Question

Suppose you deposit 1,000​$ in a savings account that pays interest at an annual rate of ​6%. If no money is added or withdrawn from the​ account, answer the following questions.
a. How much will be in the account after 4 ​years?
b. How much will be in the account after 17 ​years?
c. How many years will it take for the account to contain ​1,500$​?
d. How many years will it take for the account to contain ​2,000$​?

Answer 1

Answer:

a. $1240

b. $2020

c. 8.3 years

d. 16.7 years

Step-by-step explanation:

Using the simple interest formula, I = Prt, we solve as follows.

a.

I = 1000 × 0.06 × 4 = 240

1000 + 240 = 1240

Answer: a. $1240

b.

I = 1000 × 0.06 × 17 = 1020

1000 + 1020 = 2020

Answer: b. $2020

c.

$1500 - $1000 = $500

Interest is $500

500 = 1000 × 0.06 × t

t = 8.3

Answer: c. 8.3 years

d.

$2000 - $1000 = $1000

Interest is $1000

1000 = 1000 × 0.06 × t

t = 16.7

Answer: d. 16.7 years

Answer 2

Answer:

Please note I have assumed the interest is compounded annually.

a)  $1,262.48

b)  $2,692.77

c)  7 years

d)  12 years

Step-by-step explanation:

Assuming the interest is compounded annually.

$\boxed{\begin{minipage}{7 cm}\underline{Annual Compound Interest Formula}\\\\ A=P\left(1+r\right)^{t} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\ \phantom{ww} \bullet P = principal amount \\ \phantom{ww} \bullet r = interest rate (in decimal form) \\ \phantom{ww} \bullet t = time (in years) \\ \end{minipage}}$

Given:

• P = $1,000
• r = 6% = 0.06

Substitute the given values into the formula to create an equation for A in terms of t:

$\implies A=1000(1+0.06)^t$

$\implies A=1000(1.06)^t$

Part (a)

To calculate how much will be in the account after 4 years, substitute t = 4 into the equation:

$\implies A=1000(1.06)^4$

$\implies A=1000(1.2624769...)$

$\implies A=1262.4769...$

Therefore, there will be $1,262.48 in the account after 4 years.

Part (b)

To calculate how much will be in the account after 17 years, substitute t = 17 into the equation:

$\implies A=1000(1.06)^{17}$

$\implies A=1000(2.69277278...)$

$\implies A=2692.77278...$

Therefore, there will be $2,692.77 in the account after 17 years.

Part (c)

To calculate how many years it will take for the account to contain $1,500, substitute A = 1500 into the equation and solve for t:

$\implies 1500=1000(1.06)^t$

$\implies 1.5=(1.06)^t$

$\implies \ln 1.5=\ln (1.06)^t$

$\implies \ln 1.5=t \ln (1.06)$

$\implies t=6.95851563...$

Therefore, it would take 7 years for the account to contain $1,500.

Part (d)

To calculate how many years it will take for the account to contain $2,000, substitute A = 2000 into the equation and solve for t:

$\implies 2000=1000(1.06)^t$

$\implies 2=(1.06)^t$

$\implies \ln 2=\ln (1.06)^t$

$\implies \ln 2=t \ln (1.06)$

$\implies t=11.8956610...$

Therefore, it would take 12 years for the account to contain $2,000.