Question

Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4,500 in interest in one year?

Answer 1

Answer:

The amount invested at 5.5% was $36,000 and the amount invested at 9% was $28,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

Let

x -----> the amount invested at 5.5%

(64,000-x) -----> the amount invested at 9%

in this problem we have

$t=1\ years\\I=\ 4,500\\ P=\ 64,000\\r1=0.055\\P1=\ x\\P2=\ 64,000-\ x\\r2=0.09$

so

$I=P1(r1t)+P2(r2t)$

substitute the given values

$4,500=x(0.055*1)+(64,000-x)(0.09*1)$

$4,500=0.055x+5,760-0.09x$

$0.09x-0.055x=5760-4,500$

$0.035x=1,260$

$x=\ 36,000$

$64,000-x=64,000-36,000=\ 28,000$

therefore

The amount invested at 5.5% was $36,000 and the amount invested at 9% was $28,000