Please how genuine is the purchase of Custom Auction if impounded cars from smugglers. Legit or scam?

MPSA:


I don't understand why you add the interest of the first year on second year.

Total savings on 1rst year=24000
& total interest is: 24000*0,12=2880
Total amount p/a:
total savings+total interest
24000+2880
26880

For second year is:
Total savings+total interest
Total savings=2000*24=48000
Total interest=48000*0,12=5760
.:total amount for 2nd year is:
=48000+5760
=53760

the total amount of 20 year is:
Total savings+total interest
Total savings=2000*12*20=480000
Total interest=480000*0.12=57600
.:total amount+total interest
480000+57600
537600
Don't apply arithmatic formula to get the answer, because interest rates are different.

You need to understand that this is a Progressive compound interest problem. For the convectional compound interest problem, the man would have simply saved for one year then the bank will compound his interest to give a total amount expressed as
Amount = P x [1 + R/100]^n.
where
P=principal (savings for one year)
R=interest rate
n=number of year
I term this problem 'progressive compound interest' because the man keeps on saving every month for the 20 year period. Hence, I modified the convectional compound interest formular above to come up with the model below:

Amount=YP*(n+(n*R)+((n-1)*R^2)+((n-2)*R^3)+((n-3)*R^4)+((n-4)*R^5)+((n-5)*R^6)+((n-6)*R^7)+((n-7)*R^+((n-l*R^9)+((n-9)*R^10)+((n-10)*R^11)+((n-11)*R^12)+((n-12)*R^13)+((n-13)*R^14)+((n-14)*R^15)+((n-15)*R^16)+((n-16)*R^17)+((n-17)*R^18)+((n-18)*R^19)+(R^20))
where
YP=principal (yearly savings)
R=interest rate
n=number of year.
Actually, this is a mathematical model that will perfectly compute the total amount for anyone's savings with any bank at any yearly saving(YP) at a particular interest rate(R) at the end of some number of years(n).
A Bank should PAY ME FOR THIS MATHEMATICAL MODEL but since this is a knowledge forum, I decided to upload it at zero cost.
Any one, especially walexy30, should try this model for any set problem at his set yearly savings, at a specific interest rate, and for any desired number of years, it's only then that you'll understand the potency of this model. You might need a computer program to run this model. Just Just 5 to 6 lines of code will compute ypor money instaed of wasting time solving year by year by hand. Imagine how long and strenuous this could be. My model will do it in few seconds.
This is the last I should post on this topic, I hope to hear from the OP(walexy30).

MPSA:


The correct answer is:
20 years total savings*interest rate
(2000*12*20)(1+12/100)
(480000)(1,12)
537600

This solution is very wrong. You've simply accumulated the total savings and computed a single interest on it at the end of the 20 years period using
Amount = P(total) + [P(total)xRate/100]. The given rate(R) is annual and not for 20 years. This is wrong sir!

Check my solution pls.
This is a complex form of a Compound Interest question. (Reason: You need to keep on adding the monthly savings he makes per month.)
The mistake from many is to assume a one-time principal.
For clarity, at the end of the first year, the Principal = #2000 x 12 = #24000. 12% interest on this gives 24000 x 0.12 = 2880.
At the end of the 2nd year, the New Principal = (Principal of year 1) + (Savings/principal of year 2) + Interest from year 1 = 24000 +24000 +2880 = 50880. Interest from year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics student, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+(n*x)+((n-1)*x^2)+((n-2)*x^3)+((n-3)*x^4)+((n-4)*x^5)+((n-5)*x^6)+((n-6)*x^7)+((n-7)*x^cool+((n-cool*x^9)+((n-9)*x^10)+((n-10)*x^11)+((n-11)*x^12)+((n-12)*x^13)+((n-13)*x^14)+((n-14)*x^15)+((n-15)*x^16)+((n-16)*x^17)+((n-17)*x^18)+((n-18)*x^19)+(x^20))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 545,010.

walexy30:
I save N2000 every month in a bank with an annual interest rate of 12% for 20 years while still working. How much will i receive at the end of the period (20 years).

1) This is a complex form of a Compound Interest question. (Reason: You need to keep on adding the monthly savings he makes per month.)
The mistake from many is to assume a one-time principal.
For clarity, at the end of the first year, the Principal = #2000 x 12 = #24000. 12% interest on this gives 24000 x 0.12 = 2880.
At the end of the 2nd year, the New Principal = (Principal of year 1) + (Savings/principal of year 2) + Interest from year 1 = 24000 +24000 +2880 = 50880. Interest from year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics teacher, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+(n*x)+((n-1)*x^2)+((n-2)*x^3)+((n-3)*x^4)+((n-4)*x^5)+((n-5)*x^6)+((n-6)*x^7)+((n-7)*x^cool+((n-cool*x^9)+((n-9)*x^10)+((n-10)*x^11)+((n-11)*x^12)+((n-12)*x^13)+((n-13)*x^14)+((n-14)*x^15)+((n-15)*x^16)+((n-16)*x^17)+((n-17)*x^18)+((n-18)*x^19)+(x^20))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 545,010.
This is an adjusted solution to correct an error in the former.
This answer should be CORRECT.

DATBOI:
am i correct

no. The answer is # 541,700.

SOLUTION:

1) This is a complex form of a Compound Interest question. (Reason: You need to keep on adding the monthly savings he makes per month.)
The mistake from many is to assume a one-time principal.
For clarity, at the end of the first year, the Principal = #2000 x 12 = #24000. 12% interest on this gives 24000 x 0.12 = 2880.
At the end of the 2nd year, the New Principal = (Principal of year 1) + (Savings/principal of year 2) + Interest from year 1 = 24000 +24000 +2880 = 50880. Interest from year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics student, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+((n-1)*x)+((n-2)*x^2)+((n-3)*x^3)+((n-4)*x^4)+((n-5)*x^5)+((n-6)*x^6)+((n-7)*x^7)+((n-cool*x^cool+((n-9)*x^9)+((n-10)*x^10)+((n-11)*x^11)+((n-12)*x^12)+((n-13)*x^13)+((n-14)*x^14)+((n-15)*x^15)+((n-16)*x^16)+((n-17)*x^17)+((n-18)*x^18)+((n-19)*x^19))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 541,700.

OR

Amt = 24000 * [20 + 19(0.12) + 18(0.12)^2 + 17(0.12)^3 + 16(0.12)^4 + 15(0.12)^5 +14(0.12)^6 + 13(0.12)^7 + 12(0.12)^8 + 11(0.12)^9 + 10(0.12)^10 + 9(0.12)^11 + 8(0.12)^12 + 7(0.12)^13 + 6(0.12)^14 + 5(0.12)^15 + 4(0.12)^16 + 3(0.12)^17 + 2(0.12)^18 + (0.12)^19]
Amt = # 541,700.

inze:
[b]Interest would have to be calculated annually, added and carried over to the next year balance:

Year 1 = (2000*12)*12% = 2880(interest) + 24000(total savings) = 26,880:
Year 2 = (26,880*0.12) + 26880 = 30,105.6
Year 3 = (30,105.6 * 0.12) + 30,105.6 = 33,718.2
Year 4 = (33,718.2 * 0.12) + 33,718.2 = 37,764.16
Year 5 = (37,764.16 * 0.12) + 37,764.16 = 42,295.85
Year 6 = (42,295.85 * 0.12) + 42,295.85 = 47,371.35
Year 7 = (47,371.35 * 0.12) + 47,371.35 = 53,055.91
Year 8 = (53,055.91 * 0.12) + 53,055.91 = 59,422.61
Year 9 = (59,422.61 * 0.12) + 59,422.61 = 66,553.32
Year 10 = (66,553.32 * 0.12) + 66,553.32 = 74,539.71
Year 11 = (74,539.71 * 0.12) + 74,539.71 = 83,484.47
Year 12 = (83,484.47 * 0.12) + 83,484.47 = 93,502.60
Year 13 = (93,502.60 * 0.12) + 93,502.60 = 104,722.91
Year 14 = (104,722.91 * 0.12) + 104,722.91 = 117,289.65
Year 15 = (117,289.65 * 0.12) + 117,289.65 = 131,364.40
Year 16 = (131,364.40 * 0.12) + 131,364.40 = 147,128.12
Year 17 = (147,128.12 * 0.12) + 147,128.12 = 164,783.49
Year 18 = (164,783.49 * 0.12) + 164,783.49 = 184,557.50
Year 19 = (184,557.50 * 0.12) + 184,557.50 = 206,704.40
Year 20 = (206,704.40 * 0.12) + 206,704.40 = 231,508.92

Compounding total sum for each year gives you 1,936,753.17[/b]

I think you got your solution wrong from line 2. After getting 26,880 at the end of the first year, the new principal for computing interest for year2 is 26,880 + 24000 (second year savings) = # 50,880. now interest for year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics student, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+((n-1)*x)+((n-2)*x^2)+((n-3)*x^3)+((n-4)*x^4)+((n-5)*x^5)+((n-6)*x^6)+((n-7)*x^7)+((n-cool*x^cool+((n-9)*x^9)+((n-10)*x^10)+((n-11)*x^11)+((n-12)*x^12)+((n-13)*x^13)+((n-14)*x^14)+((n-15)*x^15)+((n-16)*x^16)+((n-17)*x^17)+((n-18)*x^18)+((n-19)*x^19))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 541,700.

OR

Amt = 24000 * [20 + 19(0.12) + 18(0.12)^2 + 17(0.12)^3 + 16(0.12)^4 + 15(0.12)^5 +14(0.12)^6 + 13(0.12)^7 + 12(0.12)^8 + 11(0.12)^9 + 10(0.12)^10 + 9(0.12)^11 + 8(0.12)^12 + 7(0.12)^13 + 6(0.12)^14 + 5(0.12)^15 + 4(0.12)^16 + 3(0.12)^17 + 2(0.12)^18 + (0.12)^19]
Amt = # 541,700.

GentleToks:


We are both wrong. We failed to carry our first year balance + interest forward to the subsequent year.

We arrived at our answers by calculating our interest on yearly savings without carrying the total figure to the following year.

Here is the solution:

1st year-

Saving #2000 every month

Total annual savings- #24,000

Annual interest on #24,000 is #2880 base on the 12% annual interest rate

Moving to a new financial year, the interest + total saving would be carried forward to another financial year and that would be #26,880


So 2nd year would begin with a balance of #26,880.........

If you continue like this to the last year, your correct and perfect answer is guaranteed

search for my post and go through the solution. I don't talk muck nor insult, I only wish to share knowledge.

GentleToks:


Sir, the man saves #2000 every month for 20 yrs with 12% interest rate per annum.

#2000 × 12=24,000
Calculate your interest on 24,000 and that would give you #2880

2nd yr saving is #48,000 with 12% annual rate and that would equally 2880 ×2=#5670

3rd................All the way to the last year which is 20 yrs.


All his interest for 20yrs comes to a total sum of #57,600

All his saving for 20yrs comes to a total sum of #480,000

#480,000(20yrs accumulated savings) + #57,600( 20yrs total accumulated interest) = 537,600.

The first three lines of your solution are correct. Your solution is however wrong from line 4. the principal for the second year = #48,000 + 2880 = # 50,880. I call this kind of problem Progressive Compound Interest because the man keeps saving until the the last month of the 20th year. You can go through my solution above.

walexy30:
I save N2000 every month in a bank with an annual interest rate of 12% for 20 years while still working. How much will i receive at the end of the period (20 years).

1) This is a complex form of a Compound Interest question. (Reason: You need to keep on adding the monthly savings he makes per month.)
The mistake from many is to assume a one-time principal.
For clarity, at the end of the first year, the Principal = #2000 x 12 = #24000. 12% interest on this gives 24000 x 0.12 = 2880.
At the end of the 2nd year, the New Principal = (Principal of year 1) + (Savings/principal of year 2) + Interest from year 1 = 24000 +24000 +2880 = 50880. Interest from year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics student, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+(n*x)+((n-1)*x^2)+((n-2)*x^3)+((n-3)*x^4)+((n-4)*x^5)+((n-5)*x^6)+((n-6)*x^7)+((n-7)*x^cool+((n-cool*x^9)+((n-9)*x^10)+((n-10)*x^11)+((n-11)*x^12)+((n-12)*x^13)+((n-13)*x^14)+((n-14)*x^15)+((n-15)*x^16)+((n-16)*x^17)+((n-17)*x^18)+((n-18)*x^19)+(x^20))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 545,010.
This is an adjusted solution to correct an error in the former.
This answer should be CORRECT.

missionmex:
After almost killing myself I got 1,960,769.9

This must b the answer

No. it's not. The answer is # 541,700

GentleToks:
If your answer is different from this #530,600, you have failed woefully.

I'm obliged to disagree with you on this. I think the answer is # 541,700. Can you show your solution pls.

[color=#770077][/color]It's interesting to still see the interest of many people at attempting to solve questions on the greatest mental tasking subject -Mathematics. I must confess that the numerous solutions I've seen gladden my heart, although I do not agree with them. Patiently follow the solution.
1) This is a complex form of a Compound Interest question. (Reason: You need to keep on adding the monthly savings he makes per month.)
The mistake from many is to assume a one-time principal.
For clarity, at the end of the first year, the Principal = #2000 x 12 = #24000. 12% interest on this gives 24000 x 0.12 = 2880.
At the end of the 2nd year, the New Principal = (Principal of year 1) + (Savings/principal of year 2) + Interest from year 1 = 24000 +24000 +2880 = 50880. Interest from year 2 = 50880 x 0.12 = 6105.6. At the end of year 3, the New Principal = (year 1 principal) + (year 2 principal) + (year 3 principal) + (Interest from year 2) = 24000 + 24000 + 24000 + 6105.6 = 78105.6. and on and on like that.
As a mathematics student, I computed an analytical equation for this 'progressive compound interest' problem thus.

Amt=yp*(n+(n*x)+((n-1)*x^2)+((n-2)*x^3)+((n-3)*x^4)+((n-4)*x^5)+((n-5)*x^6)+((n-6)*x^7)+((n-7)*x^+((n-*x^9)+((n-9)*x^10)+((n-10)*x^11)+((n-11)*x^12)+((n-12)*x^13)+((n-13)*x^14)+((n-14)*x^15)+((n-15)*x^16)+((n-16)*x^17)+((n-17)*x^18)+((n-18)*x^19)+(x^20))
yp = yearly savings/principal = #24000,
n = number of years = 20,
x = interest rate per annum = 12% = 0.12,
Amt = Amount at the end of 20 years
Amt = # 545,010.
This is an adjusted solution to correct an error in the former.
This answer should be CORRECT.