Question

The square root of $\frac{{\left(3\frac{1}{4}\right)}^4-{\left(4\frac{1}{3}\right)}^4}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}$ is ____.

• A. $5\frac{7}{12}$
• B. $5\frac{5}{12}$
• C. $5\frac{12}{13}$
• D. $5\frac{1}{6}$

Answer 1

The correct option is B $5\frac{5}{12}$

We know, $a^4-b^4=(a^2-b^2)(a^2+b^2)$

$\therefore \frac{(\frac{13}{4}{)}^4-(\frac{13}{3}{)}^4}{(\frac{13}{4}{)}^2-(\frac{13}{3}{)}^2}$

$=(\frac{13}{4}{)}^2+(\frac{13}{3}{)}^2$

$=\frac{169}{16}+\frac{169}{9}=169(\frac{25}{144})$

Square root of $\frac{169\times 25}{144}=\frac{13\times 5}{12}=\frac{65}{12}$

$=5\frac{5}{12}$