Question 11 Mark
Simplify: $7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } }$
Answer$7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } }$ We know that $a ^ { m } \cdot b ^ { m } = ( a \times b ) ^ { m }$.
So, $7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } } = ( 7 \times 8 ) ^ { \frac { 1 } { 2 } }$
$= ( 56 ) ^ { \frac { 1 } { 2 } }$
Therefore, the value of $7 ^ { \frac { 1 } { 2 } } \cdot 8 ^ { \frac { 1 } { 2 } }$will be $( 56 ) ^ { \frac { 1 } { 2 } }$.
View full question & answer→Question 21 Mark
Simplify: $\frac{11^{1 / 2}}{11^{1 / 4}}$
Answer$\quad \frac{11^{1 / 2}}{11^{/ 4}}=11^{1 / 2-1 / 4}=11^{1 / 4}$
View full question & answer→Question 31 Mark
Simplify: $\left(\frac{1}{3^{3}}\right)^{7}$
Answer$\left(\frac{1}{3^{3}}\right)^{7}=\frac{1^{7}}{\left(3^{3}\right)^{7}}$$=\frac{1}{3^{21}}=3^{-21}$
View full question & answer→Question 41 Mark
Simplify : $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
AnswerWe have. $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}$ {by $a^{m} \cdot a^{n}=a^{(m+n)}$}
$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\left(\frac{2 \times 5+1 \times 3}{15}\right)}$$=2^{\frac{13}{15}}$
View full question & answer→Question 51 Mark
Find: $125^{-1 / 3}$
Answer$125^{-1 / 3}=\left(5^3\right)^{-1 / 3}=5^3 \times{ }^{(-1 / 3)}=5^{-1}=\frac{1}{5}$
View full question & answer→Question 61 Mark
Find: $16^{3 / 4}$
Answer$\quad 16^{3 / 4}=\left(2^4\right)^{3 / 4}=2^4 \times 3 / 4=2^3=8$
View full question & answer→Question 71 Mark
Find: $32^{2 / 5}$
Answer$32^{2 / 5}=\left(2^5\right)^{2 / 5}=2^5 \times 2 / 5=2^2=4$.
View full question & answer→Question 81 Mark
Find: $9^{3 / 2}$
Answer$\quad 9^{3 / 2}=\left(9^{1 / 2}\right)^3=3^3=27$
View full question & answer→Question 91 Mark
Find: $125^{1 / 3}$
Answer$125^{1 / 3}=\left(5^3\right)^{1 / 3}$
$=5^3 \times{ }^{1 / 3}=5^1=5$
View full question & answer→Question 101 Mark
Find: $32^{1 / 5}$
Answer$32^{1 / 5}=\left(2^5\right)^{1 / 5}$
$=2^{5 \times 1 / 5}=2^1=2$
View full question & answer→Question 111 Mark
Find:${64^{\frac{1}{5}}}$
Answer${64^{\frac{1}{5}}}$
We know that ${a^{\frac{1}{n}}} = \sqrt[n]{a},{\text{ where }}a > 0.$
We conclude that ${64^{\frac{1}{2}}}$ can also be written as $\sqrt[2]{{64}} = \sqrt[2]{{8 \times 8}}=\sqrt[2]{{64}} = \sqrt[2]{{8 \times 8}} = 8.$
Therefore, the value of ${64^{\frac{1}{2}}}$ will be 8.
View full question & answer→Question 121 Mark
Rationalize the denominators of $\frac{1}{{\sqrt 7 - 2}}$
Answer$\eqalign{ & {1 \over {\sqrt 7 - 2}} \cr & = {1 \over {\sqrt 7 - 2}} \times {{\sqrt 7 + 2} \over {\sqrt 7 + 2}} \cr & = {{\sqrt 7 + 2} \over {{{(\sqrt 7 )}^2} - {{(2)}^2}}} = {{\sqrt 7 + 2} \over {7 - 4}} \cr & = {{\sqrt 7 + 2} \over 3} \cr}$
View full question & answer→Question 131 Mark
Rationalize the denominator of $\frac{1}{{\sqrt 5 + \sqrt 2 }}$
Answer$\frac{1}{{\sqrt 5 + \sqrt 2 }}$
multiply denominator and numerator by $\frac{1}{{\sqrt 5 - \sqrt 2 }}$
$=\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
$=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}$
$=\frac{\sqrt{5}-\sqrt{2}}{3}$
View full question & answer→Question 141 Mark
Rationalize the denominator of $\frac{1}{{\sqrt 7 }}$
Answer$\frac{1}{{\sqrt 7 }}$
We need to multiply the numerator and denominator $\frac{1}{{\sqrt 7 }}$ by $\sqrt 7,to get \frac{1}{{\sqrt 7 }} \times \frac{{\sqrt 7 }}{{\sqrt 7 }} = \frac{{\sqrt 7 }}{7}$
Therefore, we conclude that on rationalizing the denominator $\frac{1}{{\sqrt 7 }}$ we get $\frac{{\sqrt 7 }}{7}$
View full question & answer→Question 151 Mark
Recall, $\pi$ is defined as the ratio of the circumference $($say $c)$ of a circle to its diameter $($say $d).$ That is, $\pi = \frac{c}{d}$ This seems to contradict the fact that is irrational. How will you resolve this contradiction$?$
AnswerWe know that when we measure the length of a line or a figure by using a scale or any device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either circumference $(c)$ or diameter$(d)$ of a circle is irrational. Therefore, we can conclude that as such there is not any contradiction regarding the value $\pi$ of and we realize that the value of $\pi$ is irrational.
View full question & answer→Question 161 Mark
Classify the number as rational or irrational: $2\pi$
Answer$\because 2$ is a rational number and $\pi$ is an irrational number.
$(\because$ The product of a non-zero rational number with an irrational number is irrational$)$
$\therefore 2 \pi$ is an irrational number.
View full question & answer→Question 171 Mark
Classify the number as rational or irrational: $\frac{1}{\sqrt{2}}$
Answer$\because 1(\neq 0)$ is a rational number and$\sqrt{2}( \neq 0)$ is an irrational number.
($\because$ The quotient of a non-zero rational number with an irrational number is irrational.)
$\therefore$ $\frac{1}{\sqrt{2}}$ is an irrational number.
View full question & answer→Question 181 Mark
Classify the number as rational or irrational: $\frac{2 \sqrt{7}}{7 \sqrt{7}} $
Answer$\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}$ is a rational number
View full question & answer→Question 191 Mark
Classify the number as rational or irrational: $(3+\sqrt{23})-\sqrt{23}$
Answer$(3+\sqrt{23})-\sqrt{23}$= $3+\sqrt{23}-\sqrt{23} = 3$ is a rational number.
View full question & answer→Question 201 Mark
Classify the number as rational or irrational: $2 - \sqrt {5}$
Answer$2 - \sqrt {5}$
$\because 2 $ is a rational number and$\sqrt {5}$ is an irrational number.
$(\because$ The difference of a rational number and an irrational number is irrational.$)$
$\therefore 2 - \sqrt {5}$ is an irrational number.
View full question & answer→Question 211 Mark
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer$0.01001\ 0001\ 00001 . . . .,$
$0.20\ 2002\ 20003\ 200002 . . . .,$
$0.003000300003 . . . .,$
View full question & answer→Question 221 Mark
Express $0. \overline { 001 }$ in the form $\frac { p } { q }$, where $p$ and $q$ are integers and $q \neq 0 $
AnswerLet $x = 0. \overline { 001 } = 0.001001001 . . . .$
Multiplying both sides by $1000 ($since three digits are repeating$),$ we get
$1000x = 1.001001 . . . .$
$⇒ 1000x = 1+ 0.001001001 . . . .$
$⇒ 1000x = 1 + x$
$⇒ 1000x – x = 1$
$⇒ 999x = 1$
$⇒ x = \frac { 1 } { 999 }$
Thus $0 . \overline { 001 } = \frac { 1 } { 999 }$
Here $p = 1$
$q =999\left( { \ne 0} \right)$
View full question & answer→Question 231 Mark
Express $ 0.4 \overline{7}$ in the form $\frac {p} {q}$, where $p$ and $q$ are integers and $q \neq 0 \;$
AnswerLet $x = 0.4 \overline{7} = 0.47777 ...$
Multiplying both sides by $10 ($since one digit is repeating$),$ we get
$10x = 4.7777 ...$
$⇒ 10x = 4.3 + 0.47777 ...$
$⇒ 10x = 4.3 + x$
$⇒ 10x - x = 4.3$
$⇒ 9x = 4.3$
$\Rightarrow x=\frac{4.3}{9}=\frac{43}{90}$
Thus, $4 . \overline{7}$ = $\frac{43}{90}$
Here$ p = 43$
$q = 90(\neq 0)$
View full question & answer→Question 241 Mark
Express $0 . \overline{6}$ in the form $\frac {P} {q}$, where $p$ and $q$ are integers and $q \neq 0.$
AnswerLet $x = 0 . \overline{6} = 0.6666 ...$
Multiplying both sides by $10 ($since one digit is repeating$),$ we get
$10x = 0.666 ...$
$\Rightarrow 10x = 6 + 0.6666$
$\Rightarrow 10x = 6 + x$
$\Rightarrow 10x x = 6$
$\Rightarrow 9x = 6$
$\Rightarrow x = \frac{6}{9}$
Thus, $0 . \overline{6} =\frac{2}{3}$
Here $p = 2$
$q = 3(\neq 0)$
View full question & answer→Question 251 Mark
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
AnswerWe know that square root of every positive integer will not yield an integer.We know that $\sqrt 4 $ is $2,$ which is an integer. But, $\sqrt 7 {\text{ or }}\sqrt {10} $ will give an irrational number. Therefore, we conclude that square root of every positive integer is not an irrational number.
View full question & answer→Question 261 Mark
Show that $3.142678$ is a rational number. In other words, express $3.142678$ in the form $\frac{p}{q}$ , where $p$ and $q$ are integers and $q \ne 0.$
AnswerWe can write given number as, $3.142678 =\frac{3142678}{1000000}$, and hence is a rational number.
Now, let us consider the case when the decimal expansion is non-terminating recurring.
View full question & answer→Question 271 Mark
Simplify : $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}$
AnswerWe have, $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}=(13 \times 17)^{\frac{1}{5}}=221^{\frac{1}{5}}$
View full question & answer→Question 281 Mark
Simplify : $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}$
AnswerWe have, $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}=7^{\left(\frac{1}{5}-\frac{1}{3}\right)}=7^{\frac{3-5}{15}}=7^{\frac{-2}{15}}$
View full question & answer→Question 291 Mark
Simplify : $\left(3^{\frac{1}{5}}\right)^{4}$
Answerwe have $\left(3^{\frac{1}{5}}\right)^{4}=3^{\frac{4}{5}}$
View full question & answer→Question 301 Mark
Rationalise the denominator of $\frac{1}{7+3 \sqrt{2}}$
Answerwe have.. $\frac{1}{7+3 \sqrt{2}}$ = $\frac{1}{7+3 \sqrt{2}} \times\left(\frac{7-3 \sqrt{2}}{7-3 \sqrt{2}}\right)=\frac{7-3 \sqrt{2}}{49-18}=\frac{7-3 \sqrt{2}}{31}$
View full question & answer→Question 311 Mark
Find five rational numbers between $1$ and $2.$
AnswerRecall that to find a rational number between $r$ and $s,$ you can add $r$ and $s$ and divide the sum by $2,$ that is $\frac{r+s}{2}$ lies between $r$ and $s.$
So, $\frac{3}{2}$ is a number between $1$ and $2.$
You can proceed in this manner to find four more rational numbers between $1$ and $2.$
These four numbers are $\frac{5}{4}$, $\frac{11}{8}$, $\frac{13}{8}$ and $\frac{7}{4}$.
Thus five rational numbers between $1$ and $2$ are $\frac{5}{4}$, $\frac{11}{8}$, $\frac{3}{2}$, $\frac{13}{8}$ and $\frac{7}{4}$.
View full question & answer→Question 321 Mark
Rationalise the denominator of $\frac{1}{2+\sqrt{3}}$
Answer$\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} \text { (Rationalise the denominator) }$
$=\frac{2-\sqrt{3}}{4-3}\left[\text { using } \mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]$
$=2-\sqrt{3}$
View full question & answer→Question 331 Mark
Rationalise the denominator of $\frac{1}{\sqrt{2}}$
Answer$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} ($Rationalize the denominator$)=\frac{\sqrt{2}}{2} ($since $\sqrt{2} \cdot \sqrt{2} = 2)$
In this form, it is easy to locate $\frac{1}{\sqrt{2}}$ on the number line. It is halfway between $0$ and $\sqrt{2}$
View full question & answer→Question 341 Mark
Simplify the expression: $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$
Answerwe have.$(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7}) = (\sqrt{11})^{2}-(\sqrt{7})^{2} = 11 - 7 = 4$
View full question & answer→Question 351 Mark
Divide: $8 \sqrt{15} $ by $ 2 \sqrt{3}$
Answer$8 \sqrt{15} \div 2 \sqrt{3}=\frac{8}{2} \sqrt{\frac{15}{3}}=4 \sqrt{5}$.
View full question & answer→Question 361 Mark
Multiply $6 \sqrt{5} \text { by } 2 \sqrt{5}$
Answerwe have..$6 \sqrt{5} \times 2 \sqrt{5}=6 \times 2 \times \sqrt{5} \times \sqrt{5}=12 \times 5=60$
View full question & answer→Question 371 Mark
Add : $2 \sqrt{2}+5 \sqrt{3}$ and $\sqrt{2}-3 \sqrt{3}$
Answerwe have.$(2 \sqrt{2}+5 \sqrt{3})+(\sqrt{2}-3 \sqrt{3})=(2 \sqrt{2}+\sqrt{2})+(5 \sqrt{3}-3 \sqrt{3})$ = $(2+1) \sqrt{2}+(5-3) \sqrt{3}=3 \sqrt{2}+2 \sqrt{3}$
View full question & answer→Question 381 Mark
Simplify: $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }$
AnswerGiven, $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }$ We know that $a ^ { m } \cdot a ^ { n } = a ^ { ( m +n) }$.
So, $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } } = ( 2 ) ^ { \frac { 2 } { 3 } + \frac { 1 } { 3 } }$
$= ( 2 ) ^ { \frac { 2+1} { 3} }\\ = ( 2 ) ^ { \frac { 3 } { 3 } }\\=(2)^1\\=2$
Therefore, the value of $2 ^ { \frac { 2 } { 3 } } \cdot 2 ^ { \frac { 1 } { 3 } }$ will be $2$.
View full question & answer→Question 391 Mark
Simplify the following expressions::
$(\sqrt{3}+\sqrt{7})^2$
Answer$(\sqrt{3}+\sqrt{7})^2=(\sqrt{3})^2+2 \sqrt{3} \sqrt{7}+(\sqrt{7})^2=3+2 \sqrt{21}+7=10+2 \sqrt{21}$
View full question & answer→Question 401 Mark
Simplify the following expressions::
$(5+\sqrt{5})(5+\sqrt{5})$
Answer$(5+\sqrt{5})(5-\sqrt{5})=5^2-(\sqrt{5})^2=25-5=20$
View full question & answer→Question 411 Mark
Simplify the following expressions::
$(5+\sqrt{7})(2+\sqrt{5})$
Answer$(5+\sqrt{7})(2+\sqrt{5})=10+5 \sqrt{5}+2 \sqrt{7}+\sqrt{35}$
View full question & answer→