Last year, Deon had $10,000 to invest. He invested some of it in an account that paid 6%simple interest per year, and he invested the rest in an account that paid 9% simple interest per year. After one year, he received a total of $810 in interest. How much did he invest in each account?

Question

Last year, Deon had $10,000 to invest. He invested some of it in an account that paid 6%simple interest per year, and he invested the rest in an account that paid 9% simple interest per year. After one year, he received a total of $810 in interest. How much did he invest in each account?

I keep getting really big numbers (47000) for the first account which i know is not correct. Please help

Answer 1

.06 x + .09 (10,000 - x ) = 810

-.03 x + 900 = 810

.03 x = 90

x = 90/.03 = 3,000
10,000 - x = 7,000

Answer 2

what is the sum of 0.314,3.14,21,5 and 314?

Answer 3

Let's assume Deon invested x dollars in the account that pays 6% interest and (10000 - x) dollars in the account that pays 9% interest.

The interest earned from the first account can be calculated as 0.06x (6% of x), and the interest earned from the second account can be calculated as 0.09(10000 - x) (9% of (10000 - x)).

According to the given information, the total interest earned is $810. So we can write the equation:

0.06x + 0.09(10000 - x) = 810

Now, let's solve this equation step-by-step:

Step 1: Distribute 0.09 to (10000 - x):

0.06x + 900 - 0.09x = 810

Step 2: Combine like terms:

-0.03x + 900 = 810

Step 3: Move 900 to the other side by subtracting 900 from both sides:

-0.03x = 810 - 900
-0.03x = -90

Step 4: Divide both sides by -0.03 to solve for x:

x = -90 / -0.03
x = 3000

Step 5: Calculate the amount invested in the second account:

10000 - x = 10000 - 3000 = 7000

Therefore, Deon invested $3000 in the account that pays 6% interest and $7000 in the account that pays 9% interest.

Answer 4

To solve this problem, we need to set up a system of equations based on the given information.

Let's assume that Deon invested x amount of money at 6% interest, and the remaining amount, (10000 - x), at 9% interest.

The interest earned from the first account (at 6% interest) would be (x * 0.06), and the interest earned from the second account (at 9% interest) would be ((10000 - x) * 0.09).

According to the problem, the total interest earned is $810. So we can set up the equation:

(x * 0.06) + ((10000 - x) * 0.09) = 810

Now, let's solve this equation to find the value of x, which represents the amount invested in the first account:

0.06x + 900 - 0.09x = 810
-0.03x = -90
x = -90 / -0.03
x = 3000

Therefore, Deon invested $3000 in the first account (at 6% interest). To find out how much he invested in the second account, we subtract this amount from the total investment:

10000 - 3000 = $7000

So, Deon invested $3000 in the first account and $7000 in the second account.