Suppose that $2000 is placed in a savings account at an annual rate of 4.6%, compounded quarterly. Assuming that no withdrawals are made, how long will it take for the account to grow to $3000? Do not round any intermediate computations, and round your answer to the nearest hundredth.

If necessary, refer to the list of financial formulas.

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Answer:

It will take 35.46 quarters for the account to grow to $3000.

Step-by-step explanation:

Since the annual rate is compounded quarterly, this can be calculated using the formula for calculating the future value as follows:

FV = PV * (1 + r)^n ............................ (1)

Where;

FV = future value or the amount the deposit expected to grow to = $3,000

PV = Present value or the amount place in the savings = $2,000

r = Quarterly rate = Annual rate / 4 = 4.6% / 4 = 0.046 / 4 = 0.0115

n = number of quarters it will take for the loan to grow to $3000 = ?

Substituting the values into equation (1) and solve for n, we have:

$3,000 = $2,000 * (1 + 0.0115)^n

$3,000 / $2,000 = (1.0115)^n

1.50 = (1.0115)^n

Loglinearise both sides, we have:

log(1.50) = n log(1.0115)

0.176091259055681 = n * 0.00496588710682352

n = 0.176091259055681 / 0.00496588710682352

n = 35.4601816891322

Rounding to the nearest hundredth, which also implies to rounding to 2 decimal places, we have:

n = 35.46

Since the the annual rate is compounded quarterly, it will therefore take 35.46 quarters for the account to grow to $3000.

Suppose that $2000 is placed in a savings account at an annual rate of 4.6%, compounded quarterly. Assuming that no withdrawals are made, it will take him 8.86 years for the account to grow to $3000

From the information given:

  • The principal amount placed in the savings account (P) = 2000
  • The annual interest rate = 4.6%
  • number of times interest is compounded n = 4

By using the compound interest formula:

[tex]\mathbf{A = P( 1+\dfrac{r}{n})^{nt}}[/tex]

replacing the values from above, we have:

[tex]\mathbf{3000= 2000( 1+\dfrac{0.046}{4})^{4t}}[/tex]

[tex]\mathbf{\dfrac{3000}{2000}= ( 1+\dfrac{0.046}{4})^{4t}}[/tex]

[tex]\mathbf{\dfrac{3000}{2000}= ( 1+0.0115)^{4t}}[/tex]

[tex]\mathbf{\dfrac{3000}{2000}= ( 1.0115)^{4t}}[/tex]

[tex]\mathbf{log(\dfrac{3000}{2000})= 4t \times log ( 1.0115)}[/tex]

[tex]\mathbf{0.17609= 4t \times0.004966}[/tex]

[tex]\mathbf{t = \dfrac{0.17609}{4 \times 0.004966 }}[/tex]

t = 8.86 years to the nearest hundredth.

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