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Number Systems Test - 36

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Number Systems Test - 36
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  • Question 1
    1 / -0
    Find one irrational number between 0.2101 and 0.2222.
    Solution
    Given Numbers are $$0.2101$$ which is terminating decimal and $$0.22222$$__ which is non terminating recurring decimal.

    $$\Rightarrow$$ Irrational Numbers are non terminating and nonrecurring decimals

    In between $$0.2101$$ and $$0.2222 ...$$, we can find $$0.220$$

    Now,
       We can write the irrational number as given in the above figure $$\rightarrow $$
    REF. IMAGE

    Hence $$0.2201001000100001....$$ is an irrational number that lies in between $$0.2101$$ and $$0.2222....$$

  • Question 2
    1 / -0
    Give an example of two irrational numbers whose sum is an irrational number.
    Solution
    $$2\sqrt{5},3\sqrt{5}$$ are the irrational numbers and their sum,

    $$2\sqrt{5}+3\sqrt{5} = 5 \sqrt 5$$ is an irrational number.
  • Question 3
    1 / -0
    Give an example of two irrational numbers whose sum is a rational number.
    Solution
    $$\sqrt{5},-\sqrt{5}$$ are the irrational numbers, their sum

    $$\sqrt{5}-\sqrt{5}=0$$ is a rational number
  • Question 4
    1 / -0
    Simplify:$$\dfrac{7+3\sqrt{5}}{3+\sqrt{5}}-\dfrac{7-3\sqrt{5}}{3-\sqrt{5}}$$
    Solution
    Given,

    $$\dfrac{7+3\sqrt{5}}{3+\sqrt{5}}-\dfrac{7-3\sqrt{5}}{3-\sqrt{5}}$$

    $$=\dfrac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)-\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}$$
    $$=\dfrac{21-7\sqrt5+9\sqrt5-15-21-7\sqrt5+9\sqrt5+15}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}$$
    $$=\dfrac{4\sqrt{5}}{3^2-(\sqrt 5)^2}$$   By using $$(a+b)(a-b)=a^2-b^2$$

    $$=\dfrac{4\sqrt 5}{4}=\sqrt 5$$
  • Question 5
    1 / -0
    $$a\times a\times b\times b\times b$$ can be written as 
    Solution
    we have, $$a\times a\times b\times b\times b$$
    $$=a^2 \times b^3 =a^2b^3$$

    Hence, $$a\times a\times b\times b\times b$$ can be written as $$a^2b^3$$
  • Question 6
    1 / -0
    Which of the following rational numbers is positive?
    Solution
    $$\Rightarrow \dfrac{-3}{-4}= \dfrac{3}{4}$$
    The numerator and denominator both are negative, it will cancel out each other and gives a positive rational number.
  • Question 7
    1 / -0
    Which of the following rational numbers is negative?
    Solution
    $$(a) \Rightarrow -\left(\dfrac{-3}{7}\right)=\dfrac{3}{7}$$

    $$(b) \Rightarrow \dfrac{-5}{-8}=\dfrac{5}{8}$$

    $$(c) \Rightarrow \dfrac{9}{8}=\dfrac{9}{8}$$

    $$(d) \Rightarrow \dfrac{3}{-7}=-\dfrac{3}{7}$$

    $$\therefore -\dfrac{3}{7}$$ is a negative rational number.
  • Question 8
    1 / -0
    A number of the form $$\dfrac{p}{q}$$ is said to be a rational number if
    Solution

  • Question 9
    1 / -0
    $$\left(\dfrac {1}{10}\right)^0$$ is equal to
    Solution
    Using law of exponents, $$a^0=1$$
    where $$a\neq 1$$
    $$\Rightarrow \ \left(\dfrac {1}{10}\right)^0 =1$$
  • Question 10
    1 / -0
    By solving $$(6^0 -7^0) \times (6^0+7^0)$$, we get ________.
    Solution
    $$0$$
    $$(6^0 -7^0) \times (6^0+7^0)$$
    $$=(1-1) \times (1+1)$$
    $$=0\times 2=0$$
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