Boitumelo invests an amount of R28 000 for a period of 7 years. For the first twc years the interest was $$ 12 \% $$ p.a. compounded semi-annually. The interest was the changed to $$ 10 \% $$ p.a. compounded quarterly. Determine the value of the investment after the 7 -year period. Convert an interest rate of $$ 10 \% $$ p.a. compounded quarterly to an annual effective interest rate. A lawnmower was traded in for R236 after 10 years. The depreciation was calculated at $$ 9,2 \% $$ p.a. on a straight-line basis. Determine the cost of the lawnmower 10 years ago.

Question

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 1: Value of the Investment After 7 Years

1. **First 2 Years at 12% p.a. Compounded Semi-Annually:**
   - Principal (P) = R28,000
   - Rate (r) = 12% p.a. = 0.12
   - Time (t) = 2 years
   - Compounding frequency (n) = 2 (semi-annually)

The formula for compound interest is:
   $$
   A = P \left(1 + \frac{r}{n}\right)^{nt}
   $$

Substituting the values:
   $$
   A_1 = 28000 \left(1 + \frac{0.12}{2}\right)^{2 \times 2}
   $$

2. **Next 5 Years at 10% p.a. Compounded Quarterly:**
   - New Principal (P) = $$ A_1 $$ (the amount after the first 2 years)
   - Rate (r) = 10% p.a. = 0.10
   - Time (t) = 5 years
   - Compounding frequency (n) = 4 (quarterly)

Using the same formula:
   $$
   A_2 = A_1 \left(1 + \frac{0.10}{4}\right)^{4 \times 5}
   $$

Now, let's calculate $$ A_1 $$ and $$ A_2 $$.

### Part 2: Convert 10% p.a. Compounded Quarterly to Annual Effective Interest Rate

The formula for converting a nominal interest rate to an effective interest rate is:
$$
i_{eff} = \left(1 + \frac{r}{n}\right)^n - 1
$$
Where:
- $$ r = 0.10 $$
- $$ n = 4 $$

### Part 3: Cost of the Lawn Mower 10 Years Ago

1. **Depreciation Calculation:**
   - Final Value (V) = R236
   - Depreciation Rate (d) = 9.2% p.a. = 0.092
   - Time (t) = 10 years

The formula for straight-line depreciation is:
   $$
   V = C - d \cdot C \cdot t
   $$
   Rearranging gives:
   $$
   C = \frac{V}{1 - d \cdot t}
   $$

Now, let's perform the calculations for all parts.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{236}{\left(1-0.092\times 10\right)}$$
- step1: Remove the parentheses:
  $$\frac{236}{1-0.092\times 10}$$
- step2: Multiply the numbers:
  $$\frac{236}{1-0.92}$$
- step3: Subtract the numbers:
  $$\frac{236}{0.08}$$
- step4: Convert the expressions:
  $$\frac{236}{\frac{2}{25}}$$
- step5: Multiply by the reciprocal:
  $$236\times \frac{25}{2}$$
- step6: Reduce the numbers:
  $$118\times 25$$
- step7: Multiply the numbers:
  $$2950$$
Calculate or simplify the expression $$ (1 + 0.10/4)^4 - 1 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(1+\frac{0.1}{4}\right)^{4}-1$$
- step1: Divide the terms:
  $$\left(1+\frac{1}{40}\right)^{4}-1$$
- step2: Add the numbers:
  $$\left(\frac{41}{40}\right)^{4}-1$$
- step3: Rewrite the expression:
  $$\frac{41^{4}}{40^{4}}-1$$
- step4: Reduce fractions to a common denominator:
  $$\frac{41^{4}}{40^{4}}-\frac{40^{4}}{40^{4}}$$
- step5: Transform the expression:
  $$\frac{41^{4}-40^{4}}{40^{4}}$$
Calculate or simplify the expression $$ 28000 * (1 + 0.12/2)^(2*2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$28000\left(1+\frac{0.12}{2}\right)^{2\times 2}$$
- step1: Divide the terms:
  $$28000\left(1+\frac{3}{50}\right)^{2\times 2}$$
- step2: Add the numbers:
  $$28000\left(\frac{53}{50}\right)^{2\times 2}$$
- step3: Multiply the numbers:
  $$28000\left(\frac{53}{50}\right)^{4}$$
- step4: Simplify:
  $$28000\times \frac{53^{4}}{50^{4}}$$
- step5: Rewrite the expression:
  $$50\times 560\times \frac{53^{4}}{50^{4}}$$
- step6: Reduce the numbers:
  $$560\times \frac{53^{4}}{50^{3}}$$
- step7: Rewrite the expression:
  $$10\times 56\times \frac{53^{4}}{50^{3}}$$
- step8: Rewrite the expression:
  $$10\times 56\times \frac{53^{4}}{10^{3}\times 5^{3}}$$
- step9: Reduce the numbers:
  $$56\times \frac{53^{4}}{10^{2}\times 5^{3}}$$
- step10: Rewrite the expression:
  $$8\times 7\times \frac{53^{4}}{10^{2}\times 5^{3}}$$
- step11: Rewrite the expression:
  $$8\times 7\times \frac{53^{4}}{2^{2}\times 5^{2}\times 5^{3}}$$
- step12: Rewrite the expression:
  $$2^{3}\times 7\times \frac{53^{4}}{2^{2}\times 5^{2}\times 5^{3}}$$
- step13: Reduce the numbers:
  $$2\times 7\times \frac{53^{4}}{5^{2}\times 5^{3}}$$
- step14: Multiply:
  $$\frac{14\times 53^{4}}{5^{5}}$$
- step15: Simplify:
  $$\frac{14\times 53^{4}}{3125}$$
### Part 1: Value of the Investment After 7 Years

1. **First 2 Years at 12% p.a. Compounded Semi-Annually:**
   $$
   A_1 = 28000 \left(1 + \frac{0.12}{2}\right)^{2 \times 2} = 28000 \left(1 + 0.06\right)^{4} = 28000 \left(1.06\right)^{4}
   $$
   Calculating this gives:
   $$
   A_1 \approx 35349.35
   $$

2. **Next 5 Years at 10% p.a. Compounded Quarterly:**
   $$
   A_2 = A_1 \left(1 + \frac{0.10}{4}\right)^{4 \times 5} = 35349.35 \left(1 + 0.025\right)^{20} = 35349.35 \left(1.025\right)^{20}
   $$
   We will calculate $$ A_2 $$ using the value of $$ A_1 $$.

### Part 2: Convert 10% p.a. Compounded Quarterly to Annual Effective Interest Rate
The annual effective interest rate is approximately:
$$
i_{eff} \approx 0.103813 \text{ or } 10.38\%
$$

### Part 3: Cost of the Lawn Mower 10 Years Ago
Using the depreciation formula:
$$
C = \frac{236}{1 - 0.092 \times 10} = \frac{236}{1 - 0.92} = \frac{236}{0.08} = 2950
$$

### Summary of Results
1. **Value of the Investment After 7 Years:** 
   - We need to calculate $$ A_2 $$ to finalize this.
2. **Annual Effective Interest Rate for 10% p.a. Compounded Quarterly:** 
   - Approximately $$ 10.38\% $$
3. **Cost of the Lawn Mower 10 Years Ago:** 
   - R2950

Now, let's calculate $$ A_2 $$ to find the final value of the investment.
Calculate the value by following steps:
- step0: Calculate:
  $$35349.35488\left(1+\frac{0.1}{4}\right)^{4\times 5}$$
- step1: Divide the terms:
  $$35349.35488\left(1+\frac{1}{40}\right)^{4\times 5}$$
- step2: Add the numbers:
  $$35349.35488\left(\frac{41}{40}\right)^{4\times 5}$$
- step3: Multiply the numbers:
  $$35349.35488\left(\frac{41}{40}\right)^{20}$$
- step4: Convert the expressions:
  $$\frac{110466734}{3125}\left(\frac{41}{40}\right)^{20}$$
- step5: Multiply the numbers:
  $$\frac{55233367\times 41^{20}}{5^{25}\times 2^{59}}$$
### Final Calculation for the Investment Value

Continuing from where we left off, we calculate $$ A_2 $$:

$$
A_2 = 35349.35 \left(1 + \frac{0.10}{4}\right)^{4 \times 5} = 35349.35 \left(1.025\right)^{20}
$$

Calculating this gives:
$$
A_2 \approx 57924.03
$$

### Summary of All Results
1. **Value of the Investment After 7 Years:** 
   - Approximately **R57,924.03**
2. **Annual Effective Interest Rate for 10% p.a. Compounded Quarterly:** 
   - Approximately **10.38%**
3. **Cost of the Lawn Mower 10 Years Ago:** 
   - **R2,950**

If you have any further questions or need additional calculations, feel free to ask!
After 7 years, the investment is worth approximately R57,924.03. The annual effective interest rate for 10% compounded quarterly is about 10.38%. The cost of the lawnmower 10 years ago was R2,950.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Economics Questions