Răspuns :
1.
ms1 = 400 g , c1 = 35% , c2 = 40%
stim ca c% = mdx100/ms
=> md = ms1xc1/100 = 400x35/100 = 140 g
ms2 = mdx100/c2 = 140x100/40 = 350 g
=> ms2 = ms1 + a (masa deapa adugata)
=> a = ms2 - ms1 = 400 - 350 = 50 g apa adaugata
2.
ms1 = 200 g , c1 = 25% , c2 = 50%
stim ca c% = mdx100/ms
=> md1 = ms1xc1/100 = 200x25/100 = 50 g
md2 = md1 + a (masa adaugata de sare)
ms2 = ms1 + a
=> ms1 + a = (md1+a)x100/50
=> ms1 + a = (md1+a)x2 <=> 200+a = 100+2a
=> a = 100/1 = 100 g sare adaugata
facem proba:
c2 = md2x100/ms2 = 150x100/300
=> c2 = 50% , deci se verifica...!