Question
mary had 25000 dollars to invest. she invested part of that amount at 3% annual interest and part at 5% annual interest for one year. the amount of interest she earned for both investments was 1100. how much was invested at each rate
Answers
If x was invested at 3%, then the rest (25000-x) was invested at 5%. So, just add up the interest:
.03x + .05(25000-x) = 1100
.03x + .05(25000-x) = 1100
Continued from steeve...
0.03x + 1250 - 0.05x = 1500.
0.03x - 0.05x = 1500 - 1250
-0.02x = 250
Divide both sides by 0.02.
X = -12500
0.03x + 1250 - 0.05x = 1500.
0.03x - 0.05x = 1500 - 1250
-0.02x = 250
Divide both sides by 0.02.
X = -12500
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