CAT 2018 Slot 2 — QA Question 23

$\frac{1}{l{\mathrm{og}}_{2}100}$ - $\frac{1}{{\log }_{4}100}$ + $\frac{1}{{\log }_{5}100}$ - $\frac{1}{{\log }_{10}100}$ + $\frac{1}{{\log }_{20}100}$ - $\frac{1}{{\log }_{25}100}$ + $\frac{1}{{\log }_{50}100}$

= ${\log }_{100}2$ - ${\log }_{100}4$ + ${\log }_{100}5$ - ${\log }_{100}10$ + ${\log }_{100}20$ - ${\log }_{100}25$ + ${\log }_{100}50$

= ${\log }_{100}\left(\frac{2\times 5\times 20\times 50}{4\times 10\times 25}\right)$ = ${\log }_{100}\left(10\right)$ = $\frac{1}{2}$

Hence, option (d).