Question

Last year, Milan had $10,000 to invest. He invested some of it in an account that paid 9% simple interest per year, and he invested the rest in an account that

paid 7% simple interest per year. After one year, he received a total of $780 in interest. How much did he invest in each account?

Answer 1

Milan invested:

• $4,000 into the account earning 9% interest.
• $6,000 into the account earning 7% interest.

Explanation:

Given information:

• Total amount invested = $10,000.
• Account A = 9% simple interest per year.
• Account B = 7% simple interest per year.
• Total interest earned after one year = $780.

Let x be the amount invested in Account A.

Therefore, the amount invested in Account B is (10000 - x).

Simple Interest Formula

I = Prt

where:

• I = Interest earned.
• P = Principal invested.
• r = Interest rate (in decimal form).
• t = Time (in years).

Create two equations using the given information:

`\begin{aligned}\textsf{Interest: Account A} &= x \cdot 0.09 \cdot 1\\& = 0.09x\end{aligned}`

`\begin{aligned}\textsf{Interest: Account B} &= (10000 - x) \cdot 0.07 \cdot 1 \\& = 0.07(10000 - x)\\& = 700-0.07x\end{aligned}`

As the total interest earned was $780, set the sum of the two found equations to 780 and solve for x:

`\begin{aligned}\implies 0.09x+700-0.07x&=780\\0.02x+700&=780\\0.02x&=80\\x&=4000\end{aligned}`

Therefore, Milan invested:

• $4,000 into the account earning 9% interest.
• $6,000 into the account earning 7% interest.

Answer 2

• $4000 at 9% and $6000 at 7%

Explanation:

Let the amount invested at 9% be x.

Then the amount invested at 7% is 10000 - x.

The amount of interest after one year is $780.

Set up equation to represent this:

• 0.09x + 0.07(10000 - x) = 780
• 0.09x + 700 - 0.07x = 780
• 0.02x = 780 - 700
• 0.02x = 80
• x = 80/0.02
• x = 4000

Amount invested at 7% is:

• 10000 - 4000 =
• 6000