Question 11 Mark
Simplify the following: $4\sqrt{28}\div3\sqrt{7}\div\sqrt[3]{7}$
Answer$4\sqrt{28}\div3\sqrt{7}\div\sqrt[3]{7}$
$=4\sqrt{2\times2\times7}\times\frac{1}{3\sqrt{7}}\div\sqrt[3]{7}$
$=\frac{8\sqrt{7}}{3\sqrt{7}}\div\sqrt[3]{7}=\frac{8}{3}\div\sqrt[3]{7}$
We know that, the cube root of $7$ is $1.9129$ and
$\therefore\frac{8}{3}\div\sqrt[3]{7}=2.666\div1.9129$
$\Rightarrow\ 1.3936.$
View full question & answer→Question 21 Mark
Classify the following numbers as rational or irrational with justification: $-\sqrt{0.4}$
Answer$-\sqrt{0.4}=-\sqrt{\frac{4}{10}}=-\sqrt{\frac{2}{10}}$ Hence, it is a quotient of rational and irrational numbers, so it is an irrational number.
View full question & answer→Question 31 Mark
Find three rational numbers between: $0.1$ and $0.11$
Answer$0.101, 0.102, 0.103 ($terminating decimals$)$ are three rational numbers which lie between $0.1$ and $0.11.$
View full question & answer→Question 41 Mark
State whether the following statements are true or false$?$ Justify your answer. $\frac{\sqrt{2}}{3}$ is a rational number.
AnswerHere $\sqrt{2}$ is an irrational number and $3$ is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number.
View full question & answer→Question 51 Mark
Classify the following numbers as rational or irrational with justification: $\frac{\sqrt{12}}{\sqrt{75}}$
Answer$\frac{\sqrt{12}}{\sqrt{75}}=\frac{\sqrt{4\times3}}{\sqrt{25\times3}}=\frac{\sqrt{4}\sqrt{3}}{\sqrt{25}\sqrt{3}}=\frac{2}{5}$ Hence, it is a rational numbers.
View full question & answer→Question 61 Mark
Insert a rational number and an irrational number between the following:
$3.623623$ and $0.484848$
AnswerA rational number between $3.623623 $ and $0.484848$ is $1.$
An irrational number between $3.623623$ and $0.484848$ is $1.909009000........$
View full question & answer→Question 71 Mark
Classify the following numbers as rational or irrational with justification:
$0.5918$
Answer$0.5918,$ it is a number with terminating decimal, so it can be written in the form of $\frac{\text{p}}{\text{q}},$ where $\text{q}\neq0, p$ and $q$ are integer.Hence, it is a rational numbers.
View full question & answer→Question 81 Mark
Insert a rational number and an irrational number between the following: $\sqrt{2}$ and $\sqrt{3}$
AnswerA rational number between $\sqrt{2}$ and $\sqrt{3}$ i.e., between $1.414.....$ and $1.7320……$ is $1.5.$ An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is $1.585585558.........$
View full question & answer→Question 91 Mark
Find which of the variables $x, y, z$ and $u$ represent rational numbers and which irrational numbers:
$\text{u}^2=\frac{17}{4}$
Answer$\text{u}^2=\frac{17}{4}\Rightarrow\text{u}=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2},$ which is of the form $\frac{\text{p}}{\text{q}}$ where $\text{p}=\sqrt{17}$ is not an integer. Hence, $u$ is an irrational number.
View full question & answer→Question 101 Mark
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414 ,$ $\sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.
$\frac{4}{\sqrt{3}}$
Answer $\frac{4}{\sqrt{3}}=\frac{4}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$
$=\frac{4\times1.732}{3}=\frac{6.928}{3}=2.309$
View full question & answer→Question 111 Mark
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414 ,$ $\sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal. $\frac{1}{\sqrt{3}+\sqrt{2}}$
Answer$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{\sqrt{3}-\sqrt{2}}{3-2}$ $=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$ $=1.732-1.414=0.318$
View full question & answer→Question 121 Mark
State whether the following statements are true or false$?$ Justify your answer. There are infinitely many integers between any two integers.
AnswerFalse. Solution: Because between two consecutive integers $($likel and $2),$ there does not exist any other integer.
View full question & answer→Question 131 Mark
Classify the following numbers as rational or irrational with justification:
$\sqrt{196}$
Answer $\sqrt{196}=\sqrt{(14)^2}=14$
Hence, it is a rational number.
View full question & answer→Question 141 Mark
Insert a rational number and an irrational number between the following:
$.0001$ and $.001$
AnswerA rational number between $0.0001$ and $0.001$ is $0.00011.$
An irrational number between $0.0001$ and $0.001$ is $0.0001131331333...........$
View full question & answer→Question 151 Mark
Classify the following numbers as rational or irrational with justification: $1.010010001...$
Answer$1.010010001...$ is a number with non-terminating non-recurring decimal expansion. Hence, it is a irrational numbers.
View full question & answer→Question 161 Mark
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414 ,$ $\sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal. $\frac{\sqrt{10}-\sqrt{5}}{2}$
Answer$\frac{\sqrt{10}-\sqrt{5}}{2}=\frac{\sqrt{2}\times\sqrt{5}-\sqrt{5}}{2}$ $=\frac{\sqrt{5}(\sqrt{2}-1)}{2}=\frac{2.236(1.414-1)}{2}$ $=1.118\times0.414=0.463$
View full question & answer→Question 171 Mark
Find which of the variables $x, y, z$ and $u$ represent rational numbers and which irrational numbers:
$\text{x}^2=5$
Answer$\text{x}^2=5\Rightarrow\text{x}=\sqrt{5},$ which is an irrational number.
View full question & answer→Question 181 Mark
Insert a rational number and an irrational number between the following: $2$ and $3$
AnswerA rational number between $2$ and $3$ is $2.1.$ To find an irrational number between $2$ and $3.$ Find a number which is non-terminating non-recurring lying between them. Such number will be $2.040040004.......$
View full question & answer→Question 191 Mark
Find three rational numbers between: $\frac{5}{7}$ and $\frac{6}{7}$
Answer$\frac{5}{7}=\frac{5}{7}\times\frac{10}{10}=\frac{50}{70}$and $\frac{6}{7}=\frac{6}{7}\times\frac{10}{10}=\frac{60}{70}$ $\Rightarrow\frac{51}{70}, \frac{52}{70}, \frac{53}{70}$ are three rational numbers lying and between $\frac{50}{70}$ and $\frac{60}{70}$ and therefore lie between $\frac{5}{7}$ and $\frac{6}{7}.$
View full question & answer→Question 201 Mark
Find which of the variables $x, y, z$ and $u$ represent rational numbers and which irrational numbers:
$\text{y}^2=9$
Answer$\text{y}^2=5\Rightarrow\text{y}=\sqrt{9}=3,$ which is a rational number.
View full question & answer→Question 211 Mark
Classify the following numbers as rational or irrational with justification: $10.124124...$
Answer$10.124124...$ is a number with non-terminating decimal expansion. Hence, it is a rational numbers.
View full question & answer→Question 221 Mark
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414 ,$ $\sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.
$\frac{6}{\sqrt{6}}$
Answer $\frac{6}{\sqrt{6}}=\frac{6}{\sqrt{6}}\times\frac{\sqrt{6}}{\sqrt{6}}=\frac{6\sqrt{6}}{6}$
$=\sqrt{6}=\sqrt{2\times3}=\sqrt{2}\times\sqrt{3}$
$=1.414\times1.732=2.44909=2.448\text{ (approx.)}$
View full question & answer→Question 231 Mark
State whether the following statements are true or false? Justify your answer. $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.
AnswerFalse. Solution: $\frac{\sqrt{12}}{\sqrt{3}}=\frac{\sqrt{4\times3}}{\sqrt{3}}=\frac{\sqrt{4}\times\sqrt{3}}{\sqrt{3}}=2\times1=2,$ which is a rational number.
View full question & answer→Question 241 Mark
Find which of the variables $x, y, z$ and $u$ represent rational numbers and which irrational numbers:
$\text{z}^2=.04$
Answer$\text{z}^2=.04\Rightarrow\text{z}=\sqrt{.04}=0.2,$ which is a terminating decimal. Hence, it is rational number.
View full question & answer→Question 251 Mark
Classify the following numbers as rational or irrational with justification: $\sqrt{\frac{9}{27}}$
Answer$\sqrt{\frac{9}{27}}=\sqrt{\frac{9}{9\times3}}=\frac{1}{\sqrt{3}}$ Hence, it is an irrational number, because $\sqrt{3}$ is an irrational number.
View full question & answer→Question 261 Mark
Insert a rational number and an irrational number between the following: $6.375289$ and $6.375738$
AnswerA rational number between $6.375289$ and $6.375738$ is $6.3753.$ An irrational number between $6.375289$ and $6.375738$ is $6.375414114111.........$
View full question & answer→Question 271 Mark
State whether the following statements are true or false? Justify your answer. $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form $\frac{\text{p}}{\text{q}}, \text{ q}\neq0$ and so it is a rational number.
AnswerFalse. Solution: $\frac{\sqrt{15}}{\sqrt{3}}=\frac{\sqrt{5\times3}}{\sqrt{3}}=\frac{\sqrt{5}\times\sqrt{3}}{\sqrt{3}}=\sqrt{5},$ which is an irrational number.
View full question & answer→Question 281 Mark
Insert a rational number and an irrational number between the following: $\frac{1}{3}$ and $\frac{1}{2}$
AnswerA rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\frac{5}{12}.$ An irrational number between $\frac{1}{3}$ and $\frac{1}{2}$ i.e., between $0-3$ and $0.5$ is $0.4141141114.........$
View full question & answer→Question 291 Mark
Simplify: $(625)^{-\frac{1}{2}^{-\frac{1}{4}^{2}}}$
Answer$\Bigg[\Big((625)^{-\frac{1}{2}}\Big)^{-\frac{1}{4}}\Bigg]^{2}=(25^{-1})^{-\frac{1}{4}\times2}=[(5^2)^{-1}]^{-\frac{1}{4}\times2}$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$ $=5^{-2\times-\frac{1}{4}\times2}=5^1=5$
View full question & answer→Question 301 Mark
Find three rational numbers between: $-1$ and $-2$
Answer$-1.1, -1.2, -1.3 ($terminating decimals$)$ are three rational numbers lying between $-1$ and $-2.$
View full question & answer→Question 311 Mark
Simplify: $\big(1^3+2^3+3^3\big)^{\frac{1}{2}}$
Answer$\big(1^3+2^3+3^3\big)^{\frac{1}{2}}=(1+8+27)^{\frac{1}{2}}$ $=(36)^\frac{1}{2}=(6^2)^\frac{1}{2}=6^{2\times\frac{1}{2}}=6$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$
View full question & answer→Question 321 Mark
Classify the following numbers as rational or irrational with justification:$\Big(1+\sqrt{5}\Big)-\Big(4+\sqrt{5}\Big)$
Answer$\Big(1+\sqrt{5}\Big)-\Big(4+\sqrt{5}\Big)$ $=1-4+\sqrt{5}-\sqrt{5}=-3$ Hence, it is a rational numbers.
View full question & answer→Question 331 Mark
Find three rational numbers between: $\frac{1}{4}$ and $\frac{1}{5}$
Answer$\frac{1}{4}=\frac{1}{4}\times\frac{20}{20}=\frac{20}{80}$ and $\frac{1}{5}=\frac{1}{5}\times\frac{16}{16}=\frac{16}{80}$ Now, $\sqrt{2}\times\sqrt{3}\frac{18}{80}\Big(\frac{9}{40}\Big),\frac{19}{80}$ are three rational numbers lying between $\frac{1}{4}$ and $\frac{1}{5}.$
View full question & answer→Question 341 Mark
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414 ,$ $\sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal. $\frac{\sqrt{2}}{2+\sqrt{2}}$
Answer$\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}}$ $=\frac{\sqrt{2}(2-\sqrt{2})}{(2)^2-(\sqrt{2})^2}=\frac{\sqrt{2}(2-\sqrt{2})}{4-2}$ $=\frac{\sqrt{2}(2-\sqrt{2})}{2}=\frac{2\sqrt{2}-2}{2}$ $=\sqrt{2}-1=1.414-1=0.414$
View full question & answer→Question 351 Mark
Simplify the following: $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$
Answer$\sqrt{45}-3\sqrt{20}+4\sqrt{5}$ $=\sqrt{9\times5}-3\sqrt{4\times5}+4\sqrt{5}$ $=3\sqrt{5}-3\times2\sqrt{5}+4\sqrt{5}$ $=(3-6+4)\sqrt{5}=\sqrt{5}$
View full question & answer→Question 361 Mark
Insert a rational number and an irrational number between the following: $2.357$ and $3.121$
AnswerA rational number between $2.357$ and $3.121$ is $3.$ An irrational number between $2.357$ and $3.121$ is $3.101101110......$
View full question & answer→Question 371 Mark
State whether the following statements are true or false? Justify your answer. There are numbers which cannot be written in the form $\frac{\text{p}}{\text{q}},\text{ q}\neq0, \text{ p, q}$ both are integers.
AnswerTrue. Solution: Because there are infinitely many numbers which cannot be written in the form $\frac{\text{p}}{\text{q}},\text{ q}\neq0, \text{ p, q}$ both are integers and these numbers are called irrational numbers.
View full question & answer→Question 381 Mark
Insert a rational number and an irrational number between the following: $0$ and $0.1$
AnswerA rational number between $0$ and $0.1$ is $0.03.$ An irrational number between $0$ and $0.1$ is $0.007000700007.........$
View full question & answer→Question 391 Mark
Let $x$ and $y$ be rational and irrational numbers, respectively. Is $x + y$ necessarily an irrational number$?$ Give an example in support of your answer.
AnswerYes, $(x + y)$ is necessarily an irrational number. e.g., Let $\text{x}=2, \text{ y}=\sqrt{3}$ Then, $\text{x+y}=2+\sqrt{3}$
If possible, let $\text{x+y}=2+\sqrt{3}$ be a rational number.
Consider $\text{a}=2+\sqrt{3}$ On squaring both sides,
we get $\text{a}^2=(2+\sqrt{3})^2$
$ [\text{using identity (a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab}]$
$\Rightarrow\text{a}^2=2^2+(\sqrt{3})^2+2(2)(\sqrt{3})$
$\Rightarrow\text{a}^2=4+3+4\sqrt{3}\Rightarrow\frac{\text{a}^2-7}{4}=\sqrt{3}$
So, a is rational $\Rightarrow\frac{\text{a}^2-7}{4}$ is rational $\Rightarrow\sqrt{3}$ is rational.
View full question & answer→Question 401 Mark
Classify the following numbers as rational or irrational with justification: $\sqrt{\frac{\sqrt{28}}{\sqrt{343}}}$
Answer$\frac{\sqrt{28}}{\sqrt{343}}=\frac{\sqrt{2\times2\times7}}{\sqrt{7\times7\times7}}=\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$ Hence, it is a rational number.
View full question & answer→Question 411 Mark
Insert a rational number and an irrational number between the following: $\frac{-2}{5}$ and $\frac{1}{2}$
AnswerA rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ is $0.$ An irrational number between $\frac{-2}{5}$ and $\frac{1}{2}$ i.e., between $–0.4$ and $0.5$ is $0.151151115.........$
View full question & answer→Question 421 Mark
Simplify: $\frac{1}{27}^{\frac{-2}{3}}$
Answer$\Big(\frac{1}{27}\Big)^{\frac{-2}{3}}=\Big(\frac{1}{3^3}\Big)^{\frac{-2}{3}}=(3^{-3})^{-\frac{2}{3}}$ $\Big[\because\frac{1}{\text{a}}=\text{a}^{-1}\Big]$ $=3^{-3\times-\frac{2}{3}}=3^2=9$ $[\because(\text{a}^\text{m})^\text{n}=\text{a}^\text{mn}]$
View full question & answer→Question 431 Mark
Express the following in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0: 0.2$
AnswerLet $\text{x}=0.2=\frac{2}{10}=\frac{1}{5}$
View full question & answer→Question 441 Mark
Let $x$ be rational and $y$ be irrational. Is $xy$ necessarily irrational$?$ Justify your answer by an example.
AnswerNo, $(xy)$ is necessarily an irrational only when $x \neq 0.$
Let $x$ be a non-zero rational and $y$ be an irrational.
Then, we have to show that $xy$ be an irrational. If possible, let $xy$ be a rational number.
Since, quotient of two non-zero rational number is a rational number. So, $(xy/ x)$ is a rational number
$\Rightarrow y$ is a rational number. But, this contradicts the fact that $y$ is an irrational number.
Thus, our supposition is wrong.
Hence, $xy$ is an irrational number.
But, when $x = 0,$ then $xy = 0,$ a rational number.
View full question & answer→Question 451 Mark
Insert a rational number and an irrational number between the following:
$0.15$ and $0.16$
AnswerA rational number between $0.15$ and $0.16$ is $0.151.$
An irrational number between $0.15$ and $0.16$ is $0.1515515551..........$
View full question & answer→Question 461 Mark
Simplify the following: $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$
Answer$\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$ $=\sqrt{3}\Big(\frac{2}{3}-\frac{1}6{}\Big)=\sqrt{3}\Big(\frac{4-1}{6}\Big)$ $\sqrt{3}\times\frac{3}{6}=\frac{\sqrt{3}}{2}$
View full question & answer→Question 471 Mark
State whether the following statements are true or false$?$ Justify your answer. Number of rational numbers between $15$ and $18$ is finite.
AnswerBecause between any two rational numbers there exist infinitely many rational numbers.
View full question & answer→Question 481 Mark
Classify the following numbers as rational or irrational with justification: $\sqrt[3]{18}$
Answer$\sqrt[3]{18}=\sqrt[3]{(3)^2\times2}=3\times3\sqrt{2}=9\sqrt{2}$ Hence, it is an irrational number.
View full question & answer→Question 491 Mark
State whether the following statements are true or false? Justify your answer. The square of an irrational number is always rational.
Answer e.g., Let an irrational number be $\sqrt{2}$ and $\sqrt[4]{2}$
$a. (\sqrt{2})^2=2,$ which is a rational number.
$b. (\sqrt[4]{2})^2=\sqrt{2},$ which is not a rational number.
Hence, square of an irrational number is not always a rational number.
View full question & answer→