
Non - terminating non - repeating.

Answer

Non - terminating non - repeating.

Solution:
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).

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Question 81 Mark

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

$\frac{1}{2}(3-2\sqrt{2})$
$\frac{1}{3+2\sqrt{2}}$
$3-2\sqrt{2}$
$3+2\sqrt{2}$

Answer

$3+2\sqrt{2}$

Solution:
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ $[\because\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}]$
$[$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]$
$=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

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Question 91 Mark

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

Always an irrational number.
Always a rational number.
Always an integer.
Sometimes rational, sometimes irrational.

Answer

Sometimes rational, sometimes irrational.

Solution:
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
$(2+\sqrt{3})(1-\sqrt{3})$
$2(1-\sqrt{3})+\sqrt{3}(1-\sqrt{3})$
$2-2\sqrt{3}+\sqrt{3}-3$
$-1-\sqrt{3}$

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Question 101 Mark

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

$0.14$
$0.14\overline{16}$
$0.\overline{1416}$
$0.4014001400014...$

Answer

$0.4014001400014...$

Solution:
A number is irrational if and only of its decimal representation is non - terminating and nonrecurring.

$0.14$ 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.
$0.\overline{1416}$ is a non - terminating and recurring decimal and therefore cannot be irrational.
$0.4014001400014...$ is a non - terminating and non - recurring decimal and therefore is an irrational number.

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Question 111 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

$\frac{19}{10}$
$\frac{1999}{1000}$
$2$
$\frac{1}{9}$

Answer

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

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Question 121 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

Answer

Solution:
$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
$[$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.

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Question 131 Mark

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

A finite decimal.
1.41421.
Non - terminating recurring.
Non - terminating non - recurring.

Answer

Non - terminating non - recurring.

Solution:
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.

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Question 141 Mark

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

$\sqrt{\frac{4}{9}}$
$\frac{\sqrt{12}}{\sqrt{3}}$
$\sqrt{7}$
$\sqrt{81}$

Answer

$\sqrt{7}$

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

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Question 151 Mark

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

$\frac{\sqrt{7}+2}{3}$
$\frac{\sqrt{7}-2}{3}$
$\frac{\sqrt{7}+2}{5}$
$\frac{\sqrt{7}+2}{45}$

Answer

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

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

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Question 161 Mark

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

$\frac{\sqrt{2}+\sqrt{3}}{2}$
$\frac{\sqrt{2}\cdot\sqrt{3}}{2}$
$1.5$
$1.8$

Answer

$1.5$

Solution:
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.

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Question 171 Mark

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

$\frac{1}{9}$
$\frac{1}{3}$
$9$
$\frac{1}{81}$

Answer

$\frac{1}{9}$

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

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Question 181 Mark

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

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

Answer

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

Solution:

$\text{x}^{\frac{12}{7}}-\text{x}^{\frac{5}{7}}\neq\text{x}$
$\sqrt[12]{\text{(x}^4})^{\frac{1}{3}}=\sqrt[12]{\text{x}^{4\times\frac{1}{3}}}=\Big(\text{x}^\frac{4}{3}\Big)^\frac{1}{12}=\text{x}^{\frac{4}{3}\times\frac{1}{12}}=\text{x}^\frac{1}{9}\neq\text{x}$
$\Big((\text{x}^3)^\frac{1}{2}\Big)^\frac{2}{3}=\text{(x)}^{\frac{3}{2}\times\frac{2}{3}}=\text{x}^1=\text{x}$
$\text{x}^{\frac{12}{7}}\times\text{x}^\frac{7}{12}=\text{x}^{\frac{12}{7}+\frac{7}{12}}=\text{x}^\frac{193}{84}\neq\text{x}$

Hence, (c) is the correct answer.

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Question 191 Mark

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

$\sqrt{2}$
2
4
8

Answer

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

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Question 201 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

2.4142
5.8282
0.4142
0.1718

Answer

0.4142

Solution:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{{\sqrt{2}+1}}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}}$
$[$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]$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1=(1.4142...)-1=0.4142...$

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MCQ 211 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.

$4$

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

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