Part A
Sue
A = P*(1+r/n)^(n*t) is the compound interest formula
A = 2300*(1+0.024/1)^(1*3)
A = 2469.6061952
A = 2469.61
A - P = 2469.61-2300 = 169.61
<h3>Sue gets 169.61 pounds in interest</h3>
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Bill
A = P*(1+r/n)^(n*t)
A = 1800*(1+0.034/1)^(1*3)
A = 1989.9131472
A = 1989.91
A-P = 1989.91-1800 = 189.91
<h3>Bill earns 189.91 pounds in interest</h3>
Bill has earned more in interest.
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Part B
By year 2, Bill has 1924.48 pounds in his account based on the work shown below
A = P*(1+r/n)^(n*t)
A = 1800*(1+0.034/1)^(1*2)
A = 1924.4808
A = 1924.48
This amount is the new deposit, so to speak, when we change the interest rate. Now r = 0.034 changes to r = 0.04. We only go for one year so t = 1
A = P*(1+r/n)^(n*t)
A = 1924.48*(1+0.04/1)^(1*1)
A = 2001.4592
A = 2001.46
Bill has 2001.46 pounds in his account after 3 years if the interest rate for the 1234 account changes to 4% in the third year.
Now subtract off the original amount Bill deposited to get
2001.46-1800 = 201.46
For this scenario, Bill earns 201.46 pounds in interest.
Therefore, Bill has earned the most interest for both cases of the interest rate staying at 3.4% or changing to 4% for that third year.
<h3>Answer: Bill</h3>
