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Question 558217: Hello again:
I was very pleased with your previous answer to a word problem you sent to me. This one is a little different but would like to also have a format to work future problems of the same.
this is the problem: Larry Mitchell invested part of his 40,000 advance at 4% annual simple interest and the rest 3% annual simple interest. If both account was 1,330 Find the amount invested at each rate
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the amount invested at 4% then the rest (40,000-x) is the amount invested at 3%. The earnings on these two amounts is given as 1,330.
This situation can be expressed by the following equation, after converting the percentages to their decimal equivalents:
0.04x+0.03(40,000-x) = 1,330 Simplify.
0.04x+1200-0.03x = 1,330
0.01x+1200 = 1,330 Subtract 1200 from both sides.
0.01x = 130 Divide both sides by 0.01
x = 13000 and 40,000-x = 27,000
Larry Mitchell invested 13,000 at 4% and 27,000 at 3%.
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