Alfredo invested a total of $33,000 in 2 accounts, A and B, with annual interest rates of 5% and 3%, respectively. For the first year, the interest earned by Account A was twice the interest earned by Account B. What was the total interest earned by the 2 accounts for the first year?
A. $1,290
B. $1,320
C. $1,350
D. $1,430
E. $2,475
The correct answer and explanation is:
Let the amount invested in Account A be xx. Therefore, the amount invested in Account B is 33,000−x33,000 – x.
The interest earned from Account A is 0.05×0.05x, and the interest earned from Account B is 0.03(33,000−x)0.03(33,000 – x).
We are told that the interest earned by Account A is twice the interest earned by Account B, which gives the equation: 0.05x=2×0.03(33,000−x)0.05x = 2 \times 0.03(33,000 – x)
Now, solve for xx: 0.05x=0.06(33,000−x)0.05x = 0.06(33,000 – x) 0.05x=1,980−0.06×0.05x = 1,980 – 0.06x 0.05x+0.06x=1,9800.05x + 0.06x = 1,980 0.11x=1,9800.11x = 1,980 x=1,9800.11=18,000x = \frac{1,980}{0.11} = 18,000
So, Alfredo invested $18,000 in Account A and $33,000 – 18,000 = $15,000 in Account B.
Next, calculate the total interest earned:
- Interest from Account A: 0.05×18,000=9000.05 \times 18,000 = 900
- Interest from Account B: 0.03×15,000=4500.03 \times 15,000 = 450
The total interest is: 900+450=1,350900 + 450 = 1,350
Thus, the total interest earned by the two accounts for the first year is $1,350.
The correct answer is C. $1,350.