stlaibrowne:


Now all y'all management accountant, stockers, commercial students do you know of any share or stock one can invest such money to according to the question in real life

Alternatively, Mansard insurance has some annuity plans that are very good. But they pay interest of between 5-6% depending on the current MPR.

The MPR is currently 13%. The CBN fixes it from time to time. So, it changes.

walexy30:


Alternatively, Mansard insurance has some annuity plans that are very good. But they pay interest of between 5-6% depending on the current MPR.

The MPR is currently 13%. The CBN fixes it from time to time. So, it changes.

This is great Bro do you have any idea how cam one invest and start trading with them and also where is the company based and can you recommend the company to be excellent in term of performance and reliability

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).

1729258.62 (EQUAL SERIES PAYMENT COUMPOUND AMOUNT FACTOR)

stlaibrowne:


This is great Bro do you have any idea how cam one invest and start trading with them and also where is the company based and can you recommend the company to be excellent in term of performance and reliability

helpful links:

https://online.mansardinsurance.com/direct/mls.aspx

https://www.axamansard.com/saving-investments

walexy30:


helpful links:

https://online.mansardinsurance.com/direct/mls.aspx

https://www.axamansard.com/saving-investments

Gracias I have contacted the customer line and I am informed of a product that I can work with. I can start the investment with 10k and hence contribute monthly which I can withdraw at any particular time

stlaibrowne:

Gracias I have contacted the customer line and I am informed of a product that I can work with. I can start the investment with 10k and hence contribute monthly which I can withdraw at any particular time

You can see how applied annuity cash flow is.

mozeybee:



Hun. Thank you sir/ma. This is a really detailed solution.

No! Don't thank him; His answer is incorrect

The question is simple

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).

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

MPSA:


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

Mathematician!

walexy30:


Mathematician!

I enjoy to solve difficult problems, thanks for your support.

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.

MPSA:


No! Don't thank him; His answer is incorrect

I sarcastically called you a 'mathematician'. You are absolutely wrong.

MPSA:


I enjoy to solve difficult problems, thanks for your support.

I sarcastically called you a 'mathematician'. You are absolutely wrong!

itopicman:

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.

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.

walexy30:

I sarcastically called you a 'mathematician'. You are absolutely wrong!

It does't matter, but I know my answer is correct.

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).

N1978510.731

debbianah:
12 percent annual interest on ur savings of 2000 monthly. days 24,000 monthly. 12 percent of 24000 is 2880. so u shed get 2880 with of interest every year x 20 years is 57,600.

wat course did u study

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).

FM @chiefololade

stlaibrowne:


Now all y'all management accountant, stockers, commercial students do you know of any share or stock one can invest such money to according to the question in real life

such investment or real life situation you invest and this scenario applies include
1.annuity payment
2.term life insurance that life assurance with limited life of 5, 10, 15 years
3.Treasury bill
4.Fixed deposit with interest rollover at maturity
5.Sinking fund; fund one created with his banker that pay interest on the balance while the owner pay fix amount monthly or quarterly for meeting up a particular purposes in future

I'm late but I can't help it
Here's my understanding
N2,000*12 months= 24,000 a year
24,000*12%=2,880 interest rate per year
Then 24,000 + 2,880= 26,880 ( so he has this amount yearly in his saving account)
So for 20 years
26,880*20= N537, 600

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).

0


IT IS 78,720

Stephieehh:
I'm late but I can't help it
Here's my understanding
N2,000*12 months= 24,000 a year
24,000*12%=2,880 interest rate per year
Then 24,000 + 2,880= 26,880 ( so he has this amount yearly in his saving account)
So for 20 years
26,880*20= N537, 600

Go back to the 1st five pages, you'll find yhe solution there.

walexy30:

Go back to the 1st five pages, you'll find yhe solution there.

Yes but my calculations are different and straight forward

after my calculation i got #537600.00kb