Question Number 96390 by bemath last updated on 01/Jun/20
Answered by john santu last updated on 01/Jun/20
$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{initial}\:\mathrm{corn}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{weight} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{final}\:\mathrm{mixture}.\:\mathrm{therefore} \\ $$$$\left(\mathrm{20}\:\mathrm{bushels}\right)\frac{\mathrm{56}\:\mathrm{pounds}}{\mathrm{bushels}}\:+\:\left({x}\:\mathrm{bushels}\right)\frac{\mathrm{50}\:\mathrm{pounds}}{\mathrm{bushels}}\: \\ $$$$=\:\left[\:\left(\mathrm{20}+{x}\right)\mathrm{bushels}\:\right]\:\frac{\mathrm{54}\:\mathrm{pounds}}{\mathrm{bushels}} \\ $$$$\mathrm{20}×\mathrm{56}+\:\mathrm{50}{x}\:=\:\left(\mathrm{20}+{x}\right)×\mathrm{54} \\ $$