fc + (11-4)*b = fc + 7*b = 33
fc + (19-4)*b = fc + 15*b = 55
where fc is the fixed charge of the first 4 bushels, and b is the price per bushel of each additional bushel over 4
Use algebra to solve these two equations for b and fc:
fc + 7*b = 33
fc + 15*b = 55
An apple orchard charges a fixed amount for the first four bushels of apples purchased and an extra amount for each additional bushel purchased. If the cost of 11 bushels of apples is $33 and the cost of 19 bushels is $55, what is the fixed charge for the first four bushels of apples?
2 answers
let
cost = m(n-4) + b , where n is the number of bushels, b is the extra charge for the first 4 bushels and m is the cost per bushel
so for 11 bushels, cost is 33
33 = 7m + b
so for 19 bushels, cost = 55
55 = 15m + b
subtract them
22 = 8m
m = 22/8 = 2.75
sub into 33 = 7m+b
33 = 19.25 + b
b = 13.75
the fixed cost is $13.75
check:
our equation would be
cost = 2.75(n-4) + 13.75
if n = 11, cost = 2.75(7) + 13.75 = 33
if n=19, cost = 2.75(15) + 13.75 = 55 , YEAH
cost = m(n-4) + b , where n is the number of bushels, b is the extra charge for the first 4 bushels and m is the cost per bushel
so for 11 bushels, cost is 33
33 = 7m + b
so for 19 bushels, cost = 55
55 = 15m + b
subtract them
22 = 8m
m = 22/8 = 2.75
sub into 33 = 7m+b
33 = 19.25 + b
b = 13.75
the fixed cost is $13.75
check:
our equation would be
cost = 2.75(n-4) + 13.75
if n = 11, cost = 2.75(7) + 13.75 = 33
if n=19, cost = 2.75(15) + 13.75 = 55 , YEAH