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Question 75521This question is from textbook Elementary algebra
: A sum $2300 is invested, part of it at 10% interest and the reaminder at 12%. If the interest earned by the 12% investment is $100 more than the intrest earned by the 10% investment, find the amount invested at each rate?
Thanks!
This question is from textbook Elementary algebra
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Let x=amount invested at 12%
Then (2300-x)=amount invested at 10%
0.12x=interest earned on the 12% investment (for 1 year. (I=PRT))
0.10(2300-x)=interest earned on the 10% investment
So, we are told that if we add $100 to the interest on the 10% investment, it will equal the interest earned on 12% investment.
Then our equation to solve is:
0.10(2300-x)+100=0.12x get rid of parens
230-0.10x+100=0.12x add 0.10x to both sides
230-0.10x+0.10x+100=0.12x+0.10x collect like terms
330=0.22x divide both sides by 0.22
x=$1500----------------amount invested at 12%
$2300-x=$2300-$1500=$800--------------------------amount invested at 10%
CK
interest from 12% investment=1500*0.12=$180
interest from 10% investment=$800*0.10=$80
$80+$100=$180
$180=$180
Hope this helps----ptaylor
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