81 Determine the difference in interest earned on S7 SK for
Solution
The formula for simple interest is:
SI = P x r x t ;
and,
The formula for annual compound interest, including principal sum, is:
A = P (1 + r/n) (nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
r = the annual interest rate (decimal)
n = the number of times that interest is compounded per year
t = the number of years the money is invested or borrowed for
Note that this formula gives you the future value of an investment or loan, which is compound interest plus the principal. Should you wish to calculate the compound interest only, you need this:
Total compounded interest = P (1 + r/n) (nt) - P
8.1 Here, P= $7500 , t =15 yrs, r= 6%
Simple interest = $ 7500 x 15 x 6/100 = $6750
Compound Interest = $ 7500(1+0.06/1)15 - $7500 = $ 10474
Difference = $ (10474-6750) = $3724
8.2 Here, amount after 5 yrs is $10,000
So, A = 10,000, t= 5 yrs, P= ?, r= 5%, 15% and 25%
Since, it is not given whther it is compound interest or simple interest, we can assume it as simple interest. However, if it is compounded, the P can be found out easily usng the above formula(in 8.1)
Now, for 5% rate,
(P x 5/100 x 5)+ P = 10,000 (interest + presen value)
1.25P = 10,000
P = $ 8000
Now, for 15% rate,
(P x 15/100 x 5)+ P = 10,000 (interest + presen value)
1.75P = 10,000
P = $ 5714.29
Now, for 25% rate,
(P x 25/100 x 5)+ P = 10,000 (interest + presen value)
2.25P = 10,000
P = $ 4444.44
8.3 A = $ 30,000, t= 7 yrs, r=4%
When compounded i) annually, n=1 in the formula,thus
P(1+r/1)(1xt) = P x (1+0.04)7 = 30,000
or, P = 30,000/1.316 = $22796.35
when compounded semiannually, n=2 (2 times in a year) in the formula, thus
P(1+r/2)(tx2) = P x (1+0.04/2) 14 = 30,000
or, P = 30000/1.319 = $22744.5
when compounded monthly, n =12
P(1+r/12)(tx12) = P x (1+0.04/12)(7x12)
or, P = 30,000/1.319 = $22744.5
8.4 Diagrams are not available
8.5 Amount to be paid on 1 April = $3400, and on 1 sep = $3400
rate = 5.25%, t =15 months = 15/12 yrs, compounding monthly, so n=12
P (1+0.0525/12)12x15/12= P(1.004375)15 = P x 1.054 (for payment on April 1),
Now,remaining amount after paying 3400 on April 1 = P x1.054 -3400
for sep1 payment,
P\' = 1.054P-3400
A = (1.054P-3400)x (1+0.0525/12)6/12x12 (April to oct =6 months)
(1.054P-3400 )x (1.027)
Now this shoul be equal to 3400
(1.054P-3400)x1.027 = 3400
1.0824P - 3491.8 = 3400
P = $ 6367.15
9.1
Analysis for foam,
initial cost = $37000,r=7%, t=12 yrs
interest for 12 yrs = 37000 x 7/100 x 12 = $ 3108
painting cost (per 3 months) for 12 months = 4*3000 = 12000
total = 37000+3108+12000 = 52108
net savings = 6500*12 - 52108 = $25892
for fiberglass analysis,
installation = 14000, no maintainance, r=7%
interest for 12 yrs = 14000x7/100x12 = 1176
savings for 12 yrs = 2600 x12 = $31200
net savings = 31200 - 14000+1176 = $18376
savings for foam is more
9.2
machine A;
cost = $ 24000
maintainance for 8 yrs = 2600x8 = 20800
interest at 5 % for 8 yrs = 24000x5/100x8 = 9600
total cost in 8 yrs = 24000+20800+9600 = $ 54400
annual cost = 54400/12 = 4533.33
macihne B;
cost = 32000
maintainance for 8 yrs = 1200x8 = 9600
interest for 8 yrs = 32000 x 5/100 x8 = 12800
total cost = 32000+9600+12800 = $54400
annual cost = 54400/12 = $4533.33
Thus, both cost the same and either can be a better buy.


