A man sells different brands of an items. $^1/_9$ of the items he has in his shop are from Brand A, $^5/_8$ of the remainder are from Brand B and the rest are from Brand C. If the total number of Brand C items in the man's shop is 81, how many more Brand B items than Brand C does the shop has?

243

108

135

Correct answer is C

Let the total number of items in the man's shop = $y$

Number of Brand A's items in the man's shop = $\frac {1}{9} y$

Remaining items = 1 - $\frac {1}{9} y = \frac {8}{9} y$

Number of Brand B's items in The man's shop = $\frac{5}{8} of \frac{8}{9}y = \frac{5}{9}y$

Total of Brand A and Brand B's items = $\frac{1}{9}y + \frac{5}{9}y = \frac{2}{3}y$

Number of Brand C's items in the man's shop = 1 - $\frac{2}{3}y = \frac{1}{3}y$

$\implies\frac{1}{3}y$ = 81 (Given)

$\implies y$ = 81 x 3 = 243

∴ The total number of items in the man's shop = 243

∴ Number of Brand B's items in the man's shop = $\frac{5}{9}$ x 243 = 135

∴ The number of more Brand B items than Brand C = 135 - 81 =54

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A man sells different brands of an items. 1/9^1/_9 of the items he has in his shop are from Brand A, 5/8^5/_8 of the remainder are from Brand B and the rest are from Brand C. If the total number of Brand C items in the man's shop is 81, how many more Brand B items than Brand C does the shop has?