a - zarobki pani Amelii
b - zarobki pana Barnaby
c - zarobki pana Cypriana
b=a+15\%a=115\%a=1,15a\\ c=b-20\%b=80\%b=0,80b=0,80\cdot 1,15a=0,92a
a+b+c=9517\\ a+1,15a+0,92a=9517\\ 3.07a=9517\\ a=3100
b=1.15a=1,15\cdot 3100=3565
c=0,92a=0,92\cdot 3100=2852
Sprawdzenie:
3100+3565+2852=9517
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I. F
3100-2852=248
248>200
II. P
III. F
2852<2900
IV. P
\frac{3100-2852}{3100}\cdot 100\%=\frac{248}{3100}\cdot 100\%=\frac{24800}{3100}\cdot 100\%=8\%