Question

$\sqrt[4]{43},\sqrt[6]{610},\sqrt[12]{1225}$, when arranged in descending order will be

• A. $\sqrt[4]{43},\sqrt[6]{610},\sqrt[12]{1225}$
• B. $\sqrt[6]{610},\sqrt[4]{43},\sqrt[12]{1225}$
• C. $\sqrt[6]{610},\sqrt[12]{1225},\sqrt[4]{43}$
• D. $\sqrt[4]{43},\sqrt[12]{1225},\sqrt[6]{610}$

Answer 1

The correct option is B $\sqrt[6]{610},\sqrt[4]{43},\sqrt[12]{1225}$

LCM of $4,6$ and $12=12$.
$\therefore$ Raising each of the given number to power $12$, we have
$(3^{1/4}{)}^{12},({10}^{1/6}{)}^{12},({25}^{1/12}{)}^{12}$
$=3^3,{10}^2,25$
$=27,100,25$
Arranging in descending order, the numbers are $100,27,25$
$\Rightarrow \sqrt[6]{610},\sqrt[4]{43},\sqrt[12]{1225}$