Math Assignment
1. Original Condo fees for bigger condos = $500 per year
Reduction by $200 per year.
Owners of big Condos share costs by 4.5% = 4.5% of $300 = $13.5
Fee paid by large condo owners = $313.5 which is comparatively greater than small condo owners. Hence, the challenger is telling the truth.
2. Distance travelled by the car = 1.1 miles
Time taken = 57.51 sec. = 57.51/3600 hours = 0.015975 hours
Speed of the car = Distance travelled/Time taken = 1.1/0.015975
= 68.85758998435055miles per hour.
There are 16 significant digits in determining the speed of the car.
3. Time taken to reach Mom’s house = 22 min.
Time stuck in traffic = 3.18 min.
Time to be taken to reach Mom’s house = 22 + 3.18 = 25.18 min.
4. P = $30,000, r = 8/100 = 0.08, n = 12, t = 9 years.
A = P(1+r/n)nt = 30,000(1+0.08/12)12*9 = $61,485.91
In 9 years, the money will accumulate to $61,485.91
5. P = $30,000, R = 8/1200, n = 108.
Monthly payment = [ P * R * (1+R)n]/(1+R)n-1= 30,000 * 8/1200 * (1+8/1200)108 = $201.34
Monthly payment to be made = $201.34
6. A = $200,000, r = 9.5% = 0.095, n = 12, t = 25years.
A = P (1+r/n)nt
200,000 = P (1+0.095/12)300
P * 10.651 = 200,000
P = $18,777.58
Monthly payment = [P * R * (1+R)n](1+R)n-1 = 18,777.58 * 9.5/1200 * (1+9.5/1200)]/1+0.5/1200
= $149.83
Monthly payment = $149.83
7. P = $190, r = 6.25% = 0.0625, t = 40 years.
A = P(1+r/n)nt = 190 * (1+0.0625/12)12*40 * 40(Time period) = $91.992
Money received in 40 years = $91.992
8. P = $400,000, r = 7% = 0.07. n = 1
A = P(1+r/n)nt = 400,000(1+0.07)1 = $428,000.
I = A – P = 428,000 – 400,000 = $28,000.
Interest earned every year = $28,000.
9. P = $30,000, R = 8.5% = 0.0825, n = 12, t = 15 years.
Monthly payment = [P * R * (1+R)n]/(1+R)n-1 = [30,000 * 0.0825/12 * (1+0.0825/12)180](1+0.0825/12)
= $211.15
Monthly payment of student loan = $211.15
10. Monthly payment = $211.15
Amount paid after 15 years = 211.5 *12 * 15 = $38,007
I = A – P = 38,007 – 30,000 = $8,007
Interest paid after 15 years = $8,007.
11. Monthly payment = $230
Option A:
A = $230, r = 0%, n = 12, t = 3 years
A = P(1+r/n)nt
230 = P(1+o/12)36
230 = P * 1
P = $230.
Option B:
R = 0.5%, t = 4 years
230 = P(1+0.005/12)48
230 = P(1+0.005/12)48
P = $225.44
Option C:
r = 9.9%, t = 5 years
A = P(1+r/n)nt
230 = P(1+0.099/12)60
230 = P * 12.099)60
P = $24.92
Option A will allow me to borrow for $230 per month.
12. Monthly payment =[ P * R * (1+R)n]/(1+R)n-1
= [240,000 * 0,03375/12 * (1+0.03375/12)]/(1+0.03375/12) = $676.89
Monthly payment for Fixed Rated Option = $676.89
13. P = $240,000, R = 2.625% = 0.02625, n = 12, t = 30 years.
Monthly payment = [P * R * (1+R)n]/(1+R)n-1
=[240,000 * 0.02625/12 * (1+0.02625/12)360](1+0,02625/12)359
= $526.15
Monthly payment for ARM Option = $526.15
14. I would choose the Fixed Rate Option. If I had to choose the ARM Option, then the interest rate would increase by 1% each year. If I had to live in the home for 5 years, then the interest rate would go up to 7,625% which is much higher than the Fixed Rate Option.
15. E) Keep on plugging – it’s the American way.
