queenzoe67
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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 :

mhanifa

Answer:

  • $4000 at 9% and $6000 at 7%

Step-by-step 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
semsee45

Answer:

Milan invested:

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

Step-by-step 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:

[tex]\begin{aligned}\textsf{Interest: Account A} &= x \cdot 0.09 \cdot 1\\& = 0.09x\end{aligned}[/tex]

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

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

[tex]\begin{aligned}\implies 0.09x+700-0.07x&=780\\0.02x+700&=780\\0.02x&=80\\x&=4000\end{aligned}[/tex]

Therefore, Milan invested:

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

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