Question 13 Marks
Prove that if $x$ and $y$ are both odd positive integers then $x^2+y^2$ is even but not divisible by $4.$
AnswerLet the two odd positive numbers be $\mathrm{x}=2 \mathrm{k}+1$ and $\mathrm{y}=2 \mathrm{p}+1$
$ \text { Hence } x^2+y^2=(2 k+1)^2+(2 p+1)^2 $
$ =4 k^2+4 k+1+4 p^2+4 p+1 $
$ =4 k^2+4 p^2+4 k+4 p+2$
$=4\left(k^2+p^2+k+p\right)+2$
Clearly notice that the sum of square is even the number is not divisible by $4$
Hence if $x$ and $y$ are odd positive integers, then $x^2+y^2$ is even but not divisible by $4$
View full question & answer→Question 23 Marks
Express the following as a fraction in simplest form:
$2.2\bar4$
AnswerLet $\text{x}=2.2\bar4$
$\therefore\text{x}=2.2444\dots(1)$
$\text{10x}=22.2444\dots(2)$
$\text{100x}=224.444\dots(3)$
On subtracting equation $(2)$ from $(3),$ we get
$\text{90x}=202$
$\Rightarrow\text{x}=\frac{202}{90}$
$=\frac{101}{45}$
Hence, $2.24=\frac{\overline{101}}{45}$
View full question & answer→Question 33 Marks
Express the following as a fraction in simplest form:
$2.\bar4$
AnswerLet $\text{x}=2. \bar4$
$\therefore\text{x}=2.444\dots(1)$
$\text{10x}=24.444\dots(2)$
On subtracting equation $(1)$ from $(2),$ we get
$\text{9x}=22$
$\Rightarrow\text{x}=\frac{22}{9}$
$\therefore2.4=\frac{\overline{22}}{9}$
View full question & answer→Question 43 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$396, 1080$
Answer$ 396=2 \times 2 \times 3 \times 3 \times 11=2^2 \times 3^2 \times 11$
$1080=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5=2^3 \times 3^3 \times 5$
$\operatorname{HCF}(396,1080)=2^2 \times 3^2=36$
$\operatorname{LCM}(396,1080)=2^3 \times 3^3 \times 11 \times 5=11880$
$\operatorname{HCF} \times \operatorname{LCM}=427680$
$144 \times 198=427680$
$⇒ HCF × LCM =$ product of given numbers
Hence verified.
View full question & answer→Question 53 Marks
Show that any positive odd integer is of the form $(6m + 1)$ or $(6m + 3)$ or $(6m + 5),$ where $m$ is some integer.
AnswerLet $'a'$ be a given positive integer.
On dividing $'a'$ by $6,$ let $'q'$ be the quotient and $'r'$ be the remainder.
Then, by Euclid's algorithm, we have
$a = 6q + r,$ where $0 ≤ r < 6 $
$⇒ a = 6q + r,$ where $r = 0, 1, 2, 3, 4, 5, 6 $
$⇒ a = 6q$ or $a + 6q + 1$ or $a + 6q + 2$ or $a + 6q + 3$ or $a + 6q + 4$ or $a + 6q + 5$
But, $a = 6q, a + 6q + 2, a + 6q + 4$ gives even values of $'a'.$
Thus, when $'a'$ is odd, it is of the form $6q + 1, 6q + 3$ or $6q + 5$ for some integer $'q'.$
View full question & answer→Question 63 Marks
Express the following as a fraction in simplest form:
$0.1\bar2$
AnswerLet $\text{x}=0.1\bar2$
$\therefore\text{x}=0.1222\dots(1)$
$\text{10x}=12.222\dots(2)$
On subtracting equation $(1)$ from $(2),$ we get
$\text{9x}=12$
$\Rightarrow\text{x}=\frac{12}{9}$
$\therefore0.12=\frac{\overline{12}}{9}$
View full question & answer→Question 73 Marks
Find the smallest number which when divided by $28$ and $32$ leaves remainders $8$ and $12$ respectively.
AnswerThe smallest number which when divided by $28$ and $32$ can be determined by finding the $LCM$ of $28$ and $32$
$28=2^2 \times 7$
$ 32=2^5$
$ \text { LCM }=2^5 \times 7 $
$ =224$
The smallest number that when divided by $28$ and $32$ leaves a reminder $8$ and $12$
$= 224 - 8 - 12 = 204$
View full question & answer→Question 83 Marks
Short-Answer Questions:
Express $0.\overline{23}$ as a rational number in simplest form.
AnswerLet $\text{x}=0.\overline{23}$ then,
$\text{x}=0.232323\dots\ \ \dots(\text{i})$
$\therefore\text{100x}=23.2323\dots\ \ \dots(\text{ii})$
On subtracting $(i)$ from $(ii),$ we get
$\text{99x}=23$
$\Rightarrow\text{x}=\frac{23}{99}$
Hence, $0.\overline{23}=\frac{23}{99}$
View full question & answer→Question 93 Marks
Using Euclid's algortihm, find the $HCF$ of:
$960$ and $1575$
AnswerDividing $1575$ by $960,$ we get
Quotient $= 1,$ Remainder $= 615$
$\therefore 1575 = 960 × 1 + 615$
Dividing $960$ by $615,$ we get
Quotient $= 1,$ Remainder $= 345$
$\therefore 960 = 615 × 1 + 345$
Dividing $615$ by $345$
Quotient $= 1,$ Remainder $= 270$
$\therefore 615 = 345 × 1 + 270$
Dividing $345$ by $270,$ we get
Quotient $= 1,$ Remainder $= 75$
$\therefore 345 = 270 × 1 + 75$
Dividing $270$ by $75,$ we get
Quotient $= 3,$ Remainder $=45$
$\therefore 270 = 75 × 3 + 45$
Dividing $75$ by $45,$ we get
Quotient $= 1,$ Remainder $= 30$
$\therefore 75 = 45 × 1 + 30$
Dividing $45$ by $30,$ we get
Remainder $= 15,$ Quotient $= 1$
$\therefore 45 = 30 × 1 + 15$
Dividing $30$ by $15,$ we get
Quotient $= 2,$ Remainder $= 0$
$\therefore H.C.F.$ of $1575$ and $960$ is $15$
View full question & answer→Question 103 Marks
Express the following as a fraction in simplest form:
$0.\bar8$
AnswerLet
$\text{x}=0. \bar8$
$\therefore\text{x}=0.888\dots(1)$
$\text{10x}=8.888\dots(2)$
On subtracting equation $(1)$ from $(2),$ we get
$\text{9x}=8$
$\Rightarrow\text{x}=\frac{8}{9}$
$\therefore0.8=\frac{\bar8}{9}$
View full question & answer→Question 113 Marks
Show that any positive odd integer is of the form $(4m + 1)$ or $(4m + 3),$ where m is some integer.
AnswerLet $'a'$ be a given positive odd integer.
On dividing $'a'$ by $4,$ let $'q'$ be the quotient and $'r'$ be the remainder.
Then, by Euclid's algorithm, we have
$a = 4m + r,$ where $0 ≤ r < 4$
$⇒ a = 4m + r,$ where $r = 0, 1, 2, 3$
$⇒ a = 4m$ or $a = 4m + 1$ or $a = 4m +2$ or $a = 4m + 3$
But, $a = 4m$ and $a = 4m + 2 = 2(2m +1)$ are clearly even.
Thus, when $'a'$ is odd, it is of the form $a = (4m + 1)$ or $(4m + 3)$ for some integer $m.$
View full question & answer→Question 123 Marks
Give prime factorisation of $4620.$
Answer$4620 = 2 × 2 × 3 × 5 × 7 × 11$
$= 2^2× 3 × 5 × 7 × 11$ is the prime factorisation of $4620$
View full question & answer→Question 133 Marks
The $HCF$ of two numbers is $145$ and their $LCM$ is $2175.$ If one of the numbers is $725,$ find the other.
AnswerLet the two numbers be $a$ and $b.$
$HCF × LCM = ab$
$⇒ 145 × 2175 = 725 × b$
$⇒ b = 435$
So, the other number is $435$
View full question & answer→Question 143 Marks
Using Euclid's algortihm, find the $HCF$ of:
$504$ and $1188$
AnswerOn dividing $1188$ by $504,$ we get
Quotient $= 2,$ Remainder $= 180$
$\therefore 1188 = 504 × 2+ 180$
Dividing $504$ by $180$
Quotient $= 2,$ remainder $= 144$
$\therefore 504 = 180 × 2 + 144$
Dividing $180$ by $144,$ we get
Quotient $= 1$, Remainder $= 36$
Dividing $144$ by $36$
Quotient $= 4,$ Remainder $= 0$
$\therefore H.C.F.$ of $1188$ and $504$ is $36$
View full question & answer→Question 153 Marks
Find the $HCF$ of $1008$ and $1080$ by prime factorization method.
Answer$ 1008=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 $
$ =2^4 \times 3^2 \times 7 $
$ 1080=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$
$\text { So, the HCF }=2^3 \times 3^2=72$
View full question & answer→Question 163 Marks
The $HCF$ of two numbers is $27$ and their $LCM$ is $162.$ If one of the number is $81,$ find the other.
AnswerLet the number be a and $81$
$HCF × LCM =$ product of the two numbers
$27 × 162 = 81a$
$a = 54$
So, the other number is $54$
View full question & answer→Question 173 Marks
Express the following as a fraction in simplest form:
$0.\overline{24}$
AnswerLet $\text{x}=0. \overline{24}$
$\therefore\text{x}=0.2424\dots(1)$
$\text{100x}=24.2424\dots(2)$
On subtracting equation $(1)$ from $(2),$ we get
$\text{99x}=24$
$\Rightarrow\text{x}=\frac{8}{33}$
$\therefore0.24=\frac{\bar{8}}{33}$
View full question & answer→Question 183 Marks
Short-Answer Questions:
Explain why $0.15015001500015...$ is an irrational number.
AnswerThe numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational numbers.
Clearly, $0.15015001500015...$ is a non-terminating and non-repeating decimal.
Hence, it is irrational.
View full question & answer→Question 193 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$144, 198$
Answer$ 144=2 \times 2 \times 2 \times 2 \times 3 \times 3=2^4 \times 3^2 $
$ 198=2 \times 3 \times 3 \times 11=2 \times 3^2 \times 11 $
$ \operatorname{HCF}(144,198)=2 \times 3^2=18 $
$ \operatorname{LCM}(144,198)=2^4 \times 3^2 \times 11=1584 $
$ H C F \times \operatorname{LCM}=28512 $
$ 144 \times 198=28512$
$⇒ HCF × LCM =$ product of given numbers
Hence verified.
View full question & answer→Question 203 Marks
Short-Answer Questions:
Write a rational number between $\sqrt3$ and $2.$
AnswerWe have, $\sqrt{3}=1.732\dots$
Since $1.732... < 1.8 < 2$
So, $1.8$ is a rational number that lies between $\sqrt3$ and $2$
View full question & answer→Question 213 Marks
Show that every positive integer is either even or odd.
AnswerLet $'a'$ be a given positive integer.
On dividing $'a'$ by $2,$ let $q$ be the quotient and $r$ be the remainder.
Then, by Euclid's algorithm, we have
$a = 2q + r,$ where $0 ≤ r < 2$
$⇒ a = 2q + r,$ where $r = 0, 1$
$⇒ a = 2q$ or $a = 2q + 1$
When $a = 2q$ for some integer $q,$ then clearly $'a'$ is even.
When $a = 2q + 1$ for some integer $q,$ then clearly $'a'$ is odd.
Thus, every positive integer is either even or odd.
View full question & answer→Question 223 Marks
Find the least number which when divided by $35, 56$ and $91$ leaves the same remainder $7$ in each case.
AnswerThe smallest number which when divided by $35, 56$ and $91$ can be determined by finding the $LCM$ of $35, 56$ and $91$
$35 = 5 × 7$
$56 = 2 × 2 × 2 × 7 = 2^3× 7$
$91 = 7 × 13$
$LCM = 2^3× 5 × 7 × 13$
$= 3640$
The smallest number that when divided by $35, 56, 91$ leaves a reminder $7$ in each case
$= 3640 + 7 = 3547$
View full question & answer→Question 233 Marks
Examine whether $\frac{17}{30}$ is a terminating decimal.
AnswerA number is a terminating decimal, if the denominator is of the form $2^m × 5^n$, where $m$ and $n$ are non-negative integers.
$\frac{17}{30}=\frac{17}{2\times3\times5}$
Clearly, $\frac{17}{30}$ is a not a terminating decimal, since its denominator is not of the form $2^m× 5^n$
It has a factor of $3$ present too.
View full question & answer→Question 243 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$1152, 1664$
Answer$ 1152=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3=2^7 \times 3^2 $
$ 1664=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=2^7=128 $
$ H C F(1152,1664)=2^7=128 $
$ \text { LCM }(1152,1664)=2^7 \times 3^2 \times 13=14976 $
$ H C F \times \operatorname{LCM}=1916928 $
$ 1152 \times 1664=1916928$
$⇒ HCF × LCM =$ product of given numbers
Hence verified.
View full question & answer→Question 253 Marks
Find the least number which when divided by $20, 25, 35$ and $40$ leaves remainders $14, 19, 29$ and $34$ respectively.
Answer$LCM$ of $20, 25, 35,$ and $40 = 1400$
Now,
$20 - 6 = 14$
$25 - 6 = 19$
$35 - 6 = 29$
$40 - 6 = 34$
So, $(1400 - 6 = 1394)$ will have remainder of $14, 19, 29,$ and $34$ when divided by $20, 25, 35,$ and $40$ resp.
$= 1394$
View full question & answer→Question 263 Marks
Short-Answer Questions:
Explain why $3.\overline{1416}$ is a rational number.
AnswerThe number of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0$ are called rational numbers.
Let $\text{x}=3.\overline{1416}$
$⇒ x = 3.141614161416... ...(i)$
Since there are four repeating digits,
We multiply by $1000$
$⇒ 1000x = 31416.14161416... ...(ii)$
Subtracting $(i)$ from $(ii),$ we get
$999x = 31416$
$\Rightarrow\text{x}=\frac{31416}{999}$
Which is of the form $\frac{\text{p}}{\text{q}}.$
So, $3.\overline{1416}$ is a rational number.
View full question & answer→Question 273 Marks
The $HCF$ of two numbers is $27$ and their $LCM$ is $162.$ If one of the number is $81,$ find the other.
AnswerLet the number be a and $81$
$HCF × LCM =$ product of the two numbers
$27 × 162 = 81a$
$a = 54$
So, the other number is $54$
View full question & answer→Question 283 Marks
Find the smallest number which when increased by $17$ is exactly divisible by both $468$ and $520.$
AnswerThe smallest number which when increased by $17$ is exactly divisible by both $520$ and $468$ is obtained by subtracting $17$ from the $LCM$ of $520$ and $468$
$ 468=2^2 \times 3^2 \times 13$
$520=2^3 \times 5 \times 13$
$LCM =2^3 \times 3^2 \times 5 \times 13$
$= 4680$
Smallest number which when increased by $17$ is exactly divisible by both $520$ and $468 = 4680 - 17 = 4663$
View full question & answer→Question 293 Marks
For any positive integer $n,$ prove that $n^3- n$ is divisible by $6.$
Answer$n^3-n=n\left(n^2-1\right)=n(n-1)(n+1)$
Whenever a number is divided by $3,$ the remainder obtained is either $0$ or $1$ or $2.$
$\therefore$ $n = 3p$ or $3p + 1$ or $3p + 2,$ where $p$ is some integer.
If $n = 3p$, then n is divisible by $3.$
If $n = 3p + 1,$ then $n - 1 = 3p + 1 - 1 = 3p$ is divisible by $3.$
If $n = 3p + 2,$ then $n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1)$ is divisible by $3.$
So, we can say that one of the numbers among $n, n - 1$ and $n + 1$ is always divisible by $3.$
$⇒ n (n - 1)(n + 1)$ is divisible by $3.$
Similarly, whenever a number is divided $2,$ the remainder obtained is $0$ or $1.$
$\therefore$ $n = 2q$ or $2q + 1,$ where $q$ is some integer.
If $n = 2q$, then n is divisible by $2.$
If $n = 2q + 1,$ then $n - 1 = 2q + 1 - 1 = 2q$ is divisible by $2$ and $n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1)$ is divisible by $2.$
So, we can say that one of the numbers among $n, n - 1$ and $n + 1$ is always divisible by $2.$
$⇒ n (n - 1)(n + 1)$ is divisible by $2.$
Since, $n (n - 1)(n + 1)$ is divisible by $2$ and $3.$
$\therefore$ $n(n - 1)(n + 1) = n^3-n $ is divisible by $6. ($If a number is divisible by both $2$ and $3,$ then it is divisible by $6)$
View full question & answer→Question 303 Marks
Short-Answer Questions:
Show that $\frac{\sqrt2}{3}$ is irrational.
AnswerIf possible, let $\frac{\sqrt2}{3}$ be rational.
$\Rightarrow\frac{1}{3}\sqrt2$ is rational.
Now, 3 is rational, $\frac{1}{3}\sqrt2$ is rational.
$\Rightarrow\Big(3\times\frac{1}{3}\sqrt2\Big)$ is rational.
$\Rightarrow\sqrt2$ is rational.
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\frac{\sqrt2}{3}$ is rational.
Hence, $\frac{\sqrt2}{3}$ is irrational.
View full question & answer→Question 313 Marks
Using prime factorization, find the $HCF$ and $LCM$ of:
$96, 404$
Answer$96 = 2 × 2 × 2 × 2 × 2 × 3$
$404 = 2 × 2 × 101$
$LCM (96$ and $404) = 25 × 3 × 101 = 9696$
$HCF (96$ and $404)= 22=4$
Verification:
$HCF × LCM = 9696 × 4 = 38784$
product of given numbers $= 96 × 404 = 38784$
$⇒ HCF × LCM =$ Product of given numbers verified.
View full question & answer→Question 323 Marks
Using Euclid's algortihm, find the $HCF$ of:
$405$ and $2520$
AnswerOn dividing $2520$ by $405,$ we get
Quotient $= 6,$ Remainder $= 90$
$\therefore 2520 = (405 × 6) + 90$
Dividing $405$ by $90,$ we get
Quotient $= 4,$
Remainder $= 45$
$\therefore 405 = 90 × 4 + 45$
Dividing $90$ by $45$
Quotient $= 2,$ Remainder$ = 0$
$\therefore$ $90 = 45 × 2$
$\therefore$ $H.C.F$. of $405$ and $2520$ is $45$
View full question & answer→Question 333 Marks
Express the following as a fraction in simplest form:
$0.\overline{365}$
AnswerLet $\text{x}=0.\overline{365}$
$\therefore\text{x}=0.3656565\dots(1)$
$\text{10x}=3.656565\dots(2)$
$\text{1000x}=365.656565\dots(3)$
On subtracting equation $(2)$ from $(3),$ we get
$\text{990x}=362$
$\Rightarrow\text{x}=\frac{362}{990}$
$=\frac{181}{495}$
Hence, $0.365=\frac{\overline{181}}{495}$
View full question & answer→Question 343 Marks
Very-Short-Answer Questions:
Simplify: $\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}$
Answer $\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}$
$=\frac{\big(2\sqrt{45}+3\sqrt{20}\big)}{2\sqrt{5}}\times\frac{2\sqrt5}{2\sqrt5}$ ...(Rationlising the denominator)
$=\frac{2\sqrt5\big(2\sqrt{45}+3\sqrt{20}\big)}{20}$
$=\frac{4\sqrt{45\times5}+6\sqrt{20\times5}}{20}$
$=\frac{4\sqrt{3^2\times5^2}+6\sqrt{2^2\times5^2}}{20}$
$=\frac{4(3\times5)+6(2\times5)}{20}$
$=\frac{60+60}{20}$
$=6$
View full question & answer→Question 353 Marks
Three sets of English, Mathematics and Science books containing $336, 240$ and $96$ books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there$?$
AnswerLet us find the $HCF$ of $336, 240$ and $96$ through prime factorization:
$\begin{array}{c|c} 2 & 336 \\ \hline 2 & 168\\ \hline2&84\\ \hline2&42\\ \hline3&21\\\hline&7 \end{array}$ $\begin{array}{c|c} 2 & 240 \\ \hline 2 & 120\\ \hline2&60\\ \hline2&30\\ \hline3&15\\\hline&5 \end{array}$ $\begin{array}{c|c} 2 & 96 \\ \hline 2 & 48\\ \hline2&24\\ \hline2&12\\ \hline2&6\\\hline&3 \end{array}$
$ 336=2 \times 2 \times 2 \times 2 \times 3 \times 7=2^4 \times 3 \times 7$
$ 240=2 \times 2 \times 2 \times 2 \times 3 \times 5=2^4 \times 3 \times 5 $
$96=2 \times 2 \times 2 \times 2 \times 2 \times 3=2^5 \times 3$
Each stack of book will contain $48$ books
Number of stacks of the same height
$=\frac{240}{48}+\frac{336}{48}+\frac{96}{48}$
$=5+7+2=14$
View full question & answer→