Question

1. find the compound amount if \\(p20,000.00\\) is invested at \\(10\\%\\) compounded quarterly for 5 years and 6 months.
2. if \\(p18,500.00\\) is invested for 6 years and 3 months at \\(9\\%\\) compounded semi-annually, find the compound amount.
3. accumulate \\(p15,000.00\\) for 7 years and 8 months at \\(7\\%\\) compounded quarterly.
4. if \\(p45,000.00\\) is invested for 8 years and 4 months at the rate of \\(9\frac{1}{2}\\%\\) converted annually, find the compound amount.
5. on june 15, 2015, maria borrowed \\(p25,000.00\\). she agreed to pay the principal and the interest at \\(12\\%\\) compounded semi-annually on august 15, 2015. how much will she pay then?
6. find the compound amount of \\(p12,000.00\\) at the end of 6 years and 8 months if invested at \\(10\frac{1}{2}\\%\\) compounded quarterly.
7. john borrowed \\(p40,000.00\\) from philippine national bank, which charges \\(10\\%\\) compounded quarterly. if he intends to pay the accumulated amount 4 years and 8 months from now, how much will he pay then?
8. find the compound amount of \\(p25,000.00\\) if it is invested at \\(15\\%\\) converted semi-annually for 7 years and 6 months.
9. find the compound amount of \\(p35,000.00\\) at the end of \\(5\frac{1}{2}\\%\\) years at \\(8\\%\\) compounded annually.
10. find the compound amount of \\(p32,000.00\\) if invested at \\(7\frac{1}{2}\\%\\) compounded semi-annually for \\(6\frac{1}{4}\\) years.

Explanation:

Identify parameters for each question

We use the compound interest formula:

$$S = P(1 + \frac{r}{m})^{mt}$$

where \(P\) is the principal, \(r\) is the nominal annual interest rate, \(m\) is the compounding frequency per year, and \(t\) is the time in years.

1. \(P = 20,000.00\), \(r = 0.10\), \(m = 4\) (quarterly), \(t = 5.5\) years (5 years and 6 months).
2. \(P = 18,500.00\), \(r = 0.09\), \(m = 2\) (semi-annually), \(t = 6.25\) years (6 years and 3 months).
3. \(P = 15,000.00\), \(r = 0.07\), \(m = 4\) (quarterly), \(t = 7\frac{2}{3}\) years (7 years and 8 months).
4. \(P = 45,000.00\), \(r = 0.095\), \(m = 1\) (annually), \(t = 8\frac{1}{3}\) years (8 years and 4 months).
5. \(P = 25,000.00\), \(r = 0.12\), \(m = 2\) (semi-annually), \(t = \frac{61}{365}\) years (June 15 to August 15 is 61 days).
6. \(P = 12,000.00\), \(r = 0.105\), \(m = 4\) (quarterly), \(t = 6\frac{2}{3}\) years (6 years and 8 months).
7. \(P = 40,000.00\), \(r = 0.10\), \(m = 4\) (quarterly), \(t = 4\frac{2}{3}\) years (4 years and 8 months).
8. \(P = 25,000.00\), \(r = 0.15\), \(m = 2\) (semi-annually), \(t = 7.5\) years (7 years and 6 months).
9. \(P = 35,000.00\), \(r = 0.08\), \(m = 1\) (annually), \(t = 5.5\) years.
10. \(P = 32,000.00\), \(r = 0.075\), \(m = 2\) (semi-annually), \(t = 6.25\) years.

Calculate compound amounts

1. \(S = 20000(1 + \frac{0.10}{4})^{4 \times 5.5} = 20000(1.025)^{22} \approx 34,432.19\)
2. \(S = 18500(1 + \frac{0.09}{2})^{2 \times 6.25} = 18500(1.045)^{12.5} \approx 31,859.69\)
3. \(S = 15000(1 + \frac{0.07}{4})^{4 \times 7.6667} = 15000(1.0175)^{30.6667} \approx 25,516.48\)
4. \(S = 45000(1 + 0.095)^{8.3333} \approx 95,734.42\)
5. \(S = 25000(1 + \frac{0.12}{2})^{2 \times \frac{61}{365}} = 25000(1.06)^{0.3342} \approx 25,489.10\)
6. \(S = 12000(1 + \frac{0.105}{4})^{4 \times 6.6667} = 12000(1.02625)^{26.6667} \approx 23,956.19\)
7. \(S = 40000(1 + \frac{0.10}{4})^{4 \times 4.6667} = 40000(1.025)^{18.6667} \approx 63,353.60\)
8. \(S = 25000(1 + \frac{0.15}{2})^{2 \times 7.5} = 25000(1.075)^{15} \approx 73,971.44\)
9. \(S = 35000(1 + 0.08)^{5.5} \approx 53,281.82\)
10. \(S = 32000(1 + \frac{0.075}{2})^{2 \times 6.25} = 32000(1.0375)^{12.5} \approx 50,774.02\)

Answer:

No.	Answer
2	₱31,859.69
3	₱25,516.48
4	₱95,734.42
5	₱25,489.10
6	₱23,956.19
7	₱63,353.60
8	₱73,971.44
9	₱53,281.82
10	₱50,774.02