What is the number of decimal places in a cube root of a number having 27 decimal places ?
Let the number be $$N_1$$ which consists of 27 decimal places.
Now, $$N_1=\frac{N}{10^{27}}$$
Cube root of this number will be = $$\sqrt[3]{\frac{N}{10^{27}}}$$
= $$\frac{N^{\frac{1}{3}}}{10^9}=\frac{x}{10^9}$$
Here, $$x$$ is cube root of $$n$$, and will have 9 decimal places.
=> Ans - (C)
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