Question

Suppose that $250 is deposited into an account that pays 4.5% interest compoundedquarterly. Using A = P(1 +r/n)nt where t is the number of years, r the interestrate as a decimal, and n the number of times interest is compounded per year, find outhow many years it takes (to the nearest whole year) to reach $1000, and type youranswer into the box.

Answer 1

31 years

Step-by-step explanation:

We would apply the compound interest forula:

`A=P\mleft(1+(r)/(n)\mright)^(nt)`

A = future amount = $1000

P = principal = $250

r = rate = 4.5% = 0.045

n = compounded quarterly = 4 times

n = 4

t = time = ?

Inserting the values into the formula:

`\begin{gathered} 1000\text{ = 250(1 + }(0.045)/(4))^(4* t) \\ 1000=250(1+0.01125)^(4t) \\ \text{divide through by 250} \\ (1000)/(250)=\text{ }(1+0.01125)^(4t) \\ 4\text{ = (1}.01125)^(4t) \end{gathered}`
`\begin{gathered} \text{Taking log of both sides:} \\ \log 4=log(1.01125)^(4t) \\ \log 4=4t\lbrack log(1.01125)\rbrack \\ 0.6021\text{ = 4t(}0.0049) \end{gathered}`
`\begin{gathered} 0.6021\text{ = }0.0196t \\ \text{divide both sides by 0.0196} \\ (0.6021)/(0.0196)=(0.0196t)/(0.0196) \\ 30.72\text{ = t} \\ To\text{ the nearest whole number, t = 31 years} \end{gathered}`

It takes 31 years to reach $1000.