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).