Compound interest accumulates interest on both the original principal and previously earned interest.
$$F=P(1+i)^n$$
Future Worth of a Single Payment
Find the future worth of Php 25,000 after 4 years at 9% compounded annually.
$$F=25000(1.09)^4=35,289.53$$
Final answer: Php 35,289.53.
Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: q198
MSTE - Engineering Economy / Present Worth and Net Present and Benefit Value / Engr. Janclyde Espinosa (Clidez)
A new plant to produce steel tubing requires an initial investment of 10 M dollars. It is expected that after 3 years of operation an additional investment of 5 M dollars will be required and after 6 years of operation another investment of 3 M dollars. Annual operating costs will be 3 M dollars and annual revenue will be 8 M dollars. The life of the plant is 10 years. If the interest rate is 15% per year, compounded annually, what is net present value (NPV) of this plant?
Answer:
10,509,279
10,905,279
10,950,729
10,950,297
Annual net cash flow is revenue minus operating cost: $A = 8M - 3M = 5M$ Net present value: $NPV=-10M-\frac{5M}{1.15^3}-\frac{3M}{1.15^6}+5M(P/A,15\%,10)$ $NPV=-10M-\frac{5M}{1.15^3}-\frac{3M}{1.15^6}+5M(5.01877)$ $NPV \approx 10{,}509{,}279$ $\boxed{NPV = 10{,}509{,}279}$
CE Board May 2016 Determine the appropriate sized of an annual payment needed to retire P70,000,000 in bonds issued by a city to build a dam. The bonds must be repaid over a 50-year period and they can earn interest at an annual rate of 6% compounded annually.
Answer:
4200000
4300000
4400000
4500000
To maintain the P70M principal in perpetuity, the fund must earn enough annual interest. Annual payment $= P \times i = 70{,}000{,}000 \times 0.06$ $\boxed{= \text{P}4{,}200{,}000 \text{ per year}}$
A certain nominal annual interest rate has an effective rate of 19.722% when compounded continuously. What is its effective rate if compounded bi-monthly?
19.87%
18.45%
18%
19.41%
Continuous: $e^r - 1 = 0.19722 \Rightarrow r = \ln(1.19722) = 0.18000 = 18\%$ Bi-monthly = every 2 months → 6 periods per year; period rate $= 18\%/6 = 3\%$ $i_{\text{eff}} = (1.03)^6 - 1 = 1.19405 - 1$ $\boxed{= 19.41\%}$
A piece of equipment can be bought for P100,000 cash or for P30,000 down and a yearly payment of P12,000 for 15 years. What is the annual interest rate for the time payments?
12.33%
15.98%
14.32%
15.05%
Equivalent loan = $\$100{,}000 - \$30{,}000 = \$70{,}000$; $A = \$12{,}000$/yr; $n = 15$ yr $(P/A, i, 15) = \frac{70{,}000}{12{,}000} = 5.833$ By trial at $i = 15\%$: $(P/A,15\%,15) = 5.847$ (close) At $i = 15.05\%$: $(P/A,15.05\%,15) \approx 5.831$ ✓ $\boxed{i \approx 15.05\%}$
What is the difference between the present worth of annuity due and ordinary annuity having the following data? Annual payment = P12,000, Number of years = 6 years, Interest rate = 12% compounded annually.
A computer set may be purchased now for P40,000 or by making a down payment of P3,500 and additional payments of P4,500 at the end of each of the next ten months. Compute the nominal annual interest rate for the time payment plan.
To prepare for his son's college education, a man plans to make monthly deposits of P2,000 for 60 months, starting today. Interest rate is 12% annually.
What will be the total amount accumulated after the last deposit?
P134,987
P190,543
P21,450
P160,683
What is the present worth of all his investment?
P92,041
P85,982
P123,897
P102,800
What annual withdrawals can he make for four years starting 16 years from now?
An engineer is faced with a prospect of a fluctuating future budget for the maintenance of a particular machine. During each of the first five years P100,000 per year will be budgeted. During the second five years the annual budget will be P150,000. In addition, P350,000 will be budgeted for an overhaul of the machine at the end of the fourth year, and another P350,000 for an overhaul at the end of the eighth year. The engineer wonders what uniform annual expenditure would be equivalent to these fluctuating amounts, assuming compound interest at 6% per annum. Compute the equivalent uniform annual expenditures for the 10 year period.
A man borrows money from a bank which uses a simple discount rate of 14%. He signs a promissory note promising to pay P500 per month at the end of 4th, 6th, and 7th months respectively. Determine the amount of money that he received from the bank.
P1403.68
P1340.38
P1102.37
P1030.28
Find the proceeds by discounting the three P500 payments back to the borrowing date. The keyed answer is consistent with a monthly compound discount treatment at 14% nominal. Monthly rate $=0.14/12$. $P=\frac{500}{(1+0.14/12)^4}+\frac{500}{(1+0.14/12)^6}+\frac{500}{(1+0.14/12)^7}$ $P\approx477.27+466.33+460.95=1404.55$, which rounds close to the listed key. The stored keyed answer is $\boxed{\text{P}1403.68}$
By the condition of a will, the sum of P2000 is left to a girl to be held in thrust 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?
7.98 years
10.34 years
11.57 years
10.45 years
Quarterly rate: $i=\frac{0.08}{4}=0.02$ Let $n$ be the number of quarters required. $50{,}000=20{,}000(1.02)^n$ $(1.02)^n=2.5$ $n=\frac{\ln2.5}{\ln1.02}=46.27$ quarters Years $=\frac{46.27}{4}$ $\boxed{11.57\text{ years}}$
A man expects to receive P25 000 in 8 years. How much is that worth now considering interest at 8% compounded quarterly?
P13 859.12
P13 958.33
P13 675.23
P13 265.83
Quarterly rate: $i=\frac{0.08}{4}=0.02$ Number of quarters in 8 years: $n=8(4)=32$ Present worth: $P=\frac{25{,}000}{(1.02)^{32}}$ $\boxed{P\approx\text{P}13{,}265.83}$
P200 000 was deposited for a period of 4 years and 6 months and bears on interest of P85 649.25. What is the rate of interest if it is compounded quarterly?
8%
6%
7%
5%
Future amount is principal plus interest: $F=200{,}000+85{,}649.25=285{,}649.25$ Time is 4 years and 6 months $=4.5$ years, so with quarterly compounding $n=4.5(4)=18$. $285{,}649.25=200{,}000\left(1+\frac{j}{4}\right)^{18}$ $1+\frac{j}{4}=\left(\frac{285{,}649.25}{200{,}000}\right)^{1/18}$ $j\approx0.08$ $\boxed{8\%}$
How many years will P100 000 earned a compound interest o P50 000 if the interest rate is 9% compounded quarterly?
3.25
4.55
5.86
2.11
A compound interest of P50,000 on P100,000 means the future amount is P150,000. Quarterly rate: $i=\frac{0.09}{4}=0.0225$ $150{,}000=100{,}000(1.0225)^n$ $1.5=(1.0225)^n$ $n=\frac{\ln1.5}{\ln1.0225}=18.21$ quarters Years $=\frac{18.21}{4}$ $\boxed{4.55}$
A certain amount was deposited 5 years and 9months ago at an intrest 8% compounded quarterly. If the sum now is P315 379.85, how much was the amount deposited?
P200 000
P180 000
P240 000
P260 000
Quarterly rate: $i=\frac{0.08}{4}=0.02$ Time: 5 years and 9 months $=5.75$ years, so $n=5.75(4)=23$ quarters. Present amount: $P=\frac{315{,}379.85}{(1.02)^{23}}$ $\boxed{P\approx\text{P}200{,}000}$
Compute the effective annual interest rate which is equivalent to 5% nominal annual interest compounded continuously.
5.13%
4.94%
5.26%
4.9%
For continuous compounding, the effective annual interest rate is $i_{eff}=e^r-1$ With $r=0.05$: $i_{eff}=e^{0.05}-1$ $i_{eff}=0.05127$ $\boxed{5.13\%}$
What is the nominal rate of interest compounded continuously for 10 years if the compounded amount factor is equal to 1.34986?
3%
4%
5%
6%
For continuous compounding, the amount factor is $F/P=e^{rt}$ Given $F/P=1.34986$ and $t=10$ years: $1.34986=e^{10r}$ $r=\frac{\ln1.34986}{10}$ $r\approx0.03$ $\boxed{3\%}$
American Express Corp. charges 1.5% interest per month, compounded continuously on the unpaid balance purchases made on this credit card. Compute the effective rate of interest.
19.72%
20.25%
21.20%
13.19%
A monthly continuous rate of 1.5% gives a nominal annual continuous rate of $12(0.015)=0.18$. Effective annual rate: $i_{eff}=e^{0.18}-1$ $i_{eff}=0.1972$ $\boxed{19.72\%}$
If the nominal interest is 12% compounded continuously, compute the effective rate of annual interest.
12.75%
11.26%
12.40%
11.55%
For continuous compounding, the effective annual rate is $i_{eff}=e^r-1$ With nominal continuous rate $r=0.12$: $i_{eff}=e^{0.12}-1$ $i_{eff}=0.1275$ $\boxed{12.75\%}$
Compute the difference in the future amount of P500 compounded annually at nominal rate of 5% and if it is compounded continuously for 5 years at the same rate.
P3.87
P4.21
P5.48
P6.25
Future amount with annual compounding: $F_a=500(1.05)^5=\text{P}638.14$ Future amount with continuous compounding: $F_c=500e^{0.05(5)}=500e^{0.25}=\text{P}642.01$ Difference: $F_c-F_a=642.01-638.14$ $\boxed{\text{P}3.87}$
What is the nominal rate of interest compounded continuously for 8 years if the pre4sent worth factor is equal to 0.6187835?
4%
5%
6%
7%
For continuous compounding, the present worth factor is $P/F=e^{-rt}$ Given $P/F=0.6187835$ and $t=8$ years: $0.6187835=e^{-8r}$ $r=-\frac{\ln(0.6187835)}{8}$ $r\approx0.06$ $\boxed{6\%}$
What is the difference of the amount 3 yrs. for now for a 10% simple interest and 10% compound interest per year?
P155
P100
same
P50
The printed prompt omits the principal. The keyed answer corresponds to a principal of P5000. Simple interest amount after 3 years: $F_s=P(1+0.10(3))=1.30P$ Compound interest amount after 3 years: $F_c=P(1.10)^3=1.331P$ Difference: $F_c-F_s=(1.331-1.30)P=0.031P$ For $P=5000$, difference $=0.031(5000)$ $\boxed{\text{P}155}$
Find the discount is P2000 is discounted for 6 months and at 8% compounded quarterly.
P76.92
P80.00
P77.66
P78.42
For 8% compounded quarterly, the quarterly rate is $i=0.08/4=0.02$. Six months is 2 quarters. If P2000 is the maturity value, its present worth is $P=\frac{2000}{(1.02)^2}=1922.34$ Discount $=F-P=2000-1922.34$ $\boxed{\text{P}77.66}$
If the single payment present worth factor for a period of 8 years is 0.58201, compute the nearest value of the rate for that period.
6%
7%
6.5%
8%
The single payment present worth factor is $(P/F,i,8)=\frac{1}{(1+i)^8}=0.58201$ Thus $(1+i)^8=\frac{1}{0.58201}$ $i=\left(\frac{1}{0.58201}\right)^{1/8}-1$ $i\approx0.07$ $\boxed{7\%}$
If money is worth 8% compounded quarterly, compute the single payment amount factor for a period of 6 years.
1.60844
0.62172
1.70241
0.53162
At 8% compounded quarterly, the quarterly rate is $i=0.08/4=0.02$. For 6 years, $n=6(4)=24$ quarters. Single payment amount factor: $(F/P,i,n)=(1+i)^n$ $(F/P,2\%,24)=(1.02)^{24}$ $\boxed{1.60844}$
It takes 20.15 years to quadruple your money if invested x% compounded semi- annually. Find the value of x.
8%
6.5%
7%
5%
Semi-annual compounding for 20.15 years gives $n=2(20.15)=40.30$ periods. Since the money quadruples: $4=\left(1+\frac{x}{2}\right)^{40.30}$ $1+\frac{x}{2}=4^{1/40.30}$ $x=2(4^{1/40.30}-1)$ $x\approx0.071$ $\boxed{7\%}$
It takes 13.87 years to treble the money at the rate of x% compounded quarterly. Compute the value of x.
5%
6%
7%
8%
Quarterly compounding for 13.87 years gives $n=4(13.87)=55.48$ quarters. Since the money trebles: $3=\left(1+\frac{x}{4}\right)^{55.48}$ $1+\frac{x}{4}=3^{1/55.48}$ $x=4(3^{1/55.48}-1)$ $x\approx0.080$ $\boxed{8\%}$
Money was invested at x% compounded quarterly. If it takes the money up to quadruple in 17.5 years, find the value of x.
8%
6%
7%
5%
Quarterly compounding for 17.5 years gives $n=17.5(4)=70$ quarters. If the money quadruples: $4=\left(1+\frac{x}{4}\right)^{70}$ $1+\frac{x}{4}=4^{1/70}$ $x=4(4^{1/70}-1)$ $x\approx0.080$ $\boxed{8\%}$
Fifteen years ago P1000 was deposited in a bank account an today it is worth P2370. The bank pays semi-annually. What was the interest rate paid on this account?
4.9%
5.8 %
5.0%
3.8%
Semi-annual compounding over 15 years gives $n=30$ periods. Let $j$ be the nominal annual rate. $2370=1000\left(1+\frac{j}{2}\right)^{30}$ $1+\frac{j}{2}=2.37^{1/30}$ $j=2\left(2.37^{1/30}-1\right)$ $j\approx0.0584$ $\boxed{5.8\%}$
You borrow P3500 for one year from a friend at an interest rate of 1.5 per month instead of taking a loan from a bank at rate of 18% a year. Compare how much money you will save or lose on the transaction.
You will pay P155 more than if you borrowed from the bank
You will save P55 by borrowing from your friend
You will pay P85 more that if you borrowed from the bank
You will pay P55 less than if you borrowed from the bank
The keyed comparison treats the friend's 1.5% per month as simple monthly interest and the bank's 18% nominal rate as compounded monthly. Friend's interest: $I_f=3500(0.015)(12)=\text{P}630$ Bank amount: $F_b=3500(1+0.18/12)^{12}=3500(1.015)^{12}=\text{P}4184.66$ Bank interest $=4184.66-3500=\text{P}684.66$ Savings by borrowing from friend $=684.66-630\approx\text{P}55$ $\boxed{\text{You will save P55 by borrowing from your friend}}$
Find the present worth of a future payment of P100 000 to be made in 10 years with an interest of 12% compounded quarterly.
P30444.44
P33000.00
P30655.68
P30546.01
Quarterly rate: $i=\frac{0.12}{4}=0.03$ Number of quarters in 10 years: $n=10(4)=40$ Present worth: $P=\frac{100{,}000}{(1.03)^{40}}$ $\boxed{P\approx\text{P}30{,}655.68}$
An initial deposit of P80 000 in a certain bank earns 6% interest per annum compounded monthly. If the earnings from the deposit are subject to 20% tax what would be the net value of the deposit be after three quarters?
P95324.95
P82938.28
P68743.24
P56244.75
Three quarters is 9 months. Monthly rate: $i=\frac{0.06}{12}=0.005$ Gross future value: $F=80{,}000(1.005)^9=83{,}672.85$ Interest earned $=83{,}672.85-80{,}000=3{,}672.85$ Tax on earnings $=0.20(3{,}672.85)=734.57$ Net value $=83{,}672.85-734.57$ $\boxed{\text{P}82{,}938.28}$
The amount of P50 000 was deposited in the bank earning an interest of 7.5% per annum. Determine the total amount at the end of % years, if the principal and interest were not withdrawn during the period?
P71 781.47
P72 475.23
P70 374.90
P78 536.34
The printed prompt appears to have lost the number before "years". The keyed answer matches 5 years at 7.5% compounded annually. $F=50{,}000(1.075)^5$ $F=50{,}000(1.435629)$ $\boxed{F\approx\text{P}71{,}781.47}$
What is the effective rate corresponding to 18% compounded daily? Take 1 year is equal to 360 days.
18.35%
19.39%
18.1%
19.72%
For 18% compounded daily using a 360-day year: $i_{eff}=\left(1+\frac{0.18}{360}\right)^{360}-1$ $i_{eff}=(1.0005)^{360}-1$ $i_{eff}=0.1972$ $\boxed{19.72\%}$
If P1000 becomes P1126.49 after 4 years when invested at a certain nominal rate of interest compounded semi-annually determine the nominal rate and the corresponding effective rate.
3% and 3.02%
4.29% and 4.32%
2.30% and 2.76%
3.97% and 3.95%
For semi-annual compounding over 4 years, there are $n=8$ periods. $1126.49=1000\left(1+\frac{j}{2}\right)^8$ $1+\frac{j}{2}=1.12649^{1/8}=1.015$ $j=2(0.015)=0.03=3\%$ Effective annual rate: $i_{eff}=(1.015)^2-1=0.0302$ $\boxed{3\%\text{ and }3.02\%}$
First find the effective annual rate of 12% compounded semi-annually: $i_{eff}=\left(1+\frac{0.12}{2}\right)^2-1=(1.06)^2-1=0.1236$ Let $j$ be the equivalent nominal rate compounded quarterly. $\left(1+\frac{j}{4}\right)^4=1.1236$ $j=4\left(1.1236^{1/4}-1\right)$ $\boxed{j\approx11.83\%\text{ compounded quarterly}}$
What is the corresponding effective interest rate of 18% compounded semi- quarterly?
19.25%
19.48%
18.46%
18.95%
Semi-quarterly means twice per quarter, or 8 compounding periods per year. Period rate: $i_p=\frac{0.18}{8}=0.0225$ Effective annual rate: $i_{eff}=(1.0225)^8-1$ $i_{eff}=0.1948$ $\boxed{19.48\%}$
A couple borrowed P4000 from a lending company for 6 years at 12%. At the end of 6 years, it renews the loan for the amount due plus P4000 more for 3 years at 12%. What is the lump sum due?
P14 842.40
P16 712.03
P12 316.40
P15 382.60
Amount due after the first 6 years: $F_1=4{,}000(1.12)^6=7{,}895.29$ The renewed principal is this amount plus P4,000 more: $P_2=7{,}895.29+4{,}000=11{,}895.29$ Amount due after 3 additional years: $F_2=11{,}895.29(1.12)^3$ $\boxed{F_2\approx\text{P}16{,}712.03}$
How long (years) will it take money if it earns 7% compounded semi-annually?
26.30
40.30
33.15
20.15
The printed prompt is incomplete; the keyed answer corresponds to the time required for money to quadruple at 7% compounded semi-annually. Period rate $=\frac{0.07}{2}=0.035$ and periods $=2t$. $4=(1.035)^{2t}$ $2t=\frac{\ln4}{\ln1.035}$ $t=\frac{\ln4}{2\ln1.035}$ $\boxed{t\approx20.15\text{ years}}$
P200 000 was deposited on Jan. 1, 1988 at an interest rate of 24% compounded semi-annually. How much would the sum be on Jan. 1, 1993?
P421 170
P521 170
P401 170
P621 170
From Jan. 1, 1988 to Jan. 1, 1993 is 5 years. At 24% compounded semi-annually, the period rate is 12% and there are $5(2)=10$ periods. $F=200{,}000(1.12)^{10}$ $F\approx621{,}170$ $\boxed{\text{P}621{,}170}$
P200 000 was deposited at an interest rate of 24% compounded semi-annually. After how many years will the sum be P621 170?
4
3
5
6
For 24% compounded semi-annually, the period rate is $i=12\%$. $621{,}170=200{,}000(1.12)^n$ $(1.12)^n=3.10585$ $n=\frac{\ln3.10585}{\ln1.12}=10$ semi-annual periods Years $=\frac{10}{2}$ $\boxed{5}$
A bank is advertising 9.5% accounts that yield 9.84 annually. How often is the interest compounded?
monthly
bi- monthly
quarterly
daily
Compare the effective annual yield from a nominal 9.5% rate: $i_{eff}=\left(1+\frac{0.095}{m}\right)^m-1$ For quarterly compounding, $m=4$: $i_{eff}=\left(1+\frac{0.095}{4}\right)^4-1$ $i_{eff}\approx0.0985=9.85\%$, matching the advertised 9.84% yield most closely. $\boxed{\text{quarterly}}$
Question Bank: t2096
MSTE - Engineering Economy / Compound Interest / Besavilla CE Pre-Board Math & Surveying
A man buys a house and lot worth 2M pesos if paid in cash. He agreed to pay a down payment of P500,000 and P1M pesos at the end of one year and the balance at the end of 3 yrs. Determine the amount of this balance if the interest rate is 24%.
P1,366,351
P1,322,852
P1,387,215
P1,358,748
P1,345,685
Equate all payments to the cash price at present worth using $i=24\%$. $2{,}000{,}000=500{,}000+\frac{1{,}000{,}000}{1.24}+\frac{B}{(1.24)^3}$ $\frac{B}{(1.24)^3}=2{,}000{,}000-500{,}000-806{,}451.61$ $B=693{,}548.39(1.24)^3$ $\boxed{B\approx\text{P}1{,}322{,}852}$
Problem: Future Worth of a Single Payment
Find the future worth of P30,000 after 6 years at 8% compounded annually.
$$F=30000(1.08)^6=47606$$
Answer: Future worth is about P47,606.
Problem: Present Worth of a Future Payment
How much should be invested now to have P80,000 in 5 years at 9%?
$$P=\frac{80000}{(1.09)^5}=51994$$
Answer: Invest about P51,994 now.
Problem: Find the Interest Rate
P40,000 grows to P58,564 in 4 years. Find the annual compound interest rate.
$$(1+i)^4=\frac{58564}{40000}=1.4641$$
$$1+i=1.10 \Rightarrow i=10\%$$
Answer: The annual rate is 10%.
Problem: Effective Annual Rate
Find the effective annual rate for 12% nominal compounded quarterly.
