last year, justin had 30,000 to invest. he invested some of it in an account that paid 5% simple interest per year, and he invested the rest in an account that paid 10% simple interest per year. After one year, he recieved a total of $2500 in interest. How much did he invest in each account?
just add up the interest amounts:
.05x + .10(30000-x) = 2500
To solve this problem, let's break it down step by step.
Step 1: Assign variables
Let's assume that the amount Justin invested in the account with a 5% interest rate is 'x' (in dollars). Therefore, the amount he invested in the account with a 10% interest rate will be the remaining balance, which is 30,000 - x (in dollars).
Step 2: Calculate the interest
The interest from the account with a 5% interest rate would be (0.05 * x) dollars.
The interest from the account with a 10% interest rate would be (0.10 * (30,000 - x)) dollars.
Step 3: Set up the equation
According to the problem, the total interest Justin received after one year is $2500. So, we can write the equation:
(0.05 * x) + (0.10 * (30,000 - x)) = 2500
Step 4: Solve the equation
Let's simplify the equation:
0.05x + 0.10(30,000 - x) = 2500
0.05x + 3000 - 0.10x = 2500
-0.05x = 2500 - 3000
-0.05x = -500
Now, let's solve for 'x'. Divide both sides of the equation by -0.05:
x = (-500) / (-0.05)
x = 10,000
Step 5: Calculate the investment amounts
Now that we have the value of 'x', we can calculate how much Justin invested in each account:
Amount invested in the 5% interest account = x = $10,000
Amount invested in the 10% interest account = 30,000 - x = 30,000 - 10,000 = $20,000
Therefore, Justin invested $10,000 in the account with a 5% interest rate and $20,000 in the account with a 10% interest rate.