Math Problem Statement

On the first day of January 2020, Mandy deposited $100,000 in bank A. After 3 years, she obtained a simple interest of $12,000. She withdrew the amount and then deposited it in bank B on the first day of January 2023, where the interest rate was the same as that of bank A. It is given that the interest is compounded quarterly in bank B. (a) Find the interest rate per annum. (b) Find the total interest that Mandy earned at the end of 2023, correct to the nearest dollar. (c) To further her savings, Mandy additionally deposited $y in bank B using the same plan on the first day of April 2024. Find, correct to the nearest integer, the minimum value of $y if Mandy would like to have more than $200,000 in her account at the end of 2025.

Solution

To solve the given problems, let's break down each section:

(a) Find the interest rate per annum.

Mandy deposited $100,000 in bank A with simple interest and obtained $12,000 after 3 years.

  • Simple Interest Formula:
    SI=P×r×t\text{SI} = P \times r \times t where:
    • PP is the principal amount ($100,000),
    • rr is the rate of interest per annum,
    • tt is the time in years (3 years).

Given: 12,000=100,000×r×312,000 = 100,000 \times r \times 3 r=12,000100,000×3=12,000300,000=0.04r = \frac{12,000}{100,000 \times 3} = \frac{12,000}{300,000} = 0.04

Thus, the interest rate is 4% per annum.

(b) Find the total interest that Mandy earned at the end of 2023, correct to the nearest dollar.

The deposit was moved to bank B on January 1, 2023, where it is compounded quarterly with the same interest rate (4% per annum).

  • Compounded Interest Formula:
    A=P(1+rn)n×tA = P \left(1 + \frac{r}{n}\right)^{n \times t} where:
    • AA is the amount after interest,
    • nn is the number of compounding periods per year (4 for quarterly),
    • tt is the number of years (1 year for 2023).

Substitute the values: A=112,000(1+0.044)4×1A = 112,000 \left(1 + \frac{0.04}{4}\right)^{4 \times 1} A=112,000(1+0.01)4A = 112,000 \left(1 + 0.01\right)^{4} A=112,000×(1.01)4112,000×1.04060401116,547A = 112,000 \times (1.01)^{4} \approx 112,000 \times 1.04060401 \approx 116,547

Interest earned = Amount - Principal 116,547112,000=4,547116,547 - 112,000 = 4,547

Thus, the total interest earned is $4,547.

(c) Find the minimum value of yy if Mandy would like to have more than $200,000 in her account at the end of 2025.

Calculate the future value of the $116,547 (from 2023 end to 2025 end).

For 2 years compounded quarterly at 4% per annum: A=P(1+rn)n×tA = P \left(1 + \frac{r}{n}\right)^{n \times t} A=116,547(1+0.044)4×2A = 116,547 \left(1 + \frac{0.04}{4}\right)^{4 \times 2} A=116,547×(1.01)8116,547×1.082856126,177A = 116,547 \times (1.01)^{8} \approx 116,547 \times 1.082856 \approx 126,177

To achieve more than $200,000: 126,177+y×(1.01)7>200,000126,177 + y \times (1.01)^{7} > 200,000

y×1.072136>73,823y \times 1.072136 > 73,823

y>73,8231.07213668,864y > \frac{73,823}{1.072136} \approx 68,864

Thus, the minimum value of yy is $68,865 (rounded to the nearest integer).

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Compound Interest
Exponential Growth

Formulas

Simple Interest Formula: SI = P × r × t
Compound Interest Formula: A = P(1 + r/n)^(n × t)

Theorems

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Suitable Grade Level

Grades 10-12