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MATHS M.C.Q

Presentation of Data in Tabular Form · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 75 questions.

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M.C.Q

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Question 11 Mark
On simplification, the expression $\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$ equals:
  1. $\frac{5}{3}$
  2. $-\frac{5}{3}$
  3. $\frac{3}{5}$
  4. $-\frac{3}{5}$
Answer
  1. $-\frac{5}{3}$
Solution:
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$
$=\frac{5^{\text{n}+1}(5-6)}{5^{\text{n}}(13-2\times5)}$
$=\frac{5^{\text{n}}\times5\times(-1)}{5^{\text{n}}(13-10)}$
$=-\frac{5}{3}$
Hence, the correct option is (b).
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Question 21 Mark
Which of the following is an irrational number?
  1. 3.14
  2. 3.141414...
  3. 3.14444
  4. 3.141141114...
Answer
  1. 3.141141114...
Solution:
The decimal expansion of an irrational number is non-terminating recurring non-recurring.
Hence, 3.141141114... is an irrational number.
Hence, the correct opion is (d).
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Question 31 Mark
An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is:
  1. $\frac{1}{2}\Big(\frac{1}{7}+\frac{2}{7}\Big)$
  2. $\Big(\frac{1}{7}\times\frac{2}{7}\Big)$
  3. $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
  4. None of these.
Answer
  1. $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Solution:
An irrational number between a and b is given by $\sqrt{\text{ab}}$
So, an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Hence, the correct answer is option (c).
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Question 41 Mark
The rationalisation factor of $\frac{1}{\big(2\sqrt{3}-\sqrt{5}\big)}$ is:
  1. $\sqrt{5}-2\sqrt{3}$
  2. $\sqrt{3}+2\sqrt{5}$
  3. $\big(\sqrt{3}+\sqrt{5}\big)$
  4. $\sqrt{12}+\sqrt{5}$
Answer
  1. $\sqrt{12}+\sqrt{5}$
Solution:
The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is $2\sqrt{3}+\sqrt{5},$
i.e. $\sqrt{3\times4}+\sqrt{5}$
i.e. $\sqrt{12}+\sqrt{5}$
Hence, the correct option is (d).
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Question 51 Mark
Rationalisation of the denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$ gives:
  1. $\frac{1}{\sqrt{10}}$
  2. $\sqrt{5}+\sqrt{2}$
  3. $\sqrt{5}-\sqrt{2}$
  4. $\frac{\sqrt{5}-\sqrt{2}}{3}$
Answer
  1. $\frac{\sqrt{5}-\sqrt{2}}{3}$
Solution:
$\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\\=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct option is (d).
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Question 61 Mark
 $(125)^{-\frac{1}{3}}=?$
  1. 5
  2. -5
  3. $\frac{1}{5}$
  4. $-\frac{1}{5}$ 
Answer
  1. $\frac{1}{5}$
Solution:
$(125)^{-\frac{1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$
$=\frac{1}{5^{3-\frac{1}{3}}}=\frac{1}{5}$
Hence, the correct answer is option (c).
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Question 71 Mark
 The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is:
  1. $2\sqrt{2}+3$
  2. $2\sqrt{2}+\sqrt{3}$
  3. $\sqrt{2}+\sqrt{3}$
  4. $\sqrt{2}-\sqrt{3}$
Answer
  1. $2\sqrt{2}+\sqrt{3}$
Solution:
The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is $2\sqrt{2}+\sqrt{3}$
Hence, the correct option is (b).
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Question 81 Mark
The value of $\sqrt{3-2\sqrt{2}}$ is:
  1. $\sqrt{3}+\sqrt{2}$
  2. $\sqrt{3}-\sqrt{2}$
  3. $\sqrt{2}+1$
  4. $\sqrt{2}-1$
Answer
  1. $\sqrt{2}-1$
Solution:
$\sqrt{3-2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+(1)^2-2\times\sqrt{2}\times1}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
Hence, the correct option is (d).
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Question 91 Mark
If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$
  1. 34
  2. 56
  3. 28
  4. 63
Answer
  1. 34
Solution:
$\text{x}=3+\sqrt{8}$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
Now, $\frac{1}{\text{x}}=\frac{1}{3+\sqrt{8}}$
$=\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}$
$=\frac{3-\sqrt{8}}{3^2-\big(\sqrt{8}\big)^2}$
$=\frac{3-\sqrt{8}}{9-8}$
$=3-\sqrt{8}$
$\Rightarrow\Big(\frac{1}{\text{x}}\Big)^2=\big(3-\sqrt{8}\big)^2=9+8-6\sqrt{8}=17-6\sqrt{8}$
Then, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=17+6\sqrt{8}+17-6\sqrt{8}=34$
Hence, the correct option is (a).
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Question 101 Mark
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}=?$
  1. $2$
  2. $\sqrt{2}$
  3. $2\sqrt{2}$
  4. $4\sqrt{2}$
Answer
  1. $2$
Solution:
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{2^5}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$
$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{4+3+5}{12}}$
$=2^{\frac{12}{12}}$
$=2$
Hence, the correct option is (a).
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Question 111 Mark
Which of the following is a true statment?
  1. $\pi$ and $\frac{22}{7}$ are both rationals.
  2. $\pi$ and $\frac{22}{7}$ are both irrationals
  3. $\pi$ is rational and $\frac{22}{7}$ is irrational.
  4. $\pi$ is irrational and $\frac{22}{7}$ is rational.
Answer
  1. $\pi$ is irrational and $\frac{22}{7}$ is rational.
Solution:
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.
So, $\pi=3.141592...$ is irrational.
The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.
So, $\frac{22}{7}$ is rational.
Hence, the correct option is (d).
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Question 121 Mark
If $\text{x}=2+\sqrt{3}$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ equals:
  1. $-2{\sqrt{3}}$
  2. 2
  3. 4
  4. $4-2\sqrt{3}$
Answer
  1. 4
Solution:
$\text{x}=2+\sqrt{3}$
$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$
Hence, the correct option is (c).
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Question 131 Mark
Between any two rational numbers there:
  1. is no rational number.
  2. is exactly one rational numbers.
  3. are infinitely many rational numbers.
  4. is no irrational number.
Answer
  1. are infinitely many rational numbers.
Solution:
Options (a), (b) and (d) are incorrect since between two rational numbers there are infinitely many rational and irrational numbers.
Hence, the correct opion is (c).
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Question 141 Mark
When $15\sqrt{15}$ is divided by $3\sqrt{3},$ the quotient is:
  1. $5\sqrt{3}$
  2. $3\sqrt{5}$
  3. $5\sqrt{5}$
  4. $3\sqrt{3}$
Answer
  1. $5\sqrt{5}$
Solution:
$\frac{15\sqrt{15}}{3\sqrt{3}}=\frac{5\sqrt{5\times3}}{\sqrt{3}}$
$\frac{5\sqrt{5}\times\sqrt{3}}{\sqrt{3}}=5\sqrt{5}$
Hence, the correct answer is option (c).
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Question 151 Mark
The simplest rationalisation factor of $\sqrt[3]{500}$ is:
  1. $\sqrt{5}$
  2. $\sqrt{3}$
  3. $\sqrt[3]{5}$
  4. $\sqrt[3]{2}$
Answer
  1. $\sqrt[3]{2}$
Solution:
$\sqrt[3]{500}=500^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}$ $=\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=\frac{10^{3\times\frac{1}{3}}}{2^{\frac{1}{3}}}=\frac{10}{\sqrt[3]{2}}$
Thus, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}.$
Hence, the correct option is (d).
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Question 161 Mark
Which of the following numbers is irrational?
  1. $\sqrt{\frac{4}{9}}$
  2. $\frac{\sqrt{1250}}{\sqrt{8}}$
  3. $\sqrt{8}$
  4. $\frac{\sqrt{24}}{\sqrt{6}}$
Answer
  1. $\sqrt{8}$
Solution:
The decimal expansion of $\sqrt{8}=2.82842712...,$ which is non-terminating, non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is (c).
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Question 171 Mark
How many digits are there in the repeating block of digits in the decimal expansion of $\frac{17}{7}?$
  1. 16
  2. 6
  3. 26
  4. 7
Answer
  1. 6
Solution:
$\frac{17}{7}=2.\overline{428571}$
Hence, the correct opion is (b).
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Question 181 Mark
Which of the following is a rational number?
  1. $\sqrt{5}$
  2. 0.101001000100001...
  3. $\pi$
  4. 0.853853853...
Answer
  1. 0.853853853...
Solution:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
Hence, 0.853853853... is a rational number.
Hence, the correct option is (d).
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MCQ 191 Mark
There is a number $x$ such that $x^2$ is irrational but $x^4$ is rational. Then, $x$ can be:
  • A
    $\sqrt{5}$
  • B
    $\sqrt{2}$
  • C
    $\sqrt[3]{2}$
  • ✓
    $\sqrt[4]{2}$
Answer
Correct option: D.
$\sqrt[4]{2}$
$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times2}=2^{\frac{1}{2}}=\sqrt{2},$ which is irrational.
$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times4}=2^1=2,$ which is rational.
Hence, the correct option is $(d)$.
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Question 201 Mark
Which of the following is the value of $\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)?$
  1. -4
  2. 4
  3. $\sqrt{11}$
  4. $\sqrt{7}$
Answer
  1. 4
Solution:
$\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)$
$\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2$
$11-7=4$
Hence, the correct answer is option (b).
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Question 211 Mark
The decimal expansion of $\sqrt{2}$ is:
  1. finite decimal.
  2. 1.4121
  3. nonterminating recurring.
  4. nonterminating, nonrecurring.
Answer
  1. nonterminating, nonrecurring.
Solution:
The decimal expansion of $\sqrt{2}=1.41421356...,$ which is non-terminating, nonrecurring.
Hence, the correct opion is (d).
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Question 221 Mark
A rational number between -3 and 3 is:
  1. 0
  2. -4.3
  3. -3.4
  4. 1.101100110001...
Answer
  1. 0
Solution:
Since, -4.3 < -3.4 < -3 < 0 < 1.101100110001... < 3
But 1.101100110001... is an irrational number
So, the rational number between -3 and 3 is 0.
Hence, the correct option is (a).
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Question 231 Mark
Which of the following is a rational number?
  1. $\sqrt{2}$
  2. $\sqrt{23}$
  3. $\sqrt{225}$
  4. $0.1010010001...$
Answer
  1. $\sqrt{225}$
Solution:
The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.
$\sqrt{2},$ $\sqrt{23}$ and 0.1010010001... are irrational numbers, since they cannot be expressed in the form $\frac{\text{p}}{\text{q}}.$
$\sqrt{225}=15=\frac{15}{1}$ is a rational number.
Hence, the correct opion is (c).
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Question 241 Mark
The decimal representation of a rational number is:
  1. always terminating.
  2. either terminating or repeating.
  3. either terminating or non-repeating.
  4. neither terminating nor repeating.
Answer
  1. either terminating or repeating.
Solution:
The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.
Decimal representation of a rational number is either terminating or a repeating decimal, since every decimal of this form can be expressed in the form $\frac{\text{p}}{\text{q}}.$
Hence, the correct opion is (b).
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Question 251 Mark
$\pi$ is:
  1. a rational number.
  2. an integer.
  3. an irrational number.
  4. a whole number.
Answer
  1. an irrational number.
Solution:
$\pi=3.14159265359...,$ which is non-terminating non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is (c).
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MCQ 261 Mark
$9^3 + (-3)^3 - 6^3 = ?$
  • A
    $432$
  • B
    $270$
  • ✓
    $486$
  • D
    $540$
Answer
Correct option: C.
$486$
$9^3 + (-3)^3 - 6^3 = 729 - 27 - 21 6 = 486$
Hence, the correct answer is option $(c).$
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Question 271 Mark
$\frac{1}{\big(3+2\sqrt{2}\big)}=?$
  1. $\frac{3-2\sqrt{2}}{17}$
  2. $\frac{\big(3-2\sqrt{2}\big)}{13}$
  3. $\big(3-2\sqrt{2}\big)$
  4. None of these.
Answer
  1. $\big(3-2\sqrt{2}\big)$
Solution:
$\frac{1}{\big(3+2\sqrt{2}\big)}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=\frac{3-2\sqrt{2}}{9-8}$
$=\big(3-2\sqrt{2}\big)$
Hence, the correct option is (c).
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Question 281 Mark
The product of two irrational number is:
  1. always irrational.
  2. always rational.
  3. always an integer.
  4. sometimes rational and sometimes irrational.
Answer
  1. sometimes rational and sometimes irrational.
Solution:
Consider, the irrational number, $\sqrt{3}.$
Product $=\sqrt{3}\times\sqrt{3}=3,$ which is a rational number.
Consider two irrational numbers, $\sqrt{2}$ and $\sqrt{3}.$
Product $=\sqrt{2}\times\sqrt{3}=\sqrt{6},$ which is an irrational number.
Hence, the product of two irrational numbers are sometimes rational and sometimes irrational.
Hence, the correct opion is (d).
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Question 291 Mark
If $\text{x}=\big(7+4\sqrt{3}\big)$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
  1. $8\sqrt{3}$
  2. 14
  3. 49
  4. 48
Answer
  1. 14
Solution:
$\text{x}=\big(7+4\sqrt{3}\big)$
$\therefore\frac{1}{\text{x}}=\frac{1}{\big(7+4\sqrt{3}\big)}$
$=\frac{1}{\big(7+4\sqrt{3}\big)}\times\frac{\big(7-4\sqrt{3}\big)}{\big(7-4\sqrt{3}\big)}$
$=\frac{\big(7-4\sqrt{3}\big)}{7^2-\big(4\sqrt{3}\big)^2}$
$=\frac{7-4\sqrt{3}}{49-48}$
$=7-4\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(7+4\sqrt{3}\big)+\big(7-4\sqrt{3}\big)=14$
Hence, the correct option is (b).
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Question 301 Mark
The value of $\sqrt{20}\times\sqrt{5}$ is:
  1. $10$
  2. $2\sqrt{5}$
  3. $20\sqrt{5}$
  4. $4\sqrt{5}$
Answer
  1. $10$
Solution:
$\sqrt{20}\times\sqrt{5}=\sqrt{4\times5}\times\sqrt{5}$
$=2\sqrt{5}\times\sqrt{5}=2\times5=10$
Hence, the correct answer is option (a).
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Question 311 Mark
The sum of $0.\overline{3}$ and $0.\overline{4}$ is:
  1. $\frac{7}{10}$
  2. $\frac{7}{9}$
  3. $\frac{7}{11}$
  4. $\frac{7}{99}$
Answer
  1.  $\frac{7}{9}$
Solution:
Let $\text{x}=0.\overline{3}$
i.e., $\text{x}=0.3333 \ ...(\text{i})​​$
$\Rightarrow10\text{x}=3.3333 \ ...(\text{ii})​​$
On subtracting (i) and (ii), we get
$9\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{9}$
Let $\text{y}=0.\overline{4}$
i.e., $\text{y}=0.4444 \ ...(\text{i})​​$
$\Rightarrow10\text{y}=4.4444 \ ...(\text{ii})​​$
On subtracting (i) and (ii), we get
$9\text{y}=4$
$\Rightarrow\text{y}=\frac{4}{9}$
$\therefore0.\overline{3}+0.\overline{4}=\text{x}+\text{y}=\frac{\text{3}}{\text{9}}+\frac{\text{4}}{\text{9}}=\frac{\text{7}}{\text{9}}$
Hence, the correct answer is option (b). 
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Question 321 Mark
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
  1. $\big(\sqrt{2}+\sqrt{3}\big)$
  2. $\big(\sqrt{2}\times\sqrt{3}\big)$
  3. $5^{\frac{1}{4}}$
  4. $6^{\frac{1}{4}}$
Answer
  1. $6^{\frac{1}{4}}$
Solution:
An irrational number between a and b is given by $\sqrt{\text{ab}}$
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is
$\sqrt{\sqrt{2}+\sqrt{3}}=\big(\sqrt{6}\big)^{\frac{1}{2}}$
$=6^{\frac{1}{2}\times\frac{1}{2}}$
$=6^{\frac{1}{4}}$
Hence, the correct answer is option (d).
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Question 331 Mark
Which of the following is an irrational number?
  1. $\sqrt{23}$
  2. $\sqrt{225}$
  3. $0.3799$
  4. $7.\overline{478}$
Answer
  1. $\sqrt{23}$
Solution:
The decimal expansion of $\sqrt{23}=4.79583152...,$ which is non-terminating, nonrecurring.
Hence, it is an irrational number.
Hence, the correct opion is (a).
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Question 341 Mark
After simplification, $\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$ is:
  1. $13^\frac{2}{15}$
  2. $13^\frac{8}{15}$
  3. $13^\frac{1}{3}$
  4. $13^\frac{-2}{15}$
Answer
  1. $13^\frac{-2}{15}$
Solution:
$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}=13^{\frac{1}{5}-\frac{1}{3}}=13^{\frac{3-5}{15}}=13^{\frac{-2}{15}}$
Hence, the correct answer is option (d).
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Question 351 Mark
The value of $7^{\frac{1}{2}}.8^{\frac{1}{2}}$ is:
  1. $(28)^\frac{1}{2}$
  2. $(56)^\frac{1}{2}$
  3. $(14)^\frac{1}{2}$
  4. $(42)^\frac{1}{2}$
Answer
  1. $(56)^\frac{1}{2}$
Solution:
$(7)^{\frac{1}{2}}\times(8)^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}$
Hence, the correct answer is option (b).
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Question 361 Mark
A rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is:
  1. $\frac{\big(\sqrt{2}+\sqrt{3}\big)}{2}$
  2. $\sqrt{6}$
  3. 1.6
  4. 1.9
Answer
  1. 1.6
Solution:
$\sqrt{2}=1. 414213562...$ and $\sqrt{3}=1. 7320508075...$
Hence, 1.6 lies between $\sqrt{2}$ and $\sqrt{3}$
Hence, the correct option is (c).
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Question 371 Mark
The value of $2.\overline{45}+0.\overline{36}$ is:
  1. $\frac{67}{33}$
  2. $\frac{24}{11}$
  3. $\frac{31}{11}$
  4. $\frac{167}{110}$
Answer
  1.  $\frac{31}{11}$
Solution:
Let $\text{x}=2.\overline{45}$
i.e., $\text{x}=2.4545 \ ...(\text{i})​​$
$\Rightarrow100\text{x}=245.4545 \ ...(\text{ii})​​$
On subtracting (i) and (ii), we get
$99\text{x}=243$
$\Rightarrow\text{x}=\frac{243}{99}$
Let $\text{y}=0.\overline{36}$
i.e., $\text{y}=0.3636 \ ...(\text{iii})​​$
$\Rightarrow100\text{y}=36.3636 \ ...(\text{iv})​​$
On subtracting (iii) and (iv), we get
$99\text{y}=36$
$\Rightarrow\text{y}=\frac{36}{99}$
$\therefore2.\overline{45}+0.\overline{36}=\text{x}+\text{y}=\frac{\text{243}}{\text{99}}+\frac{\text{36}}{\text{99}}=\frac{\text{279}}{\text{99}}=\frac{\text{31}}{\text{11}}$
Hence, the correct answer is option (c).
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Question 381 Mark
The simplest form of $0.12\overline{3}$ is:
  1. $\frac{41}{330}$
  2. $\frac{37}{330}$
  3. $\frac{41}{333}$
  4. None of these.
Answer
  1. None of these.
Solution:
Let $\text{x}=0.12\overline{3}$
Then, $\text{x}=0.12333 \ ...(\text{i})$
$\therefore100\text{x}=12.333... \ (\text{ii})$
and $1000\text{x}=123.333... \ (\text{iii})$
On subtracting (ii) from (iii), we get
$900\text{x}=111$
$\Rightarrow\text{x}=\frac{111}{900}=\frac{37}{300}$
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Question 391 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$ is:
  1. -1
  2. 0
  3. 1
  4. 2
Answer
  1. 1
Solution:
$\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{q}}{\text{p}}}.\sqrt{\frac{\text{r}}{\text{q}}}.\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=\sqrt{1}$
$=1$
Hence, the correct option is (c).
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Question 401 Mark
The decimal expansion that a rational number cannot have is:
  1. 0. 25
  2. $0.25\overline{28}$
  3. $0.\overline{2528}$
  4. 0.5030030003...
Answer
  1. 0.5030030003...
Solution:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
The decimal expansion of 0.5030030003... is non-terminating, non-recurring, which is not a property of a rational number. 
Hence, the correct opion is (d).
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Question 411 Mark
The product of a non-zero rational number with an irrational number is always a/an:
  1. irrational number.
  2. rational number.
  3. whole number.
  4. natural number.
Answer
  1. irrational number.
Solution:
The product if a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is (a).
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Question 421 Mark
The value of $(243)^{\frac{1}{5}}$ is:
  1. 3
  2. -3
  3. 5
  4. $\frac{1}{3}$
Answer
  1. 3
Solution:
$(243)^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{5\times\frac{1}{5}}=3$
Hence, the correct answer is option (a).
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Question 431 Mark
If $\sqrt{7}=2.646$ then $\frac{1}{\sqrt{7}}=?$
  1. 0.375
  2. 0.378
  3. 0.441
  4. None of these.
Answer
  1. 0.378
Solution:
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$
$=\frac{\sqrt{7}}{7}$
$=\frac{1}{7}\times\sqrt{7}$
$=\frac{1}{7}\times2.646$
$=0.378$
Hence, the correct option is (b).
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Question 441 Mark
Simplified value of $(16)^{-\frac{1}{4}}\times\sqrt[4]{16}$ is:
  1. 0
  2. 1
  3. 4
  4. 16
Answer
  1. 1
Solution:
$(16)^{-\frac{1}{4}}\times\sqrt[4]{16}=(2^4)^{-\frac{1}{4}}\times(2^4)^\frac{1}{4}\\=2^{-1}\times2^1=\frac{1}{2}\times2=1$
Hence, the correct answer is option (a).
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Question 451 Mark
The value of $\frac{2^\circ+7^\circ}{5^\circ}$ is:
  1. 0
  2. 2
  3. $\frac{9}{5}$
  4. $\frac{1}{5}$
Answer
  1. 2
Solution:
$\frac{2^\circ+7^\circ}{5^\circ}=\frac{1+1}{1}=\frac{2}{1}=2$
Hence, the correct answer is option (b).
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Question 461 Mark
Choose the rational number which does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
  1. $-\frac{3}{10}$
  2. $\frac{3}{10}$
  3. $-\frac{1}{4}$
  4. $-\frac{7}{20}$
Answer
  1. $\frac{3}{10}$
Solution:
Given two rational numbers are negative and $\frac{3}{10}$ is a positive rational number.
 So, it does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
Hence, the correct opion is (b).
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Question 471 Mark
Which of the following is a true statment?
  1. The sum of two irrational numbers is an irrational number.
  2. The product of two irrational numbers is an irrational number.
  3. Every real number is always rational.
  4. Every real number is either rational or irrational.
Answer
  1. Every real number is either rational or irrational.
Solution:
Consider, $\big(2+\sqrt{3}\big)$ and $\big(2-\sqrt{3}\big)$ which are two irrational numbers.
$\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which are two irrational numbers.
$\sqrt{3}\times\frac{1}{\sqrt{3}}=1,$ which is a rational number.
Every real number can either be a rational number or an irrational number.
Hence, the correct opion is (d).
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Question 481 Mark
The value of $\sqrt[4]{\sqrt[3]{2^2}}$ is:
  1. $2^{-\frac{1}{6}}$
  2. $2^{-6}$
  3. $2^{\frac{1}{6}}$
  4. $2^{6}$
Answer
  1. $2^{\frac{1}{6}}$
Solution:
$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{\sqrt[3]{4}}=\sqrt[4]{4^{\frac{1}{3}}}=4^{\frac{1}{3}\times\frac{1}{4}}=4^{\frac{1}{12}}=2^{2\times\frac{1}{12}}=2^{\frac{1}{6}}$
Hence, the correct answer is option (c).
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Question 491 Mark
The value of $\frac{4\sqrt{12}}{12\sqrt{27}}$ is:
  1. $\frac{1}{9}$
  2. $\frac{2}{9}$
  3. $\frac{4}{9}$
  4. $\frac{8}{9}$
Answer
  1. $\frac{2}{9}$
Solution:
$\frac{4\sqrt{12}}{12\sqrt{27}}=\frac{4\sqrt{4\times3}}{12\sqrt{9\times3}}$
$=\frac{2\sqrt{3}}{3\times3\sqrt{3}}=\frac{2}{9}$
Hence, the correct answer is option (b).
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Question 501 Mark
Every point on a number line represents:
  1. a rational number.
  2. a natural number.
  3. an irrational number.
  4. a unique number.
Answer
  1. a unique number.
Solution:
Every point on a number line represents a unique number.
Hence, the correct opion is (d).
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