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Squares and Square Roots Test - 16

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Squares and Square Roots Test - 16
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  • Question 1
    1 / -0
    For the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.
    $$252$$
    Solution
    By prime factorisation we get
    $$252=\underline{2\times2}\times\underline{3\times3}\times7$$
    $${ \underline { 2\mid 252 }  }\\ { \underline { 2\mid 126 }  }\\ { \underline { 3\mid 63 }  }\\ { \underline { 3\mid 21 }  }\\ { \underline { 7\mid 7 }  }\\ \; \; \mid 1$$
    It is clear that in order to get a perfect square, one more $$7$$ is required.
    So, the given number should be multiplied by $$7$$ to make the product a perfect square.
    $$\therefore\;252\times7=1764$$ is a perfect square.
    Thus, $$1764=\underline{2\times2}\times\underline{3\times3}\times\underline{7\times7}$$
    $$\therefore\;\sqrt{1764}=2\times3\times7=42$$
  • Question 2
    1 / -0
    Without doing any calculation, find the numbers which are surely not perfect squares:
    $$153$$
    $$257$$
    $$408$$
    $$441$$
  • Question 3
    1 / -0
    A gardener wants to plant $$17956$$ trees and arranges them in such a way that there are as many rows as there are trees in a row. What is the number of trees in a row?
    Solution
    $$\textbf{Step 1: Assume the no. of trees  and use the information given}$$

                    $$\text{Let the no. of trees in a row be x.}$$

                    $$\therefore x\times x = 17956$$

                    $$\Rightarrow x^2 = 17956$$

                    $$\Rightarrow x = \sqrt{17956}$$

                    $$\Rightarrow x = \sqrt{2\times 2\times 67\times 67}$$

                    $$\Rightarrow x = 2\times 67$$

                    $$\Rightarrow x = 134$$

    $$\textbf{Hence, the required no. of trees = 134, option - D is the correct answer.}$$
  • Question 4
    1 / -0
    The largest number of $$5$$ digits, which is perfect square is
    Solution
    The largest number of 5 digits $$= 99999$$
    Required number $$= (99999 - 143) = 99856 = 316^2$$
  • Question 5
    1 / -0
    What could be the possible 'one's' digit of the square root of each of the following numbers?
    $$9801$$
    $$99856$$
    Solution
    As we know that all the perfect square number ends with  $$1,4,9,6,5,00$$
    Thus, the possible unit digit of square root of 9801 is either 1 or 9 and it will odd square root and the possible unit digit of square root of 99856 is either 4 or 6 and it will have even square root.
  • Question 6
    1 / -0
    Find the square root of each of the following numbers by division method:
    $$2304$$
    $$4489$$
    $$3481$$
    Solution
    By long division, we have
    $$\therefore\;\sqrt{2304}=48$$
    $$\;\;\;\;\;\;\;\;\;\;48\\ \overline { 4\;\;\mid \;\;\bar {23 } \bar { 04 }  } \\ { { \; \; \;\;\; \mid -16 } }\\ { \overline { 88\mid\;\;\;\;704 }  }\\ \; \; \; { \underline { \;\;\mid -704 }  }\\ \; \;\;\;\mid \; \; 0$$
    By long division, we have
    $$\therefore\;\sqrt{4489}=67$$
    $$\;\;\;\;\;\;\;\;\;\;67\\ \overline { \;\;6\; \mid \;\;\bar {44 } \bar { 89 }  } \\ { { \; \; \; \; \; \;\mid -36 } }\\ { \overline { 127\mid\;\; 889 }  }\\ \; \; \; { \underline { \; \;\;\mid -889 }  }\\ \; \; \; \;\; \mid \; \; 0$$
    By long division, we have
    $$\therefore\;\sqrt{3481}=59$$
    $$\;\;\;\;\;\;\;\;\;\;59\\ \overline { \;\;5\; \mid \;\;\bar{34 } \bar{ 81 } }\\ { { \; \; \; \; \; \;\mid -25 } }\\ { \overline { 109\mid\;\; 981 }  }\\ \; \; \; { \underline { \; \;\;\mid -981 }  }\\ \; \; \; \;\; \mid \; \; 0$$
  • Question 7
    1 / -0
    The students of class VII of a school denoted $$Rs. 2401$$ in all, for Prime Minister's National Relief Fund. Each student denoted as many rupees as the number of students in the class. Find the number of students in the class.
    Solution
    Let $$x$$ be the number of students in the class.
    Therefore, total money denoted by students
    $$=Rs.\;(x\times x)=Rs.\;x^2$$. But it is given as $$Rs.\;2401$$.
    $$\therefore\;x^2=2401$$
    $$\Rightarrow\;x=\sqrt{2401}$$
    $${ \underline { 7\mid 2401 }  }\\ { \underline { 7\mid 343 }  }\\ { \underline { 7\mid 49 }  }\\ { \underline { 7\mid 7 }  }\\ \; \; \mid 1$$
    $$=\sqrt{\underline{7\times7}\times\underline{7\times7}}$$
    $$=7\times7=49$$.
  • 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  $$1,4,9,6,5,00$$
    $$\therefore$$ 143642 is not the square of natural number because it ends with 2.
  • Question 9
    1 / -0
    Find the square root of $$10$$, correct to four places of decimal.
    Solution
      
    .....................................$$3.16227$$
    ...........................
        $$3$$
    $$+3$$
    $$10$$
    $$9$$
       $$61$$
      $$+1$$
    $$100$$
    $$61$$
       $$626$$
        $$+6$$
    $$3900$$
    $$3756$$
       $$6322$$
          $$+2$$
    $$14400$$
    $$12644$$
      $$63242$$
           $$+2$$
    -------------
      $$632447$$
    $$175600$$
    $$126484$$
    ---------------
    $$4911600$$
    $$4427129$$
    $$\sqrt{10}=3.16227\simeq 3.1623$$
    $$\therefore$$ The square root of $$10$$ correct to four places of decimal is $$3.1623$$
  • Question 10
    1 / -0
    What least number must be substracted from $$176$$ to make it a perfect square?
    Solution
    $$176$$ is not a perfect square .
    $$13^{2} = 169 < 176 $$
    $$176-169=7$$
    Therefore, $$7$$ need to subtract for make perfect square.
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