MATH_1191_Review_for Test 1V2_SOLUTIONS w explantn_W2024
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
Time Allowed: 3 Hours (180 Minutes) – Actual Test will be 2 Hours 15 Minutes with Fewer Questions
1.
Owen lent Hui $5,000 at 5% p.a. simple interest for 1 year and 3 months. Using the I = Prt formula, calculate the amount of interest charged at the end of the term.
Answer:
t
= 1 year and 3 months = 1 + = 1 + = years
I = Prt
= (5000) (0.05) (
) = $312.50
Therefore, he will receive an interest of $312.50 at the end of the time period.
2.
On December 25, 2016, Amy deposited $4750 at 5.5% p.a. in her savings account. How
much interest was earned and paid into Amy’s account on April 27, 2017?
Answer:
DBD: from Dec. 25, 2016 to April 27, 2017 = 123 days
P= $4750, r=5.5% p.a. = 0.055 p.a., t=123 days =
123
365
years
Using the formula
I
=
Pr
t
, we get:
I
=(
4750
)(
0.055
)(
123
365
)
=
88.0376...
Therefore, the amount of interest earned in this period was $88.04.
Page 1
of 12
MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
3.
What principal will earn $39.96 interest from June 18, 2021 to December 15, 2021 at 9.25%?
Answer:
Number of days = 180;
4.
If $650.00 is worth $673.70 after seven months, what rate of interest was charged?
Answer:
I = S – P = 673.7 – 650.00 = $23.70
Page 2
of 12
MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
5.
At what rate of interest will $1,387 earn $63.84 in 200 days?
Answer:
6.
In how many months will $1290 earn $100.51 interest at 8 ½ %?
Answer:
7.
What amount of money will accumulate to $1241.86 in five months at 3.90%?
Answer:
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
8.
A six-month promissory note for the amount $19,300 with interest at 8% is issued on April 1, 2022. Calculate the proceeds of the note on June 20, 2022, if money is worth 7.2%.
Answer:
Due date is June 20, 2022
Page 4
of 12
MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
9.
An investor purchased a 91-day, $100,000 T-bill on its issue date for $99,326.85 After holding it for 42 days, she sold the T-bill for a yield of 2.52%.
a)
What was the original yield of the T-bill?
b)
For what price was the T-bill sold?
c)
What rate of return (per annum) did the investor realize while holding this T-bill?
Answer:
(a)
(b)
Page 5
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
(c)
10. A deposit of $2,000 earns interest at 3% compounded quarterly. After two-and-a-half years, the interest rate is changed to 2.75% compounded monthly. How much is the account worth at the end of six years?
Answer:
Balance after 2.5 years:
11.
Compute the compound discount
of $3000 due in 8 years and 8 months if money is worth 9% compounded semi-annually. Answer:
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
Please note that the answer to the above highlighted question should be:
I = FV – PV = $3000.00 - $1398.85 = $1601.15
12. Find i if an amount is invested at 6.00% compounded every 3 months.
A) 3.00%
B) 2.00%
C) 6.00%
D) 1.50%
E) 0.02
Answer: D
Diff: 2 Type: MC Page Ref: 337-343
Topic: 9.1 Basic Concepts and Computations
Objective: 9-1: Calculate interest rates and the number of compounding periods.
13. Determine n
if an amount is invested for 3.5 years at 2.25% compounded quarterly.
A) 14
B) 5
C) 10.5
D) 0.5625%
E) 56.25
Answer: A
Diff: 1 Type: MC Page Ref: 337-343
Topic: 9.1 Basic Concepts and Computations
Objective: 9-1: Calculate interest rates and the number of compounding periods.
14. A loan of $9000.00 was repaid together with interest of $3728.00. If interest was Page 7
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
9.4% compounded quarterly, how long was the loan taken out? (Give answer in years and months to the nearest tenth.) Answer: PV = 9000; FV = 9000 + 3728 = 12 728; i = = 0.0235; P/Y = C/Y = 4 n = = 14.92069192 (quarters) ÷ 4 = 3.73017298 years = 3 years + (.73017298 x 12) = 3 years and 8.8 months
15. A loan of $4500.00 was repaid together with interest of $1164.00. If interest was 12 .4% compounded quarterly, for how many months was the loan taken out? Answer: PV = 4500, FV = 4500 + 1164 = 5664, i = 0.124 ÷ 4 = 0.031
n = = = = 7.535504 quarters
Number of Years = 7.535507 ÷ 4 = 1.8839 years
Number of Months = 1.8839 x 12 = 22 .6065 = 23 months
Explanation for #15: In this question we need to provide the answer in months.
When we calculate n the answer we get is in quarters. We now have to convert the answer to months. So, first we divide 7.535504 by 4 to get the answer in years (that is, 1.8839 years). We then have to multiply the number of years by 12 to get the number of months. During class I did this in one step as follows: 7.535504 x (12/4) = 22.607 or 23 months
16. In how many months will money double at 7.45% compounded semi-annually? Answer: PV = 1, FV = 2, i = 0.0745 ÷ 2 = 0.03725
n = = = = 18.95244 semi-annual periods
Page 8
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
Number of years = (18.95244 ÷2) = 9.4762 years Number of Months = 9.4762 × 12 = 113.7146 = 114 months
Explanation for #16: In this question we need to provide the answer in months.
When we calculate n the answer we get is in semi-annual periods. We now have to convert the answer
to months. So, first we divide 18.95244 semi-annual periods by 2 to get the answer in years (that is, 9.4762 years). We then have to multiply the number of years by 12 to get the number of months. During class I did this in one step as follows: 18.95244 x (12/2) = 113.715 or 114 months
17.
Scheduled debt payments of $600 each are due three months and six months from now. If interest at 10% is allowed, what single payment today is required to settle the two scheduled payments?
Answer:
Let the size of the single payment be $
x
.
The focal date is today.
The equation of equivalence is
18.
Dan borrowed $1100 today and is to repay the loan in two equal payments, one in four months and one in six months. If interest is 8.5% on the loan, what is the size of the equal payments if a focal date of today is used?
Answer:
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MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
Let the size of the equal payments be $
x
.
The focal date is today.
The equation of equivalence is
The size of the equal payments is FORMULA SHEET FOR TEST 1 (Based on Ch. 7, 8, 9, 10.1 and 10.2 Only)
S = P (1+rt)
P
=
S
(
1
+
rt
)
Page 10
of 12
MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
Page 11
of 12
n
i
PV
FV
)
1
(
+
=
(
)
n
i
FV
PV
−
+
=
1
l
÷
=
1
l
+
i
n
PV
FV
n
n
MATH 1191
Review for Test #1 (Winter 2024) – Based on Hummelbrunner Based on Ch. 7, 8, 9 and 10.1 (Excluding Ch. 8.4, 9.4 and 9.5)
SOLUTIONS
Prof. Sangita Ghosh
George Brown College
FORMULA SHEET FOR TEST 1 (Based on Ch. 7, 8, 9, 10.1 & 10.2)
To calculate the number of days between two dates:
For example, to determine the number of days between April 23, 2018 and July 21, 2018 press the following keys in your Texas Instruments BA II Plus calculator:
2nd DATE
DT1
04.2315 Enter
↓
DT2 07.2115 Enter ↓
CPT
DBD
Result is 89
Note that the month is entered first, with one or two digits, followed by a period; the day is entered
using two digits; and then the year is entered using two digits.
If only one of the dates and the desired days between the dates are entered, it is possible to determine
the second date
.
To determine the maturity date, given the start date plus the number of days:
For example, to determine the maturity date for April 15, 2017 + 60 days, the answer will be Wednesday, June 14, 2017 as follows
2nd DATE
DT1
04.1517
Enter
↓
↓
DBD = 60
Enter ↑
CPT
WED = 6-14-2017
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