Section 6.7: solutions
#1-4: Use the compound interest formula
(
)
to answer the following.
1) An initial deposit of $1,000 earns 4% interest compounded twice per year. How much will be in the
account after 5 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the
answer.
A
Solve
P
1000
(
)
r
.04
n
2
t
5
(note: you must put the 2*5 in a parenthesis when you use your calculator)
Answer: The investment will be worth $1,218.99 in 5 years
3) An initial deposit of $15,000 earns 2% interest compounded quarterly. How much will be in the
account after 8 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the
answer.
A
Solve
P
15000
(
)
r
.02
n
4
t
8
(note: you must put the 4*9 in a parenthesis when you use your calculator)
Answer: The investment will be worth $17,595.65 in 8 years.
#5-8: Use the formula A=Pert to answer the following.
5) An initial investment of $5,000 earns 6% interest compounded continuously. What will the
investment be worth in 5 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the
answer.
A
Solve
P
5000
r
.06
t
5
A = 5000e.06*5
Answer: The investment will be worth $6,749.29 in 5 years.
7) An initial investment of $15,000 earns 3% interest compounded continuously. What will the
investment be worth in 6 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the
answer.
A
Solve
P
15000
r
.03
t
6
A = 15000e.03*6
Answer: The investment will be worth $17,958.26 in 6 years.
9) How long will it take an initial investment of $1,000 to triple if it is expected to earn 6% interest
compounded continuously? (Round to 1 decimal place)
I will use this formula because interest is compounded continuously: A=Pert
The $1,000 is the P, the A is triple this amount so the A is $3,000. I need to solve for t.
I will put these values in the formula and solve for t.
A
3000
P
1000
r
.06
t
solve
3000 = 1000e.06t (divide both sides by 1000)
3 = e.06t (take ln of each side)
Ln 3 = ln e.06t (use power to product rule)
Ln 3 = .06tln e (lne = 1, so drop it)
Ln 3 = .06t (divide by .06 then use calculator)
Answer: 18.3 years (rounded to one decimal place as required)
11) How long will it take an initial investment of $100,000 to grow to $1,000,000 if it is expected to earn
4% interest compounded continuously? (Round to 1 decimal place)
I will use this formula because interest is compounded continuously: A=Pert
The $100,000 is the P, the A is$1,000,000. I need to solve for t.
I will put these values in the formula and solve for t.
A
1,000,000
P
100,000
r
.04
t
solve
1,000,000 = 100,000e.04t (divide both sides by 10,000)
10 = e.04t (take ln of each side)
Ln 10 = ln e.04t (use power to product rule)
Ln 10 = .04tln e (lne = 1, so drop it)
Ln 10 = .04t (divide by .04 then use calculator)
Answer: 57.6 years (rounded to one decimal place as required)
13) What will a $200,000 home cost in in 5 years if the price appreciation over that period is expected
to be 3% compounded annually?
(
I will use the
)
formula because the growth is compounded annually. This tells me that
n=1
I am solving for A.
A
Solve
P
200,000
(
r
.03
)
Answer: The house will be cost $231,854.81 in 5 years.
n
1
t
5
15) I will use the formula
(
)
as the growth in compounded annually. In this case n will
equal 1.
I am solving for A:
A
Solve
(
A=
P
r
n
t
100000
.02
1
3
)
Answer: The house will cost $106,120.80 in three years
17) I will use the formula
(
)
as the growth in compounded annually. In this case n will
equal 1.
I am solving for A:
A
Solve
A=
(
P
r
n
t
80
.04
1
5
)
Answer: The tuition will be $97.33 in 5 years