3 Marks Question

Question 13 Marks

Using division algorithm find quotient and remainder dividing $x^3+13 x^2+x-2$ by $2 x+1$

Answer

The prime factorisation of $72$ and $120 ,$ respectively, is given by:

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Question 23 Marks

Find the greatest number of six digit number exactly divisible by $18,24$ and $36 .$

Answer

Greatest number of $6$ digits is $999999$ The numbers given are $18, 24$ and $36.$
Here $\text{LCM}$ of $18, 24, 36 .$
$18=2 \times 3 \times 3=2 \times 3^2$
$36=2 \times 2 \times 3 \times 3=2^2 \times 3^2$
$24=2 \times 2 \times 2 \times 3=2^3 \times 3$
Now,
The $\text{LCM}$ of $18,24$ and $36=2^3 \times 3^2=72$
Now dividing $999999$ by $72$
$\frac{999999}{72}=13888$ with remainder $63$
And, $999999-63=999936$
Thus $999936$ is the greatest number $6-$digit number divisible by $18, 24$ and $36.$

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Question 33 Marks

Explain whether the number $3 \times 5 \times 13 \times 46+23$ is a prime number or a composite number.

Answer

$3 \times 5 \times 13 \times 46+23$It can be re-written as:$
\begin{aligned}
3 \times 5 \times 13 \times 2 \times 23+23 & =23(3 \times 5 \times 13 \times 2+1) \\
& =23 \times 391=8993
\end{aligned}
$
Here 8993 is written as the product of two different numbers $23 \times 391$.
It means it has 23 and 391 as its factors other than 1 and 8993.
Hence, it is a composite number.

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Question 43 Marks

Pens are sold in pack of $8$ and notepads are sold in pack of $12$ . Find the least number of pack of each type that one should buy so that there are equal number of pens and notepads.

Answer

Pens are sold in pack of $8$ and notepads are sold in pack of $12 ,$
$\text { LCM}$ of $8$ and $12$ is:$8=2^3 \text { and } 12=2^2 \times 3$
$\text { LCM }=2^3 \times 3=8 \times 3=24$
Least number of pack of pen $=\frac{24}{8}=3$
Least number of pack of notepads $=\frac{24}{12}=2$
Hence, $3$ packs of pen and $2$ packs of notepads one should buy to get $24$ pens and notepads.

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Question 53 Marks

Find $\text{HCF}$ and $\text{LCM}$ of $180,252$ and $324.$

Answer

Consider $252$ and $324.$
Let, $a=324$ and $b=252$ by Euclid's division lemma$-$
$a=b q+r, 0 < o r=r < b$
$324=252 \times 1+72$
$252=72 \times 3+36$
$72=36 \times 2+0$
Therefore, $\text{HCF} (252,324)=36$
Now consider $36$ and $180$ , here $a=180$ and $b=36$. By Euclid's division
$a=b q+r, 0 < o r=r < b$
$180=36 \times 5+0$
Therefore, $\operatorname{HCF}(180,36)=36$

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Question 63 Marks

Using Euclid's division algorithm, find whether the pair of numbers $847,2160$ are co$-$primes or not.

Answer

$a=2160, b=847$By Euclid's lemma, given positive integers $a$ and $b$ , there exist unique integers $q$ and $r$ satisfying
$a=b q+r, 0 \leq r < b$.
As $2160>847$, we apply the division lemma to $2160$ and $847$ , to get $2160=847 \times 2+466$
Since the remainder $466 \neq 0$, we apply the division lemma to $847$ and $466$ , and continue the same process till we get remainder $0 .$
$847=466 \times 1+381$
$466=381 \times 1+85$
$381=85 \times 4+41$
$85=41 \times 2+3$
$41=3 \times 13+2$
$3=2 \times 1+1$
$2=1 \times 1+1$
$1=1+0$
As $1$ is the $\text{HCF}$ of $847$ and $2160.847$ and $2160$ are the co$-$primes.

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Question 73 Marks

Find the $\text{LCM}$ and $\text{HCF}$ of $336$ and $54$ and verify that $\text{LCM} \times \text{HCF} =$ Product of the two numbers.

Answer

Find the factors of $336$ and $54.$
$336=2 \times 2 \times 2 \times 2 \times 3 \times 7$
$54=2 \times 3 \times 3 \times 3$
$\text{HCF}$ of $336$ and $54=2 \times 3=6$
$\text{LCM}$ of $336$ and $54=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7=3024$
Product of two numbers $=336 \times 54=18144$
Hence verified.

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Question 83 Marks

Show that square of any positive integer is either of the form $3 m$ or $(3 m+1)$ for some integer $m$.

Answer

Let $c$ be any positive number and $d=3$
Then $c=3 q+r$ for $q \geq 0$
Also, $r=0,1,2$ as $0 \leq r \leq 3$
Thus, $c=3 q$ or $c=3 q+1$ or $c=3 q+2$
$\Rightarrow c^2=(3 q)^2$ or $(3 q+1)^2$ or $(3 q+2)^2$
$\Rightarrow c^2=3 \times\left(3 q^2\right)$ or $9 q^2+6 q+1$ or $9 q^2+12 q+4$
$\Rightarrow c^2=3 \times\left(3 q^2\right)$ or $3\left(3 q^2+2 q\right)+1$ or $3\left(3 q^2+4 q+1\right)+1$
$\Rightarrow c^2=3 m_1$ or $3 m_2+1$ or $3 m_3+1$ where
$m_1=3 q^2, m_2=3 q^2+2 q$ and $m_3=3 q^2+4 q+1$
Hence, square of any positive integer is either of
$3 m$ or $(3 m+1)$ for some integer m .

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Question 93 Marks

Prove that $\sqrt{2}$ is an irrational number.

Answer

Let assume on the contrary that $\sqrt{2}$ is arational number.
Then, there exists positive integer $a$ and $b$ such that $\sqrt{2}=\frac{ a }{ b }$ where, $a$ and $b$ are co primes i.e. their $\text {HCF}$ is $1.$
$\Rightarrow \sqrt{2}=\left(\frac{a}{b}\right)^2$
$\Rightarrow 2=\frac{a^2}{b^2}$
$\Rightarrow a^2=2 b^2$
$\Rightarrow a^2 \text { is multiple of } 2$
$\Rightarrow a \text { is a multiple of } 2$
$\Rightarrow a=2 c \text { for some integer } c .$
$\Rightarrow a^2=4 c^2$
$\Rightarrow 2 b^2=4 c^2$
$\Rightarrow b^2=2 c^2$
$\Rightarrow b^2 \text { is a multiple of } 2$
$b \text { is a multiple of 2...(ii) }$
From $(i)$ and $(ii)$, a and b have at least $2$ as a common factor. But this contradicts the fact that a and b are co$-$prime. This means that $\sqrt{2}$ is an irrational number.

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Question 103 Marks

Find the HCF and LCM of 26, 65 and 117, using prime factorisation.

Answer

HCF by prime Factorization methodFirst, we have to find the highest common Factor of 26,65 and 117
Now let us write the prime factors of 26, 65, and 117.
$
\begin{aligned}
26 & =2 \times 13 \\
65 & =5 \times 13 \\
117 & =3 \times 3 \times 13
\end{aligned}
$
The common factor of 26,65 , and 117 is 13
Therefore, $\operatorname{HCF}(26,65,117)=13$
LCM by prime factorization method
To calculate the LCM of 26, 65 and 117
First, list the common factors of each number
$
\begin{aligned}
26 & =2 \times 13 \\
65 & =5 \times 13 \\
117 & =3 \times 3 \times 13 \\
\text { LCM } & =2 \times 5 \times 3 \times 3 \times 13 \\
& =1170
\end{aligned}
$

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Question 113 Marks

Find $\text {HCF}$ and $\text {LCM}$ of $404$ and $96$ and verify that $HCF \times LCM =$ Product of the two given numbers.

Answer

The prime factors of:
$96=2 \times 2 \times 2 \times 2 \times 2 \times 3$
$404=2 \times 2 \times 101$
Therefore the $\text {HCF =}$ Product of smallest power of each common prime factor $=2 \times 2=4$
And LCM $=$ Product of greatest power of each prime factor $=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 101$
$=9696$
To prove:
$HCF \times LCM=101 \times 96$
$\text { Here }, HCF \times LCM=4 \times 9696$
$=38784$
$101 \times 96=38784$
Hence proved, $HCF \times LCM =$ Product of the two given numbers.

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Question 123 Marks

Prove that $\sqrt{3}$ is an irrational number.

Answer

Let $\sqrt{3}$ be arational number, then its simplest form is $\frac{a}{b} ($where $a$ and $b$ are co$-$primes$)(\sqrt{3})^2=\frac{a^2}{b^2}$
$\Rightarrow a^2=3 b^2$
$\Rightarrow 3$ divides $a^2$
$\Rightarrow 3$ divides $a$
Let $a =3 c$, for some integer $c$
Putting $a =3 c$ in $(1),3 b^2=9 c^2$
$\Rightarrow b^2=3 c^2$
$\Rightarrow 3$ divides $b^2$
$\Rightarrow 3$ divides $b$
Thus, $3$ is a common factor of $a$ and $b$
But this is not possible as $a$ and $b$ are co$-$primes
$\Rightarrow$ Our assumption that $\sqrt{3}$ is rational is wrong
$\Rightarrow \sqrt{3}$ is irrational

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Question 133 Marks

Prove that $(3+2 \sqrt{5})$ is an irrational number, given that $\sqrt{5}$ is an irrational number.

Answer

If possible, let $a=(3+2 \sqrt{5})$ be a rational number On squaring both sides, we get
$a^2=(3+2 \sqrt{5})^2$
$\Rightarrow a^2=9+20+12 \sqrt{5}$
$\Rightarrow a^2=29+12 \sqrt{5}$
$\Rightarrow \sqrt{5}=\frac{a^2-29}{12}$since $'a\ '$ is a rational number,
$\therefore \frac{ a ^2-29}{12}$ is also a rational number
$\Rightarrow \sqrt{5}$ is a rational number
but It is given that $\sqrt{5}$ is an irrational number.
Hence, it is a contradiction
So, $3+2 \sqrt{5}$ is an irrational number.

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