Answer:
He invested altogether $900
Step-by-step explanation:
* <em>Lets explain how to solve the problem</em>
- Jose invests money in two simple interests account
- He invests twice as much in an account paying 13% as he does in
an account paying 5%
- That means the amount he invested in the account paying 13% is
twice the amount he invested in the account paying 5%
- He earns $93.00 in interest in one year from both accounts combined
- We need to find how much he invested in each account
- Assume that he invested $x in the account paying 5%
∵ He invested twice as much in an account paying 13% as he does in
an account paying 5%
∴ He invested $2x in the account paying 13%
- The simple interest <em>I = PRT</em>, where P is the money invested, R is the
rate of interest in decimal and t is the time of investment
# <u><em>Account paying 13%</em></u>
∵ P = 2x , R = 13/100 = 0.13 , T = 1
∴ I = 2x(0.13)(1)
∴ I = 0.26 x ⇒ (1)
# <u><em>Account paying 5%</em></u>
∵ P = x , R = 5/100 = 0.05 , T = 1
∴ I = x(0.05)(1)
∴ I = 0.05 x ⇒ (2)
∵ He earns $93.00 in interest in one year from both accounts
- Add (1) and (2) and equate the sum by 93.00
∴ 0.26 x + 0.05 x = 93.00
- Add like terms in the left hand side
∴ 0.31 x = 93.00
- Divide both sides by 0.31
∴ x = 300
∴ 2x = 2(300)
∴ 2x = 600
∵ x represents the amount of money invested in the account paying 5%
∴ <em>The amount of money invested in the account of 5% is $300</em>
∴ <em>The amount of money invested in the account of 13% is $600</em>
- The total amount of money he invested is the sum of the money
he invested in each account [$300 + $600 = $900]
∴ He invested altogether $900
(n=number). And from there, we can solve.
