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 $6^n= (2 \times 3)^n$
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$
Product of two numbers $336$ and $54 = 336 \times 54 = 18144$
$HCF \times LCM = 6 \times 3024 = 18144$
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 \times} 3^1=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, ${LCM}(6,72,120)=2^{3 \times} \times 3^2 \times 5^1=360$

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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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