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Squares and Square Roots Test - 13

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Squares and Square Roots Test - 13
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  • Question 1
    1 / -0
    Find the square root correct upto $$2$$ decimals $$60.92$$.
    Solution
    $$\sqrt{6092}$$

    $$=\sqrt{\dfrac{6092}{100}}$$

    $$=\dfrac{78.05}{10}$$

    $$=7.81$$
  • Question 2
    1 / -0
    The smallest four digit number which is perfect square is ____.
    Solution
    Consider, 
    $$31\times 31=961$$
    $$32\times 32=1024$$
    $$\therefore$$    Smallest four-digit number which is perfect square is $$1024$$.
  • Question 3
    1 / -0
    Which of the following will have $$4$$ at the units place?
    Solution
    Unit digit of $$62 $$ is $$2$$ and the square of $$2$$ is $$ 4$$.
  • Question 4
    1 / -0
    If $$\frac{18\times 72\times x}{48\times 315} = \sqrt{81}$$, then $$x = \underline{                }$$
    Solution

  • Question 5
    1 / -0
    There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
    Solution
    Total number of children in a school = $$500$$
    For P.T. Drill number of rows = number of columns =$$ x$$
    The number of children who will be left out = $$500 - x^2$$. The square root of $$500$$ gives remainder as 16 that means, the square of $$22$$ is less than $$500$$ by $$16$$
    To make it a perfect square = $$500-16 = 484$$
    Therefore, number of children left will be $$16$$

  • Question 6
    1 / -0
    The square root of 350 lies between which pair of numbers?
  • Question 7
    1 / -0
    The least number which is a perfect square and has $$540$$ as a factor is
    Solution
    The number must be a product of $$540$$ and another number.
    $$540 = 9\times 4\times 15$$
    Now, the least number to be multiplied with $$540$$ is $$15$$ to make it a perfect square.
    So, the required number is $$540\times 15 = 8100$$
  • Question 8
    1 / -0
    Pythagorean triplet of given number $$6$$ is
    Solution
    $$2k$$, $$(k^2 + 1)$$ and $$(k^2 - 1)$$ form a Pythagorean triplet for any number, k > 1

    Let $$2k = 6$$
    Hence, $$k = 3$$

    and, $$k^2 + 1 = 3^2 + 1= 9 + 1 = 10$$
    and, $$k^2 - 1 = 3^2 - 1 = 9 - 1 = 8$$

    Check  the  answers: $$6^2 + 8^2 = 36 + 64 = 100 = 10^2$$

    Hence, the triplet is $$ 6, 8,10$$
  • Question 9
    1 / -0
    By splitting into prime factors, find the square root of $$729$$.
    Solution
    Prime factorisation of $$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$$
    $$\sqrt{729} = \sqrt{3 \times 3\times 3\times 3 \times 3\times 3} = 3 \times 3\times 3 = 27 $$
  • Question 10
    1 / -0
    By splitting into prime factors, find the square root of $$11025$$.
    Solution

    Prime factorisation of $$ 11025 = {5} ^{2} \times { 3 } ^ {2} \times { 7 } ^ {2}  $$
    $$ \sqrt {11025} = \sqrt {{5} ^{2} \times { 3 } ^ {2} \times { 7 } ^ {2}} = 5 \times 3 \times 7= 105 $$.

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