Question

A man invests a total of $9,493in two savings accounts. One account yields 9% simple interest and the other 10% simple interest. He earned a total of $930.80interest for the year. How much was invested in the 9% account?

Answer 1

$1850

Explanation:

Given

• The person invests a total of $9493
• Amount invested in 9% Simple interest account =x
• Amount invested in 10% Simple interest account= 9493-x
• Earnings in Interest=$930.8

The simple interest formula

`S.I=(P* T* R)/(100)`

where

• S.I is simple interest
• P is principle invested
• T is time period
• R is rate of interest

There fore total interest

`=930.8= (x* 1* 9)/(100) +((9493-x)* 1* 10)/(100)\\930.8=(9x+94930-10x)/(100)\\93080=94930-x\\x=94930-93080\\x=1850`

Therefore x=1850

⇒Amount invested in 9% account= $1850