Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Grade 7 & 8 Math Circles
October 22/23, 2013
Finance Solutions
Problem Set
1. (a) S =?
P = 1200
r = 0.05
t=
7
12
S = P (1 + rt)
7
= 1200 1 + (0.05)
12
= 1235
The accumulated value is $1235.
(b) S =?
P = 8000
r = 0.125
t=4
S = P (1 + rt)
= 8000(1 + (0.125)(4))
= 12000
The accumulated value is $12000.
(c) S =?
P = 500
r = 0.1
t=
99
365
S = P (1 + rt)
99
= 500 1 + (0.1)
365
= 513.56
The accumulated value is $513.56.
1
(d) S =?
P = 750
r = 0.1325
t=
15
52
S = P (1 + rt)
15
= 750 1 + (0.1325)
52
= 778.67
The accumulated value is $778.67.
2. S = 3600
P =?
r = 0.08
t = 10
S = P (1 + rt)
S
P =
1 + rt
3600
=
1 + (0.08)(10)
= 2000
The principal/present value is $2000.
3. P = 100
I = 20
r =?
t=
10
12
I = P rt
I
P rt
=
Pt
Pt
I
r=
Pt
=
20
10
(100)
12
= 0.24
The annual simple interest rate is 24%.
Using the other equation:
2
P = 100
S = 120
r =?
t=
10
12
S = P (1 + rt)
S
P (1 + rt)
=
P
P
S
−1 = 1 + rt−1
P
S
−1
rt
P
=
t
t
S
−1
r= P
t
120
−1
= 100
10
12
= 0.24
The annual simple interest rate is 24%.
4. P = 3000
I = 60
r = 0.06
t =?
I = P rt
P rt
I
=
Pr
Pr
I
t=
Pr
=
60
(3000)(0.06)
= 0.333...
1
=
3
It will take
5. (a) S =?
1
of a year (4 months) to earn $60 interest.
3
P = 2000
r = 0.05
t=4
S = P (1 + r)t
= 2000(1 + 0.05)4
= 2431.01
3
The accumulated value is $2431.01.
(b) S =?
P = 100
r = 0.075
t = 25
S = P (1 + r)t
= 100(1 + 0.075)25
= 609.83
The accumulated value is $609.83.
6. (a) S = 7500
P =?
r = 0.08
t = 10
S = P (1 + r)t
S
P =
(1 + r)t
7500
=
(1 + 0.08)10
= 3473.95
The principal/present value is $3473.95.
(b) S = 25000
P =?
r = 0.045
t = 50
S = P (1 + r)t
S
P =
(1 + r)t
25000
=
(1 + 0.045)50
= 2767.74
The principal/present value is $2767.74.
7. S = 5000
P =?
r = 0.05
t=4
4
(a)
S = P (1 + rt)
S
P =
(1 + rt)
5000
=
(1 + (0.05)(4))
= 4166.67
You must invest $4166.67.
(b)
S = P (1 + r)t
S
P =
(1 + r)t
5000
=
(1 + 0.05)4
= 4113.51
You must invest $4113.51.
8. (a) S =?
P = 1000
r(52) = 0.13
m = 52
n = 52 × 3 = 156
n
r(m)
S =P 1+
m
156
0.13
= 1000 1 +
52
= 1476.26
The accumulated value is $1476.26.
(b) S =?
P = 500
m=2
r(2) = 0.04
n = 2 × 25 = 50
n
r(m)
S =P 1+
m
50
0.04
= 500 1 +
2
= 1345.79
The accumulated value is $1345.79.
5
9. (a) S = 6000
P =?
r(4) = 0.15
m=4
S=P
r(m)
1+
m
n = 4 × 10 = 40
n
S
P =
n
r(m)
1+
m
6000
=
40
0.15
1+
4
= 1376.03
The principal/present value is $1376.03.
(b) S = 25000
P =?
r(12) = 0.12
m = 12
n = 12 × 50 = 600
n
r(m)
S =P 1+
m
S
P =
n
r(m)
1+
m
25000
=
600
0.12
1+
12
= 63.84
The principal/present value is $63.84.
10. We arrange the data on the following diagram.
By finding the present values of both debts, we can find an equivalent payment that
6
can be paid today to settle both debts.
400
The present value of the $400 debt = 1
0.12
1+
4
= 388.35
500
The value at time 6 of the $500 debt = 6
0.1
1+
12
= 475.71
475.71
The present value of the $500 debt = 2
0.12
1+
4
= 448.40
Hence the present value of both debts = 388.35 + 448.40
= 836.75
The solution may also be expressed as
400
500
Present value of both debts = 1 + 6 2
0.12
0.12
0.1
1+
1+
1+
4
12
4
= 836.75
A payment of $836.75 today will repay the debts.
11. We arrange the data on the following diagram.
There are multiple ways to calculate X.
One way is to make sure the present value of the debt is equal to the present values of
7
the four payments at time 0:
present value of payments = present value of debt
200
X
X
X
1 + 2 + 3 + 4 = 1000
0.1
0.1
0.1
0.1
1+
1+
1+
1+
4
4
4
4
195.12 + 0.951814396X + 0.928599411X + 0.905950645X = 1000
195.12 + (0.951814396 + 0.928599411 + 0.905950645)X = 1000
195.12 + 2.786364452X = 1000
195.12 + 2.786364452X−195.12 = 1000−195.12
2.786364452X = 804.88
804.88
2.786364452X
=
2.786364452
2.786364452
X = 288.86
Another way is to make sure the accumulated value of the debt is equal to the accumulated values of the four payments at time 12:
accumulated value of payments = accumulated value of debt
3
2
1
4
0.1
0.1
0.1
0.1
+X 1+
+X 1+
+ X = 1000 1 +
200 1 +
4
4
4
4
215.38 + 1.050625X + 1.025X + X = 1103.81
215.38 + (1.050625 + 1.025 + 1)X = 1103.81
215.38 + 3.075625X = 1103.81
215.38 + 3.075625X−215.38 = 1103.81−215.38
3.075625X = 888.43
3.075625X
888.43
=
3.075625
3.075625
X = 288.86
So the second, third , and fourth payments are equal to $288.86.
8
12. (a)
Accumulated value after 10 years = 1000(1 + 0.04)3 (1 + 0.06)3 (1 + 0.07)4
= 1756.11
The accumulated value is $1756.11.
(b)
0.06
Accumulated value after 2 years = 500 1 +
12
12 52
0.13
1+
52
= 604.44
The accumulated value is $604.44.
13. (a)
1000
Present value =
(1 +
0.05)10 (1
= 210.83
The principal/present value is $210.83.
9
+
0.07)10
0.08
1+
2
10
(b)
Present value = 1000
156 24
0.13
0.06
1+
1+
52
12
= 600.97
The principal/present value is $600.97.
10