ms1 = 200g, c1% = 20%
md2 = 20g, c2% = 10%
md3 = 40g
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c%finala = ?%
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stm ca c% = mdx100/ms si ca ms = md+m.apa
din sol. 1 avem
md1 = ms1xc1%/100 = 200x20/100 = 40g
din sol. 2 avem
ms2 = md2x100/c2 = 20x100/10 = 200g
in final avem
ms.final = ms1+ms2+md3 = 200+200+40 = 440g
md.final = md1+md2+md3 = 40+20+40 = 100g
=> c%.final = md.finalx100/ms.final
= 100x100/440 = 22,73%