Chap 7 Eng'g Economy

CHAPTER 7

ENGINEERING ENGINEERING ECONOMICS ECONOMICS

GENERAL ENGINEERING & APPLIED SCIENCES

I.

CASH FLOW DIAGRAMS

A cash - flow diagram diagram is a graphical representation of cash flows drawn on a time scale. Borrower’s Viewpoint Arrow Convention: directed upward ↑ − arrows represents positive cash flow or cash inflow (receipts).

↓−

arrows

directed

downward

Lender’s Viewpoint

represents negative cash flow or cash outflow ➊ Single Payment Cash Flow - can occur at the beginning of the time line ( t = 0 ), at the end of the time line ( t = n ), or any time in between.

Single Payment

t =1

➋ Unifor m Series Cash Cash Flow

t =n

Uniform Series

- consists of a series of equal payments “ A” A” st starting at t = 1 and ending at t = n .

(n-1)G

➌ Gradient Series Cash Flow G - starts with a cash flow “G” “ G” at t = 2 , and increases by “G” “G” each year until t = n , at which time the final final cash cash flow flow is (n − 1)G .

t =n

6G 5G 4G 3G 2G

t=2 Gradient Series

t =n

➍ Exponential Gradient Cash Flow t =1 t =n Exponential Gradient

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GENERAL ENGINEERING AND APPLIED SCIENCES

CHAPTER 7 7

II.

SIMPLE INTEREST

➊ Ordinary Simple Interest 



I = Pin

Future Worth, F: 

F = P + I or F = P(1 + in)

Where:

I = Interest earned r i= → rate of interest per day 360 P = Present worth (capital) F = Future worth n = Total number of interest periods in days r = = interest in one year

For ordinary simple interest, the interest is computed based on one banker’s year. 1 ba b an ke ker' s year = 12 months = 360 days Each month = 30 days Computation for “ n’ and “ i” for ordinary simple interest: An interest rate of 10% for a period of 9 months: 0.10 i= → interest per day 360 n = 9(30) = 270 days An interest of 15% for 3 years: i=

0.15 360

→ interest per day

n = 3 (12 12) ( 30 )

= 1080 1080 days days



CHAPTER 7 7

II.

SIMPLE INTEREST

➊ Ordinary Simple Interest 



I = Pin


F = P + I or F = P(1 + in)

Where:

I = Interest earned r i= → rate of interest per day 360 P = Present worth (capital) F = Future worth n = Total number of interest periods in days r = = interest in one year

For ordinary simple interest, the interest is computed based on one banker’s year. 1 ba b an ke ker' s year = 12 months = 360 days Each month = 30 days Computation for “ n’ and “ i” for ordinary simple interest: An interest rate of 10% for a period of 9 months: 0.10 i= → interest per day 360 n = 9(30) = 270 days An interest of 15% for 3 years: i=

0.15 360

→ interest per day

n = 3 (12 12) ( 30 )

= 1080 1080 days days


GEAS


➋ Exact Simple Interest 



I = Pin


F = P + I = P(1 + in)

Where:

i= i=

r 365 r 366

o rdinary ye y ear → for or

→ for a leap year

A year is a leap year if it is divisible by 4 and divisible by 400 for a centennial year. (Centennial years are: 1800, 1900, 2000, etc) Example: Determine the exact simple interest on P5,000 for the period from January 1 to March 28, 2006 at 9% interest. Solution: Given: P = 5, 000

i = 9% Solving for n: January February March

− 31 days ( leap year ) − 28 days (l − 28 days _________

n = 87 days Thus, solving for the interest: I = Pin

⎛ 0.09 ⎞ ⎟ 87 ⎝ 365 ⎠

I = 5, 000 ⎜

I = P107.26



CHAPTER 7 7

III.

COMPOUND INTEREST

➊ Future Worth , F:





n

F = P (1 + i )

⎛ ⎝

r ⎞ ⎟ m⎠

⎛ ⎝

r ⎞

or F = P ⎜ 1 +

mt

Present Worth, P:



P = F (1 + i )

−n

or P = F ⎜ 1 +

−mt

⎟

m⎠

Where: (for both cases)

F = Future worth P = Present worth i = Effective interest rate per interest period (per month, per quarter, per year, etc) n = Total number of compounding periods m = mode of compounding = specified nominal rate r = t = number of years

In compound interest formula, the quantity: n the Single (1 + i) → is known as the

Payment Compound Amount Amount

Factor (SPCAF). −n

(1 + i) → is known as the Single Payment Present Worth Factor (SPPWF) ➌ Continuous Compounding: 

F = Per t → future wo w orth



P = Fe −r t → present worth


GEAS


Values Values of “n ” and “ i” fo r different different modes modes of compounding: Annu An nu all y – (every 12 months) m =1 ;

i =r

;

n=t

Semi – annually - (every 6 months) r m =2 ; i= ; n = 2t 2t 2 Quarterly - (every 3 months) r m=4 ; i= ; n = 4t 4 Bimonthly - (every 2 months) r m =6 ; i= ; n = 6t 6t 6

Semi-quarterly - (every 1.5 months) r m =8 ; i = ; n = 8t 8

Monthly - (every month) m = 12 ; i =

r 12

; n = 12t

Semi - monthly monthly - (every 0.5 month) r m = 24 ; i = ; n = 24 t 24



CHAPTER 7

IV.

RATES OF INTEREST

➊ Nominal Rate of Interest (NRI): Nominal rate of interest specifies the rate of interest and the number of interest periods per year. 

r = im

Where:

r = nominal rate of interest i = interest rate per period m = number of periods Thus, a nominal rate of interest of 6% compounded monthly simply means that there are 12 interest periods each year. The rate per interest period being: 

i=

r m

=

6% 12

= 0.5%

For compound interest, the rate of interest usually quoted is the NRI. In order to accurately reflect time - value considerations, NRI must be converted into ERI before applying the formulas for compound interest. ➋



ERI = (1 + i)m − 1

Where:

ER = Effective Rate of Interest i = interest per interest period r

=

m m = number of periods 

Effective Interest Rate for Continuous Compounding : 

ERI = er − 1

Where:

r = Nominal rate of interest


GEAS 


To find the nominal rate when given the effective continuous rate: 

r = ln (1 + i )

Where:

i = the effective continuous rate 

Equivalent Nomin al Rates For two nominal rates to be equal, their effective rates must be equal. (Effective interest rate - Continuous compounding) Calculate the effective interest rate per month for an interest rate of 15% in a continuously compounded account. Solution:

The nominal monthly rate, r is: 15 r= = 1.25% 12 i = e0.0125 − 1 = 0.012578 i = 1.2578% (Equivalent Rates) What nominal rate, which if converted quarterly will have the same effect as 12% compounded semi - annually? Solution:

Let:

r = the unknown nominal rate For two nominal rates to have the same effect, their corresponding effective rates must be equal. ERIquarterly = ERIsemi− annualy 4

2

⎛ r⎞ ⎛ 0.12 ⎞ ⎜ 1 + 4 ⎟ − 1 = ⎜1 + 2 ⎟ − 1 ⎝ ⎠ ⎝ ⎠ 4

⎛ r ⎞ ⎜ 1 + 4 ⎟ = 1.1236 ⎝ ⎠ r = 0.11825 or 11.825%



CHAPTER 7

V.

ESTIMATING DOUBLING AND TRIPLING TIME OF AN INVESTMENT

➊ Doubling Time: The time required for an initial single amount to double in value with compound interest is: 

n=

log2 log(1 + i)

(Rule of 72) The time required for an initial single amount to double in value with compound interest is approximately equal to: 

Estimated n =

72 i

Where:

i = effective interest in percent For example, at a rate of 2% per year, it would take 72 2 = 36 years for a current amount to double in size. ➋ Tripling Time: The time required for an initial single amount to triple in value with compound interest is: 

n=

log3 log(1 + i)

➌. General Formula: The time required for an initial single amount to become k times in value is: 

n=

logk log(1 + i)

Where:

k = 2, 3, 4,...k n


GEAS VI.


DISCOUNT 

Relationship Between Rate of Interest and Rate of Discount



i=

d 1− d

Where:

i = rate of interest d = rate of discount 

Successive Discount Two successive discounts of p% and q% allowed on an item are equivalent to a single discount of:

⎛ ⎝

d = ⎜p + q −

pq ⎞ ⎟% 100 ⎠

Two discounts of 15% and 5% are equivalent to what single discount? Solution:

⎛ ⎜ ⎝

d = ⎜ 15 + 5 −

(15 )( 5 ) ⎞ 100

⎟⎟ % ⎠

d = 19.25%



CHAPTER 7

VII. ANNUITIES Annu ity is a series of equal payments “ A” made at equal intervals of time. 

Ordinary Annuity - the type of annuity where the payments are made at the end of each period

Future Worth of Ordinary Annuity:



Cash Flow Diagram (Ordinary Annuity)

⎡ (1 + i )n − 1⎤ ⎥ F= A⎢ i ⎢⎣ ⎥⎦

0

1

2

3

4

A

A

A

A

(n − 1) n

Present Worth of Ordinary Annuity:

⎡ (1 + i ) − 1⎤ ⎥ n ⎢⎣ i (1 + i) ⎥⎦

A

A

n





F

P= A⎢

P

Deferred annuity - is the type of annuity where the first payment is made later than the first or is made several periods after the beginning of the annuity.

Future Worth of Deferred Annuity:



Cash Flow Diagram (Deferred Annuity)

⎡ (1 + i )n − 1⎤ ⎥ F= A⎢ i ⎢⎣ ⎥⎦

0

1

2

0 1 m

2

3…… n

A

A

A

Present Worth of Deferred Annui ty: 

⎡ (1 + i )n − 1⎤ ⎥ P=A⎢ m +n ⎢⎣ i (1 + i) ⎥⎦

P

Where:

m + n periods

A

A

n periods

A = periodic equal payments n = number of periods (equal to the number of payments) i = interest rate per payment m = number of periods before the beginning of the first payment


F

GEAS 


Annuity Due - is the type of annuity where the payment is made at the beginning of each period

Future Worth of Annuity Due:



⎡ (1 + i )n − 1⎤ ⎥ F= A⎢ i ⎢⎣ ⎥⎦

Present Worth of Annui ty Due:



Cash Flow Diagram (Annuity Due) 1

2

3

4

5

A

A

A

A

A

A

A F

P

⎡ (1 + i )n − 1⎤ ⎥ P=A⎢ n −1 ⎢⎣ i (1 + i ) ⎥⎦

(n − 1) n

n − 1 periods

Where: (in both cases)

A = periodic equal payments n = number of periods (equal to the number of payments) i = interest rate per payment 

Perpetuity - is an annuity in which the periodic payments continue indefinitely.

Present Worth of Perpetuity: ( For payments made at the end of each period) Cash Flow Diagram (Perpetuity) 

P=

A i

0

1

2

A

A

3…. n

→∞

Where:

A = periodic payment i = interest per payment

A

A

P



CHAPTER 7

VIII. CAPITALIZED COST Capitalized Cost refers to the present worth of a property that is assumed to last forever. The capitalized cost of any property is the “ sum of the first cost and the present costs of perpetual replacement, operation and maintenance” .

CASE 1: No replacement, only maintenance. 

CC = FC +

A i

Where:

CC = Capitalized Cost FC = first cost or original cost A = Annual maintenance cost CASE 2: No maintenance, only replacement.



CC = FC +

P n (1 + i) − 1

Where:

CC = Capitalized Cost FC = first cost or original cost P = the amount needed to replace the property every n periods CASE 3: Replacement and maintenance every period. 

CC = FC +

A i

+

P n (1 + i) − 1

Where:

CC = Capitalized Cost FC = first cost or original cost A = Annual maintenance cost P = the amount needed to replace the property every n periods


GEAS IX.


DEPRECIATION Depreciation is the decrease in the value of physical property due to passage of time.

Symbols used and their meaning:

= dn =

d

annual depreciation charge depreciation charge during the nth year

Dn = FC SV L n



total depreciation after “n” years

= = = =

first cost /original cost estimated salvage value after “L” years expected depreciable life of the property number of years before L

Straight Line Method Straight line method of depreciation assumes that the loss in value of the property is directly proportional to the age of the property. 

Annual depreciation: 



FC − SV L

Depreciation after “n” years:







d=

⎛ FC − SV ⎞ ⎟n L ⎝ ⎠ Dn = d × n Dn = ⎜

Book value after “n” years:



BVn = FC − Dn



CHAPTER 7 

Sinking Fund Method 

Annual depreciation: d=





(FC − SV ) i L (1 + i) − 1

Depreciation after “n” years: d ⎡(1 + i) − 1⎤





n

⎥⎦

i

Book value after “n” years:





⎢ Dn = ⎣

BVn = FC − Dn

Declining Balance Method Also called as the constant percentage method or the Mateson Formula: 

Depreciation during the nth year: 



Salvage Value at the end of its useful life: 



n −1

dn = k (FC )(1 − k )

L

SV = FC (1 − k )

Book Value at the end of n years: n



BV = FC (1− k )



⎛ SV ⎞ L BV = FC ⎜ ⎟ ⎝ FC ⎠

n


GEAS 


Rate of depreciation:

⎛ BVn ⎞ ⎛ SV ⎞ = 1− L ⎜ ⎟ ⎟ ⎝ FC ⎠ ⎝ FC ⎠

k = 1− n ⎜ 

Double Declinin g Balance Method (DDBM) Also called as the constant percentage method or the Mateson Formula: 

Depreciation during the nth year: n −1





dn =

2 ( FC )(1 − k ) L

Salvage Value at the end of its useful life: L





⎛ 2⎞ SV = FC ⎜ 1 − ⎟ ⎝ L⎠

Book Value at the end of n years:



⎛ ⎝

BV = FC ⎜ 1 −

2⎞

n

⎟

L⎠

Note that the formulas for DDB method are obtained form the formulas for Declining Balance Method by simply replacing k with 2/L. 

Sum - of - the - Year’s - Digits (SYD) Method Depreciation charge during the nth year:  



dn = (FC − SV )

L − n +1 SYD

Total depreciation after n years



Dn = (FC − SV )

n ( 2L − n + 1)

WHERE:

SYD = sum of the year's digit SYD =

n 2

(n + 1)

2 ( SYD )



CHAPTER 7 TEST - 7

1.

It is defined to be the capacity of a commodity to satisfy human want A. necessity B. utility C. luxuries D. discount

2.

It is the stock that has prior right to dividends. It usually does not bring voting right to the owners and the dividend is fixed and cannot be higher than the specified amount. A. B. C. D.

3.

It is a amount which a willing buyer will pay to a willing seller for the property where each has equal advantage and is under no compulsion to buy or sell. A. B. C. D.

4.

Common stock Voting stock Preferred stock Non par value stock

Book value Market value Use value Fair value

_________ is the loss of value of the equipment with use over a period of time. It could mean a difference in value between a new asset and the use asset currently in a service. A. Loss B. Depreciation C. Extracted D. Gain


GEAS 5.


An economic condition in which there are so few suppliers of a particular product that one supplier’s actions significantly affect prices and supply. A. Oligopoly * B. monopsony C. monopoly D. perfect competition

6.

A market whereby there is only one buyer of an item for when there are no goods substitute. A. Monopsony B. Monopoly C. Oligopoly D. Oligopsony

7.

It is the worth of a property as recorded in the book of an enterprise. A. B. C. D.

8.

Salvage value Price Book value Scrap value

Reduction in the level of national income and output usually accompanied by a fall in the general price level. A. Devaluation B. Deflation C. Inflation D. Depreciation

9.

A formal organization of producers within industry forming a perfect collusion purposely formed to increase profit and block new comers from the industry. A. Cartel B. Monopoly C. Corporation D. Competitors



CHAPTER 7

10. A market situation where there is only one seller with many buyer. A. B. C. D.

Monopoly* Monophony Oligopoly perfect competition

11. A market situation where there is one seller and one buyer. A. Bilateral monopoly* B. Monopoly C. Oligopoly D. Bilateral Monopoly 12. Reduction in the level of national income and accompanied by a fall in the general price level.

output usually

A. Deflation B. Inflation C. Devaluation D. Depreciation 13. A series of equal payments made at equal interval of time. A. Annuity* B. Amortization C. Depreciation D. Bonds 14. The money paid for the use of borrowed capital A. interest* B. amortization C. annuity D. bonds 15. The place where buyers and sellers come together. A. B. C. D.

Market * Store Bargain center Port


GEAS


16. The value of the stock as stated on the stock certificate A. B. C. D.

stock value par value* interest maturity value

17. An market situation in which two competing buyers exert controlling influence over many sellers. A. B. C. D.

bilateral monopoly oligopoly duopsony* duopoly

18. An market situation in which two powerful groups or organizations dominate commerce in one business market or commodity. A. B. C. D.

Oligopoly Duopoly* Bilateral oligopoly Bilateral Oligopsony

19. The type of annuity where the first payment is made after several periods, after the beginning of the payment. A. B. C. D.

Perpetuity Ordinary annuity Annuity due Deferred annuity*

20. The condition in which the total income equals the total operating expenses. A. B. C. D.

tally par value Check and balance Break even *



CHAPTER 7

21. The amount which has been spend or capital invested which for some reasons cannot be retrieved. A. B. C. D.

sunk cost* fixed costs depletion cost construction cost

22. An obligation with no condition attached is called A. B. C. D.

Personal Gratuitous * Concealed Private

23. The sum of all the costs necessary to prepare a construction project for operation. A. B. C. D.

operation cost construction cost* depletion cost production cost

24. The amount received from the sale of an additional unit of a product. A. B. C. D.

marginal cost marginal revenue* extra profit prime cost

25. The amount that the property would give if sold for junk. A. B. C. D.

junk value salvage value scrap value* book value


GEAS


26. The worth of the property which is equal to the original cost less the amount which has been charged to depreciation. A. B. C. D.

scrap value salvage value book value* market value

27. The sum of the direct labor cost incurred in the factory and the direct material costs of all materials that go into production is called A. B. C. D.

net cost maintenance cost prime cost* operating cost

28. The difference between the present value and the worth of money at some time in the future is called A. B. C. D.

market value net value discount* interest

29. The additional cost of producing one more unit is A. B. C. D.

prime cost marginal cost* differential cost sunk cost

30. A written contract by a debtor to pay final redemption value on an indicated date or maturity date and to pay a certain sum periodically. A. annuity B. bond* C. amortization D. collateral



CHAPTER 7

31. Estimated value of the property at the end of the useful life. A. B. C. D.

Market value Fair value Salvage value * Book value

32. Determination of the actual quantity of the materials on hand as of a given date. A. B. C. D.

physical inventory* counting principle stock assessment periodic material update

33. This consists of cash and account receivable during the next period or any other material which will be sold. A. B. C. D.

fixed assets deferred charges current asset* liability

34. A wrongful act that causes injury to a person or property and for which the law allows a claim by the injured party to recover damages. A. fraud B. tort* C. libel D. scam 35. A series of uniform payment over an infinite period of time A. depletion B. capitalized cost C. perpetuity * D. inflation


GEAS


36. These are products or services that are required to support human life and activities that will be purchased in somewhat the same quantity event though the price varies considerably. A. B. C. D.

Commodities Necessities * Demands Luxury

37. The quantity of a certain commodity that is offered for sale at a certain price at a given place and time. A. B. C. D.

utility supply* stocks goods

38. It is sometimes called the second hand value A. B. C. D.

Scrap value Salvage value* Book value Par value

39. Decreases in the value of a physical property due to the passage of time. A. Deflation B. Depletion C. Declination D. Depreciation* 40. An association of two or more individuals for the purpose of engaging business for profit. A. B. C. D.

Single proprietorship Party Corporation Partnership*



CHAPTER 7

41. The simplest form of business organization wherein the business is own entirely by one person. A. partnership B. proprietorship* C. corporation D. joint venture 42. Parties whose consent or signature in a contract is not considered intelligent. A. B. C. D.

dummy person minors demented persons * convict

43. It is defined as the capacity of a commodity to satisfy human want. A. B. C. D.

satisfaction luxury* necessity utility

44. This occurs in a situation where a commodity or service is supplied by a number of vendors and there is nothing to prevent additional vendors entering the market . A. B. C. D.

perfect competition * monophony monopoly cartel

45. These are products or services that are desired by human and will be purchased if money is available after the required necessities have been obtained. A. B. C. D.

Commodities Necessities Luxuries * Supplies


GEAS


46. Grand total of the assets and operational capability of a corporation. A. B. C. D.

authorized capital* paid off capital subscribed capital investment

47. It is where the original record of a business transaction is recorded. A. ledger B. spreadsheet C. journal* D. logbook 48. The length of time which the property may be operated at a profit. A. B. C. D.

life span economic life* operating life profitable life

49. The right and privilege granted to an individual or corporation to do business in a certain region. A. B. C. D.

permit royalty license franchise*

50. The worth of an asset as shown in the accounting records of an enterprise. A. B. C. D.

fair value par value market value book value*



CHAPTER 7

Solved Problems

In 1.

Economics

A man expects to receive P20,000 in 10 years. How much is that money worth now considering interest at 6 %compounded quarterly?

Solution: Given: F = 20,000 t = 10 yrs r = 6% ;compounded quarterly(m = 4 ) Formula: n

F = (1 + i)

r , n = mt m Substitute the given values: Where:

i=

⎛ ⎝

20,000 = P ⎜ 1 +

.06 ⎞ ⎟ 4 ⎠

4(10)

P = 11, 025.25

2. An employee obtained a loan of P10,000 at the rate of 6% compounded annually in order to repair a house. How much must he pay monthly to amortize the loan within a period of 10 ye ars? Formula:

Solution: Given: P = 10,000

- (Annuity)

⎡ (1 + i )n − 1⎤ ⎥ P=A⎢ ⎢ i (1 + i)n ⎥ ⎣ ⎦

r = 6% m = 1 ( annually t = 10 yrs


GEAS


Solving for the interest rate per month: ERImonthly = ERIannually 12 1 (1 + i) − 1 = (1.06 ) − 1

i = 0.0048675 Substitute to the formula:

⎡

⎤ −1 (1.0048675 ) ⎥ ⎢ ( 0.0048675 ) (1.0048675)120 ⎥ ⎣ ⎦ 12(10)

10,000 = A ⎢

10, 000 = 90.72A A = 110.22

3.

What is the effective rate corresponding to 16% compounded daily? Take 1 year = 360 days. Formula - (Effective Rate) m

ERI = (1 + i) − 1

Solution: Given: r = 16% ; m = 360 (daily) Substitute:

⎛ 0.16 ⎞ ERI = ⎜1 + ⎟ 360 ⎠ ⎝ ERI = 0.1735 = 17.35% 4.

m

r ⎞ ⎛ = ⎜1 + ⎟ − 1 ⎝ m⎠ Where: r = no min al rate

360

−1

m = mod e of compounding

What is the accumulated amount after 3 years of P6,500.00 invested at the rate of 12% per year compounded semi-annually? Formula - Future

Solution: Given: P = 6,500 ; r = 12%

worth

n

F = P (1 + i )

mt

r ⎞ ⎛ = P ⎜1 + ⎟ ⎝ m⎠

m = 2 (semi − annually) t = 3 years Substitute:

Where: r = no min al rate

6

⎛ 0.12 ⎞ F = 6500 ⎜1 + ⎟ = 9,220.37 2 ⎠ ⎝

m = mod e of compounding t = no. of years



CHAPTER 7

5.

Fifteen percent (15%) when compounded semi-annually will have an effective rate of Formula -

Solution:

(Effective Rate) m

ERI = (1 + i) − 1

Given: r = 15%

m

r ⎞ ⎛ = ⎜1 + ⎟ − 1 ⎝ m⎠

m=2


Substitute:

⎛ ⎝

ERI = ⎜ 1 +

0.15 ⎞

2


⎟ −1 2 ⎠

ERI = 0.1556

= 15.56% 6.

What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly?

Solution: Given: r = 8%

For two or more rates to be equivalent, their corresponding effective rates must be equal.

m=4 Let:

x = unknown rate ERIannually = ERIquarterly 4

⎛ 0.08 ⎞ (1 + x ) − 1 = ⎜ 1 + ⎟ −1 4 ⎠ ⎝ x = 0.0824 x = 8.24% 1


GEAS

7.


How long will it take the money to triple itself if invested at 10% compounded semi-annually?

Solution: Formula - Future

Let:

F = P (1 + i )

P = present worth F = 3P

mt

r ⎞ ⎛ = P ⎜1+ ⎟ ⎝ m⎠

Substitute to the formula:

⎛ 0.10 ⎞ 3P = P ⎜1 + ⎟ 2 ⎠ ⎝ 3 = (1.05 )

worth (CI)

n

Where:

2t

r = no min al rate m = mod e of compounding

2t

t = no. of years

ln 3 = 2t(ln 1.05) t = 11.3 years

8. A man wishes his son to receive P500,000.00 ten years from now. What amount should he invest now if it will earn interest of 12% compounded annually during the first 5 years and 15% compounded quarterly during the next 5 years?

Solution: Given: F = 500, 000 r = 12% m = 1 ( annually )

0

5

10

P

From:

F5

n

F = (1 + i)

F10

For the first 5 years: r = 12% ; m = 1 (annually) 5

F5 = P (1.12 ) F5 = 1.76P



CHAPTER 7

For the next 5 years,: F10 = 500,000 r = 15% ;

⎛ ⎝

m = 4 (quarterly)

F10 = F5 ⎜ 1 +

0.15 ⎞

20

⎟

4 ⎠ 5

20

500,000 = (1.12 ) P (1.0375 ) P = 135, 868.19

9.

What interest rate compounded monthly is equivalent to 10% effective rate?

Solution: Formula - Effective

Given: r = 8%

⎛ ⎝

ER = ⎜ 1 +

m=4 Let:

rate

m

r ⎞ ⎟ −1 m⎠

Where:

x = unknown rate 12

x ⎞ ⎛ 0.10 = ⎜ 1 + ⎟ ⎝ 12 ⎠

−1

r = no min al rate m = mod e of compounding t = no. of years

12

x ⎞ ⎛ 1.10 = ⎜ 1 + ⎟ ⎝ 12 ⎠ x = 0.0957 x = 9.57%


GEAS


10. A machine costs P8,000.00 and an estimated life of 10 years with a salvage value of P500.00. What is its book value after 8 years using straight-line method?

Solution: Given: Co = 8,000 ;

n = 8 years

L = 10 years (life) SV = 500 Solving for the total depreciation: Dn = d × n → (d = annual depreciation)

⎛ C − SV ⎞ = ⎜ o ⎟n L ⎝ ⎠ ⎛ 8000 − 500 ⎞ =⎜ ⎟ 8 = 6000 10 ⎝ ⎠ Thus, the Book Value (BV) is: BV = Co − Dn

= 8000 − 6000 = 2000 11. By the condition of a will, the sum of P20,000 is left to a girl to be held in trust fund by her guardian until it amounts to P50,000. When will the girl receive the money if the fund is invested at 8% compounded quarterly?


F = 50,000

r = 8%; m = 4 (quarterly) Substitute given values:

⎛ ⎝

50, 000 = 20, 000 ⎜ 1 + 2.5 = (1.02 )

.08 ⎞

Formula -

4t

⎟

(Compound Interest) mt

r ⎞ ⎛ F = P (1 + i ) = P ⎜1 + ⎟ ⎝ m⎠

4 ⎠

n

4t

Take ln both sides: ln 2.5 = 4t (ln1.02 ) t = 11.57 years



CHAPTER 7

12. The amount of P12,800 in 4 years at 5% compounded quarterly is


Formula -

t = 4 years


r ⎞ ⎛ F = P (1 + i ) = P ⎜ 1 + ⎟ ⎝ m⎠ n

r = 5% m = 4 ( quarterly )

Substitute the given values to the formula:

⎛ ⎝

F = 12,800 ⎜ 1 +

0.05 ⎞

4(4)

4 ⎟⎠

F = 15, 614.59

13. How much money must you invest today in order to withdraw P2000 annually for 10 years if the interest rate is 9%?

Solution: Given: A = 2000

Formula:

t = 10 years

⎡ (1 + i )n − 1⎤ ⎥ P = A⎢ ⎢ i (1 + i )n ⎥ ⎣ ⎦

r = 9% m = 1 ( annually ) P=? Substitute the given values:

⎡ (1.09 )10 − 1 ⎤ ⎥ 10 ⎢⎣ 0.09(1.09) ⎥⎦

- (Annuity)

Where: A = annual payment / widrawal P = present worth n = number of payment / widrawal

P = 2000 ⎢

P = 12, 835.32


GEAS


14. Money Borrowed today is to be paid in 6 equal payments at the end of 6 quarters. If the interest is 12% compounded quarterly, how much was initially borrowed if quarterly payments is P2,000.00?

Solution: Given: A = 2000

Formula:

⎡ (1 + i )n − 1⎤ ⎥ P = A⎢ ⎢ i (1 + i )n ⎥ ⎣ ⎦

r = 12% m = 4 (quarterly ) i=

r m

=

0.12 4

= 0.03

(Present worth -Annuity)

Where: A = periodic payment P = present worth n = number of payments

Substitute:

⎡ (1.03 )6 − 1 ⎤ ⎥ P = 2000 ⎢ 6 ⎢⎣ ( 0.03 )(1.03 ) ⎥⎦ P = 10, 834.38

i=

r m

→ int erest per period

15. The effective rate of 14% compounded semi-annually is


Given: r = 14%

⎛ ⎝

ER = ⎜ 1 +

m = 2 (semi − annually) ER = ?

rate

m

r ⎞

⎟ −1

m⎠


From the formula: 2

⎛ 0.14 ⎞ ER = ⎜ 1 + ⎟ −1 2 ⎠ ⎝ = 0.1449 = 14.49%




CHAPTER 7

16. A man expects to receive P25,000 in 8 years. How much is that money worth now considering interest at 8% compounded quarterly?

Solution: Given: F = 25,000

Formula -

r = 8%

n

⎛ ⎝

mt

F = P (1 + i ) = P ⎜ 1 +

m = 4 ( quarterly ) t = 8 years

r ⎞

⎟

m⎠

Where; F = future worth

P=?

P = present worth

From the formula:

⎛ 0.08 ⎞ 25,000 = P ⎜ 1 + ⎟ 4 ⎠ ⎝ P = 13, 262.83

(Compound Interest)

r = no min al rate of int e rest

4(8)


17. What is the accumulated amount of the five-year annuity paying P6,000 at the end of each year with interest at 15% compounded annually?

Solution: Formula:

⎡ (1 + i )n − 1⎤ ⎥ i ⎢ ⎥ ⎣ ⎦

Given:

F= A⎢

A = 6000 r = 15% m = 1 ( annually )

(Future worth -Annuity)

Where: A = periodic payment P = present worth

t = 5 years

n = number of payments

F=? Substitute:

i=

⎡ (1.15 )5 − 1⎤ ⎥ F = 6000 ⎢ ⎢⎣ 0.15 ⎥⎦ F = 40, 454.29

r m



GEAS


18. ABC Corporation makes its policy that for every new equipment purchased, the annual depreciation cost should not exceed 20% of the first cost at any time without salvage value. Determine the length of service life necessary if the depreciation used is the SYD method.

Solution : Using SYD method:

⎛ reversed digit ⎞ ⎟ SYD ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ n 0.2Co = ( Co − 0 ) ⎜ ⎟ ⎜⎜ n ( n + 1) ⎟⎟ ⎝2 ⎠ dn = ( Co − CL ) ⎜

0.20 =

2

n +1 n = 9 years

19. At an interest rate of 10% compounded annually, how much will a deposit of P1500 in 15 years?

Solution: Given: r = 10% m = 1 ( annually ) P = 1500 t = 15 years F=? From: mt

r ⎞ ⎛ F = P ⎜1 + ⎟ ⎝ m⎠ Substitute:

15

F = 1500 (1.10 ) F = 6, 265.87



CHAPTER 7

20. A debt of P10,000 with 10% interest compounded semi-annually is to be amortized by semi-annual payments over the next 5 years. The first due is 6 months. Determine the semi-annual payments.

Solution:

Formula:

Given: P = 10,000

⎡ (1 + i )n − 1⎤ ⎥ ⎢ i (1 + i )n ⎥ ⎣ ⎦

P=A⎢

r = 10% m = 2 (semi − annually) i=

r


0.10

= = 0.05 m 2 n = mt = 2(5) = 10

Where: A = periodic payment P = present worth n = number of payments

A = ?

i=

Substitute:

r m


⎡ (1.05 )10 − 1 ⎤ ⎥ 10,000 = A ⎢ ⎢⎣ 0.05(1.05)10 ⎥⎦ A = 1,295.05 21. If you borrowed money from your friend with simple interest of 12%, find the present worth of P50,000.00 which is due at the end of 7 months.

Solution: Given: r = 12% i=

Formula:

(Ordinary Simple

Interest)

0.12 F = P (1 + in )

360 F = 50,000

1month = 30 days Where: P = present worth

n = 7(30) = 210 days Substitute to the formula:

⎡

⎤ ⎛ 0.12 ⎞ 210⎥ ⎟ ⎝ 360 ⎠ ⎦

i = rate of int erest / day

50,000 = P ⎢1+ ⎜

⎣

n = no. of int e rest periods

P = 46, 728.97


GEAS


22. The amount of P50,000.00 was deposited in the bank earning an interest of 7.5% per annum. Determine the total amount at the end of 5years if the principal and interest were not withdrawn during the period.

Solution:

Formula -

Given:


r ⎞ ⎛ F = P (1 + i ) = P ⎜ 1 + ⎟ ⎝ m⎠ n

P = 50,000


r = 7.5% m = 1 (per annum)

P = present worth

t = 5 years


F=? Substitute:


5

F = 50,000 (1 + 0.075 ) F = 71, 781.47

23. What is the corresponding effective rate of 18% compounded semi quarterly?


Given: r = 18% m = 8 (semi − quarterly) Substitute:

⎛ ⎝

ER = ⎜ 1 +

rate

m

0.18 ⎞

8

r ⎞ ⎛ ER = ⎜ 1 + ⎟ − 1 m

⎝

⎠


⎟ −1


8 ⎠ ER = 0.1948

t = no. of years

ER = 19.48%



CHAPTER 7

24. A telephone company purchased a microwave radio equipment for P6 million. Freight and installation charges amounted to 4% of the purchased price. If the equipment will be depreciated over a period of 10 years with a salvage value of 8%, determine the depreciated cost during th the 5 year using SYD.

Solution: Co = 6M + 0.04(6M) Co = 6.24M SV = 0.08(Co ) SV = 0.08(6.24M) SV = 499, 200 n (n + 1) 2 10 SYD = (10 + 1) = 55 2 SYD =

Formula :Depreciation

dm = (FC − SV )

(SYD Method) n−m+1

SYD Where: F = First Cost SV = Salvage Value n = life of the property in years m = number of years used before n

th

Solving for the depreciation during the 5 year: (10 − 5 + 1) d5 = ( 6,240,000 − 499,200 ) 55 d5 = 626, 269.10

25. In how many years is required for P2,000 to increase by P3,000 if interest at 12% is compounded semi-annually? Solution: Formula - (Compound Interest) Given: mt P = 2000 ; F = 5000 r ⎞ n ⎛ F = P (1 + i) = P ⎜ 1 + ⎟ r = 12% ; m = 2 (semi − annually) ⎝ m⎠ Where; Substitute: 2t F = future worth ⎛ 0.12 ⎞ 5000 = 2000 ⎜ 1 + ⎟ P = present worth 2 ⎠ ⎝ r = no min al rate of int e rest 2t 2.5 = (1.06 ) m = mod e of compounding ln2.5 = 2t(ln1.06) t = no. of years t = 7.86 (say 8 years) 261 Loading Next Page

GEAS


26. A VOM has a current selling price of P400. If its selling price is expected to decline at a rate of 10% per annum due to obsolence, what will be its selling price after 5 years? Formula -

Solution:

Mateson’s Formula m

Cm = Co (1 − k )

Given: Co = 400 ; k = 10%

Where: Cm = Book Value at the

m = 5 years

end of m years k = rate of depreciation Co = first cost

Substitute: 5

Cm = 400 (1 − 0.1) Cm = 236.20

27. Find the nominal rate which if converted quarterly could be used instead of 12% compounded semi-annually.

Solution: Given: r = 12% ; m = 2 (semi − annually) Let: x = the unknown nominal rate mx = 4 (mode of compounding of x)


Equate Effective rates of interest: ERquarterly = ERsemi− annually 4

2

x⎞ ⎛ ⎛ 0.12 ⎞ ⎜ 1 + 4 ⎟ − 1 = ⎜1 + 2 ⎟ − 1 ⎝ ⎠ ⎝ ⎠ x = 0.1183 x = 11.83%

28. Find the present worth of a future payment of a P100,000 to be made in 10 years with an interest of 12% compounded quarterly.

Solution : See Formula in Problem 25:

⎛ ⎝

100,000 = P = ⎜ 1+

0.12 ⎞

4(10)

⎟

4 ⎠ P = 30, 655.68



CHAPTER 7

29. What nominal rate, compounded semi-annually, yields the same amount as 16% compounded quarterly?

Solution:


Given: r = 16% m = 4 ( quarterly ) Let:

x = the unknown nominal rate mx = (mode of compounding of x)

Equate Effective rates of interest: ERsemi− annually = ERquarterly 2

4

⎛ x⎞ ⎛ 0.16 ⎞ ⎜ 1 + 2 ⎟ − 1 = ⎜1 + 4 ⎟ − 1 ⎝ ⎠ ⎝ ⎠ x = 0.1632 x = 16.32%

30. What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly?

Solution: Equate Effective rates of interest: ERannually = ERquarterly 4

⎛ 0.08 ⎞ (1 + x ) − 1 = ⎜ 1+ ⎟ −1 4 ⎠ ⎝ x = 0.0824 x = 8.24%


GEAS


31. You loan from a loan firm an amount of P100,000 with the rate of simple interest of 20%, but interest was deducted from the loan at the time the money was borrowed. If at the end of one year you have to pay the full amount of P100,000, what is the actual rate of interest?

Solution: Given: n = 1 year P = 100,000 − 20% advance interest P = 100,000 − 0.20(100,000) P = 80,000 F = 100, 000 (to be paid at the end of 1 year ) From: F = P (1 + in ) Substitute: 100, 000 = 80, 000 [1 + i(1)] i = 0.25 i = 25%

Think of the 20% advance interest as a rate of discount: Relationship between rate of interest i and rate of discount d: d i= 1− d Then, 0.20 i= 1 − 0.20 i = 0.25 i = 25%



CHAPTER 7

32. A loan of P5,000 is made for a period of 15 months at a simple interest rate of 15%. What future amount is due at the end of a loan period?

Solution: Formula:

Given: P = 5000

(Ordinary Simple

Interest)

n = 15(30) = 450 days

F = P (1 + in )

0.15 i= 360 Substitute:

⎡


⎤ ⎛ 0.15 ⎞ 450⎥ ⎟ ⎝ 360 ⎠ ⎦

i = rate of int e rest / day

F = 5000 ⎢1 + ⎜

⎣


F = 5, 937.5

33. Mr. J. Reyes borrowed money from a bank. He received from the bank P1,842 and promise to repay P2,000 at the end of 10 months. Determine the simple interest.

Solution: Given: P = 1842 F = 2000 n = 10 ( 30 ) = 300 days i=? See formula in problem 32:

⎡

⎤ ⎛ i ⎞ 300⎥ ⎟ ⎝ 360 ⎠ ⎦

2000 = 1842 ⎢1 + ⎜

⎣

i = 0.1029 i = 10.29%


GEAS


34. What is the effective rate corresponding to 18% compounded daily? Take 1 year is equal to 360 days. Solution: Formula - Effective

Given: r = 18%

⎛ ⎝

ER = ⎜ 1 +

m = 360 (daily) Substitute:

⎛ 0.18 ⎞ ER = ⎜ 1 + ⎟ 360 ⎠ ⎝ ER = 0.1972 ER = 19.72%

rate

m

r ⎞

⎟ −1

m⎠


360


−1

t = no. of years

35. Find the annual payment to extinguish a debt of P 10,000 payable for 6 years at 12% interest annually.

Solution: Formula:

Given: P = 10,000

⎡ (1 + i )n − 1⎤ ⎥ ⎢ i (1 + i )n ⎥ ⎣ ⎦

t = 6 years

P= A⎢

r = 12% m = 1 ( annually )

Where: A = periodic payment

n = mt = (1)(6) = 6 i=


P = present worth

r 0.12 = = 0.12 m 1

n = number of payments i=

Substitute:

⎡ (1 + 0.12)6 − 1 ⎤ ⎥ 10,000 = A ⎢ ⎢⎣ 0.12 (1 + 0.12 )6 ⎥⎦ A = 2,432.257

r m




CHAPTER 7

36. What annuity is required over 12 years to equate with a future amount of P 20,000? Assume i = 6% annually. Solution:

Formula:

Given: F = 20,000

⎡ (1 + i )n − 1⎤ ⎥ F= A⎢ i ⎢ ⎥ ⎣ ⎦

t = 12 years r = 6%


m = 1 ( annually ) i=

r

=

0.06

m 1 n = mt = 12


F = Future worth

= 0.06

n = number of payments i=

A = ? Substitute:

r


⎡ (1 + 0.06 )12 − 1⎤ ⎥ 20,000 = A ⎢ 0.06 ⎢⎣ ⎥⎦ A = 1,185.54

37. Find the present worth of a future payment of 80,000 to be made in six years with an interest of 12% compounded annually.

Solution:

Formula -

Given: F = 80,000

(Compound Interest) n

⎛ ⎝

mt

F = P (1 + i ) = P ⎜ 1 +

t = 6 years

r ⎞ ⎟ m⎠

r = 12%


m = 1 ( annually )

P = present worth

P=?

r = no min al rate of int e rest m = mod e of compounding

Substitute: 80,000 = P (1 + 0.12 )

t = no. of years

6

P = 40, 530.48


GEAS


38. The amount of P 20,000 was deposited in a bank earning an interest of 6.5% per annum. Determine the total amount at the end of 7 years if the principal and interest were not withdrawn during this period. Solution:

Formula -


mt

⎛ ⎝

F = P (1 + i ) = P ⎜1 +

Given: P = 20,000

r ⎞ ⎟ m⎠


r = 6.5% m = 1 (per annum)

P = present worth

t = 7 years

r = no min al rate of int erest

F=? Substitute:


7

F = 20,000 (1 + 0.065 ) F = 31, 079.7

39. Today a businessman borrowed money to be paid in 10 equal payments for 10 quarters. If the interest rate is 10% compounded quarterly and the quarterly payment is 2,000 pesos, how much did he borrowed? Formula:

Solution:

Given: r = 10%

⎡ (1 + i )n − 1⎤ ⎥ P=A⎢ ⎢ i (1 + i )n ⎥ ⎣ ⎦

m = 4 (quarterly ) i=

r

=


0.10

= 0.025 m 4 n = 10 equal payments


A = 2000


P=?

i=

Substitute:

r m


⎡ (1 + 0.025 )10 − 1 ⎤ ⎥ P = 2000 ⎢ ⎢⎣ 0.025 (1 + 0.025 )10 ⎥⎦ P = 17, 504.13



CHAPTER 7

40. What is the present worth of a P 500 annuity starting at the end of the third year and continuing to the end of fourth year, if the annual interest rate is 10%.

Solution: Formula:

Given:

(Present worth of deferred Annuity)

A = 500

⎡ (1 + i)n − 1⎤ ⎥ P = A⎢ m +n ⎢⎣ i (1 + i ) ⎥⎦

m=2 n=2 r = 10% (annually )


Substitute:


⎡ (1 + 0.10 )2 − 1 ⎤ ⎥ P = 500 ⎢ 2+ 2 ⎢⎣ 0.10 (1 + 0.10 ) ⎥⎦ P = 717.16

m=number of periods before the beginning of the first payment

Alternate Solution:

P1 = P1 = P2 =

0

F

1

2

3

4

n

(1 + i)

500

500 3

= 375.66

4

= 341.51

(1.10 ) 500

(1.10 )

500

P1 P2

P = 375.66 + 341.51

= 717.17


GEAS


41. A copying machine has a useful life of 3 years and a salvage value of P 20,000 was bought for P 135,000. If the owner decides to sell after using it for 2 years, how much should the selling price be so that he will not lose or gain if the interest is 5%. (Hint: apply the Sinking Fund method).

Solution: Selling Price = Book Value after n years From Sinking Fund Formula: d=

( Co − SV ) → annual depreciation cost L (1 + i) − 1

Where: d = annual depreciation cost Co = first cost or original cost SV = salvage value at the end of useful life L = life Solving for d: (135, 000 − 20, 000)( 0.05 ) d= 3 (1.05 ) − 1 d = 36,479 Total depreciation at the end of 2 years:

⎡ (1 + i )n − 1⎤ ⎥ i ⎢⎣ ⎥⎦ ⎡ (1.05 )2 − 1⎤ ⎥ D2 = 36,479 ⎢ ⎢⎣ 0.05 ⎥⎦ D2 = 74,782 D2 = d ⎢

Solving for the Book Value after 2 years: BV2 = Co − D2

= 135,000 − 74,782 = 60, 218 → selling price



CHAPTER 7

42. A loan for P 50,000 is to be paid in 3 years at the amount of P65, 000. What is the effective rate of money?

Solution: Given: P = 50,000 F = 65,000 n = 3 years From: I = Pin 65,000 − 50,000 = 50,000i(3) i = 0.10 i = 10%

43. Today an investor withdraws P50,000.00 representing the accrued amount of his investment that matured. If he invested at 10% compounded semi-annually for 10 years, how much did he invest in pesos? Solution: Formula -

Given: F = 50,000

n

⎛ ⎝

mt

F = P (1 + i ) = P ⎜ 1 +

r = 10%

r ⎞ m ⎟⎠


m = 2 (semi − annually) t = 10 years

P = present worth

Substitute:

⎛ 0.10 ⎞ 50,000 = P ⎜ 1 + 2 ⎟⎠ ⎝ P = 18, 844.47

(Compound Interest)


2(10)



GEAS


44. A man invested part of P 20,000 at 18% and the rest at 16%. The annual income from 16% investment was P 620 less than three times the annual income from 18% investment. How much did he invest at 18%? Solution:

Let:

x = amount invested at 18% interest 20,000 − x = the amount invested at 16% interest Then, 0.16 ( 20,000 − x ) = 3 ( 0.18x ) − 620 3200 − 0.16x = 0.54x − 620 x = 5, 457.14

45. What is the accumulated amount after three (3) years of P 6,500 invested at the rate of 12% per year compounded semi-annually? Solution: Formula -

Given:


r ⎞ ⎛ F = P (1 + i ) = P ⎜ 1 + ⎟ ⎝ m⎠ n

P = 6,500 t = 3 years r = 12%


m = 2 (semi − annually)

P = present worth

F=?


Substitute:

⎛ 0.12 ⎞ F = 6,500 ⎜ 1 + ⎟ 2 ⎠ ⎝ F = 9220.37

t = no. of years

2(3)



CHAPTER 7

46. If the authorize capital stock of corporation is P 2,000,000 how much must the paid-up capital be? Solution:

Formula: Authorized Capital Stock = 16(Paid − up Capital) Hence, Paid − up Capital =

2,000,000

16 = 125,000

47. In how many years will the amount of P 10,000 triple if invested at an interest rate of 10% compounded per year?

Solution: Formula -

Given:


⎛ ⎝

mt

F = P (1 + i ) = P ⎜ 1 +

P = 10,000

r ⎞ ⎟ m⎠

F = 3P = 30000


r = 10%

P = present worth

m = 1 (per year)


t=?


Substitute: 30, 000 = 10, 000 (1 + 0.10 ) 3 = (1.10 )

t

t

ln 3 = t (ln1.10 ) → taking ln both sides t = 11.53


GEAS


48. What is the amount of an annuity of P5000 per year at the end of each year for 7 years at 5% interest compounded annually?

Solution:

⎡ (1 + i)n − 1⎤ ⎡ (1 + 0.05 )7 − 1⎤ ⎥ = 5000 ⎢ ⎥ i 0.05 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦

F = A⎢

F = 40,710

50. What is the book value of an equipment purchased three years ago for P15,000 if it is depreciated using the sum of the years digit method? The expected life is 5 years.

Solution: Using SYD method :

∑ year =

n ( n + 1) 2

(5 )( 6 )

=

2

= 15

⎛ n ⎞ ⎜ ∑ year ⎟⎟ ⎝ ⎠

d1 = ( Co − Cn ) ⎜

d1 = (15,000 − 0 ) d2 = (15,000 − 0 ) d3 = (15,000 − 0 )

5 15 4

= 5000

15 3

15 Total depreciation :

= 4000 = 3000

Dn = 5000 + 4000 + 3000

= 12,000 Book Value = 15, 000 − 12, 000 Book Value = 3, 000



CHAPTER 7

51. The parents planned for their son to receive P50,000 ten years from now. What amount in pesos should they invest now if it will earn interest of 12% compounded annually for the first five years and 15% compounded quarterly during the next five years?

Solution:

F10 = 50,000

For the first 5 years: 5

F5 = P (1 + 0.12 )

0

5

10

5

F5 = (1.12 ) P

F5

For the next 5 years: r ⎞ ⎛ F10 = F5 ⎜ 1 + ⎟ ⎝ m⎠

mt

P

5

⎛ ⎝

50,000 = (1.12) P ⎜ 1 +

0.15 ⎞

4(5)

⎟

4 ⎠

P = 13, 586.82

52. A customer buys an electric fan from a store that charges P1,500 at the end of 90 days. The customer wishes to pay cash. What is the cash price if the money is worth 10% simple interest?

Solution:

Formula: (Ordinary Simple Interest) F = P (1 + in )

F = 1,500 i = 10%


P=?

⎡

⎛ 0.10 ⎞ ⎤ ⎟ 90⎥ ⎝ 360 ⎠ ⎦

1500 = P ⎢1 + ⎜

⎣

i = rate of int e rest / day n = no. of int e rest periods

P = 1, 463.41


GEAS


53. In how many years will P1000 double if interest is 10% compounded quarterly? Solution:

Given: P = 1000

Formula - (Compound

F = 2P = 2000

n

r = 10%

r ⎞

⎟

m⎠


t=? Substitute:

2 = (1.025 )

mt

F = P (1 + i ) = P ⎜ 1 +

m = 4 ( quarterly )

⎛ 0.10 ⎞ 2000 = 1000 ⎜ 1 + 4 ⎟⎠ ⎝

⎛ ⎝

Interest)

P = present worth

4t


4t

t = no. of years

ln 2 = 4 t(ln 1.025) t = 7 years

54. Mr. Reyes borrowed money from a bank. He received from the bank P1842 and promised to pay P2,000 at the end of 10 months. Determine the simple interest rate. Solution:

(Ordinary Simple Interest) F = P (1 + in ) Formula:

Given: P = 1842 F = 2000


n = 10(30) = 300 days Substitute:

⎡

⎤ ⎛ i ⎞ ( 300)⎥ ⎟ ⎝ 360 ⎠ ⎦

2000 = 1842 ⎢1+ ⎜

⎣

i = rate of int erest / day n = no. of int e rest periods

i = 0.1029 i = 10.29%



CHAPTER 7

55. A man borrows P15,000 from Hong Kong Bank. The rate of simple interest is 12%, but the interest is to be deducted from the loan at the time the money is borrowed. At the end of one year he has to pay back P15,000. What is the actual rate of interest? Solution: Formula:

Given: P = 15, 000 − 12%(15, 000)

(Ordinary Simple

Interest) F = P (1 + in )

P = 13,200


F = 15,000 n = 360 days

i = rate of int erest / day Substitute:

⎡ ⎛ i ⎞ ⎤ 15,000 = 13,200 ⎢1 + ⎜ 360⎥ ⎟ ⎣ ⎝ 360 ⎠ ⎦ i = 0.1364 i = 13.64%


56. In how many years will it take money to quadruple if invested at 8% compounded semi-annually?

Solution:

Formula - (Compound

Let:

⎛ ⎝

n

Interest)

F = P (1 + i ) = P ⎜1 +

P = present worth F = 3P ( quadruple) r = 8% ; m = 2 (semi − annually)

mt

r ⎞ ⎟ m⎠

Where; F = future worth P = present worth

Substitute:

⎛ 0.08 ⎞ 4P = P ⎜1 + ⎟ 2 ⎠ ⎝ 4 = (1.04 )


2t


2t

ln 4 = 2t(ln 1.04) t = 17.67 years


GEAS


57. A brand new machine is estimated to have a salvage value of P 10,000 after ten years and a book value of P 30,000 after 5 years, what is the initial cost of the machine? (Assume straight line depreciation). Solution:

Given: SV = 10,000

Formula: Depreciation

BV = 30,000

(Straight Line

Method)-Book Value

L = 10 years

BV = Co − Dn

n = 5 years

;

⎛ Co − SV ⎞ ⎟n L ⎝ ⎠

Dn = ⎜

Where: Substitute:

⎡ ( Co − SV ) ⎤ ⎥n L ⎣ ⎦ ⎡ C − 10,000 ⎤ 30,000 = Co − ⎢ o ⎥5 10 ⎣ ⎦ BV = Co − ⎢

30,000 = Co − 25,000 =

Co 2

+ 5,000

BV = Book Value SV = Salvage Value Co = first cos t Dn = total depreciation after n years L = life n = no. of years before L

Co

2 Co = 50,000

58. A machine was brought for P 50,000 with an estimated useful life of 10 years and a salvage value of P 5,000. What is its annual depreciation assuming straight line trend? Solution:

Substitute: 50,000 − 5,000 10 d = 4,500 d=

Formula: Depreciation

(Straight Line Method)-Book Value C − SV d= o L Where: d = annual depreciation SV = Salvage Value Co = first cos t L = life



CHAPTER 7

59. What annuity is required over 10 years to equate with a future amount of P20,000. Assume i = 8% per year.

Solution: Formula:

Given: F = 20,000


⎡ (1 + i)n − 1⎤ ⎥ i ⎢ ⎥ ⎣ ⎦

r = 8%

F= A⎢

m =1 A = ?


Substitute:

F = Future worth

⎡ (1 + .08 )10 − 1⎤ ⎥ 20,000 = A ⎢ 0.08 ⎢⎣ ⎥⎦ A = 1,380.59


60. A merchant loaned P 500,000 payable in 10 years at an interest rate of 12 percent compounded annually. What is the monthly amortization for 10 years?

Solution: Solving for the interest per month: 12 1 (1 + i) − 1 = (1.12 ) − 1

i = 0.009489 or 0.9489% A ⎡(1.009489 )

120

500,000 =

⎣

− 1⎤ ⎦ 120

( 0.009489)(1.009489)

A = 6997.43


GEAS


61. 10 years ago a businessman purchased a machine for P50,000 with an expected life of 20 years based on a straight line depreciation. Today, he decided to replace it with a modern one that costs P120,000. If the salvage value of the old unit is P20,000, how much more will he raise to buy the new machine? Solution:

Given: Co = 50,000 SV = 20,000 L = 20 n = 10 Let: x = amount needed to buy the new equipment Solving for the Book Value:

⎡ ( Co − SV ) ⎤ ⎥n L ⎣ ⎦ ⎡ ( 50,000 − 20,000 ) ⎤ BV = 50,000 − ⎢ ⎥ 10 20 ⎣ ⎦ BV = 45,000 BV = Co − ⎢

Solving for x: x = Co − BV x = 120,000 − 45, 000 x = 75,000



CHAPTER 7

62. A heavy duty copying machine was procured for P100, 000 with an estimated salvage value of P10, 000 after 10 years. What is the book value after 5 years? (Assume straight line depreciation). Solution:

Given: Co = 100,000 SV = 10,000 L = 10 years n = 5 years Substitute:

⎡ ( Co − SV ) ⎤ ⎥n L ⎣ ⎦ ⎡ (100,000 − 10,000 ) ⎤ BV5 = 100,000 − ⎢ ⎥5 10 ⎣ ⎦ BV5 = 55,000 BV5 = Co − ⎢

63. Based on its purchased price a machine is expected to depreciate at a uniform rate of 18 percent annually until it has zero salvage value. Approximate the useful life of the machine in years using the SYD method?

Solution: Given:

⎛ ⎞ ⎜ ⎟ n 0.18C0 = ( Co − 0 ) ⎜ ⎟ n ⎜⎜ ( n + 1) ⎟⎟ ⎝2 ⎠ 0.18Co = C0

2

n +1 0.18n + 0.18 = 2 n = 10


Chap 7 Eng'g Economy

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