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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?

Respuesta :

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

[tex]S.I=\frac{P\times T\times R}{100}[/tex]

where

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

There fore total interest

[tex]=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[/tex]

Therefore x=1850

⇒Amount invested in 9% account= $1850

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