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

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

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