Financial Algebra Cengage

Financial Algebra 1st Edition Chapter 4 Consumer Credit

Page 202 Problem 1 Answer

Given a spreadsheet

To do: Write the spreadsheet formula to compute the new balance in cell F2

The given spreadsheet is empty there is no entry in any cell

so, we can use zero amount in the calculation.

The formula to compute the new balance we use

A₂−B₂+C₂+D₂+₂=F₂

Therefore,the formula to find the new balance F2 is A2−B2+C2+D2+E2

Page 203 Problem 2 Answer

Given Rhonda balance

To do: Write an algebraic expression that represents her current available credit

Given that there is a previous balance of 567.91 $ and on-time credit card payment is also the same for the last month, there is no balance then the expression will be the subtraction from credit line from purchases then we get x−y as the expression

Therefore,

The algebraic expression that represents her current available credit x−y

Page 203 Problem 3 Answer

Given a summary statement

To do: Find the error in the summary

From the given summary the new balance will be  850−560=290 $ and 290+300+3+4.78=597.78 $ but given as 504.78 $ this is the mistake in the summary

Therefore, the error in the summary is in the cell of new balance given as 504.78 $ but the actual new balance should be 597.78 $

Page 204 Problem 4 Answer

Given quote is Credit card companies pay college students generously to stand outside dining halls, dorms, and academic buildings and encourage their fellow students to apply for credits cards.– Louise Slaughter, American Congresswoman

To do: Write how might the quote apply to what you have learned

From the given quote we understand that the credit card companies will pay for college students which will generate curiosity among them and apply for credit cards as the students will be more interested to listen to their peers

This will be an example for advertising especially for college students

Therefore, the given quote is applied for advertising for college students

Page 202 Problem 5 Answer

Given the flashcard statement

To do: Find the sum of all purchases made

In the given table we sum of all purchases are given under new purchases i.e.,1,227.24 $

Therefore, the sum of all purchases is 1,227.24 $

Page 204 Problem 6 Answer

Given –

The Flash Card : We will determine when is the payment for this statement due.

The data at which the payment for this statement is due is given in the top right corner of the Flash Card Statement.

Payment Due =8Jun

The payment for this statement is due till 8 June.

Page 204 Problem 7 Answer

Given –

We will find the minimum amount that can be paid.

From the image, the minimum amount that can be paid=$30.

Minimum amount that can be paid=$30.

Page 204 Problem 8 Answer

Given –

We will find the number of days which are in the billing cycle.

The number of days in the billing cycle is given at the bottom of the FlashCard statement underneath “* Days in Billing Cycle”

Number of Days in Billing Cycle=30

The number of Days in Billing Cycle are 30.

Page 204 Problem 9 Answer

Given –

We will determine the previous balance.

We will see the first entry in the summary.

Hence, the previous balance=$420.50

The previous balance is=$420.50

Page 204 Problem 10 Answer

Given – Rebecca has a credit line of $6,500 on her credit card. She had a previous balance of $398.54 and made a $250 payment.

The total of her purchases is$1,257.89.

We will find Rebecca’s available credit.

Available credit= Credit line – previous balance−total amount of purchases made+payment

Available credit =6500−398.54−1257.89+250

Available credit =$5,093.57

The available credit is$5,093.57.

Page 204 Problem 11 Answer

Given -The APR on Leslie’s credit card is currently 21.6%.

We will find the monthly periodic rate.

Given: APR=21.6%

=0.216

​The APR is the annual periodic rate.

The monthly periodic rate is the annual periodic rate divided by 12 , because there are12 months in a year.

Monthly periodic rate= APR/Number of months in a year

=21.6%/12

=1.8%

Page 204 Problem 12 Answer

Given – Sheldon’s monthly periodic rate is 1.95%.

We will determine the APR.

Given: Monthly periodic rate=1.95%

The APR is the annual periodic rate.

The annual periodic rate is the monthly periodic rate multiplied by 12, because there are 12months in a year.

APR=Monthly periodic rate×Number of months in a year

​=1.95%×12

=23.4%

The ARN is 23.4%.

Page 205 Problem 13 Answer

Given –

The credit card summary :

We need to express the sum of the cycle’s daily balances algebraically.

We note that the sum of the cycle’s daily balances=Average daily balance×No. of days in billing cycle

Hence, Sum of the cycle′s daily balances = WX

We conclude that : Sum of the cycle′s daily balances = WX

Page 205 Problem 14 Answer

Given –

The credit card summary :

We need to express the monthly periodic rate as an equivalent decimal without the% symbol.

Given: APR=Y%

The APR is the annual periodic rate.

The monthly periodic rate is the annual periodic rate divided by 12, because there are 12 months in a year.

Monthly periodic rate= APR/Number of months in a year

=Y%/12

=Y/100/12

=Y/12×100

=Y/1200

Monthly periodic rate =Y/1200

Page 205 Problem 15 Answer

Given – The credit card summary :

We need to express the monthly periodic rate as an equivalent decimal without the % symbol.

Given: APR=Y%

The APR is the annual periodic rate.

The monthly periodic rate is the annual periodic rate divided by 12, because there are 12 months in a year.

Monthly periodic rate = APR/ Number of months in a year

=Y%/12

=Y/100/12

=Y/12×100

=Y/1200

Monthly periodic rate =Y/1200

Page 205 Problem 16 Answer

Given –

We will find the amounts a,b,c, and d.

The previous balance is given in the SUMMARY of the FlashCard statement underneath “Previous balance”.

Previous balance=$215.88

The finance charge is given in the SUMMARY of the FlashCard statement underneath “Finance charge”.

Finance charge=$6.70

The total credit line is given in the bottom left corner of the FlashCard statement next to “Total Credit Line”.

Total credit line=$5,000.00

The amounts of the transactions (debits and credits) are given in the column “DEBITS/CREDITS” (where the debits are the positive amounts and the credits are the negative amounts).

The new balance is then the previous balance increased by the debits, decreased by the credits and increased by the finance charge.

New balance =Previous balance +Debits−Credits+Finance charge

=215.88+(85.63+47.60+855.00+370.50)−(63.00+137.00)+6.70

=215.88+1358.73−200.00+6.70

=1381.81

​The total available credit is the total credit line decreased by the new balance:

Total available credit=Total credit line−New balance​

=5000−1381.81

=3618.19

​The payments/credits are the negative amounts in the column “DEBITS/CREDITS” of the FlashCard summary.

First payment: $63.00

Second payment:$137.00

The summary should contain the sum of all payments/credits underneath “Payments/Credits”.

Payments/Credits=First payment+ Second payment​

=63.00+137.00

=200.00

​The new purchases are the positive amounts in the column “DEBITS/CREDITS” of the FlashCard summary.

First purchase:$85.63

Second purchase:$47.60

Third purchase: $855.00

Fourth purchase: $370.50

The summary should contain the sum of all purchases underneath “New Purchases”.

New Purchases= First purchase+Second purchase+Third purchase+ Fourth purchase

​=85.63+47.60+855.00+370.50

=1358.73

​The previous balance is given in the SUMMARY of the FlashCard statement underneath “Previous balance”.

Previous balance=$215.88

The finance charge is given in the SUMMARY of the FlashCard statement underneath “Finance charge”.

Finance charge=$6.70

The total credit line is given in the bottom left corner of the FlashCard statement next to “Total Credit Line”.

Total credit line=$5,000.00

The amounts of the transactions (debits and credits) are given in the column “DEBITS/CREDITS” (where the debits are the positive amounts and the credits are the negative amounts).

The new balance is then the previous balance increased by the debits, decreased by the credits and increased by the finance charge.

New balance= Previous balance + Debits − Credits + Finance charge

=215.88+(85.63+47.60+855.00+370.50)−(63.00+137.00)+6.70

=215.88+1358.73−200.00+6.70

=1381.81

​We obtain :

a=3618.19

b=200.00

c=1358.73

d=1381.81

​Page 205 Problem 17 Answer

Given –

We need to check the new balance entry on the monthly statement below by using the first five entries.

If the new balance is incorrect, we will write the correct amount.

The previous balance is given in the SUMMARY of the FlashCard statement underneath “Previous balance”.

Previous balance=$424.41

The total payments is given in the SUMMARY of the FlashCard statement underneath “Payments/Credits”.

Total payments=$104.41

The total of the new purchses is given in the SUMMARY of the FlashCard statement underneath “New Purchases”.

New Purchases=$103.38

The late charge is given in the SUMMARY of the FlashCard statement underneath “Late charge”.

Late charge=$23.00

The finance charge is given in the SUMMARY of the FlashCard statement underneath “Finance charge”.

Finance charge=$7.77

The amounts of the transactions (debits and credits) are given in the column “DEBITS/CREDITS” (where the debits are the positive amounts and the credits are the negative amounts).

The new balance is then the previous balance increased by the new purchases, decreased by the total payments and increased by the finance charge.

New balance=Previous balance+New purchases−Total payments+Late charge+Finance charge

​=424.41+103.38−104.41+23.00+7.77

=454.15

We then note that the new balance is $454.15 as claimed by the summary of the credit card statement.

The new balance is $454.15 as claimed by the summary of the credit card statement.

Page 205 Problem 18 Answer

Given –

A credit card statement is modeled using the following spreadsheet. Entries are made in columns A–F.

We will write the formula to calculate the available credit in cell G₂.

The previous balance is given underneath “Previous balance”, which is thus cell A₂.

Previous balance=A₂

The total payments are given underneath “Payments/Credits”, which is thus cell B2.

Total payments=B₂

The total of the new purchases is given underneath “New Purchases”, which is thus cell C₂.

New Purchases=C₂

The late charge is given underneath “Late charge”, which is thus cell D2.

Late charge=D₂

The finance charge is given underneath “Finance charge”, which is thus cell E₂.

Finance charge=E₂

The credit line is given underneath “Credit line”, which is thus cell F₂.

Credit line=F₂

The available credit needs to be given underneath “Available Credit”, which is thus ce llG₂

Available credit=G2

The new balance is then the previous balance increased by the new purchases, decreased by the total payments and increased by the finance charge

New balance = Previous balance − Total payments + New purchases + Late charge + Finance charge

=A₂−B₂+C₂+D₂+E₂

​The available credit is then the credit line decreased by the new balance:

Available= Credit line−New balance

G₂ G₂=F₂−(A₂−B₂+C₂+D₂+E2)

=F₂−A₂+B₂−C₂−D₂−E₂

​We conclude that :

G₂=F₂−A₂+B₂−C₂−D₂−E₂

Page 205 Problem 19 Answer

Given –

We need to determine the amount

The previous balance is given in the SUMMARY of the FlashCard statement underneath “Previous balance”.

Previous balance=$939.81

The total payments is given in the SUMMARY of the FlashCard statement underneath “Payments/Credits”. Let us name the total payments ASAP.

Total payments=x

The total of the new purchases is given in the SUMMARY of the FlashCard statement underneath “New Purchases”.

New Purchases=$125.25

The late charge is given in the SUMMARY of the FlashCard statement underneath “Late charge”.

Late charge=$3.00

The finance charge is given in the SUMMARY of the FlashCard statement underneath “Finance charge”.

Finance charge=$15.38

The New Balance is given in the SUMMARY of the FlashCard statement underneath “new balance”.

New balance=$833.44

The new balance is then the previous balance increased by the new purchases, decreased by the total payments and increased by the finance charge.

New balance=Previous balance+New purchases−Total payments+ Late charge+Finance charge

833.44+x/x

x=939.81+125.25−x+3.00+15.38

=1083.44−x

=1083.44

=1083.44−833.44

=250

​We then note that the total payment is$250.

The amount of the payment made on this credit card is$250.

Page 205 Problem 20 Answer

Given – The previous balance after the last billing cycle is represented by A, recent purchases by B, payments by C, finance charge by D, and late charge by E.

We need to express the relationship among the variables that must be true for the new balance to be zero.

Given: Previous balance =A

Total payments =C

New Purchases =B

Late charge=E

Finance charge=D

New balance=0

The new balance is then the previous balance increased by the new purchases, decreased by the total payments, and increased by the finance charge.

New balance=Previous balance+New purchases−Total payments+ Late charge+ Finance charge

0=A+B−C+E+D

C=A+B+D+E

Thus we then note that the equation is0=A+B−C+E+D needs to be true when the balance is zero, or equivalently C=A+B+D+E

The relationship among the variables is: C=A+B+D+E

Financial Algebra 1st Edition Chapter 4 Consumer Credit

Page 195 Problem 1 Answer

Given about Carrie’s card

To do: Find how much is Carrie responsible for paying

Under the act of truth lending, there is a maximum liability of 50 $

so, Carrie is responsible for paying 50 $

Therefore, Carrie is responsible for paying 50 $

Page 196 Problem 2 Answer

Given about the paul last month payments

To do: Find the average daily balance

From the given data we have x dollars for 6 days means 6x

y dollars for 12 days means 12y

w dollars for q days means qw

d dollars for 2 days means 2d then the average will be 6x+12y+qw+2d/q+20

Therefore,the average daily balance is 6x+12y+qw+2d/q+20

Page 196 Problem 3 Answer

Given about steve finance

To do: Express his finance charge

From the given data APR is p % so, p/12 and here average daily balance is d $

Then the finance charge will be (p/100/12)d

Therefore, the finance charge is (p/100/12)d

Page 197 Problem 4 Answer

Given that Life was a lot simpler when what we honored was father and mother rather than all major credit cards. – Robert Orben, American Comedy Writer

To do: Interpret the quote

In the given quote we have horned by parents i.e., mother and father which means that while using credit cards we should have enough money in the cards as stores won’t honor your cards and the credit card makes easy for life

Therefore, the quote says that While using credit cards we should have enough money in the cards as stores won’t honor your cards and the credit card makes easy for life

Page 197 Problem 5 Answer

Given that Janine’s credit card was stolen, and the thief charged a $44 meal using it before she reported it stolen

To do: Find how much of this is Janine responsible for paying

Given that the card was stolen and before it was charged $44 as we know that the maximum liability will be $50 then Janine have to pay the entire amount i.e.,44 $

Therefore, Janine has to pay  $44

Page 197 Problem 6 Answer

Given that Dan’s credit card was lost on a vacation. He immediately reported it missing.

The person who found it days later used it and charged $x worth of merchandise on the card, where x> $200

To do: Find how much of the  $x is Dan responsible for paying

From the given question we get to know that Dan reported as soon it was lost i.e. before used so that he is responsible for paying $0

Therefore, Dan is responsible to pay  $0

Page 197 Problem 7 Answer

Given that Felix was given a card with an APR of 12 %

To do: Find his monthly percentage

Given APR=12 % and we know that there are 12 months in a year then we get monthly percentage rate​=12/12

=1 %

​Therefore, His monthly percentage rate is 1 %

Page 197 Problem 8 Answer

Given that Oscar was given a card with an APR of 15 %

To do: Find his monthly payment

Given that APR=15 % and we know that number of months in a year are 12

Then Monthly percentage rate​=15/12

=1.25 %

​Therefore, the monthly percentage rate is 1.25 %

Page 197 Problem 9 Answer

Given the credits of Felix and oscar

To do: Find how much more would Oscar pay than Felix

Solving for Felix From part a we know that the monthly percentage rate is 1 % then Finance charge​=1 %×800 $

=8 $

​Solving for Oscar From part b we know that the monthly percentage rate is 1.25 % then Finance charge​=1.25 %× $800

=10 $ ​the difference between them is 2 $

Therefore, Oscar has to pay 2 $ more than Felix

Page 197 Problem 10 Answer

Given all the finance charges per days

To do: Find the average daily balance

The sum of the daily balances is ​
9(778.12)=7003.08

8(1876)=15008

4(2112.50)=8450

10(1544.31)=15443.1 ​is 45904.18 $ and the total number of days given is 9+8+4+10=31

Now the average daily balance will be 45904.18/31

=1480.78 $

Therefore, the average daily balance is 1480.78 $

Page 197 Problem 11 Answer

Given the finance charges with respect to the days

To do: Find the finance charge

From part a we know that the average daily balance is 1,480.78 $

Now we have to find monthly percentage rate given APR=19.2 $

Monthly percentage rate ​= APR/ Number of months in a year

=19.2/12

=1.6 % then finance charge​=1.6×1,480.78

=23.69 $​

Therefore,The finance charge=23.69 $

Page 197 Problem 12 Answer

Given the set of daily balances

To do: Find the average  daily balance

The sum of daily balances =number of days at balance×balance

so we get​

​=x×y+r×q+w×d+m×p

=xy+rq+wd+mp and number of billing days will be x+r+w+m

Then the average daily balance will be =xy+rq+wd+mp/x+r+w+m

Therefore, the average daily balance=xy+rq+wd+mp/x+r+w+m

Page 197 Problem 13 Answer

Given Suzanne’s average daily balance and finance charge

To do: Find the monthly percentage rate

Given average daily balance=x

finance charge=y then monthly percentage rate=y/x

Therefore,The monthly percentage rate=y/x

Page 197 Problem 14 Answer

Given the data on Suzzanne’s income

To do: Find APR From the previous part, we know that the monthly percentage rate=y/x and

we know that the annual percentage will consider 12 months multiplying with it we get12y/x which is the APR

Therefore,the APR=12y/x

Page 197 Problem 15 Answer

Given Jared’s income details

To do: Find the monthly percentage rate

From the given data we know that the average daily basis is 560 $ the finance charge is 8.12 $

Then monthly percentage rate= finance charge/average daily basis

on substituting  we get

=8.12/560

=1.45 %

​Therefore, the monthly percentage rate is 1.45 %

Page 197 Problem 16 Answer

Given Jared’s income details

To do: Find the APR

The annual percentage income is the multiplication of the monthly percentage rate and the number of months in a year

we know that the monthly percentage rate=1.45 % and we also know that there are 12

months in a year then

APR=1.45 %×12

=17.4 %

​Therefore, APR=17.4 %

Page 198 Problem 17 Answer

Given Helene’s credit details

To do: Find her daily balance if she puts the down payment on the credit card today

Given that Today’s balance=712.0 $

down payment=5,000 $ then the daily balance will be712.0+5,000=5,712.0 $

Therefore,the daily balance is 5,712.0 $

Page 198 Problem 18 Answer

Given Helene’s credit details

To do: Find her average daily balance

We know that daily balance is 5,712.04 $ for 30 days it will be 5,712.04 $×30 then average daily balance=30×5,712.04 $/30

=5,712.04 $

​Therefore,the average daily balance is 5,712.04 $

Page 198 Problem 19 Answer

Given Helene’s credit details

To do: Find the finance charge

We know that the daily balance is 5,712.04 $ and given APR=16.8 %

with this monthly rate will be 0.618/12

now finance charge​=5,712.04×0.618/12

=79.97 $

Therefore,

Finance charge=79.97 $

Page 198 Problem 20 Answer

Given Helene’s credit card details

To do: Find her average daily balance

Given that for 29 days we have  $712.04 and for a day we have  $5712.04 then average daily balance is 29×$712.04+1×$5,712.04/30

=$878.71

Therefore,the average daily balance is  $878.71

Page 198 Problem 21 Answer

Given Helene’s credit card details

To do: Find the finance charge

We know that the average daily balance= $878.71 and APR=0.168/12 then

the finance charge will be $878.71×0.168/12

=$12.30

Therefore,the finance charge is  $12.30

Page 198 Problem 22 Answer

Given Helene’s credit card details

To do: Find how much can she save in finance charges

We have  the finance charge for the billing period is  $79.97 and the finance charge for the average daily balance is  $12.30

we get $79.97−$12.30=$67.67

Therefore, She saves  $67.67 finance charge

Page 199 Problem 23 Answer

Given the debit card register

To do: Find the missing values

The previous balance decreased by the payment amount will be the new balance

Following this we get

a.m−x

b.m−x−z

c.m−x−z−y

d.m−x−z−y−v

e.m−x−z−y−v+r

f.m−x−z−y−v+r−g

​Therefore, the missing values are m−x,m−x−z,m−x−z−y,m−x−z−y−v,m−x−z−y−v+r,m−x−z−y−v+r−g

Page 199 Problem 24 Answer

Given that Jill’s credit card was stolen. The thief charged a900 $ kayak on the card before she reported it stolen

To do: Find how much of the thief’s purchase is Jill re

The maximum liability will be  $50 for which she is responsible

Therefore,Jill is responsible for 50 $

Page 199 Problem 25 Answer

Given that Jill’s average daily balance would have been 1,240 $ without the thief’s purchase

To do; Find the sum of her daily balances

The product of the number of days and average daily balance will give the daily balances then we get

30×$1,240=$37,200

Therefore, The sum of her daily balances is 37,200 $

Page 199 Problem 26 Answer

Given that Jill’s credit card was stolen. The thief charged a 900 $ kayak on the card before she reported it stolen

To do: Find the sum of Jill’s daily balances

From the data, we get that before thief purchase we have 1,240 $ after thief purchase is 2,140 $ then the sum of Jill’s balances is

20×$1,240+10×$2,140=$46,200

Therefore,the sum of Jill’s balances is=46,200 $

Page 199 Problem 27 Answer

Given that Jill’s credit card was stolen. The thief charged a900 $ kayak on the card before she reported it stolen

To do: Find the average daily balance

Using the balance which we have previously increased by the thief’s purchase is the average daily balance we get

20×$1,240+10×$2,140=$46,200 when divided by the number of days we get

$46,200/30

=$1,540

Therefore, the average daily basis is 1,540 $

Page 199 Problem 28 Answer

Given Kristin’s credit details

To do: Find the finance charge

Firstly we have x×12%/12=x×1% and after the arrival of x we get x×13.2%/12

=x×1.1% the difference between these two will be x×1.1%−x×1%=x×0.1%

=0.001x

​Therefore, the  increase in this month’s finance charge is 0.001x

Page 199 Problem 29 Answer

Given data of credit card bill

To do: Find the total amount of overchange

The product of overcharge per credit card with the number of months is the total amount of debtors then we get

6×1,000,000×12×5=$360,000,000

Therefore,the total amount of overchange is =$360,000,000

Page 199 Problem 30 Answer

Given daily balances of Naoko

To do: Find the finance charge

Tha Naoko will not need to pay finance charges for next month then the September bill will be 0 $

Therefore,the finance charge for September month is 0 $

Page 199 Problem 31 Answer

Given the daily balances of Naoko

To do: Explain why the credit card company need to calculate his average daily balance

Naoko’s September bill is 0 $ which means we don’t have to pay any finance charges then the average daily balance is not needed

Therefore,there is no need to calculate average daily balance

Page 199 Problem 32 Answer

Given Naoko’s daily balances

To do: Find average daily balance

Firstly the sum of daily balances is ​

=2×99.78+15×315.64+11×515.64+2×580.32

=11,766.84​ and the number of billing days will be 2+15+11+2=30

Now the average daily balance is =11,766.84/30

≈392.23

​Therefore, the average daily balance is 392.23 $

Page 199 Problem 33 Answer

Given: Naoko daily balances

To do: Find the mistake of average daily balance

We have to calculate as all the number of daily balances with 4 then we get

=99.78+315.64+515.64+580.32/4

=1,511.38/4

≈377.85

now we got the correct answer

Therefore, the mistake he made is by summing all the balances and not dividing with 4

Financial Algebra 1st Edition Chapter 4 Consumer Credit

Page 188 Problem 1 Answer

The given data is, ​P=18000

T=8yrs

R=4.3%

How much will she pay in interest

The interest is calculated as,

I=PTR/100

=18000×8×4⋅3/100

=$6192

Therefore, the solution is calculated as, $6192.

​Page 189 Problem 2 Answer

Considering the previous example’s values.

Calculate the impact an increase in the monthly payment of $50 has on the length of the loan.

The loan length is calculated as,

⇒ \(=\frac{\ln \left(\frac{M}{P}\right)-\ln \left(\left(\frac{M}{p}-\frac{r}{12}\right)\right)}{12 \ln \left(1+\frac{\gamma}{12}\right)}\)

Additional amount = $50

Initial = $300

⇒ \(t=\frac{\ln \left(\frac{350}{25000}\right)-\ln \left(\frac{350}{25000}-\frac{5.9}{100^* 12}\right)}{12 \ln \left(1+\frac{5.9}{12 * 100}\right)}\)

⇒ \(=\frac{-4.2687-(-4.7013)}{0.05885}\)

= 7.35

= 7 years

Therefore, the solution is calculated as, 7 years.

Page 190 Problem 3 Answer

Considering the previous example’s values.

Compare the computed loan balances when x=2

The linear equation is, y=−6777,539x+13726080,

The quadratic equation is, y=−1007797000+1006140x−251,0951x²,

The cubic equation is, y=50082020000,00001−74983100x+37423,4x²−6,22616x³

Therefore, the solution is calculated for the equations and proved using the graphs.

Page 191 Problem 4 Answer

Find how might the quote apply to what you have learned

The quote implies that the loan and debts are considered as, which worries because this makes sure that the payment of the loan occurs every month on time and thus needs to make sure that enough money is set aside each month to pay for the loan.

But if the loans are failed to pay off, it is possible to lose everything.

Therefore, the payments of the loans have to be paid on time every month, to avoid worries.

Page 191 Problem 5 Answer

The given data is,​P=32000

r=6.1 %

t=10 yrs

​Calculate the total interest.

Hence the monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{32,000\left(\frac{0.061}{12}\right)\left(1+\frac{0.061}{12}\right)^{12 \times 10}}{\left(1+\frac{0.061}{12}\right)^{12 \times 10}-1}\)

⇒ \(\Rightarrow M=\$ 356.87\)

The total paid is the product of the monthly payment as,

$356.87×12×10=$42,824.40

$42,824.40−$32,000=$10,824.40

​Therefore, the solution is calculated as, $10824.4.

Page 191 Problem 6 Answer

The given data is, ​P=15000

t=4 yrs

r=5.5 %​

Calculate how much will the person pay in interest over the life of the loan.

Hence the monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{15,000\left(\frac{0.055}{12}\right)\left(1+\frac{0.055}{12}\right)^{12 \times 4}}{\left(1+\frac{0.055}{12}\right)^{12 \times 4}-1}\)

⇒ \(M=\$ 348.85\)

The total paid is the product of the monthly payment as,

$348.85×12×4=$16,744.80

$16,744.80−$15,000=$1,744.80

​Therefore, the solution is calculated as, $1744.8.

Page 191 Problem 7 Answer

The given data is, ​P=7000

t=8 yrs

r=8.6 %

​Calculate which loan will have the lower interest Over its lifetime.

 

Hence the monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{7,000\left(\frac{0.086}{12}\right)\left(1+\frac{0.086}{12}\right)^{12 \times 8}}{\left(1+\frac{0.086}{12}\right)^{12 \times 8}-1}\)

⇒ \(M=\$ 101.10\)

The total paid is the product of the monthly payment and the number of payments,

$101.10×12×8=$9,705.60

$9,705.60−$7,000=$2,705.60

Hence the monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{7,000\left(\frac{0.10}{12}\right)\left(1+\frac{0.10}{12}\right)^{12 \times 5}}{\left(1+\frac{0.10}{12}\right)^{12 \times 5}-1}\)

⇒ \(M=\$ 148.73\)

The total paid is the product of the monthly payment and the number of payments,

$148.73×12×5=$8,923.80

$8,923.80−$7,000=$1,923.80

Therefore, the solution is considered as first national bank.

Page 191 Problem 8 Answer

The given data is, ​P=$25000

r=7.7 %

t=2 to 10 yrs

Find the monthly payment formula for this loan situation.

Let t represent the number of years from 2 to 10 inclusive.

 

Hence the monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{25,000\left(\frac{0.077}{12}\right)\left(1+\frac{0.077}{12}\right)^{12 \times t}}{\left(1+\frac{0.077}{12}\right)^{12 \times t}-1}\)

⇒ \(M=\frac{160.417 \times 1.00642^{12 \times t}}{1.00642^{12 \times t}-1}\)

Therefore, the solution is calculated as, M=160.417×1.0064212×t/1.0064212×t−1

Page 191 Problem 9 Answer

The given data is,​P=$25000

r=7.7%

t=2 to 10 yrs ​

Find the total interest formula for this loan situation.

Let t represent the number of years from 2 to 10 inclusive.

⇒ \(M=\frac{160.417 \times 1.00642^{12 \times t}}{1.00642^{12 \times t}-1}\)

⇒ \(\frac{160.417 \times 1.00642^{12 \times t}}{1.00642^{12 \times t}-1} \times 12 \times t\)

⇒ \(\frac{160.417 \times 1.00642^{12 \times t}}{1.00642^{12 \times t}-1} \times 12 \times t-25,000\)

Therefore, the solution is 160.417×1.0064212×t/1.0064212×t−1×12×t−25,000

Page 191 Problem 10 Answer

The given data is,​P=$25000

r=7.7%

t=2 to 10yrs

Construct a graph

The graph is drawn as,

Therefore, the graph is drawn as, x is horizontal and y is vertical axis.

Page 191 Problem 11 Answer

The given data is, P=$25000

r=7.7 %

t=2 to10 yrs

​Use your graph to estimate the interest for a 6×1/2-year loan.

After plotting the graph, if t is an amount of 6 and a half, then the interest is about $6900.

Therefore, the graph was plotted and the solution was $6900.

Page 191 Problem 12 Answer

The given data is, ​P=$20000

r=7.1 %

t=1/10 yr​

Find the length of the loan.

Substituting the values,ln500

20000−(ln(500/20000−0.071/12))/12ln(1+0.071/12)

t=3.8 years

​Therefore, the solution is calculated as nearly 3.8 years.

Page 191 Problem 13 Answer

The given data is, ​M=$350

p=14,000

r=6.8

​Calculate the number of years.

⇒ \(t=\frac{\ln \left(\frac{M}{p}\right)-\ln \left(\frac{M}{p}-\frac{r}{12}\right)}{12 \ln \left(1+\frac{r}{12}\right)}\)

⇒ \(=\frac{\ln \left(\frac{350}{14000}\right)-\ln \left(\frac{350}{14000}-\frac{0.068}{12}\right)}{12 \ln \left(1+\frac{0.068}{12}\right)}\)

⇒ \(≈3.8 years\)

Therefore, decreasing the monthly payments by $50, increased the length of the loan by 0.5 years.

Page 191 Problem 14 Answer

The given data is,​P=$22000

t=15 years

r=4.85 %

​Calculate how much will he pay the bank in interest over the life of the loan.

The monthly payment formula is,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{22,000\left(\frac{0.0485}{12}\right)\left(1+\frac{0.0485}{12}\right)^{12 \times 15}}{\left(1+\frac{0.0485}{12}\right)^{12 \times 15}-1}\)

M= $172.26

The total paid is the product of the monthly payment,

⇒ \(\$ 172.26 \times 12 \times 15=\$ 31,006.80\)

⇒ \(\$ 31,006.80-\$ 22,000=\$ 9,006.80\)

≈$9000.

Therefore, the solution is calculated as, $9000.

Page 192 Exercise 1 Answer

Considering the table for the given yearly payment schedule. Calculate the loan amount.

Therefore, the value under the loan balance in the payment schedule is calculated as, $10000.

Page 192 Exercise 2 Answer

Considering the table for the given yearly payment schedule. Calculate the length of the loan.

Therefore, the last value under the year column is 10 years.

Page 192 Exercise 3 Answer

Considering the table for the given yearly payment schedule. Calculate the monthly payment.

The principal is considered as, 1455.93.

Dividing it by 12 to get the monthly pay,

1455.93/12≈121.33

Therefore, the solution is 121.33.

Page 192 Exercise 4 Answer

Considering the table for the given yearly payment schedule. Calculate the total interest paid.

Therefore, the total interest is calculated as, $4559.31, added from the entries under interest paid.

Page 192 Exercise 5 Answer

Considering the table for the given yearly payment schedule. Construct a scatterplot using the data points (year, loan balance).

The graph was plotted as shown,

Therefore, the given graph was plotted as,

Page 192 Exercise 6 Answer

Given a yearly payment schedule

To do: Write a linear regression

Determining the regression using STAT> CALC>4:LinReg(ax+b)L1,L2 then we get the following

​y=ax+b

a≈−992.49

b≈10554.14

Substituting in the equation y=ax+b

we get y=−992.49x+10554.14

Therefore, the linear regression equation is y=−992.49x+10554.14

Page 192 Exercise 7 Answer

Given a yearly payment schedule

To do: Write a quadratic regression

Determining the regression using STAT> CALC>5:QuadReg L1,L2 then we get the result as follows

y=ax²+bx+c

a≈−39.35

b≈−599.01

c≈9963.93

Substituting in the form of the quadratic equation we get

y=−39.35x²−599.01x+9963.93

Therefore, the quadratic regression equation is y=−39.35x²−599.01x+9963.93

Page 192 Exercise 8 Answer

Given a yearly payment schedule

To do: Write a cubic regression

Determining the regression equation using STAT>CALC>6: CubicReg L1,L2 then we get the results as

​y=ax³+bx²+cx+d

a≈−1.04

b≈−23.71

c≈−658.64

d≈−10001.46

substituting in the equation we get

y=−1.04x³−23.71x²−658.64x+10001.46

Therefore, the cubic regression equation is y=−1.04x³−23.71x²−658.64x+10001.46

Page 192 Exercise 9 Answer

Given a payment schedule

To do: Find the loan amount

The first number given in the column of the loan balance is 35,000.00 $ then the loan amount will be this

Therefore, the loan amount is 35,000.00 $

Page 192 Exercise 10 Answer

Given the payment schedule

To do: Find the length of the loan

The starting year is 2009 and the ending year is 2027 the difference between them is 18

so, the length of the loan is 18 years

Therefore, the length of the loan is 18 years

age 192 Exercise 11 Answer

Given a payment schedule

To do: Find the approximate monthly payment

The summation of principal paid and interest paid for the value of 4065.23 then

4065.23/12≈338.77

so, the monthly payment will be 338.77 $

Therefore, the monthly payment is 338.77 $

Page 192 Exercise 12 Answer

Given a payment schedule

To do: Find the total interest paid

Summation of all the year’s interests we get​=3291.90+3215.15+3130.78+3038.04+2936.10+2824.03+2700.85+2565.44+2416.59+2252.96+2073.10+1875.39+1658.05+1419.14+1156.53+867.84+550.51+201.69

=$38,174.09

​Therefore, the total interest paid 38,174.09 $

Page 192 Exercise 13 Answer

Given a payment schedule

To do: Construct a scatter plot

The graph of the year-to-load balance is as

Therefore, the scatter plot is as

Page 192 Exercise 14 Answer

Given a payment schedule

To do: Find the linear regression

Determining the regression using STAT> CALC>4: Lin Reg(ax+b) L1,L2 then we get the result as

y=ax+b

a≈−1870.62

b≈3796896.07

Substituting in the line equation we get y=−1870.62x+3796896.07

Therefore, the linear regression is y=−1870.62x+3796896.07

Page 192 Exercise 15 Answer

Given a  payment schedule

To do: Write a quadratic regression

Determining the regression using STAT> CALC>5: QuadRegL1,L2 using these we get

​y=ax²+bx+c

a≈−76.07

b≈−305369.01

c≈306434730.37

Substituting the values in the quadratic equation we get

y=−76.07x²−305369.01x+306434730.37

Therefore, the quadratic regression is y=−76.07x²−305369.01x+306434730.37

Page 192 Exercise 16 Answer

Given a payment schedule

To do: Write a cubic regression

Determining the regression using STAT> CALC>6: CubicReg L1, L2 and the results using this are

y=ax³+bx²+cx+d

a≈−2.81

b≈16936.85

c≈−34,006,808.68

d≈2.2760898E10≈22,760,898,000​

substituting the values in the general equation we get

y=−2.81×3+16936.85x²−34,006,808.68x+22,760,898,000

Therefore, the cubic regression equation is y=−2.81×3+16936.85x²−34,006,808.68x+22,760,898,000

Financial Algebra, 1st Edition, Chapter 4: Consumer Credit

Page 183 Problem 1 Answer

We are given: p=$41,000

r=0.065

t=5 .

We have to find the monthly payment.

We will be using the formula of monthly payment :

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

We will substitute the given values in the formula

⇒ \(M=\frac{41000\left(\frac{0.065}{12}\right)\left(1+\frac{0.065}{12}\right)^{12(5)}}{\left(1+\frac{0.065}{12}\right)^{12(5)}-1}\)

⇒ \(M=\frac{41000(0.0054166666666667)(1+0.00541666666666667)^{60}}{(1+0.00541666666666667)^{60}-1}\)

⇒ \(M=\frac{222.0833333333333(1.0054166666666667)^{60}}{(1+0.0054166666666667)^{60}-1}\)

⇒ \(M=\frac{222.0833333333333(1.38281732421)}{0.38281732421}\)

⇒ \(M=\frac{307.1006807516375}{0.38281732421}\)

⇒ \(M=802.2120769622562\)

The monthly payment is $802.2120769622562, we can round off the answer to the nearest cent.

Therefore, we get $802.21

Page 1843 Problem 2 Answer

We are given: Borrowed money=x dollars

Monthly payment=y dollars

Number of years=3

Number of months=36.

We have to express the finance charge algebraically.

Firstly, we will find the total of monthly payments.

The total of monthly payments is given by: 36×y=36y dollars.

Now, we will find the finance charge.

Total amount of monthly payments−Borrowed amount

Finance charge=36y−x dollars.

We can express the finance charge algebraically as 36y−x dollars.

Page 184 Problem 3 Answer

We are given: p=$1000

r=0.075

t=1 .

We have to find the monthly payment.

We will be using the formula of monthly payment :

\(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1} \text {. }\)

We will substitute the given values in the formula

⇒ \(M=\frac{1000\left(\frac{0.075}{12}\right)\left(1+\frac{0.075}{12}\right)^{12(1)}}{\left(1+\frac{0.075}{12}\right)^{12(1)}-1}\)

⇒ \(M=\frac{1000(0.00625)(1+0.00625)^{12(1)}}{\left(1+\frac{0.075}{12}\right)^{12(1)}-1}\)

⇒ \(M=\frac{1000(0.00625)(1.00625)^{12}}{\left(1+\frac{0.075}{12}\right)^{12}-1}\)

⇒ \(M=\frac{(6.25)(1.07763259886)}{(1.07763259886)^{12}-1}\)

⇒ \(M=\frac{(6.25)(1.07763259886)}{(1.07763259886)-1}\)

⇒ \( M=\frac{6.735203742875}{0.07763259886}\)

⇒ \(M=86.75741689214363 \)

The monthly payment is$86.75741689214363.

Page 185 Problem 4 Answer

We are given that :

Monthly payments are made for a five-year loan and a two-year loan.

We have to find how many more monthly payments are made for a five-year loan than for a two-year loan.

Number of monthly payments is the product of 12 months and number of years.

Several monthly payments for a five-year loan is given by: 12×5=60.

Number of monthly payments for a five-year loan is given by:12×2=24.

Now, we will subtract the number of monthly payments for a two-year loan from the number of monthly payments for a five-year loan.

60−24=36 .

36 more monthly payments are made for a five-year loan than for a two-year loan.

Page 185 Problem 5 Answer

We are given: an A2x1/2 loan.

We have to find the number of monthly payments that must be made for a 2×1/2-year loan.

Number of monthly payments is the product of 12 months and number of years.

Number of monthly payments for a 2 x1/2 loan is given by :

​2×1/2×12=5

2×1/2

2×1/2×12=2.5×12

2×1/2×12=60 .

​60 monthly payments must be made for a 2×1/2-year loan.

Number of monthly payments for a2x1/2the loan is given by :

​2×1/2×12=5

2×1/2

2×1/2×12=2.5×12

2×1/2×12=60 .

​60 monthly payments must be made for a 2×1/2-year loan.

Page 185 Problem 6 Answer

We are given: p=$7000

r=0.0975

t=1.

We have to find the monthly payment for a one-year loan

We will be using the formula of monthly payment.

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1} .\)

We will substitute the given values in the formula

⇒ \(M=\frac{7000\left(\frac{0.0975}{12}\right)\left(1+\frac{0.0975}{12}\right)^{12(1)}}{\left(1+\frac{0.0975}{12}\right)^{12(1)}-1}\)

⇒ \(M=\frac{7000(0.008125)(1+0.008125)^{12}}{(1+0.008125)^{12}-1}\)

⇒ \(M=\frac{56.875(1.10197721973)}{1.10197721973-1}\)

⇒ \(M=\frac{62.67495437214375}{0.10197721973}\)

M = 614.5975987390625.

The monthly payment for a one-year loan is $614.5975987390625, we can round off the answer to the nearest cent.

Therefore we get $614.60.

Page 185 Problem 7 Answer

The given data is, ​p=7,000,

r=0.0975,

t=3

Calculate the monthly payment for a three-year loan.

Substituting the given data as

⇒ \(M=\frac{7000\left(\frac{0.0975}{12}\right)\left(1+\frac{0.0975}{12}\right)^{12(3)}}{\left(1+\frac{0.0975}{12}\right)^{12(3)}-1}\)

 

⇒ \(M=\frac{7000\left(\frac{0.0975}{12}\right)\left(1+\frac{0.0975}{12}\right)^{36}}{\left(1+\frac{0.0975}{12}\right)^{36}-1}\)

Using the calculator, \(\frac{7000(0.0975 / 12)(1+0.0975 / 12) 36}{(1+0.0975 / 12)^{36}-1}=225.05\)

Therefore, the solution was calculated as, $225.05

Page 185 Problem 8 Answer

The given data is, ​p=7,000,

r=0.0975,

t=3

​Calculate the monthly payment for a five-year loan.

Substituting the given data as

⇒ \(M=\frac{7000\left(\frac{0.0975}{12}\right)\left(1+\frac{0.0975}{12}\right)^{12(5)}}{\left(1+\frac{0.0975}{12}\right)^{12(5)}-1}\)

⇒ \(M=\frac{7000\left(\frac{0.0975}{12}\right)\left(1+\frac{0.0975}{12}\right)^{60}}{\left(1+\frac{0.0975}{12}\right)^{60}-1}\)

Using the calculator, \(\frac{7000(0.0975 / 12)(1+0.0975 / 12)^{60}}{(1+0.0975 / 12)^{60}-1}=147.87\)

Therefore, the solution was calculated as, $147.87.

Page 185 Problem 9 Answer

The given data is, ​​p=10,000,

r=0.0725,

t=3

​Calculate the monthly payment.

Substituting the given data as

⇒ \(M=\frac{10000\left(\frac{0.0725}{12}\right)\left(1+\frac{0.0725}{12}\right)^{12(3)}}{\left(1+\frac{0.0725}{12}\right)^{12(3)}-1}\)

⇒ \(M=\frac{10000\left(\frac{0.0725}{12}\right)\left(1+\frac{0.0725}{12}\right)^{36}}{\left(1+\frac{0.0725}{12}\right)^{36}-1}\)

Using the calculator \(\frac{10000(0.0725 / 12)(1+0.0725 / 12)^{36}}{(1+0.0725 / 12)^{36}-1}=309.92\)

Therefore, the solution was calculated as, $309.92.

Page 185 Problem 10 Answer

The given data is, ​​p=10,000,

r=0.0725

t=3

​Calculate the total amount of the monthly payments.

Considering the previous question as the Monthly payment is 309.92, The total of monthly payments as,309.92×36=11157.12.

Therefore, the solution is calculated as, 309.92.

Page 185 Problem 11 Answer

The given data is, ​ p​=10,000

r=0.0725

t=3

Calculate the finance charge.

Considering the previous question as,

The total of monthly payments is, $11,157.12, Subtracting principal from a total of monthly payments,

​​11,157.12−10,000=1,157.12.

Therefore, the solution is calculated as, $1157.12.

Page 185 Problem 12 Answer

The given data is,​​p=6,000,

r=0.1,

t=4

​Calculate the monthly payment.

Substituting the values as

⇒ \(M=\frac{6000\left(\frac{0.1}{12}\right)\left(1+\frac{0.1}{12}\right)^{12(4)}}{\left(1+\frac{0.1}{12}\right)^{12(4)}-1}\)

⇒ \(M=\frac{6000\left(\frac{0.1}{12}\right)\left(1+\frac{0.1}{12}\right)^{48}}{\left(1+\frac{0.1}{12}\right)^{48}-1}\)

Using the calculator, \(\frac{6000(0.1 / 12)(1+0.1 / 12)^{48}}{(1+0.1 / 12)^{48}-1}=152.18\)

​Therefore, the solution is calculated as, $152.18.

Page 185 Problem 13 Answer

The given data is,​​p=6,000,

r=0.1

t=4

​Find the total amount of the monthly payments.

Considering the previous question, the Monthly payment is 152.18.

The total of monthly payments is 152.18×48=7,304.64.

Therefore, the solution is calculated as, $7304.64

Page 185 Problem 14 Answer

The given data is,​​p=6,000,

r=0.1

t=4

​Calculate the finance charge.

Considering the previous question, The total of monthly payments is, 7304.64, Subtracting principal from the total of monthly payments,

7,304.64−6,000=1,304.64

Therefore, the solution is calculated as, 1304.64.

Page 185 Problem 15 Answer

The given data is,​R=$3000

N=$1200

v=35 %

​Calculate the maximum amount that could be borrowed from Broadway.

This shows that the collateral for a loan is, 3000+1200=4200 USD

Since the borrower’s value is 35 %, 4200×0.35=1470 USD.

Therefore, the solution is calculated as 1470 USD.

Page 185 Exercise 1 Answer

The given data is,​​p=8,700,

r=0.0931,

t=3.5

​Calculate the monthly payment for this loan.

Substituting the values in the formula, Using the calculator,

⇒ \(M=\frac{8700\left(\frac{0.0931}{12}\right)\left(1+\frac{0.0931}{12}\right)^{12(3.5)}}{\left(1+\frac{0.0931}{12}\right)^{12(3.5)}-1}\)

⇒ \(M=\frac{8700\left(\frac{0.0931}{12}\right)\left(1+\frac{0.0931}{12}\right)^{42}}{\left(1+\frac{0.0931}{12}\right)^{42}-1}\)

​(8700(0.0931/12)(1+0.0931/12)∧42)/((1+0.0931/12)∧42−1)​

Therefore, the solution is calculated as, $243.52.

Page 185 Exercise 2 Answer

The given data is, ​p=7500

r=6.875 %

Calculate how many monthly payments the person makes.

The monthly payments are calculated as, Multiply 6 by 12.

1 year =12 months.

So 6 years 6×12=72 months.

Therefore, the solution is calculated as 72 months.

Page 186 Exercise 3 Answer

The given data is, ​P=15320

r=10.29 %

Calculate the finance charge for this loan to the nearest dollar.

Substituting the value is substituted as,​

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{15,320\left(\frac{0.1029}{12}\right)\left(1+\frac{0.1029}{12}\right)^{12 \times 2}}{\left(1+\frac{0.1029}{12}\right)^{12 \times 2}-1}\)

⇒ \(\$ 708.99 \times 12 \times 2=\$ 17,015.76\)

⇒ \(\$ 17,015.76-\$ 15,320=\$ 1,695.76\)

Therefore, the solution is calculated as, $1695.76.

Page 186 Exercise 4 Answer

The given data is that the credit union will lend the person $8000 for three years at 8.25 % APR.

The same loan at her savings bank has an APR of 10.5 %.

Calculate how much would the given person would save in finance charges if joined the credit union.

Using the monthly payment formula,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{8,000\left(\frac{0.0825}{12}\right)\left(1+\frac{0.0825}{12}\right)^{12 \times 3}}{\left(1+\frac{0.0825}{12}\right)^{12 \times 3}-1}\)

⇒ \(M=\$ 251.61\)

⇒ \(\$ 251.61 \times 12 \times 3=\$ 9,057.96\)

⇒ \(\$ 9,057.96-\$ 8,000=\$ 1,057.96\)

Also calculating the values as,

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{8,000\left(\frac{0.105}{12}\right)\left(1+\frac{0.105}{12}\right)^{12 \times 3}}{\left(1+\frac{0.105}{12}\right)^{12 \times 3}-1}\)

⇒ \(M=\$ 260.02\)

⇒ \(\$ 260.02 \times 12 \times 3=\$ 9,360.72\)

⇒ \(\$ 9,360.72-\$ 8,000=\$ 1,360.72\)

⇒ \(\$ 1,360.72-\$ 1,057.96=\$ 302.76\)

Therefore, the solution is calculated as, $302.76.

Page 186 Exercise 5 Answer

The given data is, ​P=$5000

d=$800

n=$202.5

​Calculate the interest on this installment agreement.

The paid is the product of the monthly payments as,

​$202.50×12×2+$800=$5,660

$5,660−$5,000=$660

​Therefore, the solution is calculated as, $660.

Page 186 Exercise 6 Answer

The given data is, ​P=$5000

d=$800

n=$202.5

​Calculate the monthly payment for this loan using the table.

The monthly payment formula is as,

 

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{5,000\left(\frac{0.13}{12}\right)\left(1+\frac{0.13}{12}\right)^{12 \times 2}}{\left(1+\frac{0.13}{12}\right)^{12 \times 2}-1} \)

⇒ \(M=\$ 237.71\)

Therefore, the solution is calculated as, $237.71.

Page 186 Exercise 7 Answer

The given data is, ​P=$5000

d=$800

n=$202.5

​Calculate the interest the finance company charges.

The product of the monthly payments is,

$237.71×12×2=$5,705.04

$5,705.04−$5,000=$705.04

​Therefore, the solution was calculated as, $705.04.

Page 186 Exercise 8 Answer

The given data is, ​P=$5000

d=$800

n=$202.5

​Calculate whether Rob uses the installment plan or borrows the money from the finance company.

Since Rob has to pay less interest, the installment plan is a better choice.

Page 186 Exercise 9 Answer

The given data is, P=$8400

r=7 %

t=2 yrs

Explain what was incorrectly entered.

Since Lee missed the parentheses in the calculator, hence can’t differentiate between numerator and denominator.

Hence the BODMAS rule is applied as,

​(8400(.07/12)(1+0.07/12)∧24)/((1+0.07/12)∧24−1).

Therefore, the solution is calculated as 376.089.

Page 186 Exercise 10 Answer

The given data is,p=430,000,

r=0.08,

t=30​

Compute the monthly payment.

Substituting the value in the formula,

⇒ \(M=\frac{430,000\left(\frac{0.08}{12}\right)\left(1+\frac{0.08}{12}\right)^{12(30)}}{\left(1+\frac{0.08}{12}\right)^{12(30)}-1}\)

⇒ \(M=\frac{430,000\left(\frac{0.08}{12}\right)\left(1+\frac{0.08}{12}\right)^{360}}{\left(1+\frac{0.08}{12}\right)^{360}-1}\)

Therefore, the solution is calculated as, $3155.19.

Page 186 Exercise 11 Answer

The given data is, ​p=430,000,

r=0.08,

t=30​

Find the total of all of the monthly payments for the 30 years.

Considering the values from the previous question, The monthly payment is 3155.19

Multiple monthly payments by the month are,

3155.19×30×12=3155.19×360

=1,135,868.40

​Therefore, the total amount of monthly payments is $1135868.4

Page 186 Exercise 12 Answer

The given data is, P=430000

r=8 %

t=30 yrs

​Calculate the finance charge.

Considering the previous question as the Total of monthly payments is, $1135868.4, Subtracting principal from total monthly payments,

1,135,868.40−430,000=705,868.40

Therefore, the solution is calculated as a finance charge, $705868.4

Page 186 Exercise 13 Answer

The given data is, P=430000

r=8%

t=30yrs​

Prove which is greater, the interest or the original cost of the home.

Considering the values from the previous questions,

Hence, the interest or Finance charge is $705868.4, Compare the interest ($705868.4)

with original cost ($430000).

Therefore, the interest is more than the original cost of the home.

Page 186 Exercise 14 Answer

The given table is considered. Write the spreadsheet formula to compute cell D₂

Write the spreadsheet formula to compute cell E₂.

The cell should contain the time in month for the data in the second row as,

Time in months=D₂

The years of the second row are given in the cell,

Time in years=C₂

Time in months

D₂= Time in years ×12

=C₂×12

The formula for the monthly payment is,

 

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(M=\frac{p\left(\frac{r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}}{\left(1+\frac{r}{12}\right)^{12 t}-1}\)

⇒ \(E 2=\frac{A 2\left(\frac{B 2}{12}\right)\left(1+\frac{B 2}{12}\right)^{12 \times C 2}}{\left(1+\frac{B 2}{12}\right)^{12 \times C 2}-1}\)

⇒ \(E 2=(A 2 *(B 2 / 12) *(1+B 2 / 12) \hat{0}(12 * C 2)) /((1+B 2 / 12) \hat{0}(12 * C 2)-1)\)

Therefore, the solution is calculated as D₂=C₂⋅1₂ and E₂=(A₂x(B₂/1₂)∗(1+B₂/12)0

(12∗C₂))/((1+B₂/12)0

(12x C₂)−1).