Give step-by-step solution with explanation and final answer:Because Market conditions appeared favorable, Mitsubishi Electric's Treasurer decided to
raise funds through a debt issuance. Assume that

Question

Give step-by-step solution with explanation and final answer:Because Market conditions appeared favorable, Mitsubishi Electric's Treasurer decided to
raise funds through a debt issuance. Assume that on March 31, 2010, the company issued its
0.58% Japanese yen bods due 2013. As the company had no immediate plans for the funds
raised, the Treasurer placed the proceeds in an interest - bearing bank account. i. Assume
that the bank account pays 3% interest per year. The interest is paid annually on the
anniversary date of the deposit and is left in the account. What will the balance of the bank
account be on March 31, 2011? March 31, 20127 March 31, 20137 Use the first row in the
table on the following page to tabulate your answers. ii. Now assume that the bank account
pays 5% per year. Complete the second row of the table. iii. Now assume that interest of 5%
per year is paid semiannually (i.e... 2.5% every six months) . Complete the third row of the
table.

VIPSolutions · Accepted Answer

Let's break down each part step by step:

---

## **Given:**

- Initial deposit date: **March 31, 201**
- Periods to calculate: **2011, 2012, 2013** (so, 1, 2, and 3 years)
- **Let the principal (P) be 1 (or 100 for easier calculation), since the actual amount is not specified.**

---

## **Part (i): 3% Interest Annually (Compounded Annually)**

### **Formula:**
$$
A = P(1 + r)^n
$$

Where:  
- $ P $ = Principal (initial deposit)  
- $ r $ = Annual interest rate (3% = .03)  
- $ n $ = Number of years

---

### **Calculations:**

#### **Year 1 (March 31, 2011):**
$$
A_1 = 1 \times (1 + .03)^1 = 1.03
$$

#### **Year 2 (March 31, 2012):**
$$
A_2 = 1 \times (1 + .03)^2 = 1.0609
$$

#### **Year 3 (March 31, 2013):**
$$
A_3 = 1 \times (1 + .03)^3 = 1.092727
$$

---

## **Part (ii): 5% Interest Annually (Compounded Annually)**

### **Formula:**
$$
A = P(1 + r)^n
$$

Where $ r = .05 $.

---

### **Calculations:**

#### **Year 1 (March 31, 2011):**
$$
A_1 = 1 \times (1 + .05)^1 = 1.05
$$

#### **Year 2 (March 31, 2012):**
$$
A_2 = 1 \times (1 + .05)^2 = 1.1025
$$

#### **Year 3 (March 31, 2013):**
$$
A_3 = 1 \times (1 + .05)^3 = 1.157625
$$

---

## **Part (iii): 5% Interest Paid Semiannually (2.5% Every Six Months)**

### **Formula:**
$$
A = P \left(1 + \frac{r}{m}\right)^{n \times m}
$$

Where:  
- $ r = .05 $ (annual rate)  
- $ m = 2 $ (compounding periods per year)  
- $ n = $ number of years

---

### **Calculations:**

#### **Year 1 (March 31, 2011):**
$$
A_1 = 1 \left(1 + \frac{.05}{2}\right)^{2 \times 1} = 1 \times (1.025)^2 = 1.050625
$$

#### **Year 2 (March 31, 2012):**
$$
A_2 = 1 \left(1 + .025\right)^{4} = (1.025)^4 = 1.10381289
$$

#### **Year 3 (March 31, 2013):**
$$
A_3 = 1 \left(1 + .025\right)^{6} = (1.025)^6 = 1.160754517
$$

---

## **Summary Table**

| Year                | (i) 3% Annually | (ii) 5% Annually | (iii) 5% Semiannually |
|---------------------|:---------------:|:----------------:|:--------------------:|
| March 31, 2011 (1 yr) |   **1.03000**   |    **1.05000**   |     **1.050625**     |
| March 31, 2012 (2 yr) |   **1.06090**   |    **1.10250**   |     **1.103813**     |
| March 31, 2013 (3 yr) | **1.09273**     |   **1.15763**    |    **1.160755**      |

---

### **Final Answers**

- **(i) 3% Annually:** 1.03000, 1.06090, 1.09273
- **(ii) 5% Annually:** 1.05000, 1.10250, 1.15763
- **(iii) 5% Semiannually:** 1.050625, 1.103813, 1.160755

> **Multiply these by the initial amount deposited to get the actual balances.**

Let me know if you need the table formatted for a report!

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