SOLUTION: Zoya Lon invested part of her $25,000 advance at 8% annual simple interest and the rest at 9% annual simple interest. If her total yearly interest from both accounts was $2135, fin
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Question 67441: Zoya Lon invested part of her $25,000 advance at 8% annual simple interest and the rest at 9% annual simple interest. If her total yearly interest from both accounts was $2135, find the amount invested at each rate.
I'm having a problem setting this up as an equation using the formula
Int. = P · r · t Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Let's say that x dollars were in the 8% account and the rest ($25000-x) was in the 9% account.
So Zoya Lon received two interest ammounts that added up to $2135.00 for the year. You can set this up as follows using I = Prt: Where r1 = 8% or 0.08 and r2 = 9% or 0.09, P1 = x and P2 = $25000-x, and t = 1 year.
Note that two interest amounts I1 + I2 = $2135
I1 = x(0.08)(1) = 0.08x
I2 = ($25000-x)(0.09)(1) = $2250-0.09x
Now add I1 and I2
I1+I2 = 0.08x + ($25000-x)(0.09) = $2135 Simplify this and solve for x.
0.08x+$2250-0.09x = $2135
-0.01x+$2250 = $2135
-.01x = -$115 Divide both sides by -0.01
x = $11500 This is the amount invested at 8%
$25000-x = $25000-$11500 = $13500 This is the amount invested at 9%
Check:
0.08($11500) + 0.09($13500) = $920 + $1215 = $2135
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