Question 712081
Let a = the number of 20's
let b = the number of 50's
let c = the number of 100's
B = c + 43
A = 3c
20a + 50b + 100c = 6560
260c + 50c + 2150 + 100c = 6560
210c = 4410
c = 21
b = 21 + 43
b = 64
a = 3x21
a = 63
There were 63 20's, 64 50's and 21 100's