Question

You invested $6000 in two accounts paying 7% and 9% annual interest. At the end of the year, the total interest from these investments was $510. How much was invested at each rate?

Answer 1

The amount invested at 7% was $1,500 and the amount invested at 9% was $4,500

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

in this problem

Let

x-----> the amount invested at account paying 7%

(6,000-x) -----> the amount invested at account paying 9%

`t=1\ years\\ P1=\$x\\ P2=\$(6,000-x)\\I=\$510\\r1=0.07\\r2=0.09`

substitute in the formula above

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

`510=x(0.07*1)+(6,000-x)(0.09*1)`

Solve for x

`510=0.07x+540-0.09x`

`0.09x-0.07x=540-510`

`0.02x=30`

`x=\$1,500`

`6,000-x=6,000-1,500=\$4,500`

therefore

The amount invested at 7% was $1,500 and the amount invested at 9% was $4,500