# Cash Flow and Equivalence

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Cash Flow and Equivalence

1. Cash Flow .............................. 51-1 2. Time Value of Money .................... 51-2 3. Discount Factors and Equivalence ........ 51-2 4. Functional Notation ..................... 51-5 5. Nonannual Compounding ................ 51-5 6. Continuous Compounding ................ 51-6
Sample Problems ........................ 51-6 FE-Style Exam Problems ................ 51-8 Solutions ............................... 51-11

Nomenclature

A

annual amount or annual value

C

initial cost, or present worth (present value) of all

costs

F

future worth or future value

G

interest rate per period

m

number of compounding periods per year

n

number of compounding periods

P

present worth (present value)

r

nominal rate per year (rate per annum)

time

Subscripts

0

initial

e

annual effective rate

j

at time j

n

at time n

1. CASH FLOW
The sums of money recorded as receipts or disbursements in a project's financial records are called cash flows. Examples of cash flows are deposits to a bank, dividend interest payments, loan payments, operating and maintenance costs, and trade-in salvage on equipment. Whether the cash flow is considered to be a receipt or disbursement depends on the project under consideration. For example, interest paid on a sum in a bank account will be considered a disbursement to the bank and a receipt to the holder of the account.
Because of the time value of money, the timing of cash flows over the life of a project is an important factor. Although they are not always necessary in simple problems (and they are often unwieldy in very complex problems), cash flow diagrams can be drawn to help visualize and simplify problems that have diverse receipts and disbursements.
The following conventions are used to standardize cash flow diagrams.

• The horizontal (time) axis is marked off in equal increments, one per period, up to the duration of the project.
• Receipts are represented by arrows directed upward. Disbursements are represented by arrows ·directed downward. The arrow length is approximately proportional to the magnitude of the cash flow.
• Two or more transfers in the same period are placed end to end, and these may be combined.
• Expenses incurred before t = 0 are called sunk costs.
Sunk costs are not relevant to the problem unless they have tax consequences in an after-tax analysis.
For example, consider a mechanical device that will cost \$20,000 when purchased. Maintenance will cost \$1000 each year. The device will generate revenues of \$5000 each year for five years, after which the salvage value is expected to be \$7000. The cash flow diagram is shown in Fig. 51.1(a), and a simplified version is shown in Fig. 51.1(b).
In order to evaluate a real-world project, it is necessary to present the project's cash flows in terms of standard

Figure 51.1 Cash Flow Diagrams

\$7000

\$5000 \$5000 \$5000 I \$5000 \$5000

\$1000 \$1000 \$1000 \$1000 \$1000

\$20,000

(a) cash flow diagram

\$11,000 \$4000

\$20,000 (b) simplified cash flow diagram

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51-2 FE REVIEW MANUAL

cash flows that can be handled by engineering economic analysis techniques. The standard cash flows are single payment cash flow, uniform series cash flow, and gradient series cash flow.
A single payment cash flow can occur at the beginning of
the time line (designated as t = 0), at the end of the time
line (designated as t= n), or at any time in between.
The uniform series cash flow, illustrated in Fig. 51.2, consists of a series of equal transactions starting at t = 1 and ending at t= n. The symbol A (representing an annual amount) is typically given to the magnitude of each individual cash flow.

Figure 51.2 Uniform Series

A each

t t t t t t t t 1
t=Ot=1

t=n

Notice that the cash flows do not begin at the beginning of a year (i.e., the year 1 cash flow is at t= 1, not t=O). This convention has been established to accommodate the timing of annual maintenance and other cash flows for which the year-end convention is applicable. The year-end convention assumes that all receipts and disbursements take place at the end of the year in which they occur. The exceptions to the year-end convention are initial project cost (purchase cost), trade-in allowance, and other cash flows that are associated with the inception of the project at t= 0.
The gradient series cash flow, illustrated in Fig. 51.3, starts with a cash flow (typically given the symbol G) at t= 2 and increases by G each year until t= n, at which time the final cash flow is (n - 1) G. The value of the gradient at t= 1 is zero.

4G 3G 2G G
t=O t=2

(n- 1)G
t= n

2. TIME VALUE OF MONEY
Consider \$100 placed in a bank account that pays 5% effective annual interest at the end of each year. After the first year, the account will have grown to \$105. After the second year, the account will have grown to \$110.25.
The fact that \$100 today grows to \$105 in one year at 5% annual interest is an example of the time value of money principle. This principle states that funds placed

in a secure investment will increase in value in a wa that depends on the elapsed time and the interest rat/
The interest rate that is used in calculations is known as
the effective interest rate. If compounding is once a year it is known as the effective annual interest rate. How~ ever, effective quarterly, monthly, or daily interest rates are also used.
~~-. ~.~.!i.~.QY.~.!..F.:~~I9.~~. ~~.~. ~;~y~~A~~~~.~
Assume that you will have no need for money during the next two years, and any money you receive will immediately go into your account and earn a 5% effective annual interest rate. Which of the following options would be more desirable to you?

t= 0 option b: receive \$105 in one year

t=0 t= 1

option

c:

\$110.25

in

ttwo years \$110.25

I

t= 0

t= 2

None of the options is superior under the assumptions given. If you choose the first option, you will immediately place \$100 into a 5% account, and in two years the account will have grown to \$110.25. In fact, the account will contain \$110.25 at the end of two years regardless of which option you choose. Therefore, these alternatives are said to be equivalent.
The three options are equivalent only for money earning 5% effective annual interest rate. If a higher interest rate can be obtained, then the first option will yield the most money after two years. Thus, equivalence depends on the interest rate, and an alternative that is acceptable to one decision maker may be unacceptable to another who invests at a higher rate. The procedure for determining the equivalent amount is known as discounting. For a list of discount factors, see Table 51.1.

Single Payment Equivalence
The equivalent future amount, F, at t= n, of any present amount, P, at t = 0 is called the future worth and can be calculated from Eq. 51.1. In this equation, and for all the other discounting formulas, the interest rate used must be the effective rate per period. The basis of the

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CASH F L 0 W AND E QUI VALENCE 51-3

Table 51.1 Discount Factors for Discrete Compounding

factor name
single payment compound amount
single payment present worth
uniform series sinking fund

converts Pto F Fto P Fto A

capital recovery

Pto A

uniform series compound amount
uniform series present worth

A to F AtoP Gto P Gto F Gto A

symbol (F/P, i%, n)
(P/F, %%, n)
(A/F, i%, n)
(A/P, %%, n)
(F/ A,%%, n)
(P/A, %%, n) (P/G, %%, n) (F/G, %%, n) (A/G, %%, n)

formula
(1 +it

(1 + i)-n

i
(1 +it -1

i(1 +it (1 +it -1

(1 +it -1
i
(1 +it -1 i(l +it

(1 +it -1 n i2(1 +it - i(1 +it

(1 +it -1 --n

i2

i

-1 -
i

(1

n
+it

-1

rate (annually, monthly, etc.) must agree with the type of period used to count n. It would be incorrect to use an effective annual interest rate if n was the number of compounding periods in months.

Similarly, the present worth factor has a functional
notation of (P/F, i%, n). The present worth of a future
amount n periods from now would be symbolically writ-
ten as

F=P(l+i)"

51.1

P = F(P/F, i%, n)

51.4

lS

The factor (1 + i) n is known as the single payment com-

The discounting factors are listed in Table 51.1 in sym-

i-

pound amount factor.

bolic and formula form. Normally, it will not be neces-

te

Similarly, the equivalence of any future amount to any

sary to calculate factors from these formulas. Values of

lt

present amount is called the present worth and can be

these cash flow (discounting) factors are tabulated in

oesf calculated from Eq. 51.2.

Table 51.2 through Table 51.11 for various combinations of i and n. For intermediate values, computing

the factors from the formulas may be necessary, or linear

lg

P = F(1 + if" = (1 : i)"

51.2

interpolation can be used as an approximation.

te

st

Uniform Series Equivalence

)ll

The factor (1 + i)-n is known as the single payment

to

present worth factor.

A cash flow that repeats at the end of each year for n

10

years without change in amount is known as an annual

lg

Rather than actually writing the formula for the com-

amount and is given the symbol A. (This is shown in

a

pound amount factor (which converts a present amount

Fig. 51.2.) Although the equivalent value for each of the

to a future amount), it is common convention to sub-

n annual amounts could be calculated and then

stitute the standard functional notation of (F/ P, z%, n).

summed, it is more expedient to use one of the uniform

This notation is interpreted as, "Find F, given P, using

series factors. For example, it is possible to convert from

an interest rate of z% over n years." Thus, the future

an annual amount to a future amount by using the

value in n periods of a present amount would be symbo-

{F/ A) uniform series compound amount factor.

an

lically written as

all

ehde F= P(FjP, i%, n) 51.3 F = A(F/A, i%, n) 51.5

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51-4 ~F~E~R~E~V~I~E~W~~M~A~N~U~A~L~---------------------------------------------------------

Example 51.1
Suppose you deposited \$200 at the end of every year for seven years in an account that earned 6% annual effective interest. At the end of seven years, how much would the account be worth?

Solution

Using Eq. 51.5,

F = (\$200)(FI A, 6%, 7)

1) =

(\$200)

((1

+

7
0.06)

-

0.06

= (\$200)(8.3938) = \$1678.76

(The value of 8.3938 could easily have been obtained directly from Table 51.7 at the end of this chapter.)
A sinking fund is a fund or account into which annual
deposits of A are made in order to accumulate Fat t = n
in the future. Because the annual deposit is calculated as
A= F(A/ F, z%, n), the (A/ F) factor is known as the
sinking fund factor.

Example 51.2
Suppose you want exactly \$1600 in the previous investment account at the end of the seventh year. By using the sinking fund factor, you could calculate the necessary annual amount you would need to deposit.

Solution From Table 51.1,

A= F(A/ F, 6%, 7)

= (\$1600) (

0·06 )

(1 + 0.06) 7 - 1

= (\$1600)(0.1191)

= \$190.56

An annuity is a series of equal payments, A, made over a period of time. Usually, it is necessary to "buy into" an investment (a bond, an insuran,ce policy, etc.) in order to fund the annuity. In the case of an annuity that starts at the end of the first year and continues for n years, the purchase price, P, would be

P= A(P/A, i%, n)

51.6

Example 51.3
Suppose you will retire in exactly one year and want an account that will pay you \$20,000 a year for the

next 15 years. (The fund will be depleted at the end 0f

the fifteenth year.) Assuming a 6% annual effect·

m. t erest rate, what 1.s the amount you would needlVt e

deposit now?

0

Solution Using Eq. 51.6,

P= A(P/A, 6%, 15)

= (\$20,000) ( (1 + 0.06)15 - 1 )

.

(0.06)(1 + 0.06) 15

= (\$20,000) (9. 7122)

= \$194,244

If the cash flow has the proper form (i.e., Fig. 51.3), its present worth can be determined by using the uniform
factor finds the present worth of a uniformly increasing cash flow. By definition of a uniform gradient, the cash flow starts in year 2, not in year 1.
There are three common difficulties associated with the form of the uniform gradient. The first difficulty is that the first cash flow starts at t= 2. This convention recognizes that annual costs, if they increase uniformly, begin with some value at t = 1 (due to the year-end convention), but do not begin to increase until t= 2. The tabulated values of (P/ G) have been calculated to find the present worth of only the increasing part of the annual expense. The present worth of the base expense
incurred at t = 1 must be found separately with the
(P/ A) factor.
The second difficulty is that, even though the (P/ G, z%, n) factor is used, there are only n- 1 actual cash flows. n
must be interpreted as the period number in which the last gradient cash flow occurs, not the number of gradient cash flows.
Finally, the sign convention used with gradient cash flows may seem confusing. If an expense increases each year, the gradient will be negative, since it is an expense. If a revenue increases each year, the gradient will be positive. In most cases, the sign of the gradient depends on whether the cash flow is an expense or a revenue.
Example 51.4
A bo~us package pays an employee \$1000 at the end of the first year, \$1500 at the end of the second year, and so on, for the first nine years of employment. What is the present worth of the bonus package at 6% interest?

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i of jve l to
, its
?1m
ient ling :ash the ;hat
~og­
~gin
renThe find the

CASH F L 0 W AND E QUI VALENCE 51-5

Solution The present worth of the bonus package is

P= (\$1000)(P/A, 6%, 9) + (\$500)(P/G, 6%, 9)

= (\$1000)

(

(1

+

9
0.06)

-

1 )

(0.06)(1 + 0.06)9

+ (\$500) ( i~~·o(~~,o~~' )

(0.06)(1 + 0.06)9 = (\$1000)(6.8017) + (\$500)(24.5768)

= \$19,090

4. FUNCTIONAL NOTATION
There are several ways of remembering what the functional notation means. One method of remembering which factor should be used is to think of the factors as conditional probabilities. The conditional probability of event A given that event B has occurred is written as P{AjB}, where the given event comes after the vertical bar. In the standard notational form of discounting factors, the given amount is similarly placed after the slash. What you want, A, comes before the slash. (F/P) would be a factor to find F given P.
Another method of remembering the notation is to interpret the factors algebraically. The (F/ P) factor could be thought of as the fraction F/ P. The numerical values of the discounting factors are consistent with this algebraic manipulation. The (F/ A) factor could be calculated as (F/ P)(P/A). This consistent relationship can be used to calculate other factors that might be occasionally needed, such as (F/ G) or (G/ P). For instance, the annual cash flow that would be equivalent to a uniform gradient may be found from.

A = G(P/ G, i%, n)(A/ P, i%, n)

51.7

~~---~.9.~~-~~-~~~--~.9.~F.».9.~.~~~~-~-··· ······························
If \$100 is invested at 5%, it will grow to \$105 in one year. If only the original principal accrues interest, the interest is known as simple interest and the account will grow to \$110 in the second year, \$115 in the third year, and so on. Simple interest is rarely encountered in engineering economic analyses.
More often, both the principal and the interest earned accrue interest, and this is known as compound interest. If the account is compounded yearly, then during the second year, 5% interest continues to be accrued, but on \$105, not \$100, so the value at year end will be \$110.25. The value after the third year will be \$115.76, and so on.

The interest rate used in the discount factor formulas is the interest rate per period, i (called the yield by banks). If the interest period is one year (i.e., the interest is compounded yearly), then the interest rate per period, i, is equal to the annual effective interest rate, ie. The annual effective interest rate is the rate that would yield the same accrued interest at the end of the year if the account were compounded yearly.
The term nominal interest rate, r (rate per annum), is encountered when compounding is more than once per year. The nominal rate does not include the effect of compounding and is not the same as the annual effective interest rate.
The effective interest rate can be calculated from the nominal rate if the number of compounding periods per year is known. If there are m compounding periods during the year (two for semiannual compounding, four for quarterly compounding, twelve for monthly compounding, etc.), the effective interest rate per period, i, is r/m. The effective annual interest rate, ie, can be calculated from the interest rate per period by using Eq. 51.9.

. r

z=-

51.8

m

ie = ( 1 + i) rn - 1

= (1+mr)m -1

51.9

Sometimes, only the effective rate per period (e.g., per month) is known. However, compounding for m periods at an effective interest rate per period is not affected by the definition or length of the period. For example, compounding for 365 periods (days) at an interest rate of 0.03808% is the same as compounding for 12 periods (months) at an interest rate of 1.164%, or once at an effective annual interest rate of 14.9%. In each case, the interest rate per period is different, but the effective annual interest rate is the same. If only the daily effective rate were given, the discount factor formulas could be used with i= 0.03808% and n= 365 to represent each yearly cash flow. Equation 51.9 could be used to calculate ie = 14.9% to use with n = 1 for each yearly cash flow.
Since they do not account for the effect of compounding, nominal rates cannot be compared unless the method of compounding is specified. The only practical use for a nominal rate is for calculating the effective rate.
The following rules may be used to determine what type of interest rate is given in a problem.
• Unless specifically qualified in the problem, the interest rate given is an annual rate. If the compounding period is not specified, the interest rate is the annual effective interest rate, ie.

51-6 FE REVIEW MANUAL

• If the compounding is annual, the rate given is the effective rate, ie· If compounding is not annual, the rate given is the nominal rate, r.

6. CONTINUOUS COMPOUNDING

.... ······································································-·

·····-······················

Discount factors for continuous compounding are different from those for discrete compounding. The discounting factors can be calculated directly from the nominal interest rate, r, and number of years, n, without having to find the effective interest rate per period.

(F / P, r%, n) =ern
(P/F,r%,n)=e-rn
er -1 (A/ F, r%, n) = - ern- - 1
ern -1 (F/A, r%, n) = - -
er- 1
(A/ P, r%, n) = 1e-r -e1-rn
1 -rn
(P/A, r%, n) = - e
er- 1

51.10 51.11 51.12 51.13 51.14 51.15

The effective annual interest rate determined on a daily compounding basis will not be significantly different than if continuous compounding is assumed.

SAMPLE PROBLEMS
Problem 1 If a credit union pays 4.125% interest compounded quarterly, what is the effective annual interest rate?
(A) 4.189% (B) 8.250% (C) 12.89% (D) 17.55%
Solution 4.125% is the nominal annual rate, r.
ie = (1 + i) m - 1 = ( 1 + 0.0!125f - 1
= 0.04189 (4.189%)

Problem 2 The national debt is approximately \$4 trillion. What is the required payment per year to completely pay off the debt in 20 years, assuming an interest rate of 6%?
(A) \$315 billion (B) \$325 billion (C) \$350 billion (D) \$415 billion
Solution Create a cash flow diagram.
\$4trillion

t= 1 t= 0

···-nt=A20

Use the capital recovery discount factor from the tables.

(A/ P, 6%, 20) = 0.0872
A= P(A/ P, 6%, 20) = (\$4,000,000,000,000) (0.0872)
= \$348,800,000,000 (\$350 billion)

Problem 3 The president of a growing engineering firm wishes to give each of 50 employees a holiday bonus. How much is needed to invest monthly for a year at 12% nominal interest rate, compounded monthly, so that each employee will receive a \$1000 bonus?
(A) \$2070 (B) \$3840
(C) \$3940 (D) \$4170
Solution
The total holiday bonus is
(50)(\$1000) = \$50,000

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C A 5 H F L 0 W AND E QUI VALENCE 51-7

\$50,000

r-.-.-.-.-.-.-.-.-.-.-~t=12
t= 0 A

Use the uniform series sinking fund discount factor. The
interest period is one month, there are 12 compounding
periods, and the effective interest rate per interest per-
iod is 12%/12 = 1%.

(A/ F, 1%, 12) =

0·0\

= 0.0788

(1+0.01) 2 -1

A= F(Aj F, 1%, 12)

= (\$50,000)(0.0788)

= \$3940

Problem 4 If the nominal interest rate is 3%, how much is \$5000 worth in 10 years in a continuously compounded account?
(A) \$3180 (B) \$4490 (C) \$5420 (D) \$6750
Solution Create a cash flow diagram.
F

t=o.-r-r-~+-+-+-+-+-4-~t=10
\$5000

Use the single payment compound amount factor for continuous compounding.
n= 10 r=3% (FjP, r%, n) =ern
(F/ P, 3%, 10) = e(0·03)(IO) = 1.34986
F= P(FjP, 3%, 10)
= (\$5000)(1.34986) = \$6749

Problem 5
An engineering graduate plans to buy a home. She has been advised that her monthly house and property tax payment should not exceed 35% of her disposable monthly income. After researching the market, she determines she can obtain a 30-year home loan for 6.95% annual interest per year, compounded monthly. Her monthly property tax payment will be approximately \$150. What is the maximum amount she can pay for a house if her disposable monthly income is \$2000?
(A) \$80,000
(B) \$83,100
(C) \$85,200
(D) \$90,500

Solution Create a cash flow diagram.

A each

tt=Ot

t
1

t

t

t

t

t

t

t

t

...

l

Jt~30

The amount available for monthly house payments, A, is

(\$2000)(0.35) - \$150 = \$550

Use the uniform series present worth discount factor. The effective rate per period is

i = 0·0695hs = 0.00579 per month
12 mont

n =

( 30

years) ( 12

months) year

= 360

months

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51-8 FE REVIEW MANUAL

There are no tables for this interest rate.

(PIA 0.579% 360) = (1 + it - 1

'

'

i(1 +it

(1 + 0.00579) 360 - 1 (0.00579)(1 + 0.00579)360

= 151.10

P =A(PI A, 0.579%, 360)

= (\$550)(151.10)

= \$83,105

Problem 6
The designer of the penstock for a small hydroelectric co~eneration station has the option of using steel pipe, wh1ch costs \$150,000 installed and requires \$5000 yearly for painting and leak-checking maintenance, or DSR4.3 (heavy-duty plastic) pipe, which costs \$180,000 installed and requires \$1200 yearly for leak-checking maintenance. Both options have an expected life of 25 years. If the interest rate is 8%, which choice has the lower present equivalent cost and how much lower is it?
(A) DSR4.3 costs less by \$10,600.
(B) Steel pipe costs less by \$10,600.
(C) DSR4.3 costs less by \$65,000.
(D) Steel pipe costs less by \$65,000.

Solution

The problem requires a comparison of the uniform series present worth of each alternative.

F

lor stee t= 0 t

l

p
t

i

pe
t

,

t

t

t

t t t

t

...

t

lt =

25 A

=

\$

5

0

0

0

\$150,000
For DSR4.3,

I 1

- t= 0
lt t t t t l

t t l

t

...

t lt= 25 A

=

\$

1

2

0

0

\$180,000
P(steel pipe)= \$150,000 + A(PI A, 8%, 25) = \$150,000 + (\$5000)(10.6748)
= \$203,374
P(DSR4.3) = \$180,000 + A(PI A, 8%, 25) = \$180,000 + (\$1200)(10.6748)
= \$192,810

Using DSR4.3 is less expensive by

\$203,374- \$192,810 = \$10,564 (\$10,600)

r=.~.~~T.!.~~...I::~.fttt. ~~~.~.~.~.e.:~~. . . . . . . . .
1. If the interest rate on an account is 11.5% compounded yearly, approximately how many years will it take to triple the amount?
(A) 8 years
(B) 9 years
(C) 10 years
(D) 11 years

2. Fifteen years ago \$1000 was deposited in a bank account, and today it is worth \$2370. The bank pays interest semi-annually. What was the nominal annual interest rate paid on this account?
(A) 2.9%
(B) 4.4%
(C) 5.0%
(D) 5.8%

3. Mr. Jones plans to deposit \$500 at the end of each month for 10 years at 12% annual interest, compounded monthly. The amount that will be available in two years is
(A) \$13,000
(B) \$13,500
(C) \$14,000
(D) \$14,500

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C A S H F L 0 W A N D E Q U I V A L E N C E 51-9

4. The purchase price of a car is \$25,000. Mr. Smith makes a down payment of \$5000 and borrows the balance from a bank at 6% annual interest, compounded monthly for five years. Calculate the nearest value of the required monthly payments to pay off the loan.
(A) \$350 (B) \$400
(C) \$450 (D) \$500
5. A piece of machinery can be bought for \$10,000 cash or for \$2000 down and payments of \$750 per year for 15 years. What is the annual interest rate for the time payments?
(A) 1.51% (B) 4.61% (C) 7.71% (D) 12.0%
For the following problems, use the NCEES Handbook as your only reference.
6. You have borrowed \$5000 and must pay it off in five equal annual payments. Your annual interest rate is 10%. How much interest will you pay in the first two years?
(A) \$855 (B) \$868 (C) \$875 (D) \$918
7. A company puts \$25,000 down and will pay \$5000 every year for the life of a machine (10 years). If the salvage value is zero and the interest rate is 10% compounded annually, what is the present value of the machine?
(A) \$55,700 (B) \$61,400 (C) \$75,500 (D) \$82,500

8. You borrow \$3500 for one year from a friend at an interest rate of 1.5% per month instead of taking a loan from a bank at a rate of 18% per year. Compare how much money you will save or lose on the transaction.
(A) You will pay \$55 more than if you borrowed from the bank.
(B) You will pay \$630 more than if you borrowed from the bank.
(C) You will pay \$685 more than if you borrowed from the bank.
(D) You will save \$55 by borrowing from your friend.
9. If you invest \$25,000 at 8% interest compounded annually, approximately how much money will be in the account at the end of 10 years?
(A) \$31,000
(B) \$46,000
(C) \$54,000
(D) \$75,000
10. A college student borrows \$10,000 today at 10% interest compounded annually. Four years later, the student makes the first repayment of \$3000. Approximately how much money will the student still owe on the loan after the first payment?
(A) \$7700
(B) \$8300
(C) \$11,000
(D) \$11,700
11. A 40-year-old consulting engineer wants to set up a retirement fund to be used starting at age 65. \$20,000 is invested now at 6% compounded annually. Approximately how much money will be in the fund at retirement?
(A) \$84,000
(B) \$86,000
(C) \$88,000
(D) \$92,000

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51-10 FE REVIEW MANUAL

1 2. The maintenance cost for a car this y~ar is expected to be \$500. The cost will increase \$50 each year for the subsequent 9 years. The interest is 8% compounded annually. What is the approximate present worth of maintenance for the car over the full 10 years?
(A) \$4300
(B) \$4700
(C) \$5300
(D) \$5500
13. A house is expected to have a maintenance cost of \$1000 tlie first year. It is believed that the maintenance cost will increase \$500 per year. The interest rate is 6% compounded annually. Over a 10-year period, what will be the approximate effective annual maintenance cost?
(A) \$1900
(B) \$3000
(C) \$3500
(D) \$3800
14. You deposited \$10,000 in a savings account five years ago. The account has earned 5.25% interest compounded continuously since then. How much money is in the account today?
(A) \$12,800
(B) \$12,900
(C) \$13,000
(D) \$13,600
15. A young engineer wants to surprise her husband with a European vacation for their tenth anniversary, which is five years away. She determines that the trip will cost \$5000. Assuming an interest rate of 5.50% compounded daily, approximately how much money does she need to deposit today for the trip?
(A) \$3790
(B) \$3800
(C) \$3880
(D) \$3930

17. A deposit of \$1000 is made in a bank account that pays 8% interest compounded annually. Approximately how much money will be in the account after 10 years?
(A) \$1890 (B) \$2000 (C) \$2160 (D) \$2240
18. A deposit of \$1000 is made in a bank account that pays 24% interest per year compounded quarterly. Approximately how much money will be in the account after 10 years?
(A) \$7000 (B) \$7200 (C) \$8600 (D) \$10,000
19. A machine costs \$20,000 today and has an estimated scrap cash value of \$2000 after eight years. Inflation is 8% per year. The effective annual interest rate earned on money invested is 6%. How much money needs to be set aside each year to replace the machine with an identical model eight years from now?
(A) \$2970 (B) \$3000 (C) \$3290 (D) \$3540
20. At what rate of annual interest will an investment quadruple itself in 12 years?
(A) 10.1% (B) 11.2% (C) 12.2% (D) 13.1%

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