Honors MAT111 |
Test II - Quantitative |
Fall 2002 |
A = 10,000(1 + .06)5 = $13382.26
A = $10,000(1 + .05875/12)60 = $13,404.87
A = $10,000(1 + .0579/365)5 x 365 = $13,357.29
The best choice is b.
A = $200[(1 + .065/12)72 – 1] / (.065 / 12) = $17,554.23
i. 6% annual rate?
A1 = $200[(1 + .06/12)12 x 49 – 1] / (.06 / 12) = $711,111.31
A2 = $17,554.23(1 + .06/12) 12 x 49 = $329,629.52
Total = $1,040,740.83
ii. 7% annual rate?
A1 = $200[(1 + .07/12)12 x 49 – 1] / (.07 / 12) = $1,013,844.75
A2 = $17,554.23(1 + .07/12) 12 x 49 = $536,641.09
Total = $1,550,485.84
iii. 8% annual rate?
A1 = $200[(1 + .08/12)12 x 49 – 1] / (.08/ 12) = $1,462,471.13
A2 = $17,554.23(1 + .08-/12) 12 x 49 = $873,306.05
Total = $2,335,777.18
i. Five years
Pmt = $5,000(.0999/12) / (1 – (1 + .0999/12)-60) = $106.21
ii. Three years
Pmt = $5,000(.0999/12) / (1 – (1 + .0999/12)-60) = $161.31
iii. One year
Pmt = $5,000(.0999/12) / (1 – (1 + .0999/12)-60) = $439.56
i. $6,372.60
ii. $5,807.16
iii. $5,274.67
.33 x 2200 = $726
i. 7%
P = $726[1 – (1 + .07/12)-360)] / (.07/12) = $109,123.29
ii. 6.5%
P = $726[1 – (1 + .065/12)-360)] / (.065/12) = $114,861.05
iii. 6%
P = $726[1 – (1 + .06/12)-360)] / (.06/12) = $121,090.71
$261,360
$3,000 x .33 = $990
i. P = $990[1 – (1 + .07/12)-360)] / (.07/12) = $148,804.49
ii. P = $990[1 – (1 + .065/12)-360)] / (.065/12) = $156,628.71
iii. P = $990[1 – (1 + .06/12)-360)] / (.06/12) = $165,123.70
What amount of your payment on a $100,000 mortgage at 6.5% goes towards principal in the following cases?
Pmt = $100,000(.065/12) / (1 – (1 + .065/12)-360) = $632.07
Pmt = $100,000(.065/12) / (1 – (1 + .065/12)-180) = $871.11
iv. First payment, 30-year loan?
Int $632.07 / (1+.065/12) = $628.66
Prin. $632.07 - $628.66 = $3.40
v. First payment, 15-year loan?
Int $871.11 / (1+.065/12) = $866.42
Prin. $871.11 - $866.42= $4.69
vi. 120th payment, 30-year loan?
Int $632.07 / (1+.065/12)120 = $330.55
Prin. $632.07 - $330.55 = $301.52
vii. 120th payment, 15-year loan?
Int $871.11 / (1+.065/12)120 = $455.56
Prin. $871.11 - $455.56 = $415.55
Rolling two dies and getting a sum of 0 or 13, 14, 15, etc.
Rolling two dies and getting a sum between 2 and 12 inclusive
Tossing a fair coin twice and getting exactly two heads, or getting exactly two tails
i. Exactly two heads. What is the probability of getting exactly 2 heads?
ii. Exactly three heads. What is the probability of getting exactly three heads?
Look at the chart to find 6/36 possibilities.
Sum of 3 or 11 = 2/36
Sum of 4 or 10 = 3/36
Sum of 5 or 8 = 5/36
Sum of 7 = 6/36
i. Two pieces?
1 – (15/20)2 = .4375
ii. Three pieces?
1 – (15/20)3 = .578125
1 – (.9)31 = .9618
P(Q or ©) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13
From the above solution, 4/13 = 30.77%. So, you’d expect to get a Queen or Heart 30 times out of 100, but looking at figure 7.6 on pg. 429 you shouldn’t be that confident about it.
The expected value of each bet is E.V.(Black) = (-$5)(18/38) + ($5)(16/38) = 26¢
(26¢ ) x (40 Bets) = $10.40
The expected value of each bet is E.V.(BlackJack) = (-$5)(.507) + ($5)(.493) = 7¢
(7¢ ) x (40 Bets) = $2.80
You would expect only about 600 cancer deaths. Or you could note that 1500/300,000 = .5%