2 Marks Questions

Question 12 Marks

Prove that $6+\sqrt { 2 }$ is irrational.

Answer

Let us assume that $6 + \sqrt2$ is a rational number.
So we can write this number as
$6 + \sqrt2 = a/b$
Here a and b are two co-prime numbers and b is not equal to $0$
Subtract 6 both side we get
$\sqrt2 = a/b – 6$
$\sqrt2 = (a-6b)/b$
Here $a$ and $b$ are integers so (a-6b)/b is a rational number. So $\sqrt2$ should be a rational number. But $\sqrt2$ is an irrational number. It is a contradiction.
Hence result is $6 + \sqrt2$ is a irrational number

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Question 22 Marks

Prove that $7 \sqrt { 5 }$ is irrational.

Answer

We can prove $7 \sqrt { 5 }$ irrational by contradiction.
Let us suppose that $7 \sqrt { 5 }$ is rational.
It means we have some co-prime integers $a$ and $b (b≠ 0)$ such that
$7 \sqrt { 5 } = \frac { a } { b }$
$\Rightarrow \sqrt { 5 } = \frac { a } { 7 b }$ $.......(1)$
$R.H.S$ of $(1)$ is rational but we know that $\sqrt { 5 }$ is irrational.
It is not possible which means our supposition is wrong.
Therefore, $7 \sqrt { 5 }$ cannot be rational.
Hence, it is irrational.

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Question 32 Marks

Check whether $6^n$ can end with the digit $0$ for any natural number $n$.

Answer

If any number ends with the digit $0$, it should be divisible by $10$ or in other words, it will also be divisible by $2$ and 5 as $10 = 2 \times 5$ Prime factorisation of $ 6^n= (2 ×3)^n$ It can be observed that $5$ is not in the prime factorisation of $6^n$.
Hence, for any value of n, $6^n$ will not be divisible by $5$.
Therefore, $6^n$ cannot end with the digit $0$ for any natural number $n$.

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Question 42 Marks

Find the $LCM$ and $HCF$ of $336$ and $54$ pairs of integers and verify that $LCM$ $\times$ $HCF =$ product of the two numbers.

Answer

$336$ and $54$
$ 336=2 \times 2 \times 2 \times 2 \times 3 \times 7=2^4 \times 3 \times 7 $
$ 54=2 \times 3 \times 3 \times 3=2 \times 3^3 $
$ H C F=2 \times 3=6 $
$ L C M=2^4 \times 3^3 \times 7=3024$
$\text { Product of two numbers } 336 \text { and } 54=336 \times 54=18144$
$\mathrm{HCF} \times \mathrm{LCM}=6 \times 3024=18144$
Hence, product of two numbers $= HCF \times LCM$

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Question 52 Marks

Find the $HCF$ and $LCM$ of $6, 72$ and $120$, using the prime factorisation method.

Answer

We have: $6=2 \times 3,72=2^{3} \times 3^{2}, 120=2^{3} \times 3 \times 5$
Here, $2^1$ and $3^1$ are the smallest powers of the common factors $2$ and $3$ respectively.
So, $HCF$ $(6, 72, 120) = 2^1\times3^1= 2 \times 3 = 6$
$2^3, 3^2$ and $5^1$ are the greatest powers of prime factors $2, 3$ and $5$ respectively involved in the three numbers.
So, $LCM$ $(6, 72, 120) = 2^3 \times 3^2 \times 5^1= 360$

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Question 62 Marks

Find the $HCF$ of $96$ and 404 by prime factorisation method. Hence, find their $LCM$.

Answer

We have,
$96 = 2 ^ { 5 } \times 3$ and $404 = 2 ^ { 2 } \times 101$
$\therefore \quad \mathrm { HCF } = 2 ^ { 2 } = 4$
Now, $\mathrm { HCF } \times \mathrm { LCM } = 96 \times 404$
$\Rightarrow \quad \mathrm { LCM } = \frac { 96 \times 404 } { \mathrm { HCF } } = \frac { 96 \times 404 } { 4 }$ $= 96 \times 101 = 9696$

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Question 72 Marks

Find the $LCM$ and $HCF$ of $6$ and $20$ by the prime factorisation method.

Answer

We have: $6=2^{1} \times 3^{1}$ and $20 =2 \times 2 \times 5=2^{2} \times 5^{1}$
Now $HCF$$(6, 20) = 2^1= 2 =$ Product of the smallest power of each common prime factor
and $LCM$ $(6, 20) = 2^{2} \times 3^{1} \times 5^{1}$ $= 60 =$ Product of the greatest power of each prime factor

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Question 82 Marks

Prove that $2+3 \sqrt{5}$ is irrational.

Answer

Let's assume, for the sake of contradiction, that $2+3 \sqrt{5}$ is a rational number.
By definition, we can express a rational number as a fraction $\frac{a}{b}$,where a and b are integers and $b \neq 0$.
So, we have:
$2+3 \sqrt{5}=\frac{a}{b}$
Now, we isolate the $\sqrt{5}$ term:
$\begin{array}{l}3 \sqrt{5}=\frac{a}{b}-2 \\ 3 \sqrt{5}=\frac{a-2 b}{b} \\ \sqrt{5}=\frac{a-2 b}{3 b}\end{array}$
Since a and b are integers, the expression $\frac{a-2 b}{3 b}$ is a rational number. This would mean that $\sqrt{5}$ is also a rational number.
However, we know that $\sqrt{5}$ is an irrational number. This contradicts our initial assumption.
Therefore, our assumption that $2+3 \sqrt{5}$ is rational must be false.
Hence,$2+3 \sqrt{5}$ is an irrational number.

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