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Cubes and Cube Roots Test - 12

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Cubes and Cube Roots Test - 12
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  • Question 1
    1 / -0
    The smallest natural number by which 3636 must be multiplied to get a perfect cube is _____.
    Solution

    Prime factorising 3636, we get,

    36=2×2×3×3=22×3236=2 \times 2 \times 3 \times 3=2^2 \times 3^2.

    We know, a perfect cube has multiples of 33 as powers of prime factors.

    Here, number of 22's is 22 and number of 33's is 22.

    So we need to multiply another 22 and 33 in the factorization to make 3636 a perfect cube.

    Hence, the smallest number by which 3636 must be multiplied to obtain a perfect cube is 2×3=62\times 3 =6.

    Hence, option AA is correct.
  • Question 2
    1 / -0
    Choose the correct option:
    There is no perfect cube which ends in 44.
    Solution
    We know,
    cube of 44, i.e. 43=644^3=64, which is a perfect cube.
    That is, there exists a perfect cube which ends in 44.
    Therefore, the given statement is false and option BB is correct.
  • Question 3
    1 / -0
    Value of 0.0083\sqrt [3]{0.008} is: 
    Solution
    We know,
    0.008=8×103=8×110×110×1100.008 = 8 \times { 10 }^{ -3 }=8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}.

    Also, on prime factorising 88, we get,
    8=2×2×2=238=2\times 2\times 2=2^3.

    Then, 0.0083\sqrt [ 3 ]{ 0.008 }  =8×110×110×1103=\sqrt[3]{8 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}
                             =23×110×110×1103=\sqrt[3]{2^3 \times \dfrac {1}{10} \times \dfrac{1}{10} \times \dfrac {1}{10}}
                             =2×110=0.2=2\times \dfrac {1}{10}=0.2.

    Hence, option AA is correct.
  • Question 4
    1 / -0
    The cube of  3.1-3.1 is:
    Solution
    Cube of the number 3.1-3.1:
    (3.1)3=3.1×3.1×3.1=29.791(-3.1)^{ 3 } = -3.1\times-3.1\times -3.1 = -29.791.

    Hence, option AA is correct.
  • Question 5
    1 / -0
    Find the cube of 0.60.6.
    Solution
    Cube of 0.60.6 is:
    (0.6)3=0.6×0.6×0.6(0.6)^3=0.6\times 0.6\times 0.6
                =0.216=0.216.

    Hence, option AA is correct.
  • Question 6
    1 / -0
    Find the smallest number by which the following number must be multiplied to obtain a perfect cube:
    243243
    Solution
    Prime factorizing 243243, we get,
    243=3×3×3×3×3=35243=3\times 3\times 3\times 3\times 3=3^5.

    We know, a perfect cube has multiples of 33 as powers of prime factors.
    Here, number of 33's is 55.
    So we need to multiply another 33 in the factorization to make 243243 a perfect cube.

    Hence, the smallest number by which 243243 must be multiplied to obtain a perfect cube is 33.

    Therefore, option AA is correct.

  • Question 7
    1 / -0
    Find 1253.\sqrt[3]{125}. 
    Solution

    On prime factorising, we get,

    125=5×5×5125=5 \times 5\times 5 =53= 5^3.

    Then, cube root of 125125 is:

    1253=533=5\sqrt[3]{125}=\sqrt[3]{5^3}= 5.

    Therefore, option BB is correct.

  • Question 8
    1 / -0
    Find the smallest number by which the following number must be divided to obtain a perfect cube.
    192
    Solution
    Factorizationof192Factorization\quad of\quad 192

    192=2×2×2×2×2×2×3192\quad =\quad 2\times2\times2\times2\times2\times2\times3

                =26×3=\quad 2^{ 6 }\times3

    Tomakeaperfectcube,weneedtohavemultiplesof3apowersofprimefactors,i.e,wedividethenumberby3To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 3
  • Question 9
    1 / -0
    Value of 3433 \displaystyle \sqrt[3]{343} is:
    Solution
    On prime factorising, we get,
    343{343}=7×7×7= {7\times 7\times 7} =73=7^3.

    Then,
    3433\sqrt [3] {343}=733=\sqrt [3] {7^3} =7=7.

    Hence, option AA is correct.
  • Question 10
    1 / -0
    Find the smallest number by which the following number must be divided to obtain a perfect cube.
    704
    Solution
    Factorizationof704Factorization\quad of\quad 704

    704=2×2×2×2×2×2×11 704\quad =\quad 2\times2\times2\times2\times2\times2\times11

                =26×11=\quad 2^{ 6 }\times11

    Tomakeaperfectcube,weneedtohavemultiplesof3apowersofprimefactors,i.e,wedividethenumberby11To\quad make\quad a\quad perfect\quad cube,\quad we\quad need\quad to\quad have\quad multiples\quad of\quad 3\quad a\quad powers\quad of\quad prime\quad factors,\\ i.e,\quad we\quad divide\quad the\quad number\quad by\quad 11
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