Question

Adele deposited $25,000 into her savings account on September 17, 2018. (a) If Adele's account earns an effective annual interest rate of 3.5%, how much money will Adele have in her account on September 17, 2025? account to accumulate to $40,000? effective annual interest rate she earns on her deposit? (b) If Adele's account earns an effective annual interest rate of 3.5%, how long will it take her (c) Suppose that Adele will have $30,000 in her account by September 17,2023. What is the

Answer 1

a. FV = PV (1+r)^n = 25000(1+0.035)^7 = 25000*1.272279 = $31806.981569

b. 40000 = 25000(1.035)^n

40000/25000 = 1.035^n

1.6 = 1.035^n

if we try trial and error then 1.035^13 = 1.56395606 and 1.035^14 = 1.618694522

If we see from above then it would take around 14 years to accumate $40000 but to be exact we will calculate as below

If no of year decreases by 1 i.e from 14 to 13 then annuity factor decreases by 0.05474 i.e. 1.618694522 - 1.56395606 so if annuity factor decreases by 0.0186964522 (1.618694522-1.6) then no of years decreases by = 0.0186964522/0.05474 = 0.341524

Therefore exact no of years = 14-0.341524 = 13.65848 i.e 13.66

c. 30000 = 25000(1+r)^5

30000/25000 = (1+r)^5

1.2 = (1+r)^5

5th root of 1.2 = 1+r

By trail and error,

Multiplying 1.038 five times we get 1.204999225

Multiplying 1.037 five times we get 1.19920597

So by reducing multiplier by 0.001 discounting factor is reduced by 0.00579325 thus to reduce the same by 0.00499922 (1.20499922-1.2) multiplier needs to be reduced by = 0.00499922*0.001/0.00579325 =0.0008625

Thus multiplying 1.0371375 (1.038-0.0008625) five times = 1.2

1.0371375 = 1+r

r = 1.0371375-1 = 0.0371375 i.e 3.71375%

So if we multiply 0.654389 (0.7-0.045611) 5 times we get = 0.12

Therefore,

0.654389 = 1+r