document.write( "Question 953743: ERNEST INVESTED A PART OF HIS MONEY AT 16% ANNUAL INTEREST. ANOTHER PART WHICH IS P50,000 MORE WAS INVESTED AT 18% ANNUAL INTEREST. IF THE TOTAL AMOUNT INCOME FROM THE TWO INVESTMENTS WAS P94,000, HOW MUCH WAS INVESTED AT EACH RATE?
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Algebra.Com's Answer #582487 by macston(5194)![]() ![]() You can put this solution on YOUR website! S=amount at 16%; E=amount invested at 18%=S+P50000 \n" ); document.write( "0.16S+0.18E=P94000 Substitute for E \n" ); document.write( "0.16S+0.18(S+P50000)=P94000 \n" ); document.write( "0.16S+0.18S+P9000=P94000 Subtract P9000 from each side. \n" ); document.write( "0.34S=P85000 Divide each side by 0.34 \n" ); document.write( "S=P250000 ANSWER 1: The amount invested at 16% was P250000. \n" ); document.write( "E=S+P50000=P250000+P50000=P300000 ANSWER 2: The amount invested at 18% was P300000. \n" ); document.write( "CHECK: \n" ); document.write( "0.16S+0.18E=P94000 \n" ); document.write( "0.16(P250000)+0.18(P300000)=P94000 \n" ); document.write( "P40000+P54000=P94000 \n" ); document.write( "P94000=P94000 \n" ); document.write( " |