A total of $8000 is deposited into two simple interest accounts. On one account, the annual interest rate is 10%, while on the second account the annual simple interest rate is 12%. How much should be invested in the 10% account so that the total annual interest earned is $900?
.10x + .12(8000-x) = 900
my answer 3000
mine, too
An investment club invested a part of $14,000 in a 9% annual simple interest account and the remainder in a 15% annual simple interest account. The amount of interest earned for one year was $1800. How much was invested in the 9% account?
my answer =2400
.09x + .15(14000-x) = 1800
better check your work
To solve this problem, let's break it down step by step:
Step 1: Assign variables
Let's assign a variable to represent the amount of money invested in the 10% account. Let's call this amount "x".
Step 2: Determine the amount invested in the 12% account
Since the total amount deposited is $8000 and the amount invested in the 10% account is "x", the amount invested in the 12% account would be (8000 - x).
Step 3: Calculate the interest earned on each account
To calculate the interest earned on the 10% account, use the formula:
Interest10% = (amount invested) * (interest rate)
The interest earned on the 10% account would be: 0.10x.
Similarly, the interest earned on the 12% account would be: 0.12(8000 - x).
Step 4: Set up and solve the equation
Since the total annual interest earned is $900, we can set up the equation:
0.10x + 0.12(8000 - x) = 900
Step 5: Solve the equation
Let's solve the equation and find the value of "x".
0.10x + 0.12(8000 - x) = 900
0.10x + 960 - 0.12x = 900
(0.10x - 0.12x) + 960 = 900
-0.02x + 960 = 900
-0.02x = 900 - 960
-0.02x = -60
x = (-60) / (-0.02)
x = 3000
Therefore, $3000 should be invested in the 10% account so that the total annual interest earned is $900.