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

Answer:

$1850

Step-by-step 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

$S.I=\frac{P\times T\times R}{100}$

where

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

There fore total interest

$=930.8= \frac{x\times 1\times 9}{100} +\frac{(9493-x)\times 1\times 10}{100}\\930.8=\frac{9x+94930-10x}{100}\\93080=94930-x\\x=94930-93080\\x=1850$

Therefore x=1850

⇒Amount invested in 9% account= $1850