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Question Number 180490 by Shrinava last updated on 12/Nov/22

Answered by a.lgnaoui last updated on 13/Nov/22

$$m=60\mathrm{\%}M=Masse\left(final\right)$$$$m+xx=massesugar$$$$76\mathrm{\%}(60\mathrm{\%}M+x)+x=100\mathrm{\%}M$$$$76(60\mathrm{\%}M+x)=100M$$$$76\times 60\mathrm{\%}M+76x+x=100M$$$$x=\frac{100-\left(76\times 60\right)\mathrm{\%}}{77}=\frac{100-45,60}{77}=\frac{53,4}{77}=0,7/kg?$$$$$$

Commented by Shrinava last updated on 13/Nov/22

$$\mathrm{Dear}\mathrm{professor},$$$$\mathrm{A}\mathrm{n}\mathrm{s}\mathrm{w}\mathrm{ers}:\mathrm{a})8\mathrm{b})18\mathrm{c})10\mathrm{d})24\mathrm{e})36$$

Answered by mr W last updated on 13/Nov/22

$$Behmez=bkg$$$$Sugaradded=skg$$$$massofsugaris24\mathrm{\%}ofmassof$$$$behmezinfinalproduct.$$$$s=0.24\times 0.6b=0.144b=\frac{18b}{125}$$$$s,bareinteger.$$$$\Rightarrow b=125k,s=18kwithk\in \mathbb{N}$$$$thatmeansatleast125kgbehmez$$$$andatleast18kgsugar.$$

Commented by Shrinava last updated on 17/Nov/22

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

Answered by manxsol last updated on 13/Nov/22

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