MCQ 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:
- A
$\frac{5}{3}$
- ✓
$-\frac{5}{3}$
- C
$\frac{3}{5}$
- D
$-\frac{3}{5}$
AnswerCorrect option: B. $-\frac{5}{3}$
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$
$=\frac{5^{\text{n}+1}(5-6)}{5^{\text{n}}(13-2\times5)}$
$=\frac{5^{\text{n}}\times5\times(-1)}{5^{\text{n}}(13-10)}$
$=-\frac{5}{3}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 21 Mark
Which of the following is an irrational number?
- A
$3.14$
- B
$3.141414...$
- C
$3.14444$
- ✓
$3.141141114...$
AnswerCorrect option: D. $3.141141114...$
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 $(d)$.
View full question & answer→MCQ 31 Mark
An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is:
- A
$\frac{1}{2}\Big(\frac{1}{7}+\frac{2}{7}\Big)$
- B
$\Big(\frac{1}{7}\times\frac{2}{7}\Big)$
- ✓
$\sqrt{\frac{1}{7}\times\frac{2}{7}}$
- D
AnswerCorrect option: C. $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
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 $(c)$.
View full question & answer→MCQ 41 Mark
The rationalisation factor of $\frac{1}{\big(2\sqrt{3}-\sqrt{5}\big)}$ is:
AnswerCorrect option: D. $\sqrt{12}+\sqrt{5}$
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 $(d)$.
View full question & answer→MCQ 51 Mark
Rationalisation of the denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$ gives:
AnswerCorrect option: D. $\frac{\sqrt{5}-\sqrt{2}}{3}$
$\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\\=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 61 Mark
If $\text{x}=2+\sqrt{3}$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ equals:
- A
$-2{\sqrt{3}}$
- B
$2$
- ✓
$4$
- D
$4-2\sqrt{3}$
Answer$\text{x}=2+\sqrt{3}$
$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 71 Mark
Between any two rational numbers there:
- A
- B
is exactly one rational numbers.
- ✓
are infinitely many rational numbers.
- D
AnswerCorrect option: C. are infinitely many rational numbers.
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 $(c)$.
View full question & answer→MCQ 81 Mark
When $15\sqrt{15}$ is divided by $3\sqrt{3},$ the quotient is:
- A
$5\sqrt{3}$
- B
$3\sqrt{5}$
- ✓
$5\sqrt{5}$
- D
$3\sqrt{3}$
AnswerCorrect option: C. $5\sqrt{5}$
$\frac{15\sqrt{15}}{3\sqrt{3}}=\frac{5\sqrt{5\times3}}{\sqrt{3}}$
$\frac{5\sqrt{5}\times\sqrt{3}}{\sqrt{3}}=5\sqrt{5}$
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 91 Mark
The simplest rationalisation factor of $\sqrt[3]{500}$ is:
- A
$\sqrt{5}$
- B
$\sqrt{3}$
- C
$\sqrt[3]{5}$
- ✓
$\sqrt[3]{2}$
AnswerCorrect option: D. $\sqrt[3]{2}$
$\sqrt[3]{500}=500^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}$ $=\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=\frac{10^{3\times\frac{1}{3}}}{2^{\frac{1}{3}}}=\frac{10}{\sqrt[3]{2}}$
Thus, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}.$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 101 Mark
Which of the following numbers is irrational?
AnswerCorrect option: C. $\sqrt{8}$
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 $(c)$.
View full question & answer→MCQ 111 Mark
How many digits are there in the repeating block of digits in the decimal expansion of $\frac{17}{7}?$
Answer$\frac{17}{7}=2.\overline{428571}$
Hence, the correct opion is $(b)$.
View full question & answer→MCQ 121 Mark
Which of the following is a rational number?
- A
$\sqrt{5}$
- B
$0.101001000100001...$
- C
$\pi$
- ✓
$0.853853853...$
AnswerCorrect option: D. $0.853853853...$
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 $(d)$.
View full question & answer→MCQ 131 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
$\sqrt[3]{2}$
- ✓
$\sqrt[4]{2}$
AnswerCorrect 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 $(d)$.
View full question & answer→MCQ 141 Mark
Two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are:
- A
$\frac{1}{6}$ and $\frac{2}{6}$
- B
$\frac{1}{2}$ and $\frac{2}{1}$
- ✓
$\frac{5}{6}$ and $\frac{7}{6}$
- D
$\frac{2}{3}$ and $\frac{4}{3}$
AnswerCorrect option: C. $\frac{5}{6}$ and $\frac{7}{6}$
We have,
$\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{5}{3}=\frac{5\times2}{3\times2}=\frac{10}{6}$
And, $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$ and $\frac{2}{1}=\frac{2\times6}{1\times6}=\frac{12}{6}$
Also, $\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}$
Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\Big(\frac{1}{2}\Big)<\frac{4}{6}\Big(=\frac{2}{3}\Big)<\frac{5}{6}<\frac{7}{6}\\<\frac{8}{6}\Big(=\frac{4}{3}\Big)<\frac{10}{6}\Big(=\frac{5}{3}\Big)<\frac{12}{6}\Big(=\frac{2}{1}\Big)$
So, the two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are $\frac{5}{6}$ and $\frac{7}{6}.$
Hence, the correct opion is (c).
View full question & answer→MCQ 151 Mark
Which of the following is the value of $\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)?$
- A
$-4$
- ✓
$4$
- C
$\sqrt{11}$
- D
$\sqrt{7}$
Answer$\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)$
$\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2$
$11-7=4$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 161 Mark
The decimal expansion of $\sqrt{2}$ is:
- A
- B
$1.4121$
- C
nonterminating recurring.
- ✓
nonterminating, nonrecurring.
AnswerCorrect option: D. nonterminating, nonrecurring.
The decimal expansion of $\sqrt{2}=1.41421356...,$ which is non-terminating, nonrecurring.
Hence, the correct opion is $(d)$.
View full question & answer→MCQ 171 Mark
A rational number between $-3$ and $3$ is:
- ✓
$0$
- B
$-4.3$
- C
$-3.4$
- D
$1.101100110001...$
AnswerSince, $-4.3 < -3.4 < -3 < 0 < 1.101100110001... < 3$
But $1.101100110001...$ is an irrational number
So, the rational number between $-3$ and $3$ is $0$.
Hence, the correct option is $(a)$.
View full question & answer→MCQ 181 Mark
Which of the following is a rational number?
- A
$\sqrt{2}$
- B
$\sqrt{23}$
- ✓
$\sqrt{225}$
- D
$0.1010010001...$
AnswerCorrect option: C. $\sqrt{225}$
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 $(c)$.
View full question & answer→MCQ 191 Mark
The decimal representation of a rational number is:
- A
- ✓
either terminating or repeating.
- C
either terminating or non-repeating.
- D
neither terminating nor repeating.
AnswerCorrect option: B. either terminating or repeating.
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 $(b)$.
View full question & answer→MCQ 201 Mark
$\pi$ is:
Answer$\pi=3.14159265359...,$ which is non-terminating non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is $(c)$.
View full question & answer→MCQ 211 Mark
$9^3+(-3)^3-6^3 =$ ?
Answer$9^3+(-3)^3-6^3=729-27-216=486$
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 221 Mark
$\frac{1}{\big(3+2\sqrt{2}\big)}=?$
AnswerCorrect option: C. $\big(3-2\sqrt{2}\big)$
$\frac{1}{\big(3+2\sqrt{2}\big)}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$
$=\frac{3-2\sqrt{2}}{3^2-\big(2\sqrt{2}\big)^2}$
$=\frac{3-2\sqrt{2}}{9-8}$
$=\big(3-2\sqrt{2}\big)$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 231 Mark
The product of two irrational number is:
- A
- B
- C
- ✓
sometimes rational and sometimes irrational.
AnswerCorrect option: D. sometimes rational and sometimes irrational.
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 $(d)$.
View full question & answer→MCQ 241 Mark
If $\text{x}=\big(7+4\sqrt{3}\big)$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
- A
$8\sqrt{3}$
- ✓
$14$
- C
$49$
- D
$48$
Answer$\text{x}=\big(7+4\sqrt{3}\big)$
$\therefore\frac{1}{\text{x}}=\frac{1}{\big(7+4\sqrt{3}\big)}$
$=\frac{1}{\big(7+4\sqrt{3}\big)}\times\frac{\big(7-4\sqrt{3}\big)}{\big(7-4\sqrt{3}\big)}$
$=\frac{\big(7-4\sqrt{3}\big)}{7^2-\big(4\sqrt{3}\big)^2}$
$=\frac{7-4\sqrt{3}}{49-48}$
$=7-4\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(7+4\sqrt{3}\big)+\big(7-4\sqrt{3}\big)=14$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 251 Mark
The value of $\sqrt{20}\times\sqrt{5}$ is:
- ✓
$10$
- B
$2\sqrt{5}$
- C
$20\sqrt{5}$
- D
$4\sqrt{5}$
Answer$\sqrt{20}\times\sqrt{5}=\sqrt{4\times5}\times\sqrt{5}$
$=2\sqrt{5}\times\sqrt{5}=2\times5=10$
Hence, the correct answer is option $(a)$.
View full question & answer→MCQ 261 Mark
The sum of $0.\overline{3}$ and $0.\overline{4}$ is:
- A
$\frac{7}{10}$
- ✓
$\frac{7}{9}$
- C
$\frac{7}{11}$
- D
$\frac{7}{99}$
AnswerCorrect option: B. $\frac{7}{9}$
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 $(b)$.
View full question & answer→MCQ 271 Mark
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
AnswerCorrect option: D. $6^{\frac{1}{4}}$
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 $(d)$.
View full question & answer→MCQ 281 Mark
Which of the following is an irrational number?
- ✓
$\sqrt{23}$
- B
$\sqrt{225}$
- C
$0.3799$
- D
$7.\overline{478}$
AnswerCorrect option: A. $\sqrt{23}$
The decimal expansion of $\sqrt{23}=4.79583152...,$ which is non-terminating, nonrecurring.
Hence, it is an irrational number.
Hence, the correct opion is $(a)$.
View full question & answer→MCQ 291 Mark
After simplification, $\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$ is:
- A
$13^\frac{2}{15}$
- B
$13^\frac{8}{15}$
- C
$13^\frac{1}{3}$
- ✓
$13^\frac{-2}{15}$
AnswerCorrect option: D. $13^\frac{-2}{15}$
$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}=13^{\frac{1}{5}-\frac{1}{3}}=13^{\frac{3-5}{15}}=13^{\frac{-2}{15}}$
Hence, the correct answer is option $(d)$.
View full question & answer→MCQ 301 Mark
The value of $7^{\frac{1}{2}}.8^{\frac{1}{2}}$ is:
- A
$(28)^\frac{1}{2}$
- ✓
$(56)^\frac{1}{2}$
- C
$(14)^\frac{1}{2}$
- D
$(42)^\frac{1}{2}$
AnswerCorrect option: B. $(56)^\frac{1}{2}$
$(7)^{\frac{1}{2}}\times(8)^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 311 Mark
A rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is:
Answer$\sqrt{2}=1. 414213562...$ and $\sqrt{3}=1. 7320508075...$
Hence, 1.6 lies between $\sqrt{2}$ and $\sqrt{3}$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 321 Mark
The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is:
- A
$2\sqrt{2}+3$
- ✓
$2\sqrt{2}+\sqrt{3}$
- C
$\sqrt{2}+\sqrt{3}$
- D
$\sqrt{2}-\sqrt{3}$
AnswerCorrect option: B. $2\sqrt{2}+\sqrt{3}$
The simplest rationalisation factor of $\big(2\sqrt{2}-\sqrt{3}\big)$ is $2\sqrt{2}+\sqrt{3}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 331 Mark
The value of $2.\overline{45}+0.\overline{36}$ is:
- A
$\frac{67}{33}$
- B
$\frac{24}{11}$
- ✓
$\frac{31}{11}$
- D
$\frac{167}{110}$
AnswerCorrect option: C. $\frac{31}{11}$
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 $(c)$.
View full question & answer→MCQ 341 Mark
If $\sqrt{7}=2.646$ then $\frac{1}{\sqrt{7}}=?$
- A
$0.375$
- ✓
$0.378$
- C
$0.441$
- D
AnswerCorrect option: B. $0.378$
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$
$=\frac{\sqrt{7}}{7}$
$=\frac{1}{7}\times\sqrt{7}$
$=\frac{1}{7}\times2.646$
$=0.378$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 351 Mark
The simplest form of $0.12\overline{3}$ is:
- A
$\frac{41}{330}$
- B
$\frac{37}{330}$
- C
$\frac{41}{333}$
- ✓
AnswerLet $\text{x}=0.12\overline{3}$
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
$900\text{x}=111$
$\Rightarrow\text{x}=\frac{111}{900}=\frac{37}{300}$
View full question & answer→MCQ 361 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$ is:
Answer$\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{q}}{\text{p}}}.\sqrt{\frac{\text{r}}{\text{q}}}.\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=\sqrt{1}$
$=1$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 371 Mark
The decimal expansion that a rational number cannot have is:
- A
$0. 25$
- B
$0.25\overline{28}$
- C
$0.\overline{2528}$
- ✓
$0.5030030003...$
AnswerCorrect option: D. $0.5030030003...$
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 $(d)$.
View full question & answer→MCQ 381 Mark
The product of a non-zero rational number with an irrational number is always a/an:
AnswerThe product if a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is $(a)$.
View full question & answer→MCQ 391 Mark
The value of $(243)^{\frac{1}{5}}$ is:
- ✓
$3$
- B
$-3$
- C
$5$
- D
$\frac{1}{3}$
Answer$(243)^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{5\times\frac{1}{5}}=3$
Hence, the correct answer is option $(a)$.
View full question & answer→MCQ 401 Mark
Simplified value of $(16)^{-\frac{1}{4}}\times\sqrt[4]{16}$ is:
Answer$(16)^{-\frac{1}{4}}\times\sqrt[4]{16}=(2^4)^{-\frac{1}{4}}\times(2^4)^\frac{1}{4}\\=2^{-1}\times2^1=\frac{1}{2}\times2=1$
Hence, the correct answer is option $(a)$.
View full question & answer→MCQ 411 Mark
The value of $\frac{2^\circ+7^\circ}{5^\circ}$ is:
- A
$0$
- ✓
$2$
- C
$\frac{9}{5}$
- D
$\frac{1}{5}$
Answer$\frac{2^\circ+7^\circ}{5^\circ}=\frac{1+1}{1}=\frac{2}{1}=2$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 421 Mark
Choose the rational number which does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
- A
$-\frac{3}{10}$
- ✓
$\frac{3}{10}$
- C
$-\frac{1}{4}$
- D
$-\frac{7}{20}$
AnswerCorrect option: B. $\frac{3}{10}$
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 $(b)$.
View full question & answer→MCQ 431 Mark
Which of the following is a true statment?
- A
The sum of two irrational numbers is an irrational number.
- B
The product of two irrational numbers is an irrational number.
- C
Every real number is always rational.
- ✓
Every real number is either rational or irrational.
AnswerCorrect option: D. Every real number is either rational or irrational.
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 $(d)$.
View full question & answer→MCQ 441 Mark
The value of $\sqrt[4]{\sqrt[3]{2^2}}$ is:
- A
$2^{-\frac{1}{6}}$
- B
$2^{-6}$
- ✓
$2^{\frac{1}{6}}$
- D
$2^{6}$
AnswerCorrect option: C. $2^{\frac{1}{6}}$
$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{\sqrt[3]{4}}=\sqrt[4]{4^{\frac{1}{3}}}=4^{\frac{1}{3}\times\frac{1}{4}}=4^{\frac{1}{12}}=2^{2\times\frac{1}{12}}=2^{\frac{1}{6}}$
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 451 Mark
The value of $\frac{4\sqrt{12}}{12\sqrt{27}}$ is:
- A
$\frac{1}{9}$
- ✓
$\frac{2}{9}$
- C
$\frac{4}{9}$
- D
$\frac{8}{9}$
AnswerCorrect option: B. $\frac{2}{9}$
$\frac{4\sqrt{12}}{12\sqrt{27}}=\frac{4\sqrt{4\times3}}{12\sqrt{9\times3}}$
$=\frac{2\sqrt{3}}{3\times3\sqrt{3}}=\frac{2}{9}$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 461 Mark
Every point on a number line represents:
AnswerEvery point on a number line represents a unique number.
Hence, the correct opion is $(d)$.
View full question & answer→MCQ 471 Mark
A rational number equivalent to $\frac{7}{19}$ is:
- A
$\frac{17}{119}$
- B
$\frac{14}{157}$
- C
$\frac{21}{38}$
- ✓
$\frac{21}{57}$
AnswerCorrect option: D. $\frac{21}{57}$
$\frac{7}{17}=\frac{7\times3}{17\times3}=\frac{21}{57}$
Hence, the correct opion is $(d)$.
View full question & answer→MCQ 481 Mark
Every rational number is:
AnswerA number whose square is non-negative, is called a real number.
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, a rational number is a real number since $p$ and $q$ which form a rational number are integers.
Hence, the correct opion is $(d)$.
View full question & answer→MCQ 491 Mark
The decimal representation of an irrational number is:
- A
- B
either terminating or repeating.
- C
either terminating or non-repeating.
- ✓
neither terminating nor repeating.
AnswerCorrect option: D. neither terminating nor repeating.
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.
So, decimal representation of an irrational number is neither terminating nor a repeating decimal.
Hence, the correct opion is (d).
View full question & answer→MCQ 501 Mark
If $\sqrt{2}=1.414$ then $\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=?$
- A
$0.207$
- B
$2.414$
- ✓
$0.414$
- D
$0.621$
AnswerCorrect option: C. $0.414$
$\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{1}}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
$=1.414-1$
$=0.414$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 511 Mark
The simplest for of $0.\overline{32}$ is:
- A
$\frac{16}{45}$
- B
$\frac{32}{99}$
- C
$\frac{29}{90}$
- ✓
View full question & answer→MCQ 521 Mark
The value of $x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$ is equal to:
- A
$0$
- ✓
$1$
- C
$x$
- D
$x^{\text {pqr }}$
Answer$x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$
$=x^{p-q-q-r+r-p}$
$=x^0$
$=1$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 531 Mark
If $\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$ then $x = ?$
Answer$\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{3^4}{2^4}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}+2\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\Big(\frac{3}{2}\Big)^{\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\text{x}=4$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 541 Mark
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
Answer$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 551 Mark
An irrational number between $5$ and $6$ is:
- A
$\frac{1}{2}(5+6)$
- B
$\sqrt{5+6}$
- ✓
$\sqrt{5\times6}$
- D
AnswerCorrect option: C. $\sqrt{5\times6}$
An irrational number between $a$ and $b$ is given by $\sqrt{\text{ab}}$
So, an irrational number between $5$ and $6$ is given by $\sqrt{5\times6}$
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 561 Mark
If $\sqrt{2}=1.41$ then $\frac{1}{\sqrt{2}}=?$
- A
$0.075$
- B
$0.75$
- ✓
$0.705$
- D
$7.05$
AnswerCorrect option: C. $0.705$
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 571 Mark
The value of $0.\overline{2}$ in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0,$ is:
- A
$\frac{1}{5}$
- ✓
$\frac{2}{9}$
- C
$\frac{2}{5}$
- D
$\frac{1}{8}$
AnswerCorrect option: B. $\frac{2}{9}$
$\frac{2}{9}=0.2222222...=0.\overline{2}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 581 Mark
The value of $\sqrt[4]{(64)^{-2}}$ is:
- ✓
$\frac{1}{8}$
- B
$\frac{1}{2}$
- C
$8$
- D
$\frac{1}{64}$
AnswerCorrect option: A. $\frac{1}{8}$
$\sqrt[4]{(64)^{-2}}=\sqrt[4]{(8^2)^{-2}}=8^{-4\times\frac{1}{4}}=8^{-1}=\frac{1}{8}$
Hence, the correct answer is option $(a)$.
View full question & answer→MCQ 591 Mark
The simplest for of $0.\overline{54}$ is:
- A
$\frac{27}{50}$
- ✓
$\frac{6}{11}$
- C
$\frac{4}{7}$
- D
AnswerCorrect option: B. $\frac{6}{11}$
Let $\text{x}=0.\overline{54}$
Then, $\text{x}=54.5454 \ ...(\text{i})$
$\therefore100\text{x}=54.5454 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii)$, we get
$99\text{x}=54$
$\Rightarrow\text{x}=\frac{54}{99}=\frac{6}{11}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 601 Mark
If $\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$ then the value of $p$ is:
- A
$\frac{7}{25}$
- ✓
$\frac{25}{7}$
- C
$\frac{7}{15}$
- D
$\frac{15}{7}$
AnswerCorrect option: B. $\frac{25}{7}$
$\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{5}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{25}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\text{p}=\frac{25}{\sqrt{7}\times\sqrt{7}}=\frac{25}{7}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 611 Mark
The value of $\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}$ is:
- A
$\frac{3\text{y}}{8\text{x}^4}$
- ✓
$\frac{3\text{y}}{4\text{x}^4}$
- C
$\frac{4\text{y}}{5\text{x}^4}$
- D
$\frac{4\text{x}^4}{3\text{y}}$
AnswerCorrect option: B. $\frac{3\text{y}}{4\text{x}^4}$
$\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\Big(\frac{3^4\text{y}^{4}}{4^4(\text{x}^4)^4}\Big)^{\frac{1}{4}}=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{4\times\frac{1}{4}}=\frac{3\text{y}}{4\text{x}^4}$
Hence, the correct option is $(b)$.
View full question & answer→MCQ 621 Mark
If $\left(3^3\right)^2=9^x$ then $5^x=$ ?
Answer$\left(3^3\right)^2=9^x$
$\Rightarrow\left(3^2\right)^3=\left(3^2\right)^x$
$\Rightarrow x=3$
Then $5^x=5^3=125$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 631 Mark
$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$ when simplified is:
Answer$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$
$=6+\sqrt{27}-3-\sqrt{3}-2\sqrt{3}$
$=6-3+3\sqrt{3}-\sqrt{3}-2\sqrt{3}$
$=3$
which is positive and rational number.
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 641 Mark
The simplest for of $1.\overline{6}$ is:
- A
$\frac{833}{500}$
- B
$\frac{8}{5}$
- ✓
$\frac{5}{3}$
- D
AnswerCorrect option: C. $\frac{5}{3}$
Let $\text{x}=1.\overline{6}$
$\Rightarrow\text{x}=1.666 \ ...(\text{i})$
$\therefore10\text{x}=16.666 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii)$, we get
$9\text{x}=15$
$\Rightarrow\text{x}=\frac{15}{9}=\frac{5}{3}$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 651 Mark
The value of $\sqrt{5+2\sqrt{6}}$ is:
- A
$\sqrt{5}+\sqrt{6}$
- B
$\sqrt{5}-\sqrt{6}$
- ✓
$\sqrt{3}+\sqrt{2}$
- D
$\sqrt{3}-\sqrt{2}$
AnswerCorrect option: C. $\sqrt{3}+\sqrt{2}$
$\sqrt{5+2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}}$
$=\sqrt{\big(\sqrt{2}+\sqrt{3}\big)^2}$
$=\sqrt{2}+\sqrt{3}$
Hence, the correct option is $(c)$.
View full question & answer→MCQ 661 Mark
$(125)^{-\frac{1}{3}}=?$
Answer$(125)^{-\frac{1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$
$=\frac{1}{5^{3-\frac{1}{3}}}=\frac{1}{5}$
Hence, the correct answer is option $(c)$.
View full question & answer→MCQ 671 Mark
Simplified value of $(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}$ is:
Answer$(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}=5^{2\times\frac{1}{3}}\times5^\frac{1}{3}$
$=5^\frac{2}{3}\times5^\frac{1}{3}=5^{\frac{2}{3}+\frac{1}{3}}=5^{\frac{3}{3}}=5$
View full question & answer→MCQ 681 Mark
Which of the following is a rational number?
- A
$1+\sqrt{3}$
- B
$\pi$
- C
$2\sqrt{3}$
- ✓
$0$
AnswerSince, the sum and product of a rational and an irrational is always irrational.
So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.
Also, $\pi$ is an irrational number.
And, $0$ is an integer.
So, $0$ is a rational number.
Hence, the correct option is $(d)$.
View full question & answer→MCQ 691 Mark
The value of $\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$ is:
- ✓
$3$
- B
$-3$
- C
$9$
- D
$\frac{1}{3}$
Answer$\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}=\Big[\big(9^2\big)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$
$=9^{2\times\frac{1}{2}\times\frac{1}{2}}=9^{\frac{1}{2}}=3^{2\times\frac{1}{2}}=3$
Hence, the correct option is $(a)$.
View full question & answer→MCQ 701 Mark
$\sqrt{10}\times\sqrt{15}=?$
- A
$\sqrt{25}$
- ✓
$5\sqrt{6}$
- C
$6\sqrt{5}$
- D
AnswerCorrect option: B. $5\sqrt{6}$
$\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt{3}=5\sqrt{6}$
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 711 Mark
Which of the following is a true statment?
- A
$\pi$ and $\frac{22}{7}$ are both rationals.
- B
$\pi$ and $\frac{22}{7}$ are both irrationals
- C
$\pi$ is rational and $\frac{22}{7}$ is irrational.
- ✓
$\pi$ is irrational and $\frac{22}{7}$ is rational.
AnswerCorrect option: D. $\pi$ is irrational and $\frac{22}{7}$ is rational.
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 $(d$).
View full question & answer→MCQ 721 Mark
$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$ when simplified is:
Answer$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$
$\big(-2\big)^2-\big(\sqrt{3}\big)^2$
$=4-3=1,$ which is positive and rational number.
Hence, the correct answer is option $(b)$.
View full question & answer→MCQ 731 Mark
The value of $\sqrt{3-2\sqrt{2}}$ is:
- A
$\sqrt{3}+\sqrt{2}$
- B
$\sqrt{3}-\sqrt{2}$
- C
$\sqrt{2}+1$
- ✓
$\sqrt{2}-1$
AnswerCorrect option: D. $\sqrt{2}-1$
$\sqrt{3-2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+(1)^2-2\times\sqrt{2}\times1}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
Hence, the correct option is $(d)$.
View full question & answer→MCQ 741 Mark
If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$
Answer$\text{x}=3+\sqrt{8}$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
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 $(a)$.
View full question & answer→MCQ 751 Mark
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}=?$
- ✓
$2$
- B
$\sqrt{2}$
- C
$2\sqrt{2}$
- D
$4\sqrt{2}$
Answer$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{2^5}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$
$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{4+3+5}{12}}$
$=2^{\frac{12}{12}}$
$=2$
Hence, the correct option is $(a)$.
View full question & answer→