Math Problem Statement

The XYZ Bank pays Betty, simple interest at the rate of 5% per annum, on all funds in her account. Interest is calculated and credited to her account quarterly on March 31%, June 30", September 30th and December 31st, based on the minimum quarterly balance.

Betty has $ 80,000 in her account on January 1%. Given the following transactions to her account, determine (with all appropriate calculations), how much interest will be credited to the account on December 31st

  • [9]

Date

Deposit

Jan 10

$ 2,500

Withdrawal

Mar 5

$ 4,000

Nov. 1

$ 13,000

Solution

To determine how much interest will be credited to Betty's account on December 31st, we will apply the simple interest formula on a quarterly basis:

Interest=P×r×t100\text{Interest} = \frac{P \times r \times t}{100} Where:

  • PP is the principal or minimum balance for the quarter,
  • rr is the annual interest rate (5% in this case),
  • tt is the time period in years (each quarter represents 3 months or 312=14\frac{3}{12} = \frac{1}{4} of a year).

Steps:

  1. January 1 to March 31:

    • Starting balance on January 1: $80,000.
    • January 10: Deposit of $2,500, new balance = $82,500.
    • March 5: Withdrawal of $4,000, new balance = $78,500.
    • Minimum balance during the quarter: After March 5, the balance was $78,500, so this is the minimum balance for this quarter.

    Interest for this quarter: I1=78,500×5×14100=78,500×1.25100=981.25I_1 = \frac{78,500 \times 5 \times \frac{1}{4}}{100} = \frac{78,500 \times 1.25}{100} = 981.25

  2. April 1 to June 30:

    • Starting balance: $78,500 (no changes during this period).
    • Minimum balance for the quarter: $78,500.

    Interest for this quarter: I2=78,500×5×14100=981.25I_2 = \frac{78,500 \times 5 \times \frac{1}{4}}{100} = 981.25

  3. July 1 to September 30:

    • Starting balance: $78,500 (no changes during this period).
    • Minimum balance for the quarter: $78,500.

    Interest for this quarter: I3=78,500×5×14100=981.25I_3 = \frac{78,500 \times 5 \times \frac{1}{4}}{100} = 981.25

  4. October 1 to December 31:

    • Starting balance: $78,500.
    • November 1: Deposit of $13,000, new balance = $91,500.
    • Minimum balance for the quarter: The balance before November 1 was $78,500, so this is the minimum balance for this quarter.

    Interest for this quarter: I4=78,500×5×14100=981.25I_4 = \frac{78,500 \times 5 \times \frac{1}{4}}{100} = 981.25

Total Interest for the Year:

Total Interest=I1+I2+I3+I4=981.25+981.25+981.25+981.25=3,925\text{Total Interest} = I_1 + I_2 + I_3 + I_4 = 981.25 + 981.25 + 981.25 + 981.25 = 3,925

Conclusion:

The total interest credited to Betty's account on December 31st will be $3,925.

Would you like further details or have any questions?


Here are 5 related questions:

  1. How does simple interest differ from compound interest in long-term savings?
  2. What would happen if Betty made a withdrawal in July? How would that affect the interest?
  3. How could increasing the deposit frequency impact Betty's total interest?
  4. How is the quarterly interest calculation different from annual interest?
  5. What if Betty’s bank offered a higher interest rate? How would the final amount change?

Tip: Always track minimum balances when dealing with simple interest calculated over time, especially when transactions occur in the middle of interest periods!

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Minimum Balance
Quarterly Interest Calculation

Formulas

Interest = (P × r × t) / 100
Where P is the principal (minimum balance), r is the interest rate, and t is the time in years (quarter is 1/4 year)

Theorems

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Suitable Grade Level

Grades 9-12