We are presented with 10 finance-related math problems covering simple interest, compound interest, annuities, and loan amortization. We will solve each problem step-by-step.

1. Problem Description

We are presented with 10 finance-related math problems covering simple interest, compound interest, annuities, and loan amortization. We will solve each problem step-by-step.

2. Solution Steps

Problem 1: Simple Interest

John invests $5000 in a savings account that pays an annual simple interest rate of 6%. Calculate the interest earned and the total amount after 3 years.

The formula for simple interest is:

$I = P * r * t$

Where:

$I$ = Interest earned

$P$ = Principal amount

$r$ = Annual interest rate

$t$ = Time in years

$I = 5000 * 0.06 * 3 = 900$

Total amount after 3 years = Principal + Interest

$A = P + I = 5000 + 900 = 5900$

Problem 2: Simple Interest Loan

A loan of $12,000 is taken at an annual simple interest rate of 8% for 5 years. Find the total interest paid and the final amount to be repaid.

Using the same simple interest formula:

$I = P * r * t$

$I = 12000 * 0.08 * 5 = 4800$

Total amount to be repaid = Principal + Interest

$A = P + I = 12000 + 4800 = 16800$

Problem 3: Compound Interest

Mary deposits $3000 in a bank account with a compounded 5% annual interest rate. Calculate the total amount after 4 years.

The formula for compound interest is:

$A = P(1 + r)^n$

Where:

$A$ = Total amount after $n$ years

$P$ = Principal amount

$r$ = Annual interest rate

$n$ = Number of years

$A = 3000(1 + 0.05)^4$

$A = 3000(1.05)^4$

$A = 3000 * 1.21550625 = 3646.52$

Problem 4: Compound Interest Semi-Annually

A sum of $7,500 is invested at an interest rate of 4% compounded semi-annually. Determine the amount accumulated after 6 years.

The formula for compound interest is:

$A = P(1 + \frac{r}{n})^{nt}$

Where:

$A$ = Total amount after $t$ years

$P$ = Principal amount

$r$ = Annual interest rate

$n$ = Number of times interest is compounded per year

$t$ = Number of years

$A = 7500(1 + \frac{0.04}{2})^{(2*6)}$

$A = 7500(1 + 0.02)^{12}$

$A = 7500(1.02)^{12}$

$A = 7500 * 1.268241795 = 9511.81$

Problem 5: Compound Interest Monthly and Quarterly

Mr. Asif has invested Tk.1,00,000 for 5 years at 10% rate of interest. Calculate the compound interest and total amount after 5 years if interest is paid: (i) Monthly, or (ii) Quarterly.

(i) Monthly Compounding:

$A = P(1 + \frac{r}{n})^{nt}$

$A = 100000(1 + \frac{0.10}{12})^{(12*5)}$

$A = 100000(1 + 0.00833333)^{60}$

$A = 100000(1.00833333)^{60}$

$A = 100000 * 1.645308935 = 164530.89$

Compound Interest = $164530.89 - 100000 = 64530.89$

(ii) Quarterly Compounding:

$A = P(1 + \frac{r}{n})^{nt}$

$A = 100000(1 + \frac{0.10}{4})^{(4*5)}$

$A = 100000(1 + 0.025)^{20}$

$A = 100000(1.025)^{20}$

$A = 100000 * 1.63861644 = 163861.64$

Compound Interest = $163861.64 - 100000 = 63861.64$

Problem 6: Future Value of an Ordinary Annuity

A person deposits $200 at the end of every month into a savings account that earns 6% annual interest compounded monthly. Find the amount accumulated after 5 years.

$FV = P * \frac{((1 + r)^n - 1)}{r}$

Where:

$FV$ = Future value of the annuity

$P$ = Periodic payment

$r$ = Interest rate per period

$n$ = Number of periods

$r = \frac{0.06}{12} = 0.005$

$n = 5 * 12 = 60$

$FV = 200 * \frac{((1 + 0.005)^{60} - 1)}{0.005}$

$FV = 200 * \frac{(1.005^{60} - 1)}{0.005}$

$FV = 200 * \frac{(1.348850153 - 1)}{0.005}$

$FV = 200 * \frac{0.348850153}{0.005}$

$FV = 200 * 69.7700306 = 13954.01$

Problem 7: Future Value of an Ordinary Annuity

A firm intends to invest Tk.2,50,000 at the end of each year and receive interest on the amounts at 10% per annum. What sum of money will be available at the end of the fifth year?

$FV = P * \frac{((1 + r)^n - 1)}{r}$

Where:

$FV$ = Future value of the annuity

$P$ = Periodic payment

$r$ = Interest rate per period

$n$ = Number of periods

$P = 250000$

$r = 0.10$

$n = 5$

$FV = 250000 * \frac{((1 + 0.10)^5 - 1)}{0.10}$

$FV = 250000 * \frac{(1.10^5 - 1)}{0.10}$

$FV = 250000 * \frac{(1.61051 - 1)}{0.10}$

$FV = 250000 * \frac{0.61051}{0.10}$

$FV = 250000 * 6.1051 = 1526275$

Problem 8: Compound Interest with Withdrawals

A man invests Tk. 10,000 once and withdraws Tk. 1500 at the end of each year starting at the end of the first year. How much will be left after seven years if the money is invested at 4% per annum?

Year 1:

Amount after interest: $10000 * 1.04 = 10400$

Amount after withdrawal: $10400 - 1500 = 8900$

Year 2:

Amount after interest: $8900 * 1.04 = 9256$

Amount after withdrawal: $9256 - 1500 = 7756$

Year 3:

Amount after interest: $7756 * 1.04 = 8066.24$

Amount after withdrawal: $8066.24 - 1500 = 6566.24$

Year 4:

Amount after interest: $6566.24 * 1.04 = 6828.8896$

Amount after withdrawal: $6828.8896 - 1500 = 5328.8896$

Year 5:

Amount after interest: $5328.8896 * 1.04 = 5542.045184$

Amount after withdrawal: $5542.045184 - 1500 = 4042.045184$

Year 6:

Amount after interest: $4042.045184 * 1.04 = 4203.72699136$

Amount after withdrawal: $4203.72699136 - 1500 = 2703.72699136$

Year 7:

Amount after interest: $2703.72699136 * 1.04 = 2811.8760710144$

Amount after withdrawal: $2811.8760710144 - 1500 = 1311.88$

Problem 9: Present Value of an Annuity Due

Mr. Karim can purchase a machine by paying Tk.40,000 in cash now. He can also purchase the machine by 8 equal yearly installments to be paid at the beginning of each year. If the interest rate is 12%, what should be the amount of each installment?

$PV = P * \frac{1 - (1 + r)^{-n}}{r} * (1 + r)$

Where:

$PV$ = Present value of the annuity due

$P$ = Periodic payment

$r$ = Interest rate per period

$n$ = Number of periods

$40000 = P * \frac{1 - (1 + 0.12)^{-8}}{0.12} * (1 + 0.12)$

$40000 = P * \frac{1 - (1.12)^{-8}}{0.12} * 1.12$

$40000 = P * \frac{1 - 0.403875}{0.12} * 1.12$

$40000 = P * \frac{0.596125}{0.12} * 1.12$

$40000 = P * 4.96770833 * 1.12$

$40000 = P * 5.564$

$P = \frac{40000}{5.564} = 7188.90$

Problem 10: Loan Amortization Schedule

You borrowed Tk.42,000 at 12% to be repaid over the next 8 years. Equal installment payments are required at the end of each year and these payments must be sufficient to repay the Tk.42,000 while providing the lender a 8% return. Prepare an amortization schedule based on these data.

This problem statement is contradictory. The loan is at 12%, but the lender expects an 8% return. This is unusual. I will solve it as if the loan interest rate is 12% and find the annual payment.

$P = \frac{PV * r}{1 - (1 + r)^{-n}}$

Where:

$P$ = Periodic payment

$PV$ = Present value of the loan

$r$ = Interest rate per period

$n$ = Number of periods

$P = \frac{42000 * 0.12}{1 - (1 + 0.12)^{-8}}$

$P = \frac{5040}{1 - (1.12)^{-8}}$

$P = \frac{5040}{1 - 0.403875}$

$P = \frac{5040}{0.596125}$

$P = 8454.63$

Here's the amortization schedule assuming 12% interest for the loan repayment:

Year | Beginning Balance | Payment | Interest | Principal | Ending Balance

------- | -------- | -------- | -------- | -------- | --------

1 | 42000.00 | 8454.63 | 5040.00 | 3414.63 | 38585.37

2 | 38585.37 | 8454.63 | 4630.24 | 3824.39 | 34760.98

3 | 34760.98 | 8454.63 | 4171.32 | 4283.31 | 30477.67

4 | 30477.67 | 8454.63 | 3657.32 | 4797.31 | 25680.36

5 | 25680.36 | 8454.63 | 3081.64 | 5372.99 | 20307.37

6 | 20307.37 | 8454.63 | 2436.88 | 6017.75 | 14289.62

7 | 14289.62 | 8454.63 | 1714.75 | 6739.88 | 7549.74

8 | 7549.74 | 8454.63 | 905.97 | 7548.66 | 1.08

3. Final Answer

Problem 1: Interest earned = $900, Total amount =$5900

Problem 2: Total interest = $4800, Total amount to be repaid =$16800

Problem 3: Total amount after 4 years = $3646.52

Problem 4: Amount accumulated after 6 years = $9511.81

Problem 5: (i) Monthly: Compound Interest = Tk.64530.89, Total amount = Tk.164530.89, (ii) Quarterly: Compound Interest = Tk.63861.64, Total amount = Tk.163861.64

Problem 6: Amount accumulated after 5 years = $13954.01

Problem 7: Sum of money available at the end of the fifth year = Tk.1526275

Problem 8: Amount left after seven years = Tk.1311.88

Problem 9: Amount of each installment = Tk.7188.90

Problem 10: Annual Payment = Tk.8454.63, Loan Amortization Schedule provided above.