Tag: estimation of cube roots

Questions Related to estimation of cube roots

Find the cube root of the number 514.

  1. 8.0104

  2. 8.1104

  3. 8.2104

  4. 8.3104


Correct Option: A
Explanation:
We use the Babylonian Algorithm for cube roots here
According to the algorithm, the cube root is given by the formula 
$x _{n+1}=\dfrac{(2x _n+(N/x _{n^2}))}{3}$
where
  • $N$ is the number for which cube root is to be found
  • $x _{n}$ is the initial approximation of the cube root
  • $x _{n+1}$ is the subsequent improvement on the cube root
In this case,
$N = 514$
    $x _0 =8$ since, $8^3<514 <9^3$

      $ \therefore$ $x _1 = \dfrac{((2\times8)+(514 /8^2))}{3} = \dfrac{(16+(514/64))}{3}=8.0104$

      $\Rightarrow x _2 = \dfrac{((2\times8.0104+(514/(8.0104)^2))}{3} = \dfrac{(16.02.08+((514/64.1666))}{3}=8.0104$

      We can see the value stabilizes around $8.0104$. Hence the answer is A.

      Estimate the cube root of the number $40.$

      1. $3.1$

      2. $3.2$

      3. $3.4$

      4. $3.9$


      Correct Option: C
      Explanation:

      We use the Babylonian Algorithm for cube roots here

      According to the algorithm, the cube root is given by the formula 
        $x _{n+1}=\dfrac{(2x _n+(N/x _{n^2}))}{3}$
        where,
        • $N$ is the number for which cube root is to be found
        • $x _{n}$ is the initial approximation of the cube root
        • $x _{n+1}$ is the subsequent improvement on the cube root
        • In this case 
        $N = 40$
          $x _0 =3$ since $3^3<40 <4^3$


            $ \therefore$ $x _1 = \dfrac{((2\times3)+(40 /3^2))}{3} = \dfrac{(6+(40/9))}{3}=3.4$

            $\Rightarrow x _2 = \dfrac{(2\times3.4+40/(3.4)^2)}{3} = \dfrac{(6.8+(40/11.56))}{3}= \dfrac{(6.8+3.46)}{3} = 3.4$

            We can see the value stabilizes around 3.4. Hence, the answer is C.

          Find the value of cube root of the number $1290$. (Round off your number to the nearest whole number)

          1. $9.2$

          2. $10.2$

          3. $10.96$

          4. $11$


          Correct Option: D
          Explanation:

          We need to find value of $\sqrt[3]{1290}$
          Take, $n = 1290$, choose any starting value of $x$.
          So, $10^3$ is $1000 < 1290$
          So, $x = 10$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}10+\dfrac{1290}{3\times 10^2}$
          $x _\text{next} =  10.966$
          So, the nearest whole number for the cube root $1290$ is $11$.

          Find the value of cube root of the number $6860$. (Round off your number to the nearest hundredth)

          1. $19.003$

          2. $19.00$

          3. $19.32$

          4. $19.008$


          Correct Option: B
          Explanation:

          We need to find value of $\sqrt[3]{6860}$
          Take, $n = 6860$, choose any starting value of $x$.
          So, $19^3$ is $6859 < 6860$
          So, $x = 19$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} = $ $\dfrac{2}{3}19+\dfrac{6860}{3\times 19^2}$
          $x _\text{next} = 19.00085 $    ....(1)
          So, the approximate value of $\sqrt[3]{6860}$ nearest hundredth place is $19.00$.

          Find the value of cube root of the number $823$. (Round off your number to the nearest whole number)

          1. $8$

          2. $7$

          3. $9$

          4. $10$


          Correct Option: C
          Explanation:

          We need to find value of $\sqrt[3]{823}$
          Take, $n = 823$, choose any starting value of $x$.
          So, $9^3$ is $729 < 823$
          So, $x = 9$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}9+\dfrac{823}{3\times 9^2}$
          $x _\text{next} = 9.386$
          So, the nearest whole number for the cube root $823$ is $9$.

          Find the value of cube root of the number $2486$. (Round off your number to the nearest whole number)

          1. $11$

          2. $12$

          3. $13$

          4. $14$


          Correct Option: D
          Explanation:

          We need to find value of $\sqrt[3]{2486}$
          Take, $n = 2486$, choose any starting value of $x$.
          So, $13^3$ is $2197 < 2486$
          So, $x = 13$
          $x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
          $x _\text{next} =$ $\dfrac{2}{3}13+\dfrac{2486}{3\times 13^2}$
          $x _\text{next} = 13.56$
          So, the nearest whole number for the cube root $2486$ is $14$.