Question

Mrs. Alford invested $6700 in securities. Part of the money was invested at 1% and part at 9%. The total annual income was $275. How much was invested at each rate?
$ at 1%
$ at 9%

Answer 1

Answer:

the amount invested at 1% = $4100

the amount invested at 9% = $2600

Step-by-step explanation:

Hello

let

A=amount invested at 1%

B=amount invested at 9%

the total amount invested is % 6700, so

A+B=6700 (1)

the profits generated by A= (c)

$c=A\frac{1}{100}$

the profits generated by B=(d)

$d=B\frac{9}{100}$

the total profit is $275,so

$275=\frac{A}{100}+0.09B\\ A=100*(275-0.09B)\\A=27500-9B\\\\replacing \ in \ (1)\\27500-9B+B=6700\\27500-6700=8B\\B=\frac{20800}{8}\\ B=2600$

$A=6700-B\\A=6700-2600\\A=4100$

the amount invested at 1% = $4100

the amount invested at 9% = $2600

I hope it helps