Container A contains 480 blue buttons and 590 purple buttons. Container B contains 590 blue buttons and 410 purple buttons. How many blue buttons and purple buttons must be removed from Container B to put into Container A so that 50% of the buttons in Container A are blue and 60% of the buttons in Container B are blue?
If b blue buttons are moved, then you want (in urn A)
(480+b)/(480+590+b) = 1/2
Unfortunately, this does not also make 60% of B's buttons blue...
why are you moving purple buttons?
A : 480 blue, 590 purple
B: 590 blue, 410 purple
After the Removal:
number of blues removed --- x, number of purple removed ---- y
A: 480-x blue, 590-y purple
B: 590+x blue, 410+y purple
new condition:
50% of total of A is blue
(1/2)(480-x + 590-y) = 480-x
480-x+590-y = 960-2x
x - y = -110 -----> x = y - 110
60% of the buttons in Container B are blue?
(6/10)(590+x + 410+y) = 590+x
590+x+410+y = 5/3 (590+x)
3x + 3y + 3000 = 2950 + 5x
-2x + 3y = -50
use substitution:
-2(y-110) + 3y = -50
y = -270
Uh,uh, something is wrong, either in my arithmetic or the question itself
I can't find my "error"