✔ 最佳答案
we have 2 equations.
(k/10) c = 19-----1
(k/20) c = 15-------2
a)
(k/10) - (k/20) = 4
k = 80
80/20 c = 15
c = 11
the price of a lunch box when 40 boxes are ordered
80/40 11 = $13
b) let the number of boxes ordered be x
(80/x) 11 = 21
80/x = 10
x = 8
However,it is impossible to have $19 for 10 boxes and $15 for 20 boxes.
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2008-09-20 20:34:03 補充:
we have 2 equations.
(k/10)+c = 19-----1
(k/20)+c = 15-------2
a)
(k/10) - (k/20) = 4
k = 80
80/20+c = 15
c = 11
the price of a lunch box when 40 boxes are ordered
80/40+11 = $13
b) let the number of boxes ordered be x
(80/x)+11 = 21
80/x = 10
x = 8