MATH 150 Final Exam Solution: Mortgage, Annuity, Probability Analysis

This document presents a complete solution to a MATH 150 final exam, addressing various financial and probability concepts. The solution begins with a mortgage calculation, determining the down payment, mortgage amount, and monthly installments (EMI) over a 30-year period, along with the total interest paid. The second question analyzes two options for purchasing goods, comparing down payments, interest rates, and monthly installments to determine the best financial choice. The third question involves calculating the time required for an investment to earn a specific interest amount. The fourth question focuses on the future value of annuities, calculating the total value and interest earned over a 30-year period. The fifth question involves probability calculations, including the probability of selecting a cat, the odds of selecting a car, and probabilities related to specific animal breeds from a sample. The solutions provide detailed calculations and explanations for each problem, demonstrating a strong understanding of financial mathematics and probability principles.

MATH 150
FINAL EXAM
[DATE]
Grizli777
[Company address]

Question 1
a) Total purchase price of house = $ 375,000
It is required that 20% of the purchase price needs to be made as the down payment and the
loan can be taken for the remaining amount.
Down payment required to purchase house ¿ ( 20
100 )∗375,000=$ 75,000
b) Mortgage amount = Purchase price of the house – Down payment
Purchase price = $ 375,000
Down payment as computed in part a = $ 75,000
Hence, mortgage amount = $375,000 - $ 75,000 = $ 300,000
c) The relevant formula for computation of EMI (Equal Monthly Instalment) is given below.
EMI= [ P x R x ( 1+R ) N ]
[ ( 1+ R ) N −1 ]
For the given problem, P = $300,000, R = (3.4/12) = 0.283% per month, N= 30*12 = 360
months
Hence, EMI = 300000∗0.00283∗1.00283 363
1.0028360−1 =$ 1,330.44
d) Total amount paid by the buyer over a 30 year mortgage = $1,330.44 * 360 = $478,959.73
Since, $300,000 was the principal amount, hence the remaining amount paid to the bank
would be interest only.
Amount paid as interest over the 30 year mortgage period = $478,959.73 - $300,000 =
$178,959.73

Question 2
The analysis of the two options needs to be performed so as to answer the various questions.
Buying price of goods = $ 1,600
Option I
Down payment required = (20/100)*1,600 = $ 320
Hence, amount borrowed = $1,600 - $ 320 = $1,280
Interest paid over a two year period = 1280*(6.4/100)*2 = $163.84
Total amount to be paid during two years = Principal + Interest = $1280 + $163.84 =
$1443.84
Since the repayment would be made over a period of 24 months through equal monthly
instalments, hence monthly instalment = (1443.84/24) = $60.16
Option II
As there is no down payment required under this option, hence principal borrowed would be
$ 1,600
Interest paid over a three year period ¿ 1600∗( 6.5
100 )∗3=$ 312
Total amount to be paid during two years = Principal + Interest = $1600+ $312 = $1912
Since the repayment would be made over a period of 36 months through equal monthly
instalments, hence monthly instalment ¿ ( 1912
36 )=$ 53.11
a) Option I should be chosen if the customer intends for the finance charge to be lower. This
is because option I has total interest cost of $163.84 as against interest cost of $312 for
option II.

b) Option I should be chosen for the smallest total instalment price. For Option I, this is
$1443.84 as compared to $ 1,912 for Option II.
c) Option II should be chosen for the smallest monthly payment. The option I has a monthly
instalment of $ 60.16 while this is $ 53.11 for Option II.
d) Principal invested = $ 1,600
Interest rate = 2.5% p.a.
Interest amount to be earned = $ 250
Hence, 1600*(2.5/100)*t = 250
Solving the above, t = 6.25 years
Question 3
P = $15,000
R = 2.6%
N = 18 years
(a) Money at the end of 18 years for Maria
¿ 15000∗
( 1+ ( 2.6
100 ))
18
=15000∗1.58727616=$ 23809.1424
(b) Interest earned at the end of 18 years for Maria
¿ 23809.1424−15000=$ 8809.1424
(c) Money at the end of 18 years for John
¿ 15000∗
( 1+ ( 2.6
100 ))
18
=15000∗1.5873=$ 23809. 5
(d) Interest earned at the end of 18 years for John
¿ 23809.5−1500 0=$ 8809.5
Question 4

A. a) The requisite formula for future value of annuity is shown below.
F= P∗[ 1+I ]N−1
I
Here, P = $ 200
I = (2.2/12) = 0.1833% per month
N= 30 years or (30*12) = 360 months
Hence, total value of the annuity at the end of 30 years ¿ 200∗1.00183360−1
0.001833 =$ 101.849 .84
b) Amount = Principal + Interest
Total principal deposited over 30 years = 200*360 = $ 72,000
Interest earned over 30 year period = Amount at the end of 30 years – Principal deposited
over 30 years = $101.849.84 - $ 72,000 = $29,849.84
B. a) The requisite formula for future value of annuity is shown below.
F= P∗[ 1+ I ] N−1
I
Here, P = $ 1200
I = (2.5/2) = 1.25% per six month
N= 30 years or (30*2) = 60 periods of 6 months each
Hence, total value of the annuity at the end of 30 years ¿ 1200∗1.012 560−1
0.0125 =$ 106,289.40
b) Amount = Principal + Interest
Total principal deposited over 30 years = 1200*60 = $ 72,000
Interest earned over 30 year period = Amount at the end of 30 years – Principal deposited
over 30 years = $106,289.4- $ 72,000 = $34,289.41

Question 5
Total number of animals = 8+12+6+5+11 = 42
a) Total cats = 12+ 8 = 20
Hence, probability of selecting a cat ¿ ( 20
42 )= (10
21 )
b) Odds of selecting a cat= ( 10
21 )
1− (10
21 )= ( 10
21 )
11
21
=10/11
Hence, odds of selecting a car are 10:11.
c) Number of common cats in the sample = 12
Number of Labrador Retriever in the sample = 5
Hence, requisite probability ¿ 12+5
42 =( 17
42 )
d) Total number of dogs in the sample = 6+5+11 = 22
Dogs that do not belong to mixed breed = 6+5 = 11
Hence, requisite probability ¿ 11
22= 1
2