Chloe wants to invest a total of $39000 into 2 savings account, one paying 6% per year in interest and the other paying 9% per year in interest (a more risky investment). If after 1 year she wants the total interest from both accounts to be $2730. How much should she invest in each account.
investment at the 6% simple interest?
investment at the 9% simple interest?
Po1*r1*t + Po2*r2*t = 2730.
Po1*0.06*1 + Po2*0.09*1 = 2730,
0.06Po1 + 0.09Po2 = 2730.
Eq1: 6Po1 + 9Po2 = 273,000,
Eq2: Po1 + Po2 = 39000.
Multiply Eq2 by -6 and Add:
+6Po1 + 9Po2 = 273000.
-6Po1 - 6Po2 = -234,000.
Sum: 3Po2 = 39,000, Po2 =
$13,000 Invested in acc. #2.
Po1 = 39,000-13,000 = $26,000 Invested in acc. #1.
To determine the amount Chloe should invest in each account, let's use a system of equations.
Let's assume Chloe invests x dollars in the account with a 6% interest rate and (39000 - x) dollars in the account with a 9% interest rate.
Now, we can set up the equations based on the given information:
Equation 1: 6% of the amount invested at 6% interest + 9% of the amount invested at 9% interest = total interest earned (2730).
0.06x + 0.09(39000 - x) = 2730
Simplifying the equation:
0.06x + 3510 - 0.09x = 2730
-0.03x = -780
Dividing both sides by -0.03:
x = -780 / -0.03
x = 26,000
Therefore, Chloe should invest $26,000 in the account with a 6% interest rate and the remaining $39,000 - $26,000 = $13,000 in the account with a 9% interest rate.