Question

PLEASE HELP THIS IS THE FINAL SUBMISSION

2) Ruth borrows $440 at 0.6% simple interest per month. When Ruth pays the loan back 4 years later, how much interest does Ruth pay?=

4) Ralph borrows $360 at 0.7% simple interest per month. When Ralph pays the loan back 3 months later, how much interest does Ralph pay?=

8) Craig borrows $1000 at 9% simple interest per year. When Craig pays the loan back 11 years later, what is the total amount that Craig ends up repaying?=

Answers

  1. 2) Answer:  I = $ 10.56
    Equation:
    I = Prt
    Calculation:
    First, converting R percent to r a decimal
    r = R/100 = 0.6%/100 = 0.006 per year,
    then, solving our equation
    I = 440 × 0.006 × 4 = 10.56
    I = $ 10.56
    The simple interest accumulated
    on a principal of $ 440.00
    at a rate of 0.6% per year
    for 4 years is $ 10.56.
    4) Answer: I = $ 0.63
    Equation:
    I = Prt
    Calculation:
    First, converting R percent to r a decimal
    r = R/100 = 0.7%/100 = 0.007 per year,
    putting time into years for simplicity,
    3 months ÷ 12 months/year = 0.25 years,
    then, solving our equation
    I = 360 × 0.007 × 0.25 = 0.63
    I = $ 0.63
    The simple interest accumulated
    on a principal of $ 360.00
    at a rate of 0.7% per year
    for 0.25 years (3 months) is $ 0.63.
    8) Answer: I = $ 990.00
    Equation:
    I = Prt
    Calculation:
    First, converting R percent to r a decimal
    r = R/100 = 9%/100 = 0.09 per year,
    then, solving our equation
    I = 1000 × 0.09 × 11 = 990
    I = $ 990.00
    The simple interest accumulated
    on a principal of $ 1,000.00
    at a rate of 9% per year
    for 11 years is $ 990.00.

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