Question

If the square root of a number is between $6$ and $7$,then its cube root lies between:

• A. $2$ and $3$
• B. $2.5$ and $3$
• C. $3$ and $4$
• D. $4$ and $4.5$

Answer 1

Answer: (C)

$3$ and $4$

Let $x$ be the number whose square root is between $6$ and $7$

$\therefore$ $6<\sqrt{x}<7$

$⟹$ $36<x<49$ .............. taking square

$⟹$ $(36{)}^{\frac{1}{3}}<x^{\frac{1}{3}}<(49{)}^{\frac{1}{3}}$

$⟹$ $3=(27{)}^{\frac{1}{3}}<(36{)}^{\frac{1}{3}}<x^{\frac{1}{3}}<(49{)}^{\frac{1}{3}}<(64{)}^{\frac{1}{3}}=4$

$\therefore$ $3<x^{\frac{1}{3}}<4$

Hence, option C is correct.