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?
A.
243
B.
108
C.
54
D.
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
