SOLUTION: Please help me solve this equation:
Karen Waugtal invested some money at 9% annual simple interest and $250 more than that amount at 10% annual simple interest. If her total ye
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Karen Waugtal invested some money at 9% annual simple interest and $250 more than that amount at 10% annual simple interest. If her total ye
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Question 99677This question is from textbook Beginning Algebra
: Please help me solve this equation:
Karen Waugtal invested some money at 9% annual simple interest and $250 more than that amount at 10% annual simple interest. If her total yearly interest was $101, how much was invested at each rate?
I'm stuck. This question is from textbook Beginning Algebra
You can put this solution on YOUR website! Let P=amount invested at 9%
Then P+250=amount invested at 10%
Interest(I)=Principal(P) times Rate(R) times Time(T) or I=PRT
(1) Interest at 9%=P*(0.09)*(1)=0.09P
(2) Interest at 10%=(P+250)*(0.10)*(1)=(0.10)(P+250)
Now we are told that (1)+(2)=101 so our equation to solve is:
0.09P+(0.10)(P+250)=101 get rid of parens
0.09P+0.10P+25=101 subtract 25 from both sides
0.09P+0.10P+25-25=101-25 combine like terms
0.19P=76 divide both sides by 0.19
P=$400------------------------------------ amount invested at 9%
P+250=400+250=$650-----amount invested at 10%
CK
400*0.09+650*0.10=101
36+65=101
101=101
