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//chapter 10 Ex 30
clc;
clear;
close;
solAmt=30; solPercent=2/100; incPercent=10/100;
saltAmt=solAmt*solPercent;
//let x kg of pure salt be added, thus equation will be (0.6+x)/(30+x)=10/100
pureAmt=240/90;
mprintf("The amount of pure salt that must be added is %.2f kg.",pureAmt);
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