Download Assignment 3 Solutions

Assignment 3 Solutions
Timothy Vis
January 30, 2006
3-1-6 P = 900, r = 10%, t = 9 months, I =?. Given I = P rt, we have
9
) = 67.50
I = (900)(0.10)( 12
3-1-8 I = 40, P = 400, t = 4 years, r =?. Given I = P rt, we have 40 =
40
(400)r(4), so that r = (400)(4)
= 0.025 = 2.5%.
3-1-10 P = 6000, r = 6%, t = 8 months, A =? Given A = P (1 + rt), we
8
)) = 6240.
have A = 6000(1 + (0.06)( 12
3-1-12 A = 8000, r = 12%, t = 7 months, P =? Given A = P (1 = rt), we
8000
7
)), so that P = 1+(0.12)(
= 7476.64.
have 8000 = P (1 + (0.12)( 12
7
)
12
3-1-14 Solve I = P rt for P .
I
I
rt
I
rt
= P rt
P rt
=
rt
= P
3-1-16 Solve A = P + P rt for r.
A = P + P rt
A−P
A−P
Pt
A−P
Pt
= P rt
P rt
=
Pt
= r
3-1-22 Here we are given r = 18%, P = 835, t = 2 months, and asked to
find I.
I
= P rt
1
= (835)(0.18)(
2
)
12
= 25.05
3-1-24 Here we are given P = 10, 000, t = 6 months, r = 6.5% and asked
to find A.
A = P (1 + rt)
= 10, 000(1 + (0.065)(
6
))
12
= 10, 325
3-1-26 Here we are given A = 3097.50, t = 5 months, P = 3000 and asked
to find r.
A
= P (1 + rt)
3097.50 = 3000(1 +
3097.50
3000
3097.50
−1
3000
12 3097.50
(
− 1)
5
3000
0.078
7.8%
= 1+
=
5
r)
12
5
r
12
5
r
12
= r
= r
= r
3-1-32 Here we are given t = 33 days, A = 1000, P = 996.16 and asked to
find r.
A = P (1 + rt)
1000 = 996.16(1 +
1000
996.16
1000
−1
996.16
1 1000
(
− 1)
11 996.16
0.04205
4.205%
= 1 + 11r
= 11r
= r
= r
= r
2
33
r)
363
3-1-36 This problem really takes two steps. First we need to figure out the
final value of the note after the 180 days. Then we use this final value to determine the effective rate earned by the third party. For the first part, we have
P = 10, 000, t = 180 days, r = 7%, and we must find A.
A = P (1 + rt)
A = (10, 000)(1 + (0.07)(
180
))
360
A = 10, 350
With this information we go on to the second part. We now have P = 10, 124,
t = 180 − 60 = 120 days, A = 10, 350, and we must find r.
A = P (1 + rt)
10, 350 = 10, 124(1 +
10, 350
10, 124
10, 350
−1
10, 124
10, 350
− 1)
3(
10, 124
0.06697
120
r)
360
1
= 1+ r
3
1
=
r
3
= r
= r
6.697% = r
3-1-38 The principal P is the cost of the stock plus the broker’s commission
on that purchase. The cost of the stock is (450)(21.40) = 9630, so the commission, using the commission schedule, is 37 + (0.014)(9630) = 171.82. Adding
these gives a principal P = 9630 + 171.82 = 9801.82. The final amount A is the
selling price of the stock minus the broker’s commission on that sale. The selling
price of the stock is (450)(24.60) = 11, 070, so the commission, using the commission schedule, is 107 + (0.007)(11, 070) = 184.49. This gives a final amount
of 11, 070 − 184.49 = 10, 885.51. To make things easier, we find the interest
earned as A − P = 10, 885.51 − 9801.82 = 1083.69. So, we have I = 1083.69,
t = 26 weeks, P = 9801.82 and we need to find r.
I
= P rt
1083.69 = (9801.82)r(
1083.69
9801.82
1083.69
2
9801.82
=
1
r
2
= r
3
26
)
52
0.22112 = r
22.112% = r
3-2-4 We have P = 10, 000, i = 0.08, n = 30 and we need to find A
= P (1 + i)n
A
= 10, 000(1 + 0.08)30
= 100, 626.57
3-2-6 We have A = 1000, i = 0.015, n = 60 and we need to find P
A
P
= P (1 + i)n
A
=
(1 + i)n
1000
=
(1 + 0.015)60
= 409.30
3-2-18 In all cases, we have P = 2000, r = 0.07, t = 5 years. Compounded
annually, we get i = 0.07, n = 5.
A = P (1 + i)n
= 2000(1 + 0.07)5
= 2805.10
Also, I = A − P = 2805.10 − 2000.00 = 805.10.
Compounded quarterly, we get i = 0.07
4 = 0.0175 and n = (5)(4) = 20.
A = P (1 + i)n
= 2000(1 + 0.0175)20
= 2829.56
Then, I = A − P = 2829.56 − 2000.00 = 829.56.
Compounded monthly, we get i = 0.07
12 and n = (5)(12) = 60.
A = P (1 + i)n
= 2000(1 +
= 2835.25
4
0.07 60
)
12
Then, I = A − P = 2835.25 − 2000.00 = 835.25.
3-2-28 This is just annual percentage yield. In the first case, we have r = 0.06,
m = 12.
r
AP Y = (1 + )m − 1
m
0.06 12
= (1 +
) −1
12
= 0.06168
= 6.168%
In the second case, we have r = 0.08, m = 2.
AP Y
r m
) −1
m
0.08 2
= (1 +
) −1
2
= 0.08160
= (1 +
= 8.16%
3-2-40 This is essentially the same sort of problem; here we have r = 0.052,
t = 8, A = 160, 000, and we need to find P . Since the compounding is annual,
we also have i = r and n = t.
A = P (1 + i)n
A
P =
(1 + i)n
160, 000
=
(1 + 0.052)8
= 106, 658.15
3-2-42 Here we have r = 0.0133, A = 10, P = 6, and we need to find t.
Since the compounding is annual, we have i = r and n = t. Note that we can
use billions of people rather than people, hence the 10 and 6, with the same
outcome.
A = P (1 + i)n
A
= (1 + i)n
P
A
ln
= ln(1 + i)n
P
ln A − ln P = n ln(1 + i)
5
ln A − ln P
ln(1 + i)
ln 10 − ln 6
=
ln(1 + 0.0133)
= 39
n =
3-2-44 To determine the better investment, we consider the effective annual
rates of each. Clearly 8.3% compounded annually has an annual percentage
yield of 8.3%. It remains to find the APY of 8% compounded quarterly.
r
AP Y = (1 + )m − 1
m
0.08 4
) −1
= (1 +
4
= 0.0824
= 8.24%
The annual yield is better for the annual rate of 8.3%.
3-2-48 Here we have P = 15, 000, A = 20, 000, r = 0.07, m = 4, which gives us
i = 0.0175, and we need to find n.
A
A
P
ln A − ln P
= P (1 + i)n
= (1 + i)n
= n ln(1 + i)
ln A − ln P
n =
ln(1 + i)
ln 20, 000 − ln 15, 000
=
ln(1 + 0.0175)
= 17
So 17 quarters are needed or 4.25 years for the desired growth to occur.
3-2-50 Essentially, we just plug into the various formulas with P = 1, r = 0.02,
t = 2010. With simple interest, we have:
A
= P (1 + rt)
= 1(1 + (0.02)(2010))
= 41.20
With annual compounding, we have i = 0.02, and n = 2010, so that:
A = P (1 + i)n
= 1(1 + 0.02)2 010
= 193, 350, 474, 163, 852, 184.92
6
This is about 193 quadrillion dollars.
3-2-56 To solve this, we form the appropriate equations for compounding in0.05 n
n
terest on each investment: A = 4800(1 + 0.08
12 ) and A = 5000(1 + 12 ) . Set
these equal to y1 and y2 and graph in a window. Find the intersection point of
the graphs at n = 16.42. You need to do this graphically. This corresponds to
17 months as the point when the 4800 dollar investment at the higher interest
rate becomes more valuable.
3-2-64 Here we have A = 20, 000, r = 0.04194, t = 10. Since m = 1, this
gives i = 0.04194, n = 10.
A = P (1 + i)n
A
P =
(1 + i)n
20, 000
=
(1 + 0.04194)10
= 13, 261.81
3-2-66 Here we have A = 40, 000, P = 32, 000, t = 5. Since m = 1, this
gives n = 5 and i = r.
A = P (1 + i)n
A
= (1 + i)n
P
(
A 1
)n
P
= 1+i
A 1
)n − 1
P
40, 000 1
= (
)5 − 1
32, 000
= 0.04564
= 4.564%
i = (
7