QUESTION

Find the cube root of number 120.A) 4.1B) 4.2C) 4.7D) 4.9

Hint: We will use the Babylonian Algorithm for cube roots to determine the result. The Babylonian Algorithm for cube roots is given by ${{x}_{n+1}}=\dfrac{\left( 2{{x}_{n}}+\left( \dfrac{N}{{{x}_{{{n}^{2}}}}} \right) \right)}{3}$.
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.

We have to find out the cube root of 120.
We use the Babylonian Algorithm for cube roots here,
According to the algorithm, the cube root is given by the formula
${{x}_{n+1}}=\dfrac{\left( 2{{x}_{n}}+\left( \dfrac{N}{{{x}_{{{n}^{2}}}}} \right) \right)}{3}$
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 the question of our case,
N=120
${{x}_{0}}=4$
Since, $~{{4}^{3}}<120<{{5}^{3}}$
Substituting all the above values we get the required expression as,

\begin{align} & {{x}_{1}}=\left( \dfrac{(2)(4)+\left( \dfrac{120}{{{4}^{2}}} \right)}{3} \right) \\ & \Rightarrow {{x}_{1}}=\left( \dfrac{8+\left( \dfrac{120}{16} \right)}{3} \right) \\ & \Rightarrow {{x}_{1}}=4.9 \\ \end{align}
Similarly, $x_2$ can be determined by,
\begin{align} & {{x}_{2}}=\left( \dfrac{(2)(4.9)+\left( \dfrac{120}{{{(4.9)}^{2}}} \right)}{3} \right) \\ & \Rightarrow {{x}_{2}}=\left( \dfrac{9.8+\left( \dfrac{120}{24.01} \right)}{3} \right) \\ & \Rightarrow {{x}_{2}}=\dfrac{9.8+4.99}{3} \\ & \Rightarrow {{x}_{2}}=4.9 \\ \end{align}
We can see the value stabilizes around 4.9.
Therefore, the cube root of 120, is obtained as 4.9.
Matching from the give options we get, the cube root of 120 is 4.9, that is option (D).

Note: The possibility of error in this question can be restricted to the value of only $x_1$ and not $x_2$ also which would be wrong. Always go for determining both the values then arrive at the final answer.