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.

The interest for the first year must definately be diff from that of the second year the second year diff from the third year......etc
Year 1 24000*12% = 2880
Year2. 24000 + 24000 + 2880 =50880*12% =6105.6
Year. 3 24000+50880+6105.6=80985.6*12%=9718.272

If u follow this method then u will arrive at 1960769.9 in year 20.

E long o but e sure pass.

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.

hmmm...

[quote author=DATBOI post=N1,771,984.672[/quote] check dis answer up

pls look up dis answer N1,771,984.672pls look up dis answer N1,771,984.672pls look up dis answer N1,771,984.672

DATBOI:
pls look up dis answer N1,771,984.672pls look up dis answer N1,771,984.672pls look up dis answer N1,771,984.672

am i correct

This is the surest answer 1, 729,258.8

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.

walexy30:


You have to bring it down to month because the depositor pays the money into his account every month not every year.

Walexy, I think your calculation is flawed. Bringing down to a month is correct, but the effective monthly rate is 1.0046% not 1%. 1% of the nominal monthly rate. So Pymt is -N2,000, Interest is 1.0046, # of compounding is 240, therefore future value is 1,993,284.33.

P=d(((1+r/k)^Nk)-1)/(r/k). P=balance after N years, k= number of compounding period, r=annual interest rate and d= regular deposit each period. Substituting those values will give us P=N1,978,510.730

Oya op giv us d ans, we all v tried.

itopicman:

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-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.
I love Mathematics !

Someone's dad needs a refund

12423.393

please can someone borrow me his/her CALCULATOR cos am using SCIENTIFIC CALCULATOR.

ubecharles:
. How did u get 0.01 as r

12% divided by 12

Lol, I tell you,
And he's so confident... "as a mathematics student"..."I love mathematics",
Empty vessels really are the loudest

OKeokpa:





Someone's dad needs a refund

your are so confident in your failure, cumulated annual income/savings aloneeee!!! for the 20 year period is already 480,000....go figure

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

mfm04622:


Wrong! Since the savings is monthly, you earn interest from the first month.

chigoizie7:


It should be compounded, use compound interest

Alcheringa:
PMT * (((1 + r/n)^nt - 1) / (r/n))
Where:
A = the future value of the investment/
loan, including interest
P = the principal investment amount (the
initial deposit or loan amount)
PMT = the monthly payment
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


2000*(((1+0.12/12)v12*20) - 1)/(0.12/12)) :
Equals 2000*((1.01)v240 - 1) * (12/0.12)) :
Equals 2000*(9.893 * 100) :
Equals. #1,978,510.73

N.b: v means raised to the power. for this calculation due to the sign unavailability.

I corrected it long ago.

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).

There are loopholes in this question:
1. What's the fixed deposit tenure (roll over)-is it monthly, quarterly, bi-annual or yearly? Eg . I fixed my money for about two years but it was on a 180day cycle.

2. Is the interest compounded at rollover cycles ? In my own case, it was compounded, interest was added to principal and rolled over for the next 180 days. You can elect not to rollover interest.

These things happen automatically, even if you fix for 50yrs, it's always in cycles.

These are needed to answer your question .

2,000,295.1

691200

One year=12months. Therefore, 2000*12=24000.
For 20yrs, 24000*20=480000.
Annual interest=24000*12*1/100=2880
20years interest=2880*20=57600
Total money he will receive at the end of the period =480000+57600=#537600

537,600

You have not told us where you got this money from?

Drahl:



I doubt it...this is simple annuity..
Annual savings-2000*12=24000
Then A=p(1+r/100)^n
24000(1.12)^20
231,511.034

i think the interest is re-capitalised monthly. buit he should say so. if not then i dont see why ur answer is not correct

Answer is 537600

N2,000,296 (I just used the online calculator for compound interest)

is this not amortization?!

1729258 if compounding is yearly

1978510 if compounding is monthly