10
Annuities:
Future Value and Present Value
Checkpoint Questions (Section 10.1)
1. a. False. Although the payments are equal in value, they are not being paid at regular
intervals of time. Therefore, the series of payments does not meet the definition of an
annuity.
b. False. When an annuity’s payments are made at the beginning of each period, the
annuity is classified as an annuity due.
c. True. When the payment interval and the compounding interval are equal, the annuity
is classified as a simple annuity.
d. True. When the payment interval and the compounding interval are not equal, the
annuity is classified as a general annuity.
e. False. The premiums are paid in advance, at the start of each month. The payments
are at the beginning of each payment interval. Therefore, the payments form an
annuity due.
3. The two types of annuities are distinguished by comparing the timing of the payments. If
the payments are made at the end of each payment interval, the annuity is an ordinary
annuity. If the payments are made at the beginning of each payment interval, the annuity
is an annuity due.
Exercise 10.1
Basic Problems
1. With payments at the end of every period, this is an ordinary annuity.
With monthly payments and interest compounding quarterly, this is a general annuity.
Hence, this is an ordinary general annuity.
3. With payments at the beginning of every period, this is an annuity due.
With monthly payments and interest compounding monthly, this is a simple annuity.
Hence, this is a simple annuity due.
5. Since the payments are made every quarter, the payment interval is every 3 months.
The term of the annuity is 42 payments x 3 months = 126 months. Since there are 12
months in a year, the term of the annuity is 126/12 or 10 years, 6 months.
298
Fundamentals of Business Mathematics in Canada, 2/e
Intermediate Problems
7.
First
payment
interval
(3 mths)
Loan
Advanced
PMT
#1
PMT
#2
PMT
#3
PMT
#4
PMT
#5
PMT
#6
Start of
Annuity’s
Term
PMT
#7
End of
Annuity’s
Term
Term of the annuity
7 x 3 months = 21 months, or 1 year, 9 months
9. a. Payments are made at the end of each year, therefore the annuity is an ordinary
annuity. Payments are made annually, and interest is compounded quarterly. Since
the payment interval and the compounding interval are not equal, the annuity is a
general annuity. Therefore, the series of payments represents an ordinary general
annuity.
b. First payment: December 31, 2014; Last payment: December 31, 2018
c. Start of annuity term: December 31, 2013; End of annuity term: December 31, 2018
Math Apps (Section 10.2)
Your Potential to Become a Millionaire!
1. a. The accumulated amount after 20 years will be
 1  i  n  1
 1.0075 240  1
 = $300
FV  PMT 
 = $200,366
0.0075
i




 1.0075 360  1
$300
b. The amount after 30 years will be

 = $549,223
0.0075


480
 1.0075   1
c. The amount after 40 years will be $300
 = $1,404,396
0.0075


3. If the inflation rate is 2.4% compounded monthly, you will need
FV  PV 1  i  = $1,000,000(1.002)480 = $2,609,194
40 years from now to have the same purchasing power as $1,000,000 has today.
Therefore, the amount in 1(c) will have only a little more than half of the purchasing power
of $1,000,000 today. You would/should not “feel” as wealthy as a person holding
$1,000,000 today.
n
Chapter 10: Annuities: Future Value and Present Value
299
Checkpoint Questions (Section 10.2)
1. A’s future value will be (i) double B’s future value. When we inspect the future value formula
 1  i n  1
FV  PMT 

i


we note that, for given values of i and n, the future value is proportional to PMT. Therefore,
doubling the size of the payment will double the amount of the future value.
Exercise 10.2
Basic Problems
1. Given: PMT = $100, n = 4(5.5) = 22, i =
5%
4
= 1.25%
 1  i n  1
FV  PMT 

i


 1.0125 22  1

= $100 

 0.0125 
= $2514.31
5  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
22  N 
0  PV 
100 + / – PMT
CPT  FV 
Ans: 2514.31
3. Given: PMT = $2000, n = 2(7) = 14, i =
6%
2
= 3%
6  I/Y 
 1  i n  1
FV  PMT 

i


 1.0314  1

= $2000 

 0.03 
= $34,172.65
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
14  N 
0  PV 
2000 + / –
Ans: 34,172.65
5. Given: PMT = $175, n = 12( 8 41 ) = 99, i =
 1  i n  1
FV  PMT 

i


 1.0035 99  1

= $175 

0
.
0035


= $20,662.71
PMT
CPT  FV 
4.2%
12
= 0.35%
4.2  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
99  N 
0  PV 
175 + / –
PMT
CPT  FV 
Ans: 20,662.71
300
Fundamentals of Business Mathematics in Canada, 2/e
7. Given: PMT = $160, n = 12(20) = 240, i =
7.5%
12
= 0.625%
 1  i n  1
FV  PMT 

i


240
 1.00625  1

= $160 

 0.00625 
= $88,596.92
7.5  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
240  N 
0  PV 
160 + / –
PMT
CPT  FV 
Ans: 88,596.92
Exercise 10.2 (continued)
Intermediate Problems
9.a. Given: PMT = $2(30) = $60, n = 12(3) = 36, i =
2.52%
12
= 0.21%
 1  i n  1
FV  PMT 

i


36
 1.0021  1

= $60 

 0.0021 
= $2241.30
2.52  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
36  N 
0  PV 
60 + / –
PMT
CPT  FV 
Ans: 2,241.30
b. Over the course of the three years, you will have made 36 payments of $60. This
represents a total investment of 36 x $60 = $2160 of principal. Therefore, you will
have earned $2241.30 – $2160 = $81.30 of interest.
11. Given: n = 1(25) = 25, i = 51% = 5%
a. PMT = $1000
 1  i n  1
FV  PMT 

i


 1.05 25  1

= $1000 

 0.05 
= $47,727.10
b. PMT = $2000
 1  i n  1
FV  PMT 

i


25
 1.05  1

= $2000 

 0.05 
Chapter 10: Annuities: Future Value and Present Value
5  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
25  N 
0  PV 
1000 + / –
PMT
CPT  FV 
Ans: 47,727.10
Same I/Y, C/Y, P/Y, N, PV
2000 + / – PMT
CPT  FV 
Ans: 95,454.20
301
= $95,454.20
c. PMT = $3000
 1.05 25  1

= $3000 

 0.05 
= $143,181.30
13.
a. Given: PMT = $1000, i =
5%
1
Same I/Y, C/Y, P/Y, N, PV
3000 + / – PMT
CPT  FV 
Ans: 143,181.30
= 5%, n = 5 payments
 1  i n  1
FV  PMT 

i


5
 1.05  1

= $1000 

 0.05 
= $5525.63
5  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
5  N 
0  PV 
1000 + / – PMT
CPT  FV 
Ans: 5525.63
Similarly,
b. FV = $12,577.89
c. FV = $21,578.56
d. FV = $33,065.95
e. FV = $47,727.10
f. FV = $66,438.85
for n = 10
for n = 15
for n = 20
for n = 25
for n = 30
15. For the first 2 years, PMT = $1200,
n = 2(4) = 8, i = 10%
= 2.5%.
4
10  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
8  N 
The future value after 2 years is
 1  i n  1
FV  PMT 

i


 1.025 8  1 

= $1200 
 0.025 


= $10,483.339
For the next 3 years, n = 4(3) = 12,
PV = $10,483.339, PMT = $1200, i =
0  PV 
1200 + / – PMT
CPT  FV 
Ans: 10,483.339
9%
4
= 2.25%
The combined future value of these amounts
at the end of the 3 years is
 1  i n  1
n
FV  PMT 
 + PV 1  i 
i


 1.0225 12  1
 + $10,483.339
= $1200 
 0.0225 


302
Same P/Y, C/Y, PMT
9  I/Y 
12  N 
10483.339 + / –  PV 
CPT  FV 
Ans: 30,014.43
Fundamentals of Business Mathematics in Canada, 2/e
1.0225 12
= $16,322.666 + $13,691.765
= $30,014.43
17. Given: PMT = $100, i =
6%
12
= 0.5%
a. n = 12(40) = 480
 1  i n  1
FV  PMT 

i


480
 1.005  1

= $100 

 0.005 
= $199,149.07
6  I/Y 
 P/Y  11  ENTER 
(making C/Y = P/Y = 12)
480  N 
0  PV 
100 + / –
PMT
CPT  FV 
Ans: 199,149.07
b. n = 12(35) = 420
 1  i n  1
FV  PMT 

i


420
 1.005  1

= $100 

 0.005 
= $142,471.03
Same I/Y, C/Y, P/Y, PMT, PV
420  N 
CPT  FV 
Ans: 142,471.03
c. By beginning the savings program at age 25 you will have a total of $199,149.07 –
$142,471.03 = $56,678.04 more than if you had waited until age 30 to begin saving.
$56,678.04
You will have
× 100% = 39.78% more with the early savings plan.
$142,471.03
Advanced Problems
19. At age 62, when she stops making contributions, Nina’s accumulated savings will be the
combined future value of the $10,000 already saved and the monthly $150 contributions.
That is,
4%
n = 12(32) = 384, i = 5.12
= 0.45%
 1  i n  1
n
FV  PV 1  i  + PMT 

i


 1.0045 384  1
384

= $10,000 1.0045 
+ $150 

 0.0045 
= $56,076.045 + $153,586.817
= $209,662.862
5.4  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
384  N 
10000 + / –  PV 
150 + / –
PMT
CPT  FV 
Ans: 209,662.862
(continued)
Chapter 10: Annuities: Future Value and Present Value
303
Exercise 10.2 (continued)
At age 62, Nina’s accumulated savings of $209,662.862 will accumulate interest at 5.4%
compounded monthly for three years, until age 65.
That is,
5.4  I/Y 
4%
n = 12(3) = 36, i = 5.12
= 0.45%
 P/Y  12  ENTER 
FV  PV 1  i 
n
= $209,662.862 1.0045 
= $246,444.65
36
(making C/Y = P/Y = 12)
36  N 
209662.862 + / –  PV 
0
PMT
CPT  FV 
Ans: 246,444.65
At age 65, Nina will have accumulated a total of $246,444.65.
6  I/Y 
21. After 10 years, Marika's RRSP will be worth
 1  i n  1
FV  PV 1  i n + PMT 

i


 1.03 20  1
20

= $18,000 1.03  + $2000 

0
.
03


= $32,510.002 + $53,740.749
= $86,250.751
After a further 5 years, it will be worth
 1.005 60  1

+ $300 

 0.005 
= $116,339.339 + $20,931.009
= $137,270.35
FV = $86,250.751 1.005 
60
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
20  N 
18000 + / –  PV 
2000 + / – PMT
CPT  FV 
Ans: 86,250.751
Same I/Y
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
60  N 
86250.751 + / –  PV 
300 + / – PMT
CPT  FV 
Ans: 137,270.35
Checkpoint Questions (Section 10.3)
1. A’s present value will be (i) double B’s present value. When we inspect the present
value formula
1  1  i   n 

PV  PMT 
i


we note that, for given values of i and n, the present value is proportional to PMT.
Therefore, doubling the size of the payment will double the annuity’s present value.
3. G’s present value is (iii) less than double H’s present value. The later half of G’s payments
will be discounted more heavily (and therefore contribute less to G’s present value) than the
earlier half of G’s payments.
304
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 10.3
Basic Problems
1.
Given: i =
10%
4
10  I/Y 
= 2.5%; PMT = $100, n = 4(5.5) = 22
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
22  N 
1  1  i   n 

PV  PMT 
i


 1  1.025 22 

= $100 


0
.
025


= $1676.54
3. Given: i =
100 + / – PMT
0  FV 
CPT  PV 
Ans: 1676.54
6%
4
= 1.5%; PMT = $2500, n = 4(25) = 100
1  1  i   n 

PV  PMT 
i


 1  1.015 100 

= $2500 

0
.
015


= $129,061.76
5. Given: i =
6  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
100  N 
2500 PMT
0  FV 
CPT  PV 
Ans: -129,061.76
9%
4
= 2.25%; PMT = $727.88, n = 4(7) = 28
1  1  i   n 

PV  PMT 
i


 1  1.0225 28 

= $727.88 

 0.0225 
= $15,000.03
9  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
28  N 
727.88 + / – PMT
0  FV 
CPT  PV 
Ans: 15,000.03
Rounded to the nearest dollar, the amount borrowed
was $15,000.
7. Given: i =
8.75%
12
= 0.72916 %; PMT = $350, n = 12(11
1  1  i   n 

PV  PMT 
i


 1  1.0072916 137 

= $350 

0.0072916


= $30,259.20
5
) = 137
12
8.75  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
137  N 
350 PMT
0  FV 
CPT  PV 
Ans: -30,259.20
Chapter 10: Annuities: Future Value and Present Value
305
Exercise 10.3 (continued)
9. Given: Term = 20 years; j = 4.8% compounded monthly; PMT = $1000 per month
Therefore, i = 4.8%
= 0.4% and n = 12(20) = 240
12
The amount required to purchase the annuity
4.8  I/Y 
is its present value.
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
1  1  i   n 
240  N 

PV  PMT 
i


1000 PMT
0  FV 
 1  1.004 240 
CPT  PV 

= $1000 
0.004


Ans: –154,093.30
= $154,093.30
Intermediate Problems
11.
PMT = $1000 and i = 7% in all parts.
a. For n = 5,
1  1  i   n 

PV  PMT 
i


 1  1.07  5 

= $1000 

0.07


= $4100.20
Similarly,
b. For n = 10
PV = $7023.58
c. For n = 20,
PV = $10,594.01
d. For n = 30,
PV = $12,409.04
e. For n = 100,
PV = $14,269.25
f. For n = 1000,
PV = $14,285.71
7  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
5  N 
1000 PMT
0  FV 
CPT  PV 
Ans: –4,100.20
13. The economic value of the contract at the date of termination is the
present value of the remaining payments.
5% = 0.625%
Given: PMT = $90,000, n = 12(3) = 36, i = 7.12
1  1  i   n 

PV  PMT 
i


 1  1.00625 36 

= $90,000 


0
.
00625


= $2,893,312.18
The settlement amount is $2,893,312.18.
306
7.5  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
36  N 
90000 PMT
0  FV 
CPT  PV 
Ans: –2,893,312.18
Fundamentals of Business Mathematics in Canada, 2/e
15.
i = 912% = 0.75%
The three year service contract has PMT = $75 and n = 36. The current economic value of
the three year contract is
9  I/Y 
1  1  i   n 

PV  PMT 
 P/Y  12  ENTER 
i


(making C/Y = P/Y = 12)
36  N 
 1  1.0075  36 
75
+
/
– PMT


= $75 

0
.
0075
0  FV 


= $2358.51
CPT  PV 
Since the phone is “free” with the three year
Ans: 2,358.51
contract, the total cost of the phone and three
years’ worth of service is $2358.51.
With the “no contract” provider, Marika will have PMT = $40 and n = 36
1  1  i   n 
Same I/Y, P/Y, C/Y, N, FV

PV  PMT 
40 + / – PMT
i


0  FV 
 1  1.0075  36 
CPT  PV 

= $40 

Ans:
1,257.87
0
.
0075


= $1257.87
In addition to the current economic value of the three years of service, Marika will also
have to purchase the phone, so the total cost of the phone and three years’ worth of
service is $639.99 + $1257.87 = $1897.86. In current dollars, it is $2358.51 – $1897.86 =
$460.65 cheaper to buy the phone for $639.99 and choose the “no contract” provider.
17. The appropriate price to pay is the present value of the future payments
discounted at the required rate of return. Hence,
1  1  i  n 
n
Price = PMT 
 + FV 1  i 
i


 1  1.0325 30
= $50 
0.0325

= $1332.18

 + $1000 1.0325 30


6.5  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
30  N 
50 PMT
1000  FV 
CPT  PV 
Ans: –1,332.18
Chapter 10: Annuities: Future Value and Present Value
307
Exercise 10.3 (continued)
19. Given: Term = 25 years; j = 4.8% compounded monthly;
PMT = $1000/month for the first 15 years and $1500/month for the next 10 years,
The amount Isaac will pay is the present value of the annuity.
The present value, 15 years from now, of 120 payments of $1500 is
4.8  I/Y 
 1  1.004  120 

PV = $1500 
 P/Y  12  ENTER 
0.004


(making C/Y = P/Y = 12)
120  N 
= $142,733.952
1500 PMT
0  FV 
CPT  PV 
Ans: –142,733.95
The combined present value, today, of the preceding amount & 180 payments of $1000 is
 1  1.004  180 
$142,733.952
4.8  I/Y 


+
$1000
PV =
0.004
1.004180

P/Y

12
 ENTER 


= $69,575.924 + $128,137.046
= $197,712.97
(making C/Y = P/Y = 12)
180  N 
1000 PMT
142,733.952  FV 
CPT  PV 
Ans: –197,712.97
21. a. Original loan = Present value of all payments
1  1  i  n 
Original loan = PMT 

i


 1  1.019 40 

= $808.15 

 0.019 
= $22,500.06
b. Balance = Present value of the remaining
40 – 2(8.5) = 23 payments
 1  1.019 23 

= $808.15 

 0.019 
= $14,945.45
308
3.8  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
40  N 
808.15 + / – PMT
0  FV 
CPT  PV 
Ans: 22,500.06
Same I/Y, P/Y, C/Y, PMT, FV
23  N 
CPT  PV 
Ans: 14,945.45
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 10.3 (continued)
23. Selling price = Down payment + Present value of the monthly payments
PMT = $295.88, n = 12(3.5) = 42, i = 7.5%
= 0.625%
12
1  1  i  n 
PV of monthly payments = PMT 

i


 1 1.00625 42 

= $295.88 
 0.00625 
= $10,899.99
7.5  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
42  N 
295.88 PMT
0  FV 
CPT  PV 
Ans: –10,899.99
Thus, the selling price was $2000 + $10,899.99 = $12,899.99
25. Mrs. Chan is willing to pay $100,000 at the start and a total of 16 quarterly payments of
$6250. The present value of the quarterly payments is
1  1  i   n 

PV  PMT 
i


 1  1.0125 16 

= $6250 

 0.0125 
= $90,126.83
5  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
16  N 
6250 PMT
0  FV 
CPT  PV 
Ans: -90,126.83
The total present value of Mrs. Chan’s offer is $100,000 + $90,126.83 = $190,126.83.
Mr. Seebach’s offer is worth $195,000 – $190,126.83 = $4873.17 more in current dollars.
Advanced Problems
27.
Let PMT represent the normal monthly pension payment at age 60. The choice at age 55
is between:
7.5  I/Y 
(i) receiving 0.85PMT per month for 28 years, or
 P/Y  12  ENTER 
(ii) waiting 5 years and then receiving PMT
(making C/Y = P/Y = 12)
per month for 28  5 = 23 years.
336  N 
The economic values of the two alternatives are their
1 PMT
present values at age 55. For option (i),
0
 FV 
 1  1.00625 336 


CPT

PV 
PV = 0.85(PMT)


0
.
00625
Ans: -140.2785


×
0.85 =-119.24
= 119.24(PMT)
Chapter 10: Annuities: Future Value and Present Value
309
For option (ii), the present value at
age 60 of the12(23) = 276 monthly
payments of PMT is
1  1  i   n 

PV  PMT 
i


 1  1.00625 276 

= PMT 


0
.
00625


= 131.34(PMT)
Same I/Y, P/Y, C/Y, FV
276  N 
1 PMT
CPT  PV 
Ans: –131.34
Same I/Y, P/Y, C/Y
60  N 
0 PMT
The PV at age 55 of this amount is
PV  FV 1  i 
131.34  FV 
n
CPT  PV 
Ans: –90.37
= 131.34(PMT) 1.00625 60
= 90.37(PMT)
Hence, the pension-at-age-55 option has a
119.24( PMT )  90.37( PMT )
 100%
90.37( PMT )
= 31.9% higher economic value
29. a. Given: For Annuity A, PMT = $100; i = 8%; n = 20
For Annuity B, PMT = $100; i = 8%; n = 40
PVB $1192.46
(i)
= 1.21455

PV A
$981.81
Therefore, PVB is 21.46% larger than PVA .
FVB $25,905.65
(ii)
= 5.6610

FV A
$4576.20
Therefore, FVB is 466.10% larger than FVA .
b. Given: For Annuity A, PMT = $100; n = 30; i = 8%
For Annuity B, PMT = $100; n = 30; i = 9%
PVB $1027.37
(i)
= 0.9126

PV A $1125.78
Therefore, PVB is (1 – 0.9126)100% = 8.74% smaller than PVA .
FVB $13,630.75
(ii)
= 1.2032

FV A $11,328.32
Therefore, FVB is 20.32% larger than FVA .
Checkpoint Questions (Section 10.4)
1. The payments are at the end of each payment interval (ordinary annuity) and the payment
interval is not equal to the compounding interval (general annuity).
3.
c
Number of compoundings per year
2
=
 0.16
12
Number of payments per year
i2 is the interest rate per payment interval. In this case, it is the interest rate per month. It
Nominal annual rate 6%
will be approximately equal to
= 0.5% per month.

12
12
310
Fundamentals of Business Mathematics in Canada, 2/e
The correct value will be smaller than 0.5% because i2 compounded 6 times must equal
i = 6%
= 3% per half year.
2
Exercise 10.4
Basic Problems
1. i =
7%
= 7%;
1
c
Number of compoundings per year
1
=
= 0.5
Number of payments per year
2
i2  1  i c  1 = 1.07 
0 .5
3. i =
– 1 = 0.03441 = 3.441% (per half year)
Number of compoundings per year
5%
2
= 2.5%; c 
=
= 0.16
Number of payments per year
2
12
i2  1  i c  1 = 1.0250.16 – 1 = 0.00412 = 0.412% (per month)
5. i =
c
8%
2
= 4%; n = 1(27) = 27;
Number of compoundings per year 2
=
=2
Number of payments per year
1
i2  1  i c  1 = 1.04  – 1 = 0.0816
2
 1  i n  1
a. FV  PMT 

i


 1.0816 27  1

= $1000 

 0.0816 
= $89,630.08
1  1  i  n 
b. PV  PMT 

i


 1  (1  0.0816) 27 

= $1000 

0.0816


= $10,780.86
Chapter 10: Annuities: Future Value and Present Value
8  I/Y 
 P/Y  1  ENTER 
 C/Y  2  ENTER 
27  N 
0  PV 
1000 PMT
CPT  FV 
Ans: –89,630.08
(continued)
Same I/Y, C/Y, P/Y, N, PMT
0  FV 
CPT  PV 
Ans: –10,780.86
311
Intermediate Problems
7. i =
Number of compoundings per year
4
4.5%
= 1.125%; n = 1(10) = 10; c 
=
=4
Number of payments per year
4
1
i2  1  i c  1 = 1.01125 4 – 1 = 0.045765086
 1  i n  1
FV  PMT 

i


4.5  I/Y 
 P/Y  1  ENTER 
 C/Y  4  ENTER 
 1.04576508610  1

= $3500 

 0.045765086 
10  N 
3500 + / – PMT
0  PV 
CPT  FV 
Ans: 43,162.14
= $43,162.14
5.5%
= 1.375%; n = 12(13) = 156;
4
Number of compoundings per year
4
c
=
= 0. 3
Number of payments per year
12
9. a. i =
i2  1  i c  1 = 1.01375 0.3 – 1 = 0.004562485
 1  i n  1
FV  PMT 

i


5.5  I/Y 
 P/Y  12  ENTER 
 C/Y  4  ENTER 
156  N 
 1.004562485  1

= $185 

 0.004562485 
156
= $41,936.99
185 + / – PMT
0  PV 
CPT  FV 
Ans: 41,936.99
b. Over 13 years, Larissa will have deposited a total of $185(156) = $28,860 of her own
money. Since her investment has accumulated to $41,936.99 at the end of the 13 year
period, Larissa has earned $41,936.99 – $28,860 = $13,076.99 of interest.
11. The amount required to purchase the annuity is the present value of
the payments discounted at the rate of return on the annuity.
PMT = $2500, n = 12(20) = 240, i = 6.75%,
Number of compoundings per year
1
=
= 0.08 3
c
12
Number of payments per year
i2  1  i c  1 = = 1.0675 0.083 – 1 = 0.00545813044
1  1  i  n 
PV  PMT 

i


 1  1.0054581304 4 240 

= $2500 

0
.
0054581304
4


= $333,998.96
312
6.75  I/Y 
 P/Y  12  ENTER 
 C/Y  1  ENTER 
240  N 
2500 PMT
0  FV 
CPT  PV 
Ans: 333,998.96
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 10.4 (continued)
9  I/Y 
13.
a. i =
9%
2
= 4.5% ; n = 2(8) = 16;
FV  PV 1  i 
= 5000(1+.045)16
= $10,111.85
n
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
16  N 
0
PMT
5000 + / –  PV 
9%
= 4.5%; n = 1(8) = 8;
2
Number of compoundings per year 2
c
=
=2
Number of payments per year
1
b. i =
i2  1  i c  1 = 1.045  – 1 = 0.092025
CPT  FV 
Ans: 10,111.85
2
 1  i n  1
FV  PMT 

i


 1.092025  1

= $900 

 0.092025 
8
= $9,998.73
15. a. Selling price = Down payment + Present value
of the monthly payments
Given: PMT = $259.50; n = 12(3.5) = 42.
1 = 0.08 3 , and
Then i = 7.5%
= 7.5%, c = 12
1
i2  1  i c  1 = 1.0750.083 – 1 = 0.006044919
1  1  i  n 
PV  PMT 

i


 1  1.006044919 42 

= $259.50 

 0.006044919 
= $9600.00
Thus, the selling price was $2000 + $9600.00 = $11,600.00.
9  I/Y 
 P/Y  1  ENTER 
 C/Y  2  ENTER 
8  N 
900 + / – PMT
0  PV 
CPT  FV 
Ans: 9,998.73
7.5  I/Y 
 P/Y  12  ENTER 
 C/Y  1  ENTER 
42  N 
259.50 PMT
0  FV 
CPT  PV 
Ans: –9600.00
b. Immediately after the 12th payment, there are 42 – 12 = 30 payments remaining
outstanding on the loan. The loan balance is the PV of the remaining payments.
1  1  i  n 
PV  PMT 

i


 1  1.006044919 30 

= $259.50 

 0.006044919 
= $7100.36
Chapter 10: Annuities: Future Value and Present Value
Same I/Y, P/Y, C/Y, PMT, FV
30  N 
CPT  PV 
Ans: –7100.36
313
Exercise 10.4 (continued)
17. The current balance is the present value of the remaining payments.
Given: PMT = $1167.89/month, j = 6.6% compounded semiannually, and
time remaining in the mortgage’s term = 4 years, 7 months
2
 0.16
Then i = 3.3%, n = 12(4) + 7 = 55, and c = 12
i2  1  i   1 = 1.033 
c
0.16
– 1 = 0.0054258653
6.6  I/Y 
 P/Y  12  ENTER 
 C/Y  2  ENTER 
55  N 
 1  1.0054258653  55 

Balance = $1167.89 
0.00542586
532


= $55,406.95
1167.89 + / – PMT
0  FV 
CPT  PV 
Ans: 55,406.95
19.
The fair market value of the share is the combined present value of the 30 quarterly
dividend payments of $1.63, and the $25 that will be paid in 7.5 years from now.
For the 30 quarterly dividend payments,
i=
5.5%
2
= 2.75%; PMT = $1.63; n = 30; c 
Number of compoundings per year
= 0.5
Number of payments per year
i2  1  i c  1 = 1.0275  – 1 = 0.013656747
The PV of the 30 quarterly dividend payments is
1  1  i  n 
PV  PMT 

i


 1  1.013656747 30 

= $1.63 

 0.013656747 
= $39.901
For the final $25 payment, i = 5.52% = 2.75%; n = 15
0.5
PV  FV 1  i 
= 25(1+.0275)-15
= $16.642
Hence, the fair market value of the share is $39.901 + $16.642 = $56.54.
-n
314
5.5  I/Y 
 P/Y  4  ENTER 
 C/Y  2  ENTER 
30  N 
1.63 PMT
25  FV 
CPT  PV 
Ans: – 56.54
Fundamentals of Business Mathematics in Canada, 2/e
Advanced Problems
21. Cost to purchase the annuity = PV of all payments
First calculate the PV, 15 years from now, of the last
10 years’ payments of PMT = $1500/month with
1 = 0.08 3 , n = 12(10) = 120, and
j = 5%, m = 1, c = 12
i2  1  i   1 = 1.05 
– 1 = 0.00407412378
1  1  i  n 
 1  1.00407412378 120 

PV  PMT 
 = $1500 
0.00407412
378
i




= $142,148.386
c
0.08 3
Next determine the combined PV of this amount and the
first 15 years’ payments of PMT = $1000/month
with the same i2 and n = 12(15) = 180.
 1  1.00407412378 180 
$142,148.386


PV =
+
$1000
0.00407412378
1.0515


= $68,375.804 + $127,385.158
= $195,760.96
5  I/Y 
 P/Y  12  ENTER 
 C/Y  1  ENTER 
120  N 
1500 PMT
0  FV 
CPT  PV 
Ans: –142,148.386
Same I/Y, P/Y, C/Y
180  N 
1000 PMT
142148.386  FV 
CPT  PV 
Ans: –195,760.96
23. The $43.2 million figure is simply the sum of all the monthly payments over the 7 years.
That is,
36($400,000) + 48($600,000) = $43,200,000
The current economic value of the deal is the present value of all the payments
discounted at 4% compounded annually. The corresponding interest rate per
payment interval is
i2  1  i c  1 = 1.040.083 – 1 = 0.00327374
The present value, 3 years from now, of the last 48 payments is
1  1  i  n 
PV  PMT 

i


 1  1.00327374 48 

= $600,000 

0
.
00327374


= $26,610,998.79
The present value today of the preceding amount and the
first 36 payments is
 1  1.00327374 36 
$26,610,998.79


+
$400,000
PV 
 0.00327374 
1.04 3


= $37,219,997
The current economic value of the deal is $37,220,000, rounded
to the nearest $1000.
Chapter 10: Annuities: Future Value and Present Value
4  I/Y 
 P/Y  12  ENTER 
 C/Y  1  ENTER 
48  N 
600,000 PMT
0  FV 
CPT  PV 
Ans: –26,610,998.79
Same I/Y, P/Y, C/Y
36  N 
26,610,998.79  FV 
400,000 PMT
CPT  PV 
Ans: –37,219,997
315
Checkpoint Questions (Section 10.5)
1. a. False. The focal date is at the end of the last payment interval. Since payments are at
the beginning of each payment interval, the focal date is one payment interval after the
final payment.
b. True. The focal date is at the beginning of the first payment interval. Since payments
are at the beginning of each payment interval, the focal date coincides with the first
payment.
c. False. The book value of long-term lease liability is equal to the PV of the remaining
lease payments.
Exercise 10.5
Basic Problems
1. a. Given: PMT = $100, n = 300, i =
6%
12
= 0.5%
BGN mode
 1  i n  1
FV (due) = PMT 
  1 + i 
i


300
 1.005
 1
= $100 
 (1.005)
0.005


= $69,645.89
6  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
300  N 
0  PV 
100 + / – PMT
CPT  FV 
Ans: 69,645.89
b. If instead, i =
8%
12
= 0.6 %
 1.00 6 300  1
 1.00 6
FV (due) = $100 

0.00 6 


= $95,736.66

BGN mode

3. a. Given: PMT = $1000, n = 2(25) = 50, i =
1  1  i  n 
PV (due) = PMT 
  1 + i 
i


 1  1.015  50 
 (1.015)
= $1000 

0
.
015


= $35,524.68
Same P/Y, C/Y
Same N, PV, PMT
8  I/Y 
CPT  FV 
Ans: 95,736.66
3%
2
= 1.5%
BGN mode
3  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
50  N 
1000 PMT
0  FV 
CPT  PV 
Ans: –35,524.68
(continued)
316
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 10.5 (continued)
BGN mode
b. If instead, i =
4%
2
Same P/Y, C/Y, N
Same PMT, FV
4  I/Y 
= 2%
 1  1.02  50 
 (1.02)
PV (due) = $1000 

 0.02 
= $32,052.08
Intermediate Problems
5. a. Given: PMT = $2000, n = 38, i =
8%
1
CPT  PV 
Ans: –32,052.08
= 8%
 1  i n  1
FV (due) = PMT 
  1 + i 
i


 1.08 38  1
 (1.08)
= $2000 

 0.08 
= $475,882.44
b.
If instead, i =
8%
4
=2% , c 
4
1
= 4;
and i2  1  i   1 = 1.02  – 1 = 0.082432160
c
4
 1.082432160 38  1
 1.082432160 
FV (due) = $2000 

 0.082432160 
= $506,532.59
7. Given: PMT = $2000 every 6 months.
For the first 11 years, j = 8% compounded semiannually.
We have a simple annuity due with
n = 2(11) = 22, i = 4%
 1  i n  1
FV (due) = PMT 
  1 + i 
i


 1.04 22  1
 (1.04)
= $2000 

0.
04


= $71,235.777
For the next 14 years, j = 7% compounded semiannually.
We have a simple annuity due with
n = 2(14) = 28, i = 3.5%
The overall future value is
 1.035 28  1
 (1.035)
$71,235.777( 1.035 28 ) + $2000 
 0.035 
= $186,649.985 + $95,821.599
= $282,471.58
Giorgio’s RRSP is worth $282,471.58 today.
Chapter 10: Annuities: Future Value and Present Value
BGN mode
8  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
38  N 
0  PV 
2000 + / – PMT
CPT  FV 
Ans: 475,882.44
BGN mode
Same I/Y, P/Y, N, PMT,
PV
 C/Y  4  ENTER 
CPT  FV 
Ans: 506,532.59
BGN mode
8  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
22  N 
0  PV 
2000 + / – PMT
CPT  FV 
Ans: 71,235.777
BGN mode
Same C/Y, P/Y, PMT
7  I/Y 
28  N 
71235.777 + / –  PV 
CPT  FV 
Ans: 282,471.58
317
Exercise 10.5 (continued)
9. a. Given: PMT = $500, n = 4(6.5) = 26, i =
Compoundin gs per year
c=
=
Payments per year
1
4
7.6%
1
= 7.6%,
= 0.25
i2  1  i c  1 = 1.076 0.25 – 1 = 0.0184813196
 1.0184813196  1
 (1.0184813196)
FV (due) = $500 
0.01848131
96


= $16,803.44
Amount in mutual fund will be $16,803.44.
BGN mode
7.6  I/Y 
 P/Y  4  ENTER 
 C/Y  1  ENTER 
26  N 
26
0  PV 
500 + / – PMT
CPT  FV 
Ans: 16,803.44
b. Earnings = $16,803.44  26($500) = $3803.44
11.
𝑗
Given: PMT = $75, n = 24, i = 𝑚 =
6%
12
= 0.5%
 1  i n  1
FV (due) = PMT 
  1 + i 
i


 1.005 24  1
 (1.005)
= $75 

 0.005 
= $1916.93
BGN mode
6  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
24  N 
75 + / – PMT
0  PV 
CPT  FV 
Ans: 1,916.93
Over the 24 months, you will have deposited a total
of $75 x 24 = $1800 of principal. Therefore, you will have earned
$1916.93 – $1800 = $116.93 of interest.
13. a. The initial long-term liability reported is the present value of all lease payments
discounted at the interest rate the firm would pay to borrow money. The lease
payments form a simple annuity due having
PMT = $2100, n = 4(5) = 20, and i = 84% = 2%
BGN mode
1  1  i  n 
PV (due) = PMT 
  1 + i 
i


 1  1.02  20 
 (1.02)
= $2100 

 0.02 
= $35,024.77
The initial lease liability will be $35,024.77.
8  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
20  N 
0  FV 
2100 + / – PMT
CPT  PV 
Ans: 35,024.77
b. The liability remaining at the end of the 4th year will be the present value
of the remaining 4 payments. That is,
 1  1.02  4 
BGN mode
 (1.02)
Lease liability = $2100 

Same I/Y, P/Y, C/Y, FV
 0.02 
Same PMT
= $8156.15
4  N 
CPT  PV 
Ans: 8,156.15
318
Fundamentals of Business Mathematics in Canada, 2/e
Exercise 10.5 (continued)
15. The lease consists of 36 beginning of month payments
of $575, as well as a payment of 0.20($20,000) = $4000 at
the end of the 3 year period. The initial lease liability is the present
value of all the payments, including the final lump sum payment,
discounted at the cost of borrowing.
The series of $575 payments form a general annuity due having
PMT = $575, n = 12(3) = 36, and i = 84% = 2%
c=
Compoundings per year
=
Payments per year
4
12
= 0. 3
i2  1  i c  1 = 1.020.3 – 1 = 0.00662271
BGN mode
 1  1.00662271 36 
 (1.00662271)
PV of lease payments = $575 

0
.
00662271


= $18,485.161
8  I/Y 
 P/Y  12  ENTER 
 C/Y  4  ENTER 
36  N 
575 + / – PMT
For the lump sum, PV of the $4000 lump sum
n = 4(3) = 12, and i = 84% = 2%
4000 + / –
 FV 
CPT  PV 
PV= FV 1  i 
= $4000(1+0.02)-12
=$3153.973
n
Ans: 21,639.13
Hence, the total lease liability is $18,485.161 + $3153.973 = $21,639.13. By buying rather
than leasing, one will save $21,639.13 - $20,000 = $1639.13.
17. For the monthly payment option, PMT = $38.50, n = 12(5) = 60, i =
1  1  i  n 
PV (due) = PMT 
  1 + i 
i


4.8  I/Y 
 1  1.004

 (1.004)
= $38.50 

0.004


= $2058.28
For the annual payment option, PMT = $455, n = 5,
i = 4.8%
= 0.4%,
12
Compoundings per year 12
= 1 = 12
Payments per year
= 0.4%,
BGN mode
 60
c=
4.8%
12
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
60  N 
38.50 + / – PMT
0  FV 
CPT  PV 
Ans: 2058.28
and
i2  1  i c  1 = 1.004 12 – 1 = 0.0490702075
 1  1.049070208  5 
 (1.049070208)
PV (due) = $455 

0
.
049070208


= $2071.90
Rino will save $2071.90  $2058.28 = $13.62
by choosing the monthly payment option.
BGN mode
4.8  I/Y 
 P/Y  1  ENTER 
 C/Y  12  ENTER 
5  N 
455 + / – PMT
0  FV 
CPT  PV 
Ans: 2071.90
Chapter 10: Annuities: Future Value and Present Value
319
19. The current economic value of Rosie’s offer is the present value of her payments.
The payments form a simple annuity due having
PMT = $1900, n = 5, i = 2.5%
1  1  i  n 
PV (due) = PMT 
  1 + i 
i


 1  1.025 5 
 (1.025)
= $1900 

 0.025 
= $9047.75
Rosie Senario’s offer is worth $9047.75 – $8500 =
$547.75 more in current dollars.
21. The amount of savings required is the present value of the
withdrawals.
Given: PMT = $40,000, n = 16, i = 6%
1  1  i  n 
PV (due) = PMT 
  1 + i 
i


 1 1.06 16 
 (1.06)
= $40,000 
 0.06 
= $428,489.96
Karsten must have $428,489.96 in savings at age 65.
BGN mode
5  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
5  N 
1900 PMT
0  FV 
CPT  PV 
Ans: –9,047.75
BGN mode
6  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
16  N 
40000 PMT
0  FV 
CPT  PV 
Ans: –428,489.96
Advanced Problems
23. The amount in the RRSP will be the future value of all the contributions.
4%
PMT = $2000, n = 10, i =
= 1%, c =
4
Compoundings per year
4
=
=4
Payments per year
1
i2  1  i   1 = 1.01 – 1 = 0.04060401
The amount in the RRSP after 10 years is
 1  i n  1
FV (due) = PMT 
 1 + i 
i


c
4
 1.0406040110  1
 (1.04060401)
= $2000 

0
.
04060401


= $25,057.31
BGN mode
4  I/Y 
 P/Y  1  ENTER 
 C/Y  4  ENTER 
10  N 
2000 + / – PMT
0  PV 
CPT  FV 
Ans: 25,057.31
The amount after 25 years is the future value of $25,057.31 an additional 15 years later
plus the future value of the last 15 years' contributions. That is,
320
Fundamentals of Business Mathematics in Canada, 2/e
 1.0406040115  1
 (1.04060401) + $25,057.31 1.0406040115
Amount = $4000 

0
.
04060401


= $129,243.10
BGN mode
Same I/Y, P/Y, C/Y
15  N 
4000 + / – PMT
25,057.31 + / –  PV 
CPT  FV 
Ans: 129,243.10
25. Compare the present values of the alternative payment
streams to determine the lower cost policy. Each payment
stream forms an ordinary annuity due with i = 4.8%
= 0.4%.
12
For the Sun Life option, PMT = $51.75/month, n = 120
 1  1.004 120 
 (1.004)
PV(due) = $51.75 
0.004


= $4944.02
For the Atlantic Life option,
PMT = $44.25/month for the first 5 years, and
PMT = $60.35/month for the subsequent 5 years.
The present value, 5 years from now, of the
final 5 years’ payments is
 1 1.004 60 
 (1.004)
PV(due) = $60.35 
 0.004 
= $3226.423
BGN mode
4.8  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
120  N 
51.75 PMT
0  FV 
CPT  PV 
Ans: –4944.02
BGN mode
Same I/Y, P/Y, C/Y
Same FV
60  N 
60.35 PMT
CPT  PV 
Ans: –3226.423
The present value today of all the premiums is
 1 1.004 60 
$3226.423

 (1.004)
+
$44.25
0.004
1.004 60


= $4904.90
Bram will save $39.12 by choosing the Atlantic Life policy.
BGN mode
Same I/Y, P/Y
Same C/Y, N
44.25 PMT
3226.423  FV 
CPT  PV 
Ans: –4904.897
Chapter 10: Annuities: Future Value and Present Value
321
Review Problems
Intermediate Problems
1. a. PMT = $1000, n = 2(20) = 40, i = 8.52% = 4.25%
 1  i n  1
FV  PMT 

i


 1.0425 40  1

= $1000 
 0.0425 


= $100,822.83
8.5  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
40  N 
0  PV 
1000 + / – PMT
Same I/Y, PV
 P/Y  1  ENTER 
(making C/Y = P/Y=1)
20  N 
b. PMT = $2000, n = 20, i = 8.5%
 1.085 20  1

FV = $2000 
 0.085 


= $96,754.03
CPT  FV 
Ans: 100,822.83
2000 + / – PMT
CPT  FV 
Ans: 96,754.03
3. a. Initial liability = Present value of all lease payments
With PMT = $1900, i = 8.25%
= 0.6875%, n = 12(5) = 60
12
 1  1.006875  60
Initial liability = $1900 
0.006875

= $93,794.81

 (1.006875)


BGN mode
8.25  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
60  N 
1900 + / – PMT
0  FV 
CPT  PV 
b.
Liability after the first year = Present value of the remaining
48 payments
 1  1.006875  48 
 (1.006875)
= $1900 

0.006875


= $77,987.28
The reduction in the liability during the first year will be
$93,794.81 – $77,987.28 = $15,807.53
5. a. PMT = $1000, n = 12(12) = 144, i =
6%
12
BGN mode
Same I/Y, P/Y, C/Y, PMT, FV
48  N 
CPT  PV 
Ans: 77,987.28
= 0.5%
1  1  i n 
PV  PMT 

i


 1  1.005 144 

= $1000 

0
.
005


= $102,474.74
6  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
144  N 
1000  PMT
0  FV 
CPT  PV 
(continued)
322
Ans: 93,794.81
Ans: -102,474.74
Fundamentals of Business Mathematics in Canada, 2/e
Review Problems (continued)
5. b. PMT = $1000, n = 12(12) = 144, i =
c=
6%
4
= 1.5%
Compoundings per year
4
=
= 0. 3
Payments per year
12
i2  1  i c  1 = 1.0150.3 – 1 = 0.004975206
1  1  i  n 
PV  PMT 

i


 P/Y  12  ENTER 
 C/Y  4  ENTER 
144  N 
1000 PMT
 1  1.004975206 144 

= $1000 

0.004975206


= $102,636.61
7.
6  I/Y 
PMT = $500, n = 12(20) = 240, i =
9%
12
0  FV 
CPT  PV 
Ans: - 102,636.61
= 0.75%
 1.0075  1
 (1.0075)
FV (due) = $500 

0
.
0075


= $336,448.01
240
BGN mode
9  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
240  N 
500 + / – PMT
0  PV 
CPT  FV 
Ans: 336,448.01
9. a. The value placed on half of the partnership is the present value of Dr. Wilson's
payments discounted at 7% compounded semiannually.
7
Given: PMT = $714.60, n = 12(15) = 180,
7%
2
i=
= 3.5%, c =
i2  1  i   1 = 1.035 
1  1  i  n 
PV  PMT 

i


c
0.1 6
2
12
= 0.1 6
– 1 = 0.0057500395
 1  1.0057500395 180 

= $714.60 


0
.
0057500395


= $80,000.05
The implied value of the partnership was 2($80,000) = $160,000.
 I/Y 
 P/Y  12  ENTER 
 C/Y  2  ENTER 
180  N 
714.60 + / – PMT
0  FV 
CPT  PV 
Ans: 80,000.05
b. Total interest = 180($714.60) – $80,000 = $48,628.
Chapter 10: Annuities: Future Value and Present Value
323
11. a. The original amount of the loan is the present
value of all the payments.
1  1  i  n 
Original loan = PMT 

i


 1  1.007 180 

= $587.33 


0
.
007


= $59,999.80
b. Balance after 7½ years
= Present value of the remaining payments
 1  1.007 90 

= $587.33 

 0.007 
= $39,119.37
8.4  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
180  N 
587.33 + / – PMT
0  FV 
CPT  PV 
Ans: 59,999.80
Same I/Y, P/Y, C/Y
Same PMT, FV,
90  N 
CPT  PV 
Ans: 39,119.37
13. a. The payments form an annuity due having
PMT = $1000, n = 2(20) = 40, and i = 3.52% = 1.75%
 1.0175 40  1
 (1.0175)
FV(due) = $1000 

 0.0175 
= $58,235.73
b. With PMT = $2000, n = 20, and i = 3.5%
 1.035 20  1
 (1.035)
FV(due) = $2000 

 0.035 
= $58,538.94
BGN mode
3.5  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
40  N 
1000 + / –
PMT
0  PV 
CPT  FV 
Ans: 58,235.73
BGN mode
3.5  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
20  N 
2000 + / –
PMT
0  PV 
CPT  FV 
Ans: 58,538.94
324
Fundamentals of Business Mathematics in Canada, 2/e
Review Problems (continued)
15. The payments form a general annuity due, having
PMT = $50, n = 12(19) = 228,
i = 84% = 2%, c  124  0. 3 , and
i2  1  i   1 = 1.02
c
0. 3
– 1 = 0.00662271
 1.00662271228  1
 (1.00662271)
FV(due) = $50 

 0.00662271 
= $26,630.79
17. The payments represent a simple annuity due, with:
PMT = $5000, n = 1(25) = 25, i = 61% =6%,
 1  i n  1
FV (due) = PMT 
  1 + i 
i


 1.06 25  1
 (1.06)
= $5000 

 0.06 
= $290,781.91
19. Choose the payment plan with the lower economic value.
For the monthly premiums, PMT = $33.71, n = 12,
%
and i = 3.75
= 0.3125%
12
 1  1.003125  12 
 (1.003125)
PV(due) = $33.71 

 0.003125 
= $397.66
This is $397.66 – $387.50 = $10.16 more than the single
payment falling on the focal date. Therefore, choose the
single annual payment plan.
BGN mode
8  I/Y 
 P/Y  12  ENTER 
 C/Y  4  ENTER 
228  N 
50 + / –
PMT
0  PV 
CPT  FV 
Ans: 26,630.79
BGN mode
6  I/Y 
 P/Y  1  ENTER 
(making C/Y = P/Y = 1)
25  N 
5000 + / – PMT
0  PV 
CPT  FV 
Ans: 290,781.91
BGN mode
3.75  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
12  N 
33.71 + / – PMT
0  FV 
CPT  PV 
Ans: 397.66
21. The fair market value of the contract is the combined present value of the 33 payments of
$900 plus the present value of the final lump sum payment of $37,886. The 33 payments
form an ordinary simple annuity with:
.2%
PMT = $900, n = 12(2.75) = 33, i = 712
= 0.6%,
1  1  i  n 
n
Fair market value = PMT 
 + FV 1  i 
i


 1  1.006 33 
 + $37,886 1.00633
= $900 

 0.006 
= $26,871.697 + $31,098.926
= $57,970.62
Chapter 10: Annuities: Future Value and Present Value
7.2  I/Y 
 P/Y  12  ENTER 
(making C/Y = P/Y = 12)
33  N 
900 PMT
37886  FV 
CPT  PV 
Ans: -57,970,62
325
Review Problems (continued)
23. The appropriate price to pay is the present value
of the payments discounted at the required rate
of return. The present value, 5 years from now,
of the last 7 years' payments is
1  1  i  n 
PV  PMT 

i


 1  1.0175  28 

= $1500 

0.0175


= $32,980.432
Today's present value of this amount
and the first 5 years' payments is
 1  1.015  20
 20
$32,980.432 1.015 
+ $1500 
0.015

7  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
28  N 
1500 PMT
0  FV 
CPT  PV 
Ans: 32,980.432
Same P/Y, C/Y, PMT
20  N 




= $24,486.995 + $25,752.958
= $50,239.95
6  I/Y 
32980.432  FV 
CPT  PV 
Ans:  50,239.95
The appropriate price to pay is $50,239.95.
25. The amount required to purchase the annuities is the present value, on the
purchase date, of all payments.
Step 1:
Calculate the present value, 5 years from now, of the $2500-payment
annuity. The $2500 payments form a general annuity due having
4 = 0. 3 , and
PMT = $2500, n = 12(15) = 180, i = 3.64% = 0.9%, c = 12
i2  1  i   1 = 1.009 – 1 = 0.002991045
c
0.3
 1  1.002991045  180 
 (1.002991045)
PV(due) = $2500 

0.002991045


= $348,612.482
Step 2:
Calculate today's present value of the amount from Step 1 and
the $4000-payment annuity.
The $4000 payments form an ordinary simple annuity having
PMT = $4000, n = 4(5) = 20, i = 0.9%
 1  1.009  20 
$348,612.482


PV(Today) =
+
$4000
 0.009 
1.009 20


= $291,420.1706 + $72,914.2138
BGN mode
3.6  I/Y 
 P/Y  12  ENTER 
 C/Y  4  ENTER 
180  N 
2500 PMT
0  FV 
CPT  PV 
Ans: -348,612.482
END mode
Same I/Y
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
20  N 
4000 PMT
348612.482  FV 
CPT  PV 
= $364,334.38
Ans: 364,334.38
The amount required to purchase the annuity is $364,334.38.
326
Fundamentals of Business Mathematics in Canada, 2/e
27.
The amount is the RRSP will be the future value of all payments. For the past 7 years,
PMT = $3000, n = 2(7) = 14, i = 9%
= 4.5%
2
The amount currently in Charlene's RRSP is
 1  i n  1
FV  PMT 

i


 1.045 14  1 

= $3000 
 0.045 


= $56,796.328
9  I/Y 
 P/Y  2  ENTER 
(making C/Y = P/Y = 2)
14  N 
0  PV 
3000 + / – PMT
CPT  FV 
Ans: 56,796.328
For the next 5 years,
PMT = $2000, n = 4(5) = 20, i = 7.54% = 1.875%
The amount in the RRSP 5 years from now will be
 1.01875 20  1

$56,796.328 1.01875 20 + $2000 
 0.01875 


7.5  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
20  N 
56,796.328 + / –  PV 
2000 + / – PMT
CPT  FV 
= $130,346.18
Ans: 130,346.18
Advanced Problems
29. The amount (future value) in Dr. Krawchuk’s RRSP
when he left general practice was
 1  i n  1
FV  PMT 

i


 1.0125 24  1

= $2000 

 0.0125 
= $55,576.168
After an additional 2.5 years with no
further contributions, this amount grew to
FV  PV 1  i 
n
5  I/Y 
 P/Y  4  ENTER 
(making C/Y = P/Y = 4)
24  N 
0  PV 
2000 + / – PMT
CPT  FV 
Ans: 55,576.168
Same I/Y, P/Y, C/Y
10  N 
55576.168 + / –  PV 
0 PMT
= $55,576.168 1.0125 
= $62,927.27
10
Chapter 10: Annuities: Future Value and Present Value
CPT  FV 
Ans: 62,927.27
327
31. Amount in the RRSP = Future value of all contributions.
For the first 15 years' contributions,
PMT = $2500, n = 2(15) = 30, i =
8%
4
= 2%, c =
i2  1  i c  1 = 1.02  – 1 = 0.0404
The amount in the RRSP after 15 years will be
 1  i n  1
FV (due) = PMT 
  1 + i 
i


4
=2
2
2
BGN mode
8  I/Y 
 P/Y  2  ENTER 
 C/Y  4  ENTER 
 1.0404 30  1
 (1.0404)
= $2500 

 0.0404 
= $146,855.472
30  N 
2500 + / – PMT
0  PV 
CPT  FV 
Ans: 146,855.472
For the subsequent 10 years' contributions,
PMT = $3000, n = 2(10) = 20, i = 2%, c =
i2  1  i c  1 = 1.02  – 1 = 0.0404
4
=2
2
2
The future value, 25 years from now, of all the payments will be
 1.0404 20  1
20
  1 + 0.0404 
$146,855.472 1.0404  + $3000 

 0.0404 
= $324,262.707 + $93,330.035
= $417,592.74
BGN mode
Same I/Y, P/Y, C/Y
20  N 
146,855.472 + / –  PV 
3000 + / – PMT
CPT  FV 
Ans: 417,592.74
33. The current economic value of the award is the present value, on the date of the award, of
all payments.
Step 1:
Calculate the present value, 5 years from now, of the $1000-payment
annuity. The $1000 payments form an ordinary simple annuity with
PMT = $1000, n = 12(10) = 120, i = 612% = 0.5%,
6  I/Y 
1  1  i  n 
 P/Y  12  ENTER 
PV  PMT 

(making C/Y = P/Y = 12)
i


120  N 
 1  1.005  120
= $1000 
0.005

= $90,073.453




Step 2:
Calculate today's present value of the amount from Step 1 and
the $800-payment annuity. The $800 payments form an ordinary
simple annuity having
PMT = $800, n = 12(5) = 60, i = 0.5%
The combined present value of the amount from Step 1 and
the $800-payment annuity is:
 1  1.005  60 
 60

$90,073.453 1.005 
+ $800 

 0.005 
= $66,777.954 + $41,380.449
328
1000 PMT
0  FV 
CPT  PV 
Ans: -90,073.453
Same I/Y, P/Y, C/Y
60  N 
800 PMT
90073.453  FV 
CPT  PV 
Ans: -108,158.40
Fundamentals of Business Mathematics in Canada, 2/e
= $108,158.40
The current economic value of the court award is $108,158.40.
35. The total amount in the RRSP after 30 years will be the future value of all contributions.
The future value, 10 years from now, of the $4000 contributions will be
 1.0825 10  1
BGN mode
 (1.0825)
FV(due) = $4000 
 0.0825 
8.25  I/Y 


 P/Y  1  ENTER 
= $63,476.431
(making C/Y = P/Y = 1)
10  N 
0  PV 
4000 + / – PMT
CPT  FV 
Ans: 63,476.431
The future value, 30 years from now (an additional n = 20
years), of this amount and the $6000 contributions will be
 1.0825 20  1
 (1.0825)
$63,476.431 1.082520 + 6000 
 0.0825 


= $615,447.79
BGN mode
Same I/Y, P/Y, C/Y
20  N 
63,476.431 + / –  PV 
6000 + / – PMT
CPT  FV 
Ans: 615,447.79
Chapter 10: Annuities: Future Value and Present Value
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