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Stan invested $17,000, pat at 18% and part at 4%. If the total interest at the end of the year is $1,380, how much did he invest at each rate?
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if he invested P at 18% then he invested the rest, 1700-P and 4% set \[.18P+.04(1700-P)=1380\] and solve for P
\[.18P+68-.04P=1380\] \[.12P=1380-680=1312\] \[P=\frac{700}{.12}=583\tfrac{1}{3}\]
typo on first line. should be \[.18P+680-.04P=1380\]
damn and on the second line too! should be \[.12P=1380-680=700\] answer is right tho...
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so the answer is 700 than right
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