Financial Algebra 1st Edition Chapter 3: Banking Services
Page 132 Problem 1 Answer
Given -5.51%,51/2%,55/8%,5.099%,5.6%
We need to arrange the given interest rates in descending order.
We have :
5.51%,51/2%,55/8%,5.099%,5.6%
These are equivalent to :
5.51%,5.50%,5.625%,5.099%,5.6%
Arranging these in descending order, we get :
5.625%,5.6%,5.51%,5.50%,,5.099%
We obtain :
5.625%,5.6%,5.51%,5.50%,,5.099%
Page 133 Problem 2 Answer
Given – Mae has $891 in her account. A$7 fee is charged each month the balance is below $750.
She withdraws$315. She makes no deposits or withdrawals for the next x months.
We need to express her balance algebraically.
New Balance= Prior Balance−Reduction
=891−315
=576
New Balance=Prior Balance− Reduction
=576−7
=$569.00
The new balance is: $569.00
Page 133 Problem 3 Answer
We will find the simple interest that is earned on $4,000 in 31/2 years at an interest rate of 5.2% We can calculate the simple interest as follows :
I =p⋅r⋅t
i=(4000)(0.052)(3.5)
i=812
The simple interest is $812.
Page 133 Problem 4 Answer
We will find the simple interest that $800 earn in 300 days in a non-leap year at an interest rate of 5.71%.
We will round to the nearest cent.
Here,t=300/365 years
i=p⋅r⋅t
i=800⋅0.0571⋅300/365
i=$37.55
The simple interest is: $37.55
Page 134 Problem 5 Answer
Given – A two-year simple interest account pays 31/4% interest to earn $300 in interest.
We will find the principal.We will use the formulai=p⋅r⋅t.
We have :
i=p⋅r⋅t
300=p⋅3.25⋅2
∴p=46.1538
The principal amount is$46.1538.
Page 134 Problem 6 Answer
We will determine how long will it take $10,000 to double at 11% simple interest.
We will use the formula:i=p⋅r⋅t
For the given data, we get :
i=p⋅r⋅t
20000=10000⋅0.11⋅t
t=18.1818
The time is : 18.1818 years
Page 134 Problem 7 Answer
Marcos deposited $500 into a 2.5-year simple interest account.
He wants to earn$200 interest. We will find interest rate that must the account pay.
We will use the formula :i=p⋅r⋅t.
We know :
i=p⋅r⋅t
From the given data, we get :
200=500⋅2.5⋅r 2⋅10 5⋅25
=r
r=0.16
We get r=16%
We get the rate as 16%.
Page 135 Problem 8 Answer
We will arrange the following interest rates in ascending order :
3.4%,3.039%,33/16%,3.499%,31/2%
We will first convert all the rate of interests in decimal form.
1.3.4%=3.400
2.3.039%=3.039 ( no need to make changes)
3.33/16=(16×3+3)/16→3.188
4.3.499=3.499
5.31/2=(3×2+1)/2→3.500
Comparing the decimal forms, we get,
3.039<3.188<3.400<3.499<3.500
Hence,3.039<33/16%<3.4%<3.499%<31/2%
We conclude :
3.039<33/16%<3.4%<3.499%<31/2%
Page 135 Problem 9 Answer
Given – Josh has a savings account at a bank that charges a $10 fee for every month his balance falls below$1,000.
His account has a balance of$1,203.44 and he withdraws $300.
We will determine his balance be in six months if he makes no deposits or withdrawals.
We will use the known facts.
The new balance is the previous balance decreased by the withdrawal.
$1,203.44−$300=$903.44
We note that the balance is below $1,000 and thus Josh has to pay a charge of$10
per month on each of the six months (when this balance is on the account).
Total fee: $10×6=$60
The balance after 6 months is then the previous balance decreased by the total fee of $60.
$903.44−$60=$843.44
His balance in six months if he makes no deposits or withdrawals is$843.44.
Page 135 Problem 10 Answer
Given – John, Paul, and George are having a disagreement over interest rates.
John says that 63/4% can be expressed as 6.75%.
George thinks that 63/4% can be expressed as 0.0675.
Paul remembers converting percents to equivalent decimals and thinks it can be expressed as0.0675%.
We will determine who is correct and who is incorrect.
John and George are correct. Paul’s conversion is incorrect.
63/4%=6.75%
=0.0675.
These representations are correct.
While converting percentage to decimal form, the percentage sign needs to be dropped.
Paul converted it into decimal, yet kept the percent sign.
That was his mistake!
John and George are correct. Paul’s conversion is incorrect.
Page 135 Problem 11 Answer
Given – Beth and Mark would like to put some savings in the bank.
They most likely will not need this money for4
years, so Beth wants to put it in a four-year CD.
Mark wants to put the money in a passbook savings account.
We will tell the advantage and disadvantages of a CD.
Beth and Mark would like to put some savings in the bank.
They most likely will not need this money for 4 years, so Beth wants to put it in a 4−year CD.
Mark wants to put the money in a savings account.
The disadvantage of Cd are that Cd has a penalties if money is withdrawn before maturity, and the advantage of CD is that Cd have highest interest rate.
Page 135 Problem 12 Answer
We will find the simple interest on a $2,350 principal deposited for six years at a rate of 4.77%.
Principal amount(p)=$2350
No of years/ time(t)=6
Rate of interest(r)=4.77%
=0.0477
Simple Interest (I)=ptr
=2350×6×0.0477
=$672.57
The simple interest is$672.57.
Page 135 Exercise 1 Answer
Given – Ryan depos its $775 in an account that pays 4.24%
simple interest for four years. Brian deposits $775 in an account that pays4.24%
simple interest for one year.We will determine the Ryan’s interest after the four years.
Principal(p)=$775
Rate of interest
(r)=4.24%
=0.0424
No of years/time(t)=4 years
Simple interest=p t r
=775×4×0.0424
=131.44 dollars
Hence simple interest after 4 years is$131.44.
The simple interest after 4 years is$131.44.
Page 135 Exercise 2 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits $775 in an account that pays 4.24% simple interest for one year.We will find Ryan’s balance after four years.
Principal(p)=$775
Rate of interest
(r)=4.24%
=0.0424
No of years/time(t)=4 years
Simple interest=p t r
=775×4×0.0424
=131.44 dollars
Hence simple interest after 4 years is$131.44
Amount = simple interest +principal
=775+131.44
=906.44 dollars
Hence amount after 4 years is$906.44
The amount after 4years is $906.44.
Page 135 Exercise 3 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits$775 in an account that pays 4.24% simple interest for one year.
We will determine the interest that Ryan’s account earn the first year.
Principal(p)=$775
Rate of interest
(r)=4.24%
=0.0424
No of years/time(t)=1 year
Simple interest =p t r
=775×1×0.0424
=32.86 dollars
Hence simple interest after 1 year is$32.86.
The simple interest after 1 year is$32.86.
Page 135 Exercise 4 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits$775 in an account that pays 4.24% simple interest for one year.
We will determine the interest that Ryan’s account earn the fourth year.
We will use the known facts.
We note that : i=p⋅r⋅t
From the given data, we get :
i=775⋅0.0424⋅4
i=131.44
The simple interest is$131.44.
Page 135 Exercise 5 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits $775 in an account that pays 4.24% simple interest for one year.
We will determine Brian’s interest after the first year.
Principal(p)=$775
Rate of interest
(r)=4.24%
=0.0424
No of years/time (t)=1 years
Simple interest=p t r
=775×1×0.0424
=32.86 dollars
Hence simple interest after 1 years is$32.86.
The simple interest after 1 years is$32.86.
Page 135 Exercise 6 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits $775 in an account that pays 4.24% simple interest for one year.
We will determine Brian’s balance after the first year.
Principal(p)=$775
Rate of interest
(r)=4.24%
=0.0424
No of years/time (t)=1 years
Simple interest=p t r
=775×1×0.0424
=32.86 dollars
Hence simple interest after 1 years is$32.86
Amount= simple interest+ principal =32.86+775
=807.86 dollars
Hence amount after the first year is$807.86
The amount after the first year is$807.86.
Page 135 Exercise 7 Answer
Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits $775 in an account that pays 4.24% simple interest for one year.
Here, Brian withdraws all of the principal and interest after the first year and deposits it into another one-year account at the same rate.
We will determine his interest for the second year rounded to the nearest cent.
In part (f), we determines that the balance after the first year is$807.86.
We then put the $807.86 in a one-year account at the same rate:
p=Principle =$807.86
r= Interest rate =4.24%
=0.0424
Time t=1 year
The simple interest is the product of the principal, interest rate and time expressed in years.
=p r t
=807.86×0.0424×1≈$34.25
Thus, the interest during the second year is $34.25.
The interest during the second year is $34.25.
Page 135 Exercise 8 Answer
Given – Ryan deposits $775 in an account that pays 4.24% simple interest for four years.
Brian deposits $775 in an account that pays 4.24% simple interest for one year.
We will compare the interest Brian earns with the interest Ryan earns for the second year.
We will tell who earned more interest.
We know, Simple interest=p t r
If we compare the values of p,t&r, for Ryan and Brian, we get that all the three values are equal. Hence the product p t r will also be equal for both.
Hence Ryan and Brian earn the same interest.
We conclude that Ryan and Brian earn the same interest.
Page 136 Exercise 9 Answer
Given: Principal =$2,560
Interest Rate =51/8%
Time =17 months
To find the simple interest.We will use the formula Simple Interest = Principal ⋅ Rate of interest. Time.
Substituting the given values in simple interest formula as I=p r t
=2560×0.05125×17/12
≈$185.87
The simple interest for
Principle =$2,560
Interest rate =51/8%
Time =17 months is $185.87.
Page 136 Exercise 10 Answer
Given:Interest =$450
Principle =$450
Interest rate =14%
To find the time when$450 can be doubled for the given interest rate.We will use the simple interest formula as t=I/pr.
Substituting the given values in the formula t=I/pr, we get
450/450×0.14
≈7.1429
It takes$450 approximately 7.1429 years to double at a simple interest rate of 14%.
Page 136 Exercise 11 Answer
Given:Interest =$450
Principle =$450
Interest rate =100%
To find the time such that the given interest gets doubled in the given interest rate.We will use the formula t=I/pr.
Substituting the given values in the simple interest formula as t=I/pr
=450/450×1
=1
It takes$450 one year to double at a simple interest rate of 100%.
Page 136 Exercise 12 Answer
Given:
Interest =$900
Principle =$9,500
Time =19 months
To find the rate of interest so that we can earn the given amount of interest.
We will use the formula r=I/pt.
Substituting the given values in the simple interest formula as r=I/pt
=900/9500×19/12
=108/1805≈0.0598
=5.98%
The interest rate needed for$ 9,500 to earn $900 in 19 months is 5.98%.
Page 136 Exercise 13 Answer
Given: Assume $20,000 is deposited into a savings account.
Bed ford Bank offers an annual rate of 4% simple interest for five years.
Slick Bank offers a rate of 20% simple interest for one year.
To find that which earns more interest.
We will find out the simple interest for both banks using the formulaI=prt and compare them.
For Bedford Bank, we are given
Principle =$20,000
Interest rate =4%
Time =5 years
so for them the simple interest is
I=p r t
I=20000×0.04×5
I=4000
For Slick Bank, we are given
Principle =$20,000
Interest rate =20%
Time =1 year
so for them the simple interest is
I=prt
I=20000×0.2×1
I=4000
Therefore, we get that the interest in both the cases is same and it is$4000.
Interest for both the banks is same and it is$4000.
Page 136 Exercise 14 Answer
Given: A couple is planning a savings account for a newborn baby.
They start with$3,450 received in newborn baby gifts.
If no deposits or withdrawals are made, to find the balance of the account if it earns simple interest at 5% interest for18
years. We will use the formula I=p r t.
We are given
Principle =$3,450
Interest rate =5%
Time =18 years
So using these values, we get the simple interest is
I=p r t
I=3,450×0.05×18
I=3,105
If no deposits or withdrawals are made, the balance of the account if it earns simple interest at 5% interest for18 years is$3,105.
Page 136 Exercise 15 Answer
Given: Ron estimates that it will cost $400,000 to send his daughter to a private college in18 years.
He currently has $90,000 to deposit in an account.
To find that what simple interest rate must his account have to reach a balance of $400,000 in 18 years.
We will use the formula r=I/pt.
We are given that
Interest =$400,000−$90,000
Interest =$310,000
Principle =$90,000
Time =18 years
using these values the interest rate is r=I/pt
r=310000
90000×18≈0.19
r=19%
The simple interest rate must his account have to reach a balance of $400,000 In 18 years is 19%.
Page 136 Exercise 16 Answer
Given:
For A2 to compute the interest.We will use the formulaI=prt.
From the given table, we get that for row one, we have
Interest = A2
Principal =B2
Interest rate =C2
Time =D2
So the simple interest for these values using the formulaI=prt becomes
A2=B2×C2×D2.
In a spreadsheet, this can be written as A2=B2⋅C2⋅D2.
The simple interest for A2 in the table is A2=B2⋅C2⋅D2.
Page 136 Exercise 17 Answer
Given:
For B2 to compute the principal. We will use the formula I=p r t.
We are given that
Interest = A2
Principal =B2
Interest Rate =C2
Time =D2
using these values we get that the principal formula is p=I/rt
B2=A2
C2×D2
In a spreadsheet, we write it as B2=A2/(C2⋅D2)
The value of the principal for B2 in the table is B2=A2/(C2⋅D2).
Page 136 Exercise 18 Answer
Given:
For C2 to compute the interest rate.We will use the formulaI=prt.
We are given that
Interest=A2
Principal l =B2
Interest Rate =C2
Time =D2
so using these values, the formula for interest rate is
r=I/pt
C2=A2
B2×D2
C2=A2/(B2⋅D2)
The formula of interest rate for C2 in the table is C2=A2/(B2⋅D2).
Page 136 Exercise 19 Answer
Given:
ForD2 to compute time in years, given the interest, rate, and principal. We will use the formula I=p r t.
We are given that
Interest=A2
Principal =B2
Interest Rate =C2
Time =D2 using these values the formula for time in years is
t=I/pr
D2=A2
B2×C2
In a spreadsheet, we write it as D2=A2/(B2⋅C2).
The formula of time in years for D2 in the table is D2=A2/(B2⋅C2).
Page 136 Exercise 20 Answer
Given: Zoe creates a spreadsheet to make simple interest calculations.
The user inputs values for the principal, rate, and time in years in row 2.
Write each formula.
To find: For E2 compute the time in months, given the time in years.
All values are entered or calculated in row 2.
By taking into account the column tables in row 1, we then note:
Time in years=D2
Time in months=E2
Since there are 12 months in a year, the time in month is 12 times the time in years:
E2=12×D2
In a spreadsheet, a product is most often entered using x instead of ×, and a division is entered using.
E2=12xD2
We get, E2=12⋅D2