Financial Algebra 1st Edition Chapter 3 Banking Services
Page 138 Problem 1 Answer
Given: Principal amount(P)dollars
Rate of interest(R) compounded annually. i.e.
Time(T) year
To find: Amount of interest
Solution: Interest
PRT dollars will earn dollars in one year at a rate of compounded annually
Page 139 Problem 2 Answer
Given: p=4000
r=5% compounded semi-annually
=0.05
To find: Balance after one year
Solution: Accounts that pay interest semiannually have the interest added on twice each year (every six months).
We will first find the balance after six monthsThen calculate interest on that amount
For the first six months,
t=6 months =0.5 years
I=PRT
=4000×0.05×0.5
=100
Now we will add the interest to the principal,
New principal=4000+100
=$4,100
Now we will compute interest for the next six months
I=PRT
=4100×0.05×0.5
=102.5
Now, we will add the interest to find the balance after one year
Balance after one year=4100+102.5
=4,202.5
Alex deposits$4,000 in a savings account that pays 5% interest compounded semiannually, then the balance after one year will be$4,202.5
Page 139 Problem 3 Answer
Given: p=$3000
r=4% compounded quarterly i.e.0.04
To find: Earnings in six months
Solution: Interest will be calculated every three monthsWe will find interest for the period of first three months and then again for the next three months
For the first three months
I=prt
=3000×0.04×0.25 [Convert 3 months in years]
=30
New principal=3000+30
=$3030
For the next three months
I=prt
=3030×0.04×0.25
=30.3
Total Interest=30+30.3
=$60.3
For the next three months
I=prt
=3030×0.04×0.25
=30.3
Total Interest=30+30.3
=$60.3
$3,000 earns$60.3 in six months at an interest rate of 4% compounded quarterly
Page 139 Problem 4 Answer
Given; x dollars earn in one day at an interest rate of 5 % compounded daily
To find; Express the answer algebraically.
Accounts that pay interest daily earn interest every day.There are 365 days in a year.
r=5
r=0.05
t=1 day
t=1 day =1/365 years
simple interest I=prt
I=1,000×0.05×1/365
≈0.13 interest to the principal. 1,000+0.13=1,000.13
The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.
This larger principal is used to compute the next day’s interest
Hence we conclude that The first day’s interest is approximately 13 cents, so the new balance is 1,000.13.
This larger principal is used to compute the next day’s interest
Page 140 Problem 5 Answer
Given that On January 7, Joelle opened savings account with 900. It earned 3 % interest compounded daily.
On January 8, she deposited her first paycheck of76.22.
To do: find what was her balance at the end of the day on January 8.
From the given data the starting principal amount is 900 $ and given that at an interest of 3 %
it is compounded daily then we get 900(0.03)(1/365) deposited a paycheck of 76.22 $ on adding up all of them we get
900+900(0.03)(1/365)(76.22)
=976.29 $ is the balance
Therefore, the total balance at the end of the day is 976.29 $
Page 141 Problem 6 Answer
We have to find that how might those words apply to what you learned in this lesson
In the old days, a man who would save money was called a miser, because they just saved their money and didn’t earn anything with this money.
Thus the man wouldn’t profit from saving that money and also wouldn’t use this money for some other purpose.
However, nowadays, it is very unique and a kind of miracle to find someone who saves money, because it is not profitable to start saving money at an older age (as time is crucial in making savings grow) and thus you won’t gain much on your money (we do earn some due to compound interest).
Thus if you want to profit the most of your savings, then we need to start saving at a young age, but most people don’t tend to start saving at a young age and thus it is a kind of miracle to find someone who do as such.
A miser just saved their money and didn’t earn anything with this money.
However, nowadays someone who saved money is a wonder, because they would have to start at a young age if they want to profit from their savings.
Page 141 Problem 7 Answer
Given: Jerome deposits $3,700 in a certificate of the deposit that pays 6×1/2 interest, compounded annually.
We have to find that how much interest does Jerome earn in one year
Convert rate of interest to a decimal.
r=6×1/2%=6.5%=0.065
Use the simple interest formula I=prt
Substitute I=3760×0.065×1=244.4
Jerome will earn $244.4. as interest at the end of 1 year.
Page 141 Problem 8 Answer
Sally deposits $4,000 in a certificate of deposit that pays 6×3/4% simple interest.
What is her balance after one year
Convert rate of interest to a decimal.
r=6×3/4%=6.75%=0.0675
Use the simple interest formula I=prt
Substitute I=4000×0.0675×1=270
Add interest to the principal. 4,000+270=4,270
Sally’s balance after one year will be$4,270.
Page 141 Problem 9 Answer
Given: Pierre deposits $9,000 in a certificate of deposit that pays 8% interest, compounded semiannually.
To find: How much interest does the account earn in the first six months?
What is the balance after six months
Convert rate of interest to a decimal. r=8%=0.08
Convert 6 months to years.t=6 months =0.5 years
Use the simple interest formula I=prt
By Substituting
I=9000×0.08×0.5=360
Pierre’s account will earn $360 as interest after 6 months.
The balance after 6 months will be $9,360
Add interest to the principal. 9,000+360=9,360
The balance after 6 months will be $9,360
Page 141 Problem 10 Answer
Given: Kevin has x dollars in an account that pays 2.2% interest, compounded quarterly.
We have to Express his balance after one quarter algebraically.
Convert rate of interest to a decimal. r=2.2%=0.022
Step 2 Convert 3 month to years.
t=3 months =1/4 years =0.25 years Use the simple interest formula
By Substituting
I=x×0.022×0.25=0.0055x
Kevin’s balance after one quarter will be $1.0055x.
Add interest to the principal. x+0.0055x=1.0055x
Kevin’s balance after one quarter will be $1.0055x.
Page 141 Exercise 1 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to Find the first quarter’s interest.
p=3500
r=7.5%=0.075
t=1 quarter =0.25yrs first quarter interest ∣=prt=35000.0750.25=65.625
We get, $65.625
Page 141 Exercise 2 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find the first quarter’s ending balance.
Given: Principal:$3,500
Result part (a): The first quarter interest is $65.625
First quarter interest =$65.625≈865.63
The first quarter’s ending balance if the principal increased by the first quarter interest:
First quarter’s ending balance = Principal + First quarter interest
=$3,500+$65.63
=$3,565.63
We get,
$3,565.63
Page 141 Exercise 3 Answer
Given: Liam deposits $3,500
in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find the second quarter’s interest.
Given:
p= Principal =$3,565.63 (First quarter’s ending balance found in part (b))r= Interest rate =7×1/2 %=7.5%=0.075
t= Time in year=1
Quarter =1/4 years =0.25years
The interest is the product of the principal, interest rate and time expressed in years.
I=prt=3,565.63×0.075×0.25≈$66.86
We get,
$66.86
Page 141 Exercise 4 Answer
Given: Liam deposits $3,500
in a saving account that pays7x1/2% interest, compounded quarterly.
We have to find the second quarter’s ending balance.
Given: Principal=$3,565.63 (First quarter’s ending balance found in part (b))
Result part (c): The second quarter interest is $66.86
Second quarter interest =66.86
The second quarter’s ending balance is the principal increased by the second quarter interest:
Second quarter’s ending balance = Principal + Second quarter interest
=$3,565.63+$66.86
=$3,632.49
We get,
$3,632.49
Page 141 Exercise 5 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find the third quarter’s interest.
Given: p= Principal =$3,632.49 (First quarter’s ending balance found in part (d)) r= Interest rate =7×1/2%=7.5%=0.075
t= Time in year=1 Quarter=1/4 years =0.25 years
The interest is the product of the principal, interest rate and time expressed in years.
I= prt =3,632.49×0.075×0.25≈$68.11
we get,
$68.11
Page 141 Exercise 6 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find the third quarter’s ending balance.
Given: Principal =$3,632.49 (First quarter’s ending balance found in part (d))
Result part (e): The third quarter interest is $68.11
Third quarter interest =68.11
The third quarter’s ending balance is the principal increased by the third quarter interest:
Third quarter’s ending balance = Principal + Second quarter interest
=$3,632.49+868.11
=$3,700.60
The third quarter’s ending balance is : $3,700.60
Page 141 Exercise 7 Answer
Given: Liam deposits $3,500
in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find the fourth quarter’s interest.
Given:p= Principal r= (First quarter’s ending balance found in part (f))r= Interest rate =7×1/2%=7.5%=0.075
t= Time in year =1 Quarter=1/4 years =0.25 years
The interest is the product of the principal, interest rate and time expressed in years.
I=prt=3,700.60×0.075×0.25≈$69.39
We get,
$69.39
Page 141 Exercise 8 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
To find: What is the balance at the end of one year?
Given: Principal =$3,700.60 (First quarter’s ending balance found in part (f))
Result part (g): The fourth quarter interest is $69.39
Fourth quarter interest =69.39
The fourth quarter’s ending balance is the principal increased by the fourth quarter interest:
Fourth quarter’s ending balance = Principal + Second quarter interest
=$3,700.60+$69.39
=$3,769.99
The balance at the end of one year is : $5,769.99
Page 141 Exercise 9 Answer
Given: Liam deposits $3,500 in a saving account that pays 7×1/2% interest, compounded quarterly.
We have to find that how much interest does the account earn in the first year?
Total interest in 1 st year = Balance after 1 year- original principal amount
=3769.98-3500=269.98
The amount earned in first year is : $269.98
Page 141 Exercise 10 Answer
Given : Janine opens a savings account with a deposit of $720.
The account pays3.4 percent interest, compounded daily.
We have to find first day’s interest.
We will use above formula .
Here, First convert rate of interest into decimal:r=3.4%=0.034
Convert 1 day to years:t=1day =1/365 years
Now, use simple interest formula: I=720×0.034×1/365≈0.07
Hence, the First day’s interest is approximately 7 cents.
Therefore, First day’s interest is approximately 7 cents.
Page 142 Exercise 11 Answer
We have given a table,
We have to find the missing amounts in the table.
We will find by using above information.
No principle amount initially thus
(1) Opening balance = 0
(1) Deposit on 10 August = $4,550
(3) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=0+4550−0
=4550
(4) The interest is the product of the principle, interest rate and time in years:
I=prt
=4550×0.04×1/365
≈0.50
Thus interest rate is $0.5
(5) Ending Balance = principal+interest
Ending Balance =4550+0.5=4550.50
ending balance is $4,550.50
(6) Next day opening = previous day ending balance
Opening bal aug 11=4550.50
(7) Deposit on August 11=300
(8) principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=4550.50+300−0
=4850.50
(9) The interest is the product of the principle, interest rate and time in years:
I=prt
=4850.50×0.04×1/365
≈0.53interest rate is $0.53
(10) Ending Balance = principal+interest
Ending balance =4850.5+0.53=4851.03
(11) Next day opening = previous day ending balance
Opening bal aug 12=4851.03
(12) withdrawal=900
(13) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle= Opening balance + Deposit − Withdrawal
=4851.03+0−900
=3951.03
(14) The interest is the product of the principle, interest rate and time in years:
I=prt
=3951.03×0.04×1/365
≈0.43
(15) Ending Balance = principal+interest
Ending Bal=3951.03+0.43=3951.46
The answers of the given blanks in the table are given above. All the answers are in USD.
Page 142 Exercise 12 Answer
Given: On December 18 of a leap year, Stacy opened a savings account by depositing $6,000.
The account pays 3.45% interest, compounded daily.
On December 19 she deposited $500, and on December 20 she withdrew $2,500.
By using this data we will fill all the blanks in the table.
(1) no principal amount initially
Thus opening balance is = 0
(2) Deposit = $6000
(3) Withdrawal =0
(4) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=0+6000−0
=6000
(5) The interest is the product of the principle, interest rate and time in years:
I=prt
=6000×0.0345×1/365
≈0.57
(6) Ending Balance = principal+interest
Ending Balance =6000+0.57=6000.57
(7) Next day opening = previous day ending balance
Opening bal =6000.57
(8) Deposit =500
(9) Withdraw =0
(10) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=6,000.57+500−0
=6,500.57
(11) The interest is the product of the principle, interest rate and time in years:
I=prt
=6500.57×0.0345×1/365
≈0.61
(12) Ending balance =6500.57+0.61=6501.18
(13) Next day opening = previous day ending balance
Opening bal =6501.18
(14) deposit =0
(15) withdrawal=2500
(16) Principle = Opening balance + Deposit − Withdrawal
=6,501.18+0−2,500
=4,001.18
(17) The interest is the product of the principle, interest rate and time in years:
I=prt
=4,001.18×0.0345×1/365
≈0.38
(18) Ending Balance = principal+interest
Ending Bal=4001.18+0.38=4001.56
The answers of the given blanks in the table are given above. All the answers are in USD.
Page 142 Exercise 13 Answer
Given : On May 29, Rocky had an opening balance of x dollars in an account that pays 3% interest, compounded daily. He deposits y dollars.
We have to express his ending balance on May30 algebraically.
We will use above formula .
Here, Consider that the deposit occurs on May 23
So, p=Principle
=Opening balance+Deposit
=$(x+y)
r=Interest rate
=3%
=0.03
t=Time in years
=1day
=1/365years (as there are 365 days in a year)
The interest is the product of the principle, interest rate and time in years:I=prt
=(x+y)×0.03×1/365
=x×3/100×1/365
=3(x+y)/36,500
The ending balance is then the principle increased by the interest: Ending balance
=p+I
=x+3(x+y)/36,500
=36,500x+3x+3y/36,500
=36,503x+3y/36,500
Now, p=Principle=ClosingbalanceMay29
t=$(36,503x+3y/36,500)
r=Interest rate
=3%
=0.03
The interest is the product of the principle, interest rate and time in years:I=prt
=(36,503x+3y/36,500)×0.03×1/365
=(36,503x+3y/36,500)×3/100×1/365
=(36,503x+3y/36,500)×3/36,500
=3(36,503x)+3(3y)/36,5002
=109,509x+9y/1,332,250,000
The ending balance is then the principle increased by the interest:Ending balance =p+I
=36,503x+3y/36,500+109,509x+9y/1,332,250,000
=36,500(36,503x)+36,500(3y)/1,332,250,000+109,509x+9y/1,332,250,000
=1,332,359,500x+109,500y/1,332,250,000+109,509x+9y/1,332,250,000
=1,332,359,500x+109,500y+109,509x+9y/1,332,250,000
=1,332,469,009x+109,509y/1,332,250,000
Hence, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.
Therefore, The ending balance on May 30 is 1,332,469,009x+109,509y/1,332,250,000.
Page 142 Exercise 14 Answer
Given :Linda has d dollars in an account that pays 3.4% interest, compounded weekly. She withdraws w dollars.
We have to express her first week’s interest algebraically.
We will use above formula .
Here, p=Principle
=Opening balance−Withdrawal
=d−w
r=Interest rate
=3.4%
=0.034
t=Timeinyears
=1week
=1 /52 years (as there are 52 weeks in a year)
Now, The interest is the product of the principle, interest rate and time in years: I
=prt
=(d−w)×0.034×1/52
=(d−w)×34/1000×1/52
=34(d−w)/52,000
=17(d−w)/26,000.
Hence, The first week’s interest is 17(d−w)/26,000.
Therefore, The first week’s interest is17(d−w)/26,000
Page 142 Exercise 15 Answer
Given: The table represents the compound interest calculations for an account that pays 2% interest compounded daily. Represent a–g algebraically.
Opening balance = P
Deposit = D
Withdrawal = $0
By using this data we will fill the table .
1) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=P+D−0
=P+D
(2) The interest is the product of the principle, interest rate and time in years:
I=prt
=(P+D)×0.02×1/365
=(P+D)×2/100×1/365
=2(P+D)/36,500
=P+D/18,250
(3) The ending balance is then the principle increased by the interest:
Ending balance =p+I
=(P+D)+(P+D)/18,250
=(P+D)(1+1/18,250)
=(P+D)(18,250/18,250+1/18,250)
=(P+D)(18,251/18,250)
=18,251(P+D)/18,250
(4) The opening balance on February 3 is the ending balance on the previous day and thus is the ending balance on February 2.
Opening balance on February 3= Opening balance on February 2
=18,251(P+D)/18,250
(5) The principle used to compute interest is the new balance, which is the opening balance increased by the deposit and decreased by the withdrawal.
Principle = Opening balance + Deposit − Withdrawal
=18,251(P+D)/18,250+0−W
=18,251(P+D)/18,250−W
(6) The interest is the product of the principle, interest rate and time in years:
I=prt
=(18,251(P+D)/18,250−W)×0.02×1/365
=18,251(P+D)−18,250W/18,250×2/100×1/365
=18,251(P+D)−18,250W/18,250×1/18,250
=18,251(P+D)−18,250W/18,2502
=18,251(P+D)−18,250W/333,062,500
(7) The ending balance is then the principle increased by the interest:
Ending balance =p+I
=18,251(P+D)/18,250−W+18,251(P+D)−18,250W/333,062,500
=18,251(P+D)−18,250W/18,250+18,251(P+D)−18,250W/333,062,500
=18,250(18,251)(P+D)−333,062,500W/333,062,500+18,251(P+D)−18,250W/333,062,500
=333,080,750(P+D)−333,062,500W/333,062,500/+18,251(P+D)−18,250W/333,062,500
=333,080,750(P+D)−333,062,500W+18,251(P+D)−18,250W/333,062,500
=333,099,001(P+D)−333,080,750W/333,062,500
Therefore,
1)The principle used to compute interest is the new balance is P+D
2)Interest = =P+D/18,250
3) Ending Balance =18,251(P+D)/18,250
4)Opening Balance =18,251(P+D)/18,250
5)The principle used to compute interest =18,251(P+D)/18,250−W
6)The interest is the product of the principle, interest rate and time in years: =18,251(P+D)−18,250W/333,062,500
7)Ending Balance:- 333,099,001(P+D)−333,080,750W/333,062,500
Page 142 Exercise 16 Answer
Given: George had an opening balance of m dollars in an account that pays 2.25% interest compounded monthly.
On the last day of the month, he made a deposit equal to twice his opening balance.
We have tp express his ending balance on the last day of the month algebraically.
We will use above formula .
Here, Opening balance =m
Deposit =2m (twice the opening balance)
p =Principle
= Opening balance + Deposit
=m+2m
=3m
r= Interest rate
=2.25%
=0.0225
t=Time in years
=1month
=1/12 years (as there are 12 months in a year)
Now,The interest is the product of the principle, interest rate and time in years: I
=prt
=3m×0.0225×1/12
=3m×225/10000×1/12
=3m×9/400×1/12
=(3m)(9)/400(12)
=27m/4,800
The ending balance is then the principle increased by the interest:Ending balance
=p+I
=3m+27m/4,800
=14,400m/4,800+27m/4,800
=14,400m+27m/4,800
=14,427m/4,800
=4,809m/1,600
Hence,
The ending balance on the last day of the month is 4,809m /1,600.
Therefore,The ending balance on the last day of the month is 4,809m/1,600.