Aman Bansal
Last Activity: 13 Years ago
dear doppalapudi,
the mixtures are x, y and z
3x + 4y + 5z = 240
6x + 4y + 9z = 340
eliminate y
3x + 4z = 100
3x = 100 - 4z
3x = 4*25 - 4z
assuming x, y and z are integers, x must be divisible by 4
let x = 4A where A is an integer
3*4A = 4*25 - 4z
3A = 25 - z
25 - z must be divisible by 3, possible values for z are
1, 4, 7, 10, 13, 16, 19, 22
from
6x + 4y + 9z = 340
3x + 4y + 5z = 240
multiply by 2
6x + 8y + 10z = 480
eliminate x
4y + z = 140
z = 140 - 4y
z = 4*35 - 4y
z must be divisible by 4, thus
z = 4 or 16
possible solutions for (x,y,z) are
(28,34,4) and (12,31,16)
with a ratio of 5:6:8,
5*28 + 6*34 + 8*4 = 376
5*12 + 6*31 + 8*16 = 374
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AMAN BANSAL