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 rational$?$
- A
$\sqrt{3}$
- B
$\pi$
- C
$\frac{4}{0}$
- ✓
$\frac{0}{4}$
AnswerCorrect option: D. $\frac{0}{4}$
$\sqrt{3}=1.732 \ ...=$ Non-terminating and non-repeating number, hence irrational
$\pi=3.14 \ ...$ also can not be terminated to $\frac{\text{p}}{\text{q}}$ form, and is non$-$terminating and non-repeating in nature. Hence, irrational.
$\frac{4}{0}$ is not a rational number because this is in the form $\frac{\text{p}}{\text{q}}$ where $p$ and $q$ are integers but $q = 0$
$\frac{0}{4}$ follows the defination of rational number.
Hence, correct option is $(d).$
View full question & answer→MCQ 31 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
$2^{6}$
AnswerCorrect 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 41 Mark
If $\text{a}=7-4\sqrt{3},$ then the value of $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}$ is:
AnswerLet $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=\text{x}$
Then, squaring both side, we get
$\text{a}+\frac{1}{\text{a}}+2=\text{x}^2$
$\Rightarrow\frac{\text{a}^2+1}{\text{a}}+2=\text{x}^2$
Now, put the value of $a,$
$\frac{(7-4\sqrt{3})^2+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{49+48-56\sqrt{3}+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{98-56\sqrt{3}}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow14\Big(\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\Big)+2=\text{x}^2$
$\Rightarrow16=\text{x}^2$
$\Rightarrow\text{x}=4$
So, $\text{x}=\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=4$
View full question & answer→MCQ 51 Mark
If $\text{x}=(7+4\sqrt{3})$ than $\Big(\text{x}+\frac{1}{\text{x}}\Big)=?$
- A
$49$
- ✓
$14$
- C
$48$
- D
$8\sqrt{3}$
Answer$\text{x}=(7+4\sqrt{3})$
$\frac{1}{\text{x}}=\frac{1}{7+4\sqrt{3}}=(7-4\sqrt{3})$
$\text{x}+\frac{1}{\text{x}}=(7+4\sqrt{3})+(7-4\sqrt{3})$
$=14$
View full question & answer→MCQ 61 Mark
$2\sqrt{3}+\sqrt{3}$ is equal to:
- A
$3\sqrt{6}$
- B
$3$
- ✓
$3\sqrt{3}$
- D
$2\sqrt{6}$
AnswerCorrect option: C. $3\sqrt{3}$
$(\sqrt{\text{x}^3})^{\frac{2}{3}}$
$=(\text{x}\frac{3}{2})^{\frac{2}{3}}$
$=\text{x}$
View full question & answer→MCQ 71 Mark
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to:
Answer$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$
$\Rightarrow\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$
$\Rightarrow\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}$
$\Rightarrow\frac{4}{2}(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}})$
$\Rightarrow2$
View full question & answer→MCQ 81 Mark
The number obtained on rationalising the denominator $\frac{1}{\sqrt{7}-2}$ of is:
- A
$\frac{\sqrt{7}+2}{45}$
- ✓
$\frac{\sqrt{7}+2}{3}$
- C
$\frac{\sqrt{7}-2}{1}$
- D
$\frac{\sqrt{7}+2}{5}$
AnswerCorrect option: B. $\frac{\sqrt{7}+2}{3}$
After rationalizing:
$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$
$=\frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2}$
$=\frac{\sqrt{7}+2}{7-4}$
$\frac{\sqrt{7}+2}{3}$
View full question & answer→MCQ 91 Mark
Which of the following is true statement?
- ✓
Every real number is either rational or irrational.
- B
The product of two irrational numbers is an irrational number.
- C
The sum of two irrational numbers is an irrational number.
- D
Every real number is always rational.
AnswerCorrect option: A. Every real number is either rational or irrational.
Consider, $(2+\sqrt{3})$ and $(2-\sqrt{3})$ which are two irrational number.
$(2+\sqrt{3})+(2-\sqrt{3})=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which is a rational number.
Every real number can either be a rational number or an irrational number.
View full question & answer→MCQ 101 Mark
If $\sqrt2=1.414,$ then the value of $\sqrt6-\sqrt3$ upto three place of decimal is:
- A
$0.235$
- ✓
$0.707$
- C
$1.414$
- D
$0.471$
AnswerCorrect option: B. $0.707$
$\sqrt6-\sqrt3$
$=\sqrt3\big(\sqrt2-1\big)$
Now, $\sqrt3=1.732$
$\sqrt2=1.414$
$\therefore\sqrt6-\sqrt3=1.732(1.414-1)\\ \ =1.732(0.414)=0.717$ (upto 3 decimal places)
Hence, correct option is $(b).$
View full question & answer→MCQ 111 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 121 Mark
The value of $\{5(8^{\frac{1}{4}}+27^{\frac{1}{3}})^3\}^{\frac{1}{4}}$ is:
Answer$\{5(8^{\frac{1}{4}}+27^{\frac{1}{3}})^3\}^{\frac{1}{4}}$
$\Rightarrow\{5(2^{3\times\frac{1}{3}}+3^{3\times\frac{1}{3}})^3\}^{\frac{1}{4}}$
$\Rightarrow\{5(3+2)^3\}^{\frac{1}{4}}$
$\Rightarrow\{5\times5^3\}^{\frac{1}{4}}$
$\Rightarrow5^{4\times\frac{1}{4}}$
$\Rightarrow5$
View full question & answer→MCQ 131 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 141 Mark
$\sqrt{10}\times\sqrt{15}$ is equal to:
- ✓
$5\sqrt6$
- B
$6\sqrt5$
- C
$\sqrt{30}$
- D
$\sqrt{25}$
AnswerCorrect option: A. $5\sqrt6$
$10 = 5 × 2$
$15 = 5 × 3$
$\therefore\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt3$
$=\big(\sqrt5\times\sqrt5\big)\times\sqrt2\times\sqrt3$
$=5\sqrt6$
Hence, correct option is $(a).$
View full question & answer→MCQ 151 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 161 Mark
The difference of two irrational numbers is.
- A
- B
- C
- ✓
Either irrational or rational.
AnswerCorrect option: D. Either irrational or rational.
Difference of two irrationals need not be an irrational.
Example:- each one of $(5\div\sqrt{2})$ and $(5-\sqrt{2})$ is irrational,
But, $(5+\sqrt{2})-(3+\sqrt{2})=2,$ which is rational.
View full question & answer→MCQ 171 Mark
The value of $\Big(\frac{\text{x}^{\text{l}}}{\text{x}^{\text{m}}}\Big)^{\frac{1}{\text{lm}}}\times\Big(\frac{\text{x}^{\text{m}}}{\text{x}^{\text{n}}}\Big)^{\frac{1}{\text{mn}}}\times\Big(\frac{\text{x}^{\text{n}}}{\text{x}^{\text{l}}}\Big)^{\frac{1}{\text{nl}}}$ is:
Answer$\Big(\frac{\text{x}^{\text{l}}}{\text{x}^{\text{m}}}\Big)^{\frac{1}{\text{lm}}}\times\Big(\frac{\text{x}^{\text{m}}}{\text{x}^{\text{n}}}\Big)^{\frac{1}{\text{mn}}}\times\Big(\frac{\text{x}^{\text{n}}}{\text{x}^{\text{l}}}\Big)^{\frac{1}{\text{nl}}}$
$\Rightarrow\text{x}^{\text{a}^{2}-\text{b}^2}\times\text{x}^{\text{b}^2-\text{c}^2}\times\text{x}^{\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^0=1$
View full question & answer→MCQ 181 Mark
The decimal representation of an irrational number is:
- ✓
Neither terminating nor repeating.
- B
- C
- D
Either terminating or non-repeating.
AnswerCorrect option: A. Neither terminating nor repeating.
An irrational number can not be written in the form of $\frac{\text{p}}{\text{q}}.$
And decimal representation of it is neither terminating nor repeating.
View full question & answer→MCQ 191 Mark
Between any two rational numbers there.
- A
- B
- C
Are exactly two rational numbers.
- ✓
Are many rational numbers.
AnswerCorrect option: D. Are many rational numbers.
Between any two rational number there are many rational number,
Example:- $4$ and $8$
We have $5, 6, 7, 7.5$, and many more.
View full question & answer→MCQ 201 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 211 Mark
If $\text{x}=7+4\sqrt3$ and $xy = 1,$ then $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=$
- A
$64$
- B
$134$
- ✓
$194$
- D
$\frac{1}{49}$
Answer$\text{x}=7+4\sqrt3,\ \text{xy}=1\Rightarrow\text{y}=\frac{1}{\text{x}}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
Now, $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=\frac{\text{y}^2+\text{x}^2}{\text{x}^2\text{y}^2}=\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}$
$\text{x}^2=\big(7+4\sqrt3\big)^2=49+48+56\sqrt3=97+56\sqrt3$
$\text{y}^2=\big(7-4\sqrt3\big)^2=49+48-56\sqrt3=97-56\sqrt3$
$\therefore\text{x}^2+\text{y}^2=97+56\sqrt3+97-56\sqrt3=194$
$\text{xy} = 1$
$\therefore\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}=\frac{194}{(1)^2}=194$
Hence, correct option is $(c).$
View full question & answer→MCQ 221 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 231 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 241 Mark
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then:
- ✓
$a = 2, b = 1$
- B
$a = 2, b = -1$
- C
$a = -2, b = 1$
- D
$a = b = 1$
AnswerCorrect option: A. $a = 2, b = 1$
$\frac{\sqrt3-1}{\sqrt3+1}$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get $a = 2$ and $b = 1.$
Hence, correct option is $(a).$
View full question & answer→MCQ 251 Mark
The simplified form of $16^{\frac{-1}{4}}\times\sqrt[4]{16}$ is:
Answer$16^{\frac{-1}{4}}\times\sqrt[4]{16}$
But, $16=2^4$
So,
$\Rightarrow16^{\frac{-1}{4}}\times\sqrt[4]{16}$
$\Rightarrow\{(2)^4\}^{\frac{-1}{4}}\times(2)^{4\times\frac{1}{4}}$
$\Rightarrow(2)^{4\times\frac{-1}{2}}\times2$
$\Rightarrow2^{-1}\times2$
$\Rightarrow\frac{2}{2}$
$\Rightarrow1$
View full question & answer→MCQ 261 Mark
The decimal representation of a rational number is:
AnswerCorrect option: D. Either terminating or repeating.
Rational numbers can be represented in decimal forms rather than representing infractions.
They can easily be represented as decimals by just dividing numerator $'p'$ by denominator $'q' \Big($as rational numbers is in the form of $\frac{\text{p}}{\text{q}}\Big)$
A rational number can be expressed as a terminating or no terminating, recurring decimal.
For example:
$\frac{5}{2}=2.5,$
$\frac{2}{8}=0.25,$
$7 = 7.0,$ etc., are rational numbers which are terminating decimals.
$\frac{5}{9}=0.555555555.......=0.5,$
$\frac{1}{6}=0.166666.....=0.16,$
$\frac{9}{11}=0.818181$
View full question & answer→MCQ 271 Mark
An irrational number between $2$ and $2.5$ is:
- A
$\sqrt{11}$
- ✓
$\sqrt{5}$
- C
$\sqrt{22.5}$
- D
$\sqrt{12.5}$
AnswerCorrect option: B. $\sqrt{5}$
$\sqrt{4}=2$ and $\sqrt{6.25}=2.5$
Option $(a), (c)$ and $(d): \sqrt{11},\sqrt{22.5}$ and $\sqrt{12.5},$ all are greater than $\sqrt{6.25}$
$⇒$ Out of interval $(2, 2.5)$
Option $(b): \sqrt{4}<\sqrt{5}<\sqrt{6.25}$
$\Rightarrow$ lies in the interval $(2, 2.5)$
Hence, option $(b)$ is correct.
View full question & answer→MCQ 281 Mark
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called:
AnswerAn irrational number cannot be written in the form of $\frac{\text{p}}{\text{q}}.$
Irrational number can neither be expressed as terminating decimal nor as repeating decimal.
View full question & answer→MCQ 291 Mark
If $8^{x+1} = 64$, what is the value of $3^{2x+1}?$
AnswerWe have to find the value of $3^{2 x+1}$ provided $8^{x+1}=64$
So,
$ 2^{3(x+1)}=64$
$ 2^{3 x+3}=2^6$
Equating the exponents we get
$ 3 x+3=6$
$ 3 x=6-3$
$ 3 x=3$
$x=\frac{3}{3}$
$ x=1$
By substitute in $3^{2 \mathrm{x}+1}$ we get
$ =3^{2 \times 1+1} $
$ =3^{2+1} $
$ =3^3 $
$ =27$
The real value of $3^{2 x+1}$ is $27$
Hence the correct choice is $d.$
View full question & answer→MCQ 301 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: $( 2+\sqrt2)^2=6+4\sqrt2$
Reason: $ (\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$
- ✓
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
- B
Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→MCQ 311 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 321 Mark
After simplification, $\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$ is:
- ✓
$13^{-\frac{2}{15}}$
- B
$13^{\frac{1}{3}}$
- C
$13^{\frac{2}{15}}$
- D
$13^{\frac{8}{15}}$
AnswerCorrect option: A. $13^{-\frac{2}{15}}$
$\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$
$=13^{\frac{1}{5}+\frac{1}{3}}$
$=13^{-\frac{2}{15}}$
View full question & answer→MCQ 331 Mark
If $\left(3^3\right)^2=9^x$ than $5^x= ?$
Answer$ \left(3^3\right)^2=9^x$
$ \left(3^3\right)^3=9^x$
$ 9^3=9^x $
$ \Rightarrow x=3$
$\therefore 5^3=125$
View full question & answer→MCQ 341 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
$10\sqrt{5}$
AnswerCorrect 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 351 Mark
If $\text{x}=\sqrt6+\sqrt5,$ then $\text{x}^2+\frac{1}{\text{x}^2}-2=$
- A
$2\sqrt6$
- B
$2\sqrt5$
- C
$24$
- ✓
$20$
Answer$\text{x}^2+\frac{1}{\text{x}^2}-2=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2$
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}\\ \ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
Now,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\big[\sqrt6+\sqrt5-\big(\sqrt6-\sqrt5\big]^2\\ \ =\big(2\sqrt5\big)^2=4\times5=20$
Hence, correct option is $(d).$
View full question & answer→MCQ 361 Mark
$16\sqrt{134}\div9\sqrt{52}$ is equal to:
- A
$\frac{3}{9}$
- B
$\frac{9}{8}$
- ✓
$\frac{8}{9}$
- D
AnswerCorrect option: C. $\frac{8}{9}$
$16\sqrt{134}\div9\sqrt{52}$
$\frac{16\sqrt{13}}{9\sqrt{52}}=\frac{16}{9}\times\sqrt{\frac{13}{52}}=\frac{16}{9}\times\frac{1}{2}$
$=\frac{8}{9}$
View full question & answer→MCQ 371 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 381 Mark
The simplest rationalising factor of $2\sqrt5-\sqrt3,$ is:
- A
$2\sqrt5+3$
- ✓
$2\sqrt5+\sqrt3$
- C
$\sqrt{5}+\sqrt3$
- D
$\sqrt{5}-\sqrt3$
AnswerCorrect option: B. $2\sqrt5+\sqrt3$
Rationalising factor of any number of kind $\text{a}\sqrt{\text{a}}\pm\sqrt{\text{b}}$ is $\sqrt{\text{a}}\mp\sqrt{\text{b}}$
So. for given number $2\sqrt5-\sqrt3.$ Rationalising factor would be $2\sqrt5+\sqrt3.$
Hence, correct option is $(b).$
View full question & answer→MCQ 391 Mark
The value of $\sqrt[3]{1000}$ is:
Answer$(10)^3=1000$
So, $\sqrt[3]{1000}=1000^{\frac{1}{3}}=(10^3)^{\frac{1}{3}}$
View full question & answer→MCQ 401 Mark
The value of $\sqrt[3]{1000}$ is:
Answer$(10)^3=1000$
So, $\sqrt[3]{1000}=1000^{\frac{1}{3}}=(10^3)^{\frac{1}{3}}$
$=10$
View full question & answer→MCQ 411 Mark
$(256)^{0.16} \times(256)^{0.09}$
AnswerWe have to find the value of $(256)^{0.16} \times(256)^{0.09}$. So,
By using law of rational exponents
$ a^m \times a^n=a^{m+n} \text { we get } $
$ (265)^{0.16} \times(256)^{0.09}=(256)^{0.16} \times(256)^{0.09} $
$ =(256)^{0.16+0.09} $
$ =256^{0.25} $
$=256^{\frac{25}{100}}$
$(256)^{0.16}\times(256)^{0.09 }=2^{8\times\frac{25}{100}}$
$=2^{8\times\frac{25}{100}}$
$=2^{8\times\frac{1}{4}}$
$=2^{8\times\frac{1}{4}}=4$
The value of $(256)^{0.16} \times(256)^{0.09}$ is $4$
Hence the correct choice is a.
View full question & answer→MCQ 421 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 431 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 441 Mark
The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is:
- A
$(\sqrt{3}+\sqrt{5})$
- ✓
$\sqrt{12}+\sqrt{5}$
- C
$\sqrt{5}-2\sqrt{3}$
- D
$\sqrt{3}+2\sqrt{5}$
AnswerCorrect option: B. $\sqrt{12}+\sqrt{5}$
$\frac{1}{2\sqrt{3}-\sqrt{5}}$
$=(2\sqrt{3}-\sqrt{5})(2\sqrt{3}+\sqrt{5})$
$=12-5$
$=7$
Rational number
$(2\sqrt{3}+\sqrt{5})=(\sqrt{4\times3}+\sqrt{5})=\sqrt{12}+\sqrt{5}$
View full question & answer→MCQ 451 Mark
If $\sqrt{7}=2.646$ then $\frac{1}{\sqrt{7}}=?$
- A
- ✓
$0.378$
- C
$0.441$
- D
$0.375$
AnswerCorrect option: B. $0.378$
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$
$=\frac{\sqrt{7}}{7}$
$=\frac{1}{7}\times\sqrt{7}$
$=\frac{1}{7}\times2.646$
$=0.378$
View full question & answer→MCQ 461 Mark
The value of $1.9999......$ in the from $\frac{\text{p}}{\text{q}},$ where $'p'$ and $'q'$ are integers and $\text{q}\not=0,$ is:
- A
${\frac{1}{9}}$
- B
${\frac{19999}{1000}}$
- C
${\frac{19}{10}}$
- ✓
$2$
Answer$1.9999$ can be written as $2,$
$2$ is taken as approx vlaue.
View full question & answer→MCQ 471 Mark
The value of $64^{-\frac{1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)$ is:
AnswerFind the value of $64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)$
So,
$\Rightarrow64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)=2^{6\times\frac{1}{3}}\Big(2^{6\times\frac{1}{3}}-2^{6\times\frac{2}{3}}\Big)$
$=2^{-2}(2^2-2^4)$
$=2^2(4-16)$
$\Rightarrow64^{\frac {1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)=\frac{1}{2^2}\times-12$
$=\frac{1}{4}\times-12$
$=-3$
Hence the correct statement is $c.$
View full question & answer→MCQ 481 Mark
$0.83458456……………$ is:
AnswerSome numbers cannot be written as a ratio of two integers they are called Irrational Numbers.
It is irrational because it cannot be written as a ratio (or fraction),
View full question & answer→MCQ 491 Mark
Write the correct answer in the following: The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to
Answer$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$
$=\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$
$=\frac{4}{2}=2$
Hence, $(b)$ is correct answer.
View full question & answer→MCQ 501 Mark
Which of the following is an irrational number.
- A
$0.3799$
- B
$7.\overline{478}$
- ✓
$\sqrt{23}$
- D
$\sqrt{225}$
AnswerCorrect option: C. $\sqrt{23}$
It doesn't represent the form of $\frac{\text{p}}{\text{q}}$ and $\text{q}\not=0$
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