Question 11 Mark
Without actual division, show that the following rational numbers is a terminating decimal. Express in decimal form:
$\frac{19}{3125}$
Answer$\frac{19}{3125}=\frac{19}{5^5}=\frac{19\times2^5}{5^5\times2^5}$
$=\frac{608}{100000}=0.00608$
We know either $2$ or $5$ is not a factor of $19$, so it is in its simplest form.
Moreover, it is in the form of$\left(2^m \times 5^n\right).$
Hence, the given rational is terminating.
View full question & answer→Question 21 Mark
Without actual division, show that the following rational numbers is a non-terminating repeating decimal:
$\frac{32}{147}$
Answer$\frac{32}{147}=\frac{32}{3\times7^2}$
We know either $3$ or $7$ is not a factor of $32$, so it is in its simplest form.
Moreover, $\left(3 \times 7^2\right) \neq\left(2^m \times 5^n\right)$
Hence, the given rational is non-terminating repeating decimal.
View full question & answer→Question 31 Mark
Very-Short-Answer Questions:
State fundamental theorem of arithmatic.
AnswerThe fundamental theorem of arithmetic states that, every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique except for the order in which the prime factors occur.
View full question & answer→Question 41 Mark
Without actual division, show that the following rational numbers is a non-terminating repeating decimal:
$\frac{129}{\big(2^2\times5^7\times7^5\big)}$
Answer$\frac{129}{\big(2^2\times5^7\times7^5\big)}$
We know $2, 5$ or $7$ is not a factor of $129$, so it is in its simplest form.
Moreover, $\left(2^2 \times 5^7 \times 7^5\right) \neq\left(2^m \times 5^n\right)$
Hence, the given rational is non-terminating repeating decimal.
View full question & answer→Question 51 Mark
Very-Short-Answer Questions:
Express $360$ as product of its prime factors.
Answer$360 = 2 × 2 × 2 × 3 × 3 × 5 = 2^3× 3^2× 5$
View full question & answer→Question 61 Mark
Define:
- Rational numbers.
- Irrational numbers.
- Real numbers.
Answer
- Rational numbers: The numbers of the form $\frac{\text{p}}{\text{q}}$ where $p , q$ are integers and $\text{q}\neq0$ are called rational numbers.
Example: $\frac{2}{3}$
- Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.
Example: $\sqrt{2}$
- Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.
Example: $2,\ \frac{1}{3},\ \sqrt{2},\ -3$ etc. View full question & answer→Question 71 Mark
The following numbers are irrational?
$\sqrt{2}$
Answer $\sqrt{2}$ is irrational $(\because$ if p is prime, then $\sqrt{\text{p}}$ is irrational$).$
View full question & answer→Question 81 Mark
The following numbers are irrational?
$2.\bar3$
Answer$2.\bar3$ is rational because it is a non-terminating, repeating decimal.
View full question & answer→Question 91 Mark
Very-Short-Answer Questions:
Give an example of two irrationals whose sum is rational.
AnswerConsider the irrational numbers, $\big(3+\sqrt2\big)$ and $\big(3-\sqrt2\big)$
$\big(3+\sqrt2\big)+\big(3-\sqrt2\big)=6$
which is rational number.
View full question & answer→Question 101 Mark
Classify the following number as rational or irrational:
$\sqrt[3]{3}$
Answer$\sqrt[3]{3}$ is an irrational number because $3$ is a prime number. So, $\sqrt{3}$ is an irrational number.
View full question & answer→Question 111 Mark
Classify the following number as rational or irrational:
$5.636363...$
Answer$5.636363...$ is a rational number because it is a non-terminating, repeating decimal.
View full question & answer→Question 121 Mark
Without actual division, show that the following rational numbers is a terminating decimal. Express in decimal form:
$\frac{15}{1600}$
Answer$\frac{15}{1600}=\frac{15}{2^6\times5^2}=\frac{15\times5^4}{2^6\times5^6}$
$=\frac{9375}{1000000}=0.009375$
We know either $2$ or $5$ is not a factor of $15$, so it is in its simplest form.
Moreover, it is in the form of $\left(2^m \times 5^n\right).$
Hence, the given rational is terminating.
View full question & answer→Question 131 Mark
The following numbers are irrational?
$5.27\overline{41}$
Answer$5.27\overline{41}$ is rational because it is a non-terminating, repeating decimal.
View full question & answer→Question 141 Mark
Classify the following number as rational or irrational:
$\big(5+3\sqrt{2}\big)$
AnswerLet $5+3\sqrt2$ be rational.
Hence, $$5 and $5+3\sqrt2$ are rational.
$\therefore\big(5+3\sqrt{2}-2\big)=3\sqrt2=$ rational $[\because$ Difference of two rational is rational$]$
$\therefore\frac{1}3{}\times3\sqrt{2}=\sqrt{2}=$ rational $[\because$ Product of two rational is rational$]$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming $5+3\sqrt2$ is rational.
Hence, $5+3\sqrt2$ is irrational.
View full question & answer→Question 151 Mark
State whether the given statement is true of false:
The product of a rational and an irrational is irrational.
Question 161 Mark
Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal:
$\frac{29}{343}$
Answer$\frac{29}{343}=\frac{29}{7^3}$
We know $7$ is not a factor of $29$, so it is in its simplest form.
Moreover, $7^3 ≠ \left(2^m \times 5^n\right)$
Hence, the given rational is non-terminating repeating decimal.
View full question & answer→Question 171 Mark
Very-Short-Answer Questions:
If $a$ and $b$ are two prime numbers then find $HCF (a, b)$.
AnswerIf $a$ and $b$ are prime numbers, it means they have no common factors other than $1$
So, the $HCF (a, b) = 1$
View full question & answer→Question 181 Mark
Is it possible to have two numbers whose $HCF$ is $18$ and $LCM$ is $760?$ Give reason.
HINT: $HCF$ always divides $LCM$ completely.
AnswerNo it is, not possible since the $LCM$ has to be a multiple of the $HCF$.
$760$ is not a multiple of $18$
View full question & answer→Question 191 Mark
Classify the following number as rational or irrational:
$\big(2-\sqrt{3}\big)$
AnswerLet $2-\sqrt3$ be rational.
Hence, $2$ and $2-\sqrt3$ are rational.
$\therefore\big(2-2+\sqrt{3}\big)=\sqrt3=$ rational $[\because$ Difference of two rational is rational$]$
This contradicts the fact that $\sqrt3$ is irrational.
The contradiction arises by assuming $2-\sqrt3$ is rational.
Hence, $2-\sqrt3$ is irrational.
View full question & answer→Question 201 Mark
A number when divided by $61$ gives $27$ as quotient and $32$ as remainder. Find the number.
AnswerBy Euclid's Division algorithm we have:
Dividend$ = ($divisor $\times $ quotient$) +$ remainder
$= (61 \times 27) + 32$
$= 1647 + 32$
$= 1679$
View full question & answer→Question 211 Mark
Very-Short-Answer Questions:
State Euclid's division lemma.
AnswerFor any two given positive integers $'a'$ and 'b' there exists unique whole numbers $'q'$ and $'r'$ such that $a = bq + r,$ where $0 ≤ r < b$
View full question & answer→Question 221 Mark
Classify the following number as rational or irrational:
$3.121221222...$
Answer$3.121221222...$ is an irrational number because it is a non-terminating and non-repeating decimal.
View full question & answer→Question 231 Mark
Classify the following number as rational or irrational:
$\big(3+\sqrt{2}\big)$
AnswerLet $3+\sqrt2$ be rational.
Hence, $3$ and $3+\sqrt2$ are rational.
$\therefore3+\sqrt{2}-3=\sqrt2=$ rational $[\because$ Difference of two rational is rational$]$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming $3+\sqrt2$ is rational.
Hence, $3+\sqrt2$ is irrational.
View full question & answer→Question 241 Mark
Classify the following number as rational or irrational:
$\big(\sqrt3+\sqrt5\big)$
AnswerLet $\sqrt3+\sqrt5$ be rational.
$\therefore\sqrt3+\sqrt5= \text{a},$ where aa is rational
$\therefore\sqrt3=\text{a}-\sqrt5\dots(1)$
On squaring both sides of equation $(1)$, we get
$3=\big(\text{a}-\sqrt5\big)^2$
$=\text{a}^2+5-2\sqrt5\text{a}$
$\Rightarrow\sqrt5=\frac{\text{a}^2+2}{\text{2a}}$
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, $\sqrt3+\sqrt5$ is irrational.
View full question & answer→Question 251 Mark
Without actual division, show that the following rational numbers is a non-terminating repeating decimal:
$\frac{77}{210}$
Answer$\frac{77}{210}=\frac{77\div\text{7}}{210\div7}$
$=\frac{11}{30}=\frac{11}{2\times3\times5}$
We know $2, 3$ or $5$ is not a factor of $11$, so $\frac{11}{30}$ is in its simplest form.
Moreover, $(2 × 3 × 5) ≠ \left(2^m \times 5^n\right)$
Hence, the given rational is non-terminating repeating decimal.
View full question & answer→Question 261 Mark
Very-Short-Answer Questions:
Write the decimal expansion of $\frac{73}{\big(2^4\times5^3\big)}$
AnswerThe given number is $\frac{73}{\big(2^4\times5^3\big)}$
Clearly, none of $2$ and $5$ is a factor of $73$
So, the given rational is in its simplest form.
So, the given number is a terminating decimal.
Now, $\frac{73}{\big(2^4\times5^3\big)}=\frac{73\times5}{\big(2^4\times5^3\big)}$
$=\frac{365}{10000}=0.0365$
0.0365 is the decimal expansion.
View full question & answer→Question 271 Mark
The $HCF$ of two numbers is $18$ and their product is $12960$. Find their $LCM$.
AnswerLet the two numbers be $a$ and $b$.
$HCF \times LCM = ab$
$\Rightarrow 18 \times LCM = 12960$
$\Rightarrow LCM = 720$
View full question & answer→Question 281 Mark
The following numbers are irrational?
$0.232332333...$
Answer$0.232332333...$ is irrational because it is a non-terminating, non-repeating decimal.
View full question & answer→Question 291 Mark
Without actual division, show that the following rational numbers is a terminating decimal. Express in decimal form:
$\frac{17}{320}$
Answer$\frac{17}{320}=\frac{17}{2^6\times5}=\frac{17\times5^5}{2^6\times5^6}$
$=\frac{53125}{1000000}=0.053125$
We know either $2$ or $5$ is not a factor of $17$, so it is in its simplest form.
Moreover, it is in the form of $\left(2^m \times 5^n\right)$
Hence, the given rational is terminating.
View full question & answer→Question 301 Mark
The following numbers are irrational?
$\pi$
Answer$\pi$ is irrational because it is a non-repeating, non-terminating decimal.
View full question & answer→Question 311 Mark
State whether the given statement is true of false:
The sum of two rationals is always rational.
Question 321 Mark
Classify the following number as rational or irrational:
$2.040040004...$
Answer$2.040040004...$ is an irrational number because it is a non-terminating and non-repeating decimal.
View full question & answer→Question 331 Mark
Very-Short-Answer Questions:
What is a composite number?
AnswerA whole number that can be divided evenly by numbers other than 1 or itself.
View full question & answer→Question 341 Mark
Without actual division, show that the following rational numbers is a terminating decimal. Express in decimal form:
$\frac{24}{125}$
Answer$\frac{24}{125}=\frac{24}{5^3}=\frac{24\times2^3}{5^3\times2^3}$
$=\frac{192}{1000}=0.192$
We know $5$ is not a factor of $24$, so it is in its simplest form.
Moreover, it is in the form of $(2^m× 5^n).$
Hence, the given rational is terminating.
View full question & answer→Question 351 Mark
Classify the following number as rational or irrational:
$3.\overline{142857}$
Answer $3.\overline{142857}$ is a rational number because it is a repeating decimal.
View full question & answer→Question 361 Mark
Prove that $\big(2+\sqrt3\big)$ is irrational.
AnswerIf possible, let $\big(2+\sqrt3\big)$ be rational.
Then $2$ and $\sqrt3$ are rational.
$\Rightarrow2+\sqrt3-2$ is rational $\dots(\because$ difference of two rationals is rational$)$
$\Rightarrow\sqrt3$ is rational
This contradicts the fact that $\sqrt3$ is irrational.
The contradiction arises by asuming that $\big(2+\sqrt3\big)$ is rational.
Hence, $\big(2+\sqrt3\big)$ is irrational.
View full question & answer→Question 371 Mark
The following numbers are irrational?
$3.142857$
Answer$3.142857$ is rational because it is a terminating decimal.
View full question & answer→Question 381 Mark
State whether the given statement is true of false:
The product of two irrationals is an irrational.
AnswerFalse.
Counter example:
$2\sqrt3$ and $4\sqrt3$ are two irrational numbers. But their product is $24$, which is a rational number.
View full question & answer→Question 391 Mark
Classify the following number as rational or irrational:
$\frac{3}{\sqrt{5}}$
AnswerLet $\frac{3}{\sqrt5}$ be rational.
$\therefore\frac{1}3{}\times\frac{3}{\sqrt5}=\frac{1}{\sqrt5}=$ rational $[\because$ Product of two rational is rational$]$
This contradicts the fact that $\frac{1}{\sqrt5}$ is irrational.
$\therefore\frac{1\times\sqrt5}{\sqrt5\times\sqrt5}=\frac{1}5{}\sqrt5$
So, if $\frac{1}{\sqrt5}$ is rational, then $\frac{1}{5}\sqrt5$ is rational.
$\therefore5\times\frac{1}{5}\sqrt5=\sqrt5=$ rational $[\because$ Product of two rational is rational$]$
Hence, $\frac{1}{\sqrt5}$ is irrational.
The contradiction arises by assuming $\frac{3}{\sqrt5}$ is rational.
Hence, $\frac{3}{\sqrt5}$ is irrational.
View full question & answer→Question 401 Mark
What do you mean by Euclid's division algorithm.
AnswerFor any two given positive integers $a$ and $b$ there exist unique whole numbers $q$ and $r$ such that
$\text{a}=\text{bq}+\text{r},$ where $0\le\text{r}<\text{b}$
Here, we call 'a' as dividend, $'b'$ as divisor, $'q'$ as quotient and $'r'$ as remainder.
Dividend $= ($divisor $\times $ quotient$) +$ remainder
View full question & answer→Question 411 Mark
Classify the following number as rational or irrational:
$\frac{22}{7}$
Answer$\frac{22}{7}$ is a rational number because it is of the form of $\frac{\text{p}}{\text{q}},\ \text{q}\neq0.$
View full question & answer→Question 421 Mark
Find the simplest form of:
$\frac{69}{92}$
AnswerPrime factorisation of $69$ and $92$ is:
$69 = 3 \times 23$
$92 = 2^2\times 23$
Therefore, $\frac{69}{92}=\frac{3\times\text{23}}{2^2\times23}=\frac{3}{2^2}=\frac{3}{4}$
Thus, simplest form of $\frac{69}{92}$ is $\frac{3}{4}.$
View full question & answer→Question 431 Mark
Very-Short-Answer Questions:
If the rational number $\frac{\text{a}}{\text{b}}$ has a terminating decimal expansion, what is the condition to be satisfied by $b$?
AnswerThe condition for $\frac{\text{a}}{\text{b}}$ to be a terminating decimal is that b should be of form $(2^m× 5^n)$ for some non-negative integers $m$ and $n$ .
View full question & answer→Question 441 Mark
Find the simplest form of $\frac{148}{185}.$
Answer $\frac{148}{185}=\frac{2\times2\times37}{5\times37}$
$=\frac{4}{5}$
which is in simplest form.
View full question & answer→Question 451 Mark
Without actual division, show that the following rational numbers is a non-terminating repeating decimal:
$\frac{11}{\big(2^3\times3\big)}$
Answer$\frac{11}{\big(2^3\times3\big)}$
We know either $2$ or $3$ is not a factor of $11$, so it is in its simplest form.
Moreover, $\left(2^3 \times 3\right) \neq\left(2^m \times 5^n\right)$
Hence, the given rational is non-terminating repeating decimal.
View full question & answer→Question 461 Mark
Give an examples of two irrationals whose product is rational.
HINT: Take $\big(3+\sqrt2\big)$ and $\big(3-\sqrt2\big).$
AnswerLet $2\sqrt3,\ 3\sqrt3$ be two irrationals.
$\therefore2\sqrt3\times3\sqrt3=18=$ rational number
View full question & answer→Question 471 Mark
Without actual division, show that the following rational numbers is a terminating decimal. Express in decimal form:
$\frac{171}{800}$
Answer$\frac{171}{800}=\frac{171}{2^5\times5^2}=\frac{171\times5^3}{2^5\times5^5}$
$=\frac{21375}{100000}=0.21375$
We know either $2$ or $5$ is not a factor of $171$, so it is in its simplest form.
Moreover, it is in the form of $\left(2^m \times 5^n\right)$.
Hence, the given rational is terminating.
View full question & answer→Question 481 Mark
Classify the following number as rational or irrational:
$\frac{22}{7}$
Answer$\frac{22}{7}$ is a rational number because it is of the form of $\frac{\text{p}}{\text{q}},\ \text{q}\neq0.$
View full question & answer→Question 491 Mark
Classify the following number as rational or irrational:
$\big(2+\sqrt{5}\big)$
AnswerLet $3+\sqrt2$ be rational.
Hence, $2+\sqrt5$ and $\sqrt5$ are rational.
$\therefore\big(2+\sqrt{5}\big)-2=2+\sqrt5-2=\sqrt{5}=$ rational $[\because$ Difference of two rational is rational$]$
This contradicts the fact that $\sqrt5$ is irrational.
The contradiction arises by assuming $2-\sqrt5$ is rational.
Hence, $2-\sqrt5$ is irrational.
View full question & answer→Question 501 Mark
Very-Short-Answer Questions:
If $a$ and $b$ are two prime numbers then find $LCM(a, b)$.
AnswerIf $a$ and $b$ are prime numbers, it means rthey have no common multiples other than their product.
So, the $LCM(a, b) = ab.$
View full question & answer→