Question

13. Use the appropriate percent growth to determine how much money Lyra will have in each ofthe following situations:(a) How much money will Lyra have after 10 years if she invests $5,000 at 4% interest com-pounded annually?(b) Suppose that Lyra is saving for retirement, and has saved up $20,000. If her retirementaccount earns 3% interest each year, how much will she save in 25 years?(c) How much money will Lyra have after 20 years if she $5,000 at 3.5% interest compoundedannually?(d) How much money will Lyra have after 10 years if she invests $5,000 at 1.5% interestcompounded quarterly?(e) How much money will Lyra have after 10 years if she invests $5,000 at 0.8% interestcompounded continuously?(f) Compare your answers to (a) and (c). Which one made more money, and does the resultsurprise you?

Answer 1

a) 7401.22

b) 35000

c) 9948.94

d) 5807.54

e) 5416.44

f)(c) made more money. I'm not surprised because c has more years

Step-by-step explanation:

The formula for calculating compound interest is expressed as

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

where

A is the total amount after t years

P is the principal or amount invested

r is the interest rate

n is he number of compounding periods in a year

t is the number of years

a) From the information given,

P = 5000

r = 4% = 4/100 = 0.04

t = 10

n = 1 because it was compounded once per year.

By substituting these values into the formula, we have

A = 5000(1 + 0.04/1)^1 * 10

A = 5000(1.04)^10

A = 7401.22

b) From the information given,

P = 20000

r = 3% = 3/100 = 0.03

t = 25

This is simple interest. The formula for calculating simple interest is expressed as

I = Prt

By substituting the values,

I = 20000 x 0.03 x 25 = 15000

Total amount after 25 years = P + I = 20000 + 15000

Total amount after 25 years = 35000

c) From the information given,

P = 5000

r = 3.5% = 3.5/100 = 0.035

t = 20

n = 1 because it was compounded once per year.

By substituting these values into the formula, we have

A = 5000(1 + 0.035/1)^1 * 20

A = 5000(1.035)^20

A = 9948.94

d) From the information given,

P = 5000

r = 1.5% = 1.5/100 = 0.015

t = 10

n = 4 because it was compounded quarterly.

By substituting these values into the formula, we have

A = 5000(1 + 0.015/4)^4 * 10

A = 5000(1.00375)^40

A = 5807.54

e) The formula for calculating continuously compounded interest is expressed as

A = Pe^rt

From the information given,

P = 5000

r = 0.8% = 0.8/100 = 0.008

t = 10

By substituting these values into the formula, we have

A = 5000e^(0.008 * 10)

A = 5000e^(0.08)

A = 5416.44

f) By comparing (a) and (c), (c) made more money. I'm not surprised because c has more years