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Maxwell purchased $30,000 worth of 52-week T-Bills for $29,650. Find how much money should be deposited in a bank paying interest at the rate of 8.5%/year compounded quarterly so that at the end of 7 years, the accumulated amount will be $80,000. (Round your answer to the nearest cent.)

a) $22,000.45
b) $32,780.65
c) $39,285.19
d) $40,917.83

User Osanoj
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1 Answer

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Final answer:

To find the initial deposit amount that would grow to $80,000 in 7 years in a bank account with interest compounded quarterly at 8.5%, we use the compound interest formula and solve for P. The calculated amount is approximately $37,045.94, which is not an option given, suggesting an error in the question or choices provided.

Step-by-step explanation:

The question asks us to determine the amount that must be deposited in a bank account with an interest rate of 8.5% per year, compounded quarterly, so that after 7 years the amount grows to $80,000.

To calculate this, we use the formula for compound interest:
A = P(1 + r/n)^(nt)

Where:

  • A is the amount of money accumulated after n years, including interest.
  • P is the principal amount (the initial amount of money).
  • r is the annual interest rate (decimal).
  • n is the number of times that interest is compounded per year.
  • t is the time the money is invested for, in years.

We need to solve for P, given that A is $80,000, r is 0.085 (8.5% expressed as a decimal), n is 4 (since interest is compounded quarterly), and t is 7 years.

Rearranging the formula to solve for P gives us:

P = A / (1 + r/n)^(nt)

Plugging in the values we get:

P = $80,000 / (1 + 0.085/4)^(4*7)

P = $80,000 / (1 + 0.02125)^(28)

P = $80,000 / (1.02125)^(28)

P = $80,000 / 2.159885

P ≈ $37,045.94

After calculating, we find that approximately $37,045.94 needs to be deposited to achieve the goal. However, this amount is not one of the options provided, indicating a possible error in the setup of the multiple-choice answers or in the question itself. So we need to carefully recheck the computation and the given options.

User Jll
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