Question 346293: Part of $17,000 is invested at 9% annual interest, and the rest is invested at 6%. After one year, the accounts paid $1,290 in interest. How much was invested at the lower rate? (Give your answer correct to the nearest cent.)
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! I = PRT ..... I (interest) = principal (P) × rate (R) × time (T).
Here, there are two principals (the two parts of the $17,000) and two rates, yielding two amounts of interest. T = 1 in both cases; R1 = 9% = 0.09; R2 = 6% = 0.06.
P1 + P2 = $17,000, so P1 = $17,000 - P2.
I1 = P1R1T = ($17,000 - P2)(0.09)(1) = $1,530 - 0.09P2.
I2 = P2R2T = (P2)(0.06)(1) = 0.06P2.
I1 + I2 = $1,290, so combine the above equations:
$1530 - 0.09P2 + 0.06P2 = $1,290, or
$240 - 0.09P2 + 0.06P2 = 0, or
$240 = 0.03P2, or
P2 = $8,000.
(Just for completeness: P1 = $17,000 - P2 = $17,000 - $8,000 = $9,000.)
check:
I1 = P1R1T = ($9,000)(0.09)(1) = $810.
I2 = P2R2T = ($8,000)(0.06)(1) = $480.
I1+I2 = $1,290.
|
|
|
