Question

1. helene invested a total of $1000 in two simple-interest bank accounts. one account paid 5% annual interest; the other paid 6% annual interest. the total amount of interest she earned after 1 year was $58. find the amount invested in each account.

Explanation:

Step1: Define Variables

Let $x$ be the amount invested at 5%, so the amount invested at 6% is $1000 - x$.

Step2: Interest Formula

Simple interest formula: $I = Prt$. For $t = 1$ year, $I_1 = 0.05x$ (interest at 5%), $I_2 = 0.06(1000 - x)$ (interest at 6%). Total interest $I_1 + I_2 = 58$.

Step3: Set Up Equation

$0.05x + 0.06(1000 - x) = 58$

Step4: Solve Equation

Expand: $0.05x + 60 - 0.06x = 58$
Combine like terms: $-0.01x + 60 = 58$
Subtract 60: $-0.01x = -2$
Divide by -0.01: $x = 200$

Step5: Find Both Amounts

Amount at 5%: $x = 200$. Amount at 6%: $1000 - 200 = 800$.

Answer:

Amount invested at 5%: $\$200$, Amount invested at 6%: $\$800$