2023
Mathematics
67ca00ce0c643c71d77dee1f
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: 54
Explanation
Let the total number of items in the man's shop = y Number of Brand A's items in the man's shop = 1/9y Remaining items = 1 - 1/9y=8/9y Number of Brand B's items in The man's shop = 5/8of8/9y=5/9y Total of Brand A and Brand B's items = 1/9y+5/9y=2/3y Number of Brand C's items in the man's shop = 1 - 2/3y=1/3y ⟹1/3y = 81 (Given) ⟹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 = 5/9 x 243 = 135 ∴ The number of more Brand B items than Brand C = 135 - 81 =54
Let the total number of items in the man's shop = y Number of Brand A's items in the man's shop = 1/9y Remaining items = 1 - 1/9y=8/9y Number of Brand B's items in The man's shop = 5/8of8/9y=5/9y Total of Brand A and Brand B's items = 1/9y+5/9y=2/3y Number of Brand C's items in the man's shop = 1 - 2/3y=1/3y ⟹1/3y = 81 (Given) ⟹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 = 5/9 x 243 = 135 ∴ The number of more Brand B items than Brand C = 135 - 81 =54THIS WEEK's
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