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If x books cost $5 each and y books cost $8 each, then the average (arithmetic mean) cost, in dollars per book, is equal to
A) \(\frac{5x+8y}{x+y}\)
B) \(\frac{5x+8y}{xy}\)
C) \(\frac{5x+8y}{13}\)
D) \(\frac{40xy}{x+y}\)
E) \(\frac{40xy}{13}\)
A) \(\frac{5x+8y}{x+y}\)
B) \(\frac{5x+8y}{xy}\)
C) \(\frac{5x+8y}{13}\)
D) \(\frac{40xy}{x+y}\)
E) \(\frac{40xy}{13}\)
21 May 2017, 03:44
Kudos
Bookmarks
carcass
x books cost $5 each --> total cost of x books = 5x;
y books cost $8 each --> total cost of y books = 8y.
The average cost = (total cost of x + y books)/(number of books) = \(\frac{5x + 8y}{x + y}\).
Answer: A.
21 May 2017, 03:56
Kudos
Bookmarks
Here the cost associated with x books is $5 and cost associated with y books is $8.
The total no: of books mentioned is x + y and the total cost associated with it is 5*x + 8*y
Therefore the average cost , in dollars per book is ,
(5*x + 8*y)/(x+y).
Option A is the answer.
The total no: of books mentioned is x + y and the total cost associated with it is 5*x + 8*y
Therefore the average cost , in dollars per book is ,
(5*x + 8*y)/(x+y).
Option A is the answer.
