You invest in: Bonds yielding an interest rate of 1.6 percent, Utility Stock yielding an interest rate of 5.2 percent, and a Bank Certificate yielding an interest rate of 7.5 percent. The total interest earned in all three accounts in one year was \( \$ 4,373.00 \) dollars. The amount allocated to the bonds was 500 less than the amount allocated to the bank certificate. The amount allocated to the utility stock was 4 times the amount allocated to the bonds. What was the total amount of money the person had to invest? The total amount invested was \( \$ \)
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Mind Expander
To solve for the total amount invested, we can represent the amounts allocated to each type of investment with variables. Let's define: - \( B \) as the amount invested in bonds. - \( S \) as the amount invested in utility stock. - \( C \) as the amount invested in bank certificates. From the problem, we have the following relationships: 1. \( B = C - 500 \) (The amount allocated to bonds is $500 less than that of the bank certificates.) 2. \( S = 4B \) (The amount allocated to utility stock is 4 times that of bonds.) 3. The total interest earned can be represented as: \[ 0.016B + 0.052S + 0.075C = 4,373 \] Now, substituting the expressions for \( S \) and \( B \) into the total interest equation: We start substituting \( S = 4B \) into the interest equation: \[ 0.016B + 0.052(4B) + 0.075C = 4,373 \] Simplifying that gives: \[ 0.016B + 0.208B + 0.075C = 4,373 \] \[ 0.224B + 0.075C = 4,373 \] Now substitute \( B = C - 500 \): \[ 0.224(C - 500) + 0.075C = 4,373 \] Distributing \( 0.224 \): \[ 0.224C - 112 + 0.075C = 4,373 \] Combining like terms: \[ (0.224 + 0.075)C - 112 = 4,373 \] \[ 0.299C - 112 = 4,373 \] Adding 112 to both sides: \[ 0.299C = 4,485 \] Now, divide both sides by 0.299: \[ C \approx 15,029.10 \] Now calculate \( B \) and \( S \): \[ B = C - 500 \approx 15,029.10 - 500 = 14,529.10 \] \[ S = 4B = 4(14,529.10) \approx 58,116.40 \] Finally, the total amount invested is: \[ T = B + S + C = 14,529.10 + 58,116.40 + 15,029.10 \approx 87,674.60 \] So, the total amount invested was \( \$ 87,674.60 \).