SOLUTION: two investments were made totaling $15,000. For a certain year these investments yielded $1432 in simple intrest. part of the $15,000 was invested at 9% and part at 10%. find the

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: two investments were made totaling $15,000. For a certain year these investments yielded $1432 in simple intrest. part of the $15,000 was invested at 9% and part at 10%. find the      Log On


   

Question 102266: two investments were made totaling $15,000. For a certain year these investments yielded $1432 in simple intrest. part of the $15,000 was invested at 9% and part at 10%. find the amount invested at each rate
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Interest(I)=Principal(P) times Rate(R) times Time(T) or I=PRT
Let P-amount invested at 9%
(1) Interest accrued at 9%=P*0.09*1 or 0.09P

Then 15000-P=amount invested at 10%
(2) Interest accrued at 10%=(15000-P)*0.10*1 or 0.10(15000-P)
Now we are told that (1)+(2)=1432 or
0.09P+0.10(15000-P)=1432 get rid of parens
0.09P+1500-0.10P=1432 subtract 1500 from both sides
0.09P+1500-1500-0.10P=1432-1500 collect like terms
-0.01P=-68 divide both sides by -0.01
P=$6800---------------------amount invested at 9%
15000-P=15000-6800=$8200----------------amount invested at 10%
CK
6800*0.09+8200*0.10=1432
$612+$820=$1432
$1432=$1432
Hope this helps----ptaylor