Question 14 Marks
Show that any number of the form $4^n, n \in N$ can never end with the digit $0.$
AnswerWe have $4^n=(2 \times 2)^n=2^n \times 2^n$
The prime factors of $4^n$ are $2^n$
For any $n$, the number $2^n$ would with a zero digit if number is divisible by $5$
So, the prime factorisation should contain at least one prime factor $5$
But $4 n=2^n \times 2^n$ which contains the prime factors $2$
So, by the uniqueness of the fundamental theorem, no other prime factorisation of $4^n$ exists.
Thus, any number of the form $4^n$ can never end with the digit $0$
View full question & answer→Question 24 Marks
Use Euclid's algorithm to find $HCF$ of $1190$ and $1445$. express the $HCF$ in the form $1190m + 1445n.$
AnswerUsing Euclid's algorithm
$HCF$ of ($1190$ and $1445).$
$1445 = 1190 × 1 + 255$
$1190 = 255 × 4 + 170$
$255 = 170 × 1 + 85$
$170 = 85 × 2 + 0$
$HCF = 85$
Now,
$85 = 255 - 170$
$= (1445 - 1190 × 1) - (1190 - 225 × 4)$
$= (1445 - 1190 × 1) - (1150 - (1445 - 1190 × 1) × 4)$
$= 1445 - 190 × 1 - 1190 + 1445 × 4 - 1190 × 4$
$= 1445 × 5 - 1190 × 6$
$= 85$
View full question & answer→Question 34 Marks
The traffic lights at three different road crossing change after every $48$ secons, $72$ seconds and 108 seconds respectively. If they all change simultaneously at $8$ a.m. then at what time will they again change simultaneously?
AnswerLet us find the $LCM$ of $48$, $72$ and 108 through prime factorisation:
$\begin{array}{c|c} 2 & 48 \\ \hline 2 & 24\\ \hline2&12\\ \hline2&6\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 72 \\ \hline 2 & 36\\ \hline2&18\\ \hline3&9\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 108 \\ \hline 2 & 54\\ \hline3&27\\ \hline3&9\\ \hline&3 \end{array}$
$48=2 \times 2 \times 2 \times 2 \times 3=2^4 \times 3 $
$ 72=2 \times 2 \times 2 \times 3 \times 3=2^3 \times 3^2$
$108=2 \times 2 \times 3 \times 3 \times 3=2^2 \times 3^3$
$\text { LCM of } 48,72,108 \text { is } 2^4 \times 3^3$
$= 16 × 27$sec
$= 432$sec
$=7$min $12$sec
Three bells toll together after $7$min $12$sec.
View full question & answer→Question 44 Marks
Find the leash number of square tiles required to pave the ceiling of a room $15\ m$ $17\ cm$ long and $9\ m$ $2\ cm$ broad.
AnswerLength of room $= 15\ m 17\ cm$
$= 1500cm + 17\ cm = 1517\ cm$
Breadth of room $= 9\ m 2\ cm$
$= 900\ cm + 2\ cm = 902\ cm$
Now, $H.C.F.$ of $1517$ and $902$ is $41$
Thus, area of each tile $= (41 × 41) = 1681 $ sq. cm
Hence, required number of tiles $=\frac{1517\times902}{1681}=814$
View full question & answer→Question 54 Marks
Three measuring rods are $64\ cm, 80\ cm$ and 96cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.
AnswerLet us find the $LCM$ of $64, 80$ and $96$ through prime factorization:
$\begin{array}{c|c} 2 & 64 \\ \hline 2 & 32\\ \hline2&16\\ \hline2&8\\ \hline2&4\\\hline&2 \end{array}$ $\begin{array}{c|c} 2 & 80 \\ \hline 2 & 40\\ \hline2&20\\ \hline2&10\\ \hline&5 \end{array}$ $\begin{array}{c|c} 2 & 96 \\ \hline 2 & 48\\ \hline2&24\\ \hline2&12\\ \hline2&6\\\hline&3 \end{array}$
$ 64=2 \times 2 \times 2 \times 2 \times 2 \times 2=2^6 $
$ 80=2 \times 2 \times 2 \times 2 \times 5=2^4 \times 5 $
$ 96=2 \times 2 \times 2 \times 2 \times 2 \times 3=2^5 \times 3$
$\text { L.C.M. of } 64,80 \text { and } 96$
$ =2^6 \times 5 \times 3=64 \times 15$
$=960 \mathrm{~cm}=9.6 \mathrm{~m}$
Therefore, the least length of the cloth that can be measured an exact number of times by the rods of $64\ cm, 80\ cm$ and $96\ cm = 9.6\ m$
View full question & answer→Question 64 Marks
In a seminar, the number of participants in Hindi, English and mathematics are $60, 84$ and $108$ respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.
AnswerTo find the minimum number of rooms required first find the maximum number of participants that can be accomodated in each room such that the number of participants in each room is same.
This can be determined by finding the $HCF$ of $60, 84 $and $108.$
$ 60=2^2 \times 3 \times 5 $
$ 84=2^2 \times 3 \times 7 $
$ 108=2^2 \times 3^2 $
$ H C F=2^2 \times 3 $
$ =12$
So, the minimum number of rooms required
$=\frac{\text{Total number of participants}}{12}$
$=\frac{60+84+108}{12}$
$=21$
View full question & answer→Question 74 Marks
Find the largest number which divides $546$ and $764$, leaving remainders $6$ and $8$ respectively.
Answer$546$ and $764$ are divided by the largest number leaving remainers $6$ and $8$ respectively.
$546 - 6 = 540$
$764 - 8 = 756$
So, $540$ and $756$ are exactly divisible by the required number.
Thus, the required number is the $HCF$ of $540$ and $756$
$ 540=2^2 \times 3^3 \times 5$
$756=2^2 \times 3^3 \times 7$
$\operatorname{HCF}(540,756)=2^2 \times 3^3=108$
which is the required number.
View full question & answer→Question 84 Marks
Six bells commence tolling together and toll at intervals of $2, 4, 6, 8, 10, 12$ minutes respectively. In $30$ hours, how many times do they toll together?
Answer$2 = 2 \times 1$
$4 = 2 \times 2$
$6 = 2 \times 3$
$8 = 2 \times 2 \times 2$
$10 = 2 \times 5$
$12 = 2 \times 2 \times 3$
$L.C.M. of 2, 4, 6, 8, 10, 12$ minutes
$= 2 \times 2 \times 2 \times 3 \times 5 = 120$ minutes
$= 2$ hours
After every $2$ hours they toll together,
Required number of times $=\Big(\frac{30}{2}+1\Big)$
$$= 16 times
View full question & answer→Question 94 Marks
Find the $HCF$ and $LCM$ of $\frac{8}{9},\frac{10}{27}$ and $\frac{16}{81}.$
Answer$\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{HCF of the numerators}}{\text{LCM of the denominators}}$
$=\frac{\text{HCF}(8,10,16)}{\text{LCM}(9,27,81)}$
and
$\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{LCM of the numerators}}{\text{HCF of the denominators}}$
$=\frac{\text{LCM}(8,10,16)}{\text{HCF}(9,27,81)}$
Consider,
$8=2^3 $
$ 10=2 \times 5 $
$ 16=2^4$
So, the $\operatorname{HCF}(8,10,16)=2$ and $\operatorname{LCM}(8,10,16)=2^4 \times 5=80$
$ 9=3^2 $
$ 27=3^3 $
$ 81=3^4$
So, the $\operatorname{HCF}(9,27,81)=3^2=9$ and $\operatorname{LCM}(9,27,81)=3^4=81$
$\Rightarrow\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{2}{81}$ and $\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{80}{9}$
View full question & answer→Question 104 Marks
An electronic device makes a beep after every $60$ seconds. Another device makes a beep after $62$ seconds. They beeped together at $10$ a.m. At what time will they beep together at the earliest?
AnswerInterval of beeping together $= LCM (60$ seconds, $62$ seconds)
The prime factorization of $60$ and $62:$
$60 = 30 \times 2, 62 = 31 \times 2$
$\therefore L.C.M$ of $60$ and $62$ is $30 31 \times 2 = 1860$ sec $= 31min$
$\therefore$ electronic device will beep after every $31$ minutes
After $10$ a.m., it will beep at $10$hrs $31$ minutes
View full question & answer→Question 114 Marks
Find the maximum number of students among whom $1001$ pens and $910$ pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.
AnswerTotal pens $= 1001$
Total pencils $= 910$
We need to find maximum no.of students among whom $1001$ pens and $910$ pencils can be distributed in such a way that each students get same no.of pens and pencils.
Then we need to find $HCF$ of $1001$ and $910$
Prime factorization of,
$1001 = 7 \times 11 \times 13$
$910 = 2 \times 5 \times 7 \times 13$
$HCF=$ product of commom prime factor of least power
$HCF = 7 \times 13 = 91$
Here $HCF$ of $1001$ and $910$ is $91.$
Hence among $91$ students $1001$ pens and $910$ pencils can be distributed such that each student get same no. of pens and pencils.
View full question & answer→Question 124 Marks
Show that every positive odd integer is of the form $(4 q+1)$ or $(4 q+3)$ for some integer $q$.
AnswerLet 'a' be a given positive od integer.
On dividing ' $a$ ' by $4$ , let $q$ be the quotient and $r$ be the remainder.
Then, by euclid's algorithm, we have
$a = 4q + r$, where $0 ≤ r < 4$
$⇒ a = 4q + r$, where$ r = 0, 1, 2, 3$
$⇒ a = 4q$ or$ a = 4q + 1$ or $a = 4q + 2$ or $a = 4q + 3$
But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$But, $a=4 q$ and $a=4 q+2=2(2 q+1)$ are clearly even.
Thus, when ' $a$ ' is odd, it is of the form:
$a=(4 q+1) \text { or }(4 q+3) \text { for some integer } q$
View full question & answer→Question 134 Marks
Show that $\big(4+3\sqrt2\big)$ is irrational.
AnswerIf possible, let $\big(4+3\sqrt2\big)$ be rational.
Then 4 and $3\sqrt2$ are rational.
$\Rightarrow4+3\sqrt2-4$ is rational $\dots(\because$ diference of two rationals is rational$)$
$\Rightarrow3\sqrt2$ is rational
$\Rightarrow\sqrt2$ is rational $\dots(\because 3 $ is rational$)$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\big(4+3\sqrt2\big)$ is rational.
Hence, $\big(4+3\sqrt2\big)$ is irrational.
View full question & answer→Question 144 Marks
Prove that $\frac{2}{\sqrt7}$ is an irrational number.
HINT: $\frac{2}{\sqrt7}=\Big(\frac{2}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}\Big)=\frac{2}{7}.\sqrt7$
Answer$\frac{2}{\sqrt7}=\Big(\frac{2}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}\Big)=\frac{2}{7}.\sqrt7$
Let $\frac{2}{7}\sqrt7$ is a rational number.
$\therefore\frac{2}{7}\sqrt7=\frac{\text{p}}{\text{q}},$ where p and q are some integers and $HCF (p, q) = 1 ...(1)$
$\Rightarrow2\sqrt7\text{q}=\text{7p}$
$\Rightarrow\big(2\sqrt7\text{q}\big)^2=(\text{7p})^2$
$\Rightarrow7(4\text{q}^2)=\text{49p}^2$
$\Rightarrow\text{4q}^2=\text{7p}^2$
$\Rightarrow q^2$ is divisible by $7$
$\Rightarrow q$ is divisible by $7 ...(2)$
Let $q = 7m$, where m is some integer.
$\therefore2\sqrt7\text{q}=\text{2p}$
$\Rightarrow\Big[2\sqrt7\text{q}(7\text{m})\Big]^2=\text{7p}^2$
$\Rightarrow343(4\text{m}^2)=\text{49p}^2$
$\Rightarrow7(\text{4m}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by $7$
\Rightarrow p is divisible by $7 ...(3)$
From $(2)$ and $(3), 7$ is a common factor of both $p$ and $q$, which contradicts $(1).$
Hence, our assumption is wrong.
Thus, $\frac{2}{\sqrt7}$ is irrational.
View full question & answer→Question 154 Marks
Prove that $\big(4-5\sqrt2\big)$ is an irrational number.
AnswerLet $\text{x}=4-5\sqrt2$ be a rational number.
$\text{x}=4-5\sqrt2$
$\Rightarrow\text{x}^2=\big(4-5\sqrt2\big)^2$
$\Rightarrow\text{x}^2=(4)^2+\big(2\sqrt3\big)^2-2(4)\big(5\sqrt2\big)$
$\Rightarrow\text{x}^2=16+50-40\sqrt2$
$\Rightarrow\text{x}^2-66=-40\sqrt2$
$\Rightarrow\frac{66-\text{x}^2}{40}=\sqrt2$
Since x is a rational number, $x^2$ is also a rational number.
$⇒ 66 − x^2$ is a rational number
$\Rightarrow\frac{66-\text{x}^2}{40}$ is a rational number
$\Rightarrow\sqrt2$ is a rational number
But $\sqrt2$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(4-5\sqrt2\big)$ is an irrational number.
View full question & answer→Question 164 Marks
Prove that $\big(5-2\sqrt3\big)$ is an irrational number.
AnswerLet $\text{x}=5-2\sqrt3$ be a rational number.
$\text{x}=5-2\sqrt3$
$\Rightarrow\text{x}^2=\big(5-2\sqrt3\big)^2$
$\Rightarrow\text{x}^2=(5)^2+\big(2\sqrt3\big)^2-2(5)\big(2\sqrt3\big)$
$\Rightarrow\text{x}^2=25+12-20\sqrt3$
$\Rightarrow\text{x}^2-37=-20\sqrt3$
$\Rightarrow\frac{37-\text{x}^2}{20}=\sqrt3$
Since x is a rational number, $x^2$ is also a rational number.
$⇒ 37 − x^2$ is a rational number
$\Rightarrow\frac{37-\text{x}^2}{20}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(5-2\sqrt3\big)$ is an irrational number.
View full question & answer→Question 174 Marks
Show that one and only one out of $n,(n+2)$ and $(n+4)$ is divisible by $3$ , where $n$ is any positive integer.
Answeron dividing ' $n$ ' by $3$ , let ' $q$ ' be the quotient and ' $r$ ' be the remainder.
Then, $n=3 q+r$, where $0 \leq r<3$
$\Rightarrow n=3 q+r \text {, where } r=0,1 \text { or } 2$
$\Rightarrow n=3 q \text { or } n=3 q+1 \text { or } n=3 q+2$
Case 1: If $n=3 q$, then ' $n$ ' is clearly divisible by 3
Case 2: If $n=3 q+1$, then $(n+2)=(3 q+3)=3(q+1)$, which is clearly divisible by $3$
In this case, $( n +2)$ is divisible by $3$
Case 3: If $n=3 q+2$, then $(n+4)=(3 q+6)=3(q+2)$, which is clearly divisible by $3$
In this case, $(n+4)$ is divisible by $3$
Hence, one and only one out of $n,(n+2)$ and $(n+4)$ is divisible by $3$
View full question & answer→Question 184 Marks
Prove that $\sqrt3$ is an irrational number.
AnswerIf possible, let $\sqrt3$ be rational and let its simplest form be $\frac{\text{a}}{\text{b}}$
The, $a$ and $b$ are integers having no common factor other than $1$, and $\text{b}\neq0$
Now, $\sqrt3=\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow3=\frac{\text{a}^2}{\text{b}^2}$ ...(On squaring both sides)
$\Rightarrow 3 b^2=a^2 $
$ \Rightarrow 3 \text { divides } a^2 \ldots\left[\because 3 \text { divies } 3 b^2\right] $
$ \Rightarrow 3 \text { divides a } \ldots\left[\because 3 \text { is prime and } 3 \text { divides } a^2 \Rightarrow 3 \text { divides } a\right]$
Let $\mathrm{a}=3 \mathrm{c}$ for some integers c .
Putting $a=3 c$ in (i), we get
$3 \mathrm{~b}^2=9 \mathrm{c}^2 $
$ \Rightarrow \mathrm{~b}^2=3 \mathrm{c}^2 $
$ \Rightarrow 3 \text { divides } \mathrm{b}^2 \ldots\left[\because 3 \text { divies } 3 \mathrm{c}^2\right] $
$ \Rightarrow 3 \text { divides } \mathrm{b} \ldots\left[\because 3 \text { is prime and } 3 \text { divides } \mathrm{a}^2 \Rightarrow 3 \text { divides } \mathrm{a}\right]$
Thus, $3$ is a common factor of a and $b$
But, this contradicts the fact that $a$ and $b$ have no common factor other than $1$
The contradiction arises by assuming that $\sqrt3$ is rational.
Hence, $\sqrt3$ is irrational.
View full question & answer→Question 194 Marks
Prove that $5\sqrt2$ is an irrational number.
AnswerLet $5\sqrt2$ is a rational number.
$\therefore5\sqrt2=\frac{\text{p}}{\text{q}},$ where p and q are some integers and $HCF (p, q) = 1 ...(1)$
$\Rightarrow5\sqrt2\text{q}=\text{p}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{p}^2$
$\Rightarrow2(25\text{q}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by $2$
$\Rightarrow p$ is divisible by $2 ...(2)$
Let $p = 2m$, where $m$ is some integer.
$\Rightarrow5\sqrt2\text{q}=\text{2m}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{2m}^2$
$\Rightarrow2(25\text{q}^2)=\text{4m}^2$
$\Rightarrow\text{25q}^2=\text{2m}^2$
$\Rightarrow q^2$ is divisible by $2$
\Rightarrow q is divisible by $2 ...(3)$
From $(2)$ and $(3), 2$ is a common factor of both p and q, which contradicts $(1).$
Hence, our assumption is wrong.
Thus, $5\sqrt2$ is irrational.
View full question & answer→Question 204 Marks
Prove that $\big(2\sqrt3-1\big)$ is an irrational number.
AnswerLet $\text{x}=2\sqrt3-1$ be a rational number.
$\text{x}=2\sqrt3-1$
$\Rightarrow\text{x}^2=\big(2\sqrt3-1\big)^2$
$\Rightarrow\text{x}^2=\big(2\sqrt3\big)^2+1^2-2\big(2\sqrt3\big)(1)$
$\Rightarrow\text{x}^2=12+1-4\sqrt3$
$\Rightarrow\text{x}^2-13=-4\sqrt3$
$\Rightarrow\frac{13-\text{x}^2}{4}=\sqrt3$
Since x is a rational number, $x^2$ is also a rational number.
$⇒ 13 − x^2$ is a rational number
$\Rightarrow\frac{13-\text{x}^2}{4}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(2\sqrt3-1\big)$ is an irrational number.
View full question & answer→