Question

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.