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Squares and Square Roots Test - 15

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Squares and Square Roots Test - 15
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  • Question 1
    1 / -0
    Find the square root of $$2209$$.
    Solution
    So, the square root of $$2209$$ is $$47$$.

  • Question 2
    1 / -0
    Find the square root of $$\displaystyle 6 \frac{7}{8}$$ correct to two decimal places 
    Solution
     $$\displaystyle 6 \frac{7}{8}$$
    $$=\dfrac{55}{8}$$
    $$=6.875$$

    $$2$$
    $$2$$
    $$6.875$$
    $$4$$
    $$46$$
      $$6$$
    $$287$$
    $$276$$
    $$522$$
       
    $$1150$$
    $$1044$$

    $$6$$
                     


    Therefore,
    Square root of $$\displaystyle 6 \frac{7}{8}=2.62$$
                                 
  • Question 3
    1 / -0
    Find the square root of $$\displaystyle 1 \frac{5}{16}$$  correct to two decimal places
    Solution
     $$\displaystyle 1 \dfrac{15}{16}$$
    $$=\dfrac{21}{16}$$
    $$=1.3125$$

    $$1$$
    $$1$$
    $$1.3125$$
    $$1$$
    $$21$$
      $$1$$
    $$31$$
    $$21$$
    $$224$$
       $$4$$
    $$1025$$
    $$896$$
    $$2285$$$$12900$$
    $$11425$$
                     
    $$1475$$

    Therefore,
    Square root of $$\displaystyle 1 \frac{15}{16}=1.145$$
                                       $$=1.15$$
  • Question 4
    1 / -0
    Consider the Following values of the three given number $$\displaystyle \sqrt{103},$$ $$\displaystyle \sqrt{99.35},$$ $$\displaystyle \sqrt{102.20},$$
    1.10.1489 (approx,)
    2.10.109(approx,)
    3.9.967 (approx,)
    The correct sequence of the these values matching with the above number is:
    Solution
    $$\sqrt{3}=10.14889$$
    $$\sqrt{99.35}=9.967$$
    $$\sqrt{102.30}=10.1094$$
    Hence,The correct sequence of the these values matching with the above number is:1,3,2
  • Question 5
    1 / -0
    A four digit perfect square number whose first two digits and last two digits taken seperately are also perfect squares is:
    Solution
    $$1,681$$ is the only perfect square from given options.
    $$1681=41\times 41$$
    $$1681=41^2$$
    $$\therefore$$  $$1681$$ is a square of $$41$$.

    First two digits and last two digits of $$1681$$ are $$16$$ and $$81$$
    $$16=4^2$$ and $$81=9^2$$

    We can see $$16$$ and $$81$$ are perfect squares of $$4$$ and $$9$$.

    $$\therefore$$  $$1681$$ is a four digit square number whose first two digits taken separately are also perfect square.  
  • Question 6
    1 / -0
    The value of $$\sqrt{0.9}$$ is (approximately)
    Solution
    $$\sqrt{0.9} =  {\sqrt{0.90} = \sqrt{\cfrac{90}{100}} = \cfrac{\sqrt{90}}{10}}$$ $$=  \cfrac{9.4}{10} = 0.94$$
  • Question 7
    1 / -0
    Find the square root of 784 by prime factorisation method.
    Solution
    $$784=\underline { 2\times2 } \times\underline { 2\times2 } \times\underline { 7\times7 } \\ \Rightarrow \sqrt { 784 } =2\times2\times7=28$$
  • Question 8
    1 / -0
    Which one of the following can't be the square of natural number?
    Solution
    As we know that all the perfect square number ends with  $$1,4,9,6,5,00$$
    $$2452$$ is not a perfect square because it ends in 2.
     $$\text {Also} \quad 1936= (44)^2\\1369=(37)^2\\6561=(81)^2$$
  • Question 9
    1 / -0
    The smallest number by which 396 must be multiplied so that the product becomes a perfect square is:
    Solution
    $$396\, =\, 2\, \times\, 2\, \times\, 3\, \times\, 3\, \times\, 11$$
    So 396 should be multiplied by 11 to make the product a perfect square.
  • Question 10
    1 / -0
    The least number by which 216 must be divided to make the result a perfect square, is
    Solution
    $$216 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$$
    Clearly, in order to make it a perfect square, it must be divided by $$2 \times 3$$ i.e.,$$ 6$$.
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