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Question Number 96390 by bemath last updated on 01/Jun/20

Answered by john santu last updated on 01/Jun/20

$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}$$$$\mathrm{initial}\mathrm{corn}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{weight}$$$$\mathrm{of}\mathrm{the}\mathrm{final}\mathrm{mixture}.\mathrm{therefore}$$$$\left(20\mathrm{bushels}\right)\frac{56\mathrm{pounds}}{\mathrm{bushels}}+\left(x\mathrm{bushels}\right)\frac{50\mathrm{pounds}}{\mathrm{bushels}}$$$$=\left[(20+x)\mathrm{bushels}\right]\frac{54\mathrm{pounds}}{\mathrm{bushels}}$$$$20\times 56+50x=(20+x)\times 54$$

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