Question

Alewuya invests a total of $18,000 in two accounts paying 11% and 10% simple interest, respectively. How much was invested in each account if, after one year, the total interest was $1,920.00. A) Enter an equation that uses the information as it is given that can be used to solve this problem. Use x as your variable to represent the amount of money invested in the account paying 11% simple interest. Equation: B) The answers are: $ was invested at 11% and was invested at 10%.​

Answer 1

Answer: $12,000 at 11%, $6,000 at 10%.

Explanation:

Firstly, you would identify the two interest rates as variables.

Assign variable X to the amount invested at 11% and variable Y to the amount invested at 10%. Using those two variables assigned to our interest rates, equate those to the principle of $18,000. After that, multiply the variables by the given interest rates and equate that to the total interest.

x + y = $18,000
0.11x + 0.1y = $1,920

Solve the first equation for Y, multiplying by 100 to even out decimals.

x + y = $18,000
y = $18,000 - x
11x + 10y = $192,000
11x + 10($18,000 - x) = $192,000
11x + $180,000 - 10x = $192,000
x + $180,000 = $192,000
x = $12,000

With that, we now know that the original principle for the 11% interest rate is $12,000, which when subtracted from the total principle of $18,000 gives us the interest rate for the 10% investment as well

$18,000 - $12,000 = $6,000

So, with a total Principle of $18,000, two accounts with interest rates of 11% and 10% respectively, and a total interest accumulation of $1,920, we can gather that the account with an 11% interest rate originally had $12,000 and the account with the 10% interest rate originally had $6,000.

Answer 2

$12000 was invested at 11% and $6000 was invested at 10%

Explanation:

Total invested amount is 18000

Let the amount of money invested in the account paying 11% be x

Let the amount of money invested in the account paying 10% be y

⇒ x + y = 18000

⇒ y = 18000 - x

Simple interest = PNR

where P is the amount invested, N is the number of years and R is the rate of interest

For the account paying 11%,

SI₁ = x(1)(11%) = x(11%)

For the account paying 10%,

SI₂ = y(1)(10%)

= (18000 - x)(10%)

= 18000(10%) - x(10%)

= 1800 - x(10%)

Total Interest = SI₁ + SI₂

⇒ 1920 = x(11%) + 1800 - x(10%)

⇒ 1920 - 1800 = x(11% - 10%)

⇒ 120 = x(1%)

⇒
`(x)/(100)` = 120

⇒ x = 120*100

⇒ x = 12000

y = 18000 - x

⇒ y = 18000 - 12000

⇒ y = 6000

$12000 was invested at 11% and $6000 was invested at 10%