how to solve dealing with months,years, [simple & compound interest?
457588
Simple interest has its own formula\nFV = P(1+rt)\nwhere FV is the future value(in this case the amount due on the loan), P is the principle, r is the rate and t is the time(in years)\nP = 1780, r = .12 and t = 6/12 = .5\nFV = 1780(1+.12(.5)) = 1886.8\nFor compound interest, use the formula\nFV = P(1+i)^n\nP is the principle, i is the interest rate divided by the number of times it is compounded per year, and n is the number of times it is compounded (n is in terms of months, years or whatever the problem states.\nFor example:\nIf it is compounded monthly, then\ni = .12/12 (divide by 12 because there is 12 months in one year) and r = 6 months\nFV = 1780(1+.12/12)^6 = 1889.51\nIf it is compounded annually, then\ni = .12/1 (because it is compounded once a year) and n = 6/12 = 1/2 years\nFV = 1780(1+.12/1)^(1/2) = 1883.77
749623
Simple Interest\nA = P *r * t\n\nCompound Interest (monthly)\nA = P(1 + r/12)^(12t)\n\nA1 = P(1 + .075/12)^(12*30)\nA2 = A1 * .06 * 1\n\nP(1 + .075/12)^(12*30) * .06 * 1 >= 12000\nP >= 12000 / ((1 + .075/12)^(12*30) * .06)\nP >= $21,227.97
432552
The Accumulate amount, A, = Pe^(in) where P is the investment, i is the interest rate and n is the time.  e is a mathmatical number and is on your calculator.\n\nTo find the present value of an investment n years ago you would use this formula PV = P(1+i)^n  with P the initial investment, i your interest, and n your time.\n\nyou will have to look at when your interest is compounded, if it is monthly or semi-annual make sure to adjust your interest and time accordingly
1084922
N = d + 2m + [3(m+1)/5] + y + [y/4] - [y/100] + [y/400] + 2\n\nd = day of the month\n\nm = number of the month ( for Jan and Feb use 13, 14 of the             previous year, e.g. for feb 18 2003 d=18 m=14 y=2002\n\ny = the year \n\nApril 1, 2007\n\nN = 1+2(4)+{3(4+1)/5}+2007+{2007/4}+{2007/100}-{2007/400}+2\n\nFor numbers in brackets use the whole number, drop everything after the decimal\n\nN=1+8+{3(5)/5}+2007+501+20+5+2\nN=9+15/5+2515\nN=9+3+2515\nN=2527\n\nnow divide N by 7\n\n2527/7=361.0   the number after the decimal is the day of the week. 0 in this case.\n\n0=Sun\n1=Mon\n2=Tue and so on\n\nOr for an easier way, just look on a digital calander
39368
Compound Interest= Principal Amount{1+(rate of intersest/100)^Numebr of years}\nFor you it will be\n21000=5000*(1+r/100)^5\n=>4.2=(1+r/100)^5\n=>1.33=1+r*0.01\n=>r=33%
379493
Usually, they go by average daily balance for the month and then compound it every month.  At least this is what my bank does.  I guess this is simulated by compounding it every day and added the interest once a month.\n\nThe daily interest is annual interest divided by 360.
641632
i have the same problem... just don't max it out in a day... the trick is to pay it off each month... that way you don't get charged interest.
633448
monthly installment? (sales price - deposit)/(%number.of.mths) which could be 12/18/24 +++
906582
assume $Y grows at 10% every year. after 1year\ninterest=Y*(1.1^1)-Y\n\nSo after X years\ncompound interest= Y*(1.1^X)-Y\n\nan example: $1000 grows at 10% every year for 20 year will have compound interest of\n1000(1.1^20)-1000=$5727.5   ( as suppose to $2000 which is what you will have if you earn 10% every year but not compound)\n\nAnother one: $1000 grows at 15% every year for 20 year will have compound interest of\n1000(1.15^20)-1000=$15366   ( as suppose to $3000 which is what you will have if you earn 15% every year but not compound)\n\nenjoy
749562
Use a mortgage calculator because it's very hard to figure out compound interest manually - it's compounded every month, but also reduced by the amount you pay (assuming you make monthly payments) so you can't just multiply the principal ($4,327.77) by 10%\n\nIf you're not making monthly payments, then use an investment calculator.
875073
Problem 1: If a piece of real estate purchased for $75,000 in 1998 appreciates at the rate of 6% per year, then its value t years after the purchase will be f(t)=75,000(1.06^t). According to this model, by how much will the value of this piece of property increase between the years of 2005 and 2008? \n\nf=75000*1.06^3=89326.20\nincrease=89326.20-75000=$14,326.20\n\nProblem 2: The amount A in an account after t years of an initial principle P invested at an annual rate r compounded continuously is given by A=Pe^rt where r is expressed as a decimal. What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously?\n\n500*e^.05*10=500*e^.5=$824.36\n\n\n\nProblem 3: Find the number log5(1/5). \n-1
289683
discounting factors for an interest rate of 8% \nat the end of  year 1 - 100/108 or 1/1.08 = 0.9259\nat the end of  year 2 - 1/1.1664 = 0.8573\nat the end of  year 3 - 1/1.2597 = 0.7938\n\n1.08, 1.1664, 1.2579 compound amounts at the end of year 1,2,3\n\nExpected Cash Flows\n\nDiscounting Factor.. 0.9259\nYear 1 Cash Flow .....$300...$500...$700\nYear 1 Dis Cash Flow..$278...$463...$648\nDiscounting Factor..0.8573.\nYear 2 Cash Flow .....$100...$200...$600\nYear 2 Dis Cash Flow..$86....$172...$514\nDiscounting Factor....0.7938\nYear 3 Cash Flow .....$200...$800...$900\nYear 3 Dis Cash Flow..$159...$635...$714\n\nYear 1\nDisc Cash Flow (x)..278...463...648\nProb P(x)...........0.3....0.4...0.3\nx.P(x) {px}.........83.4--185.2--194.4\n\nExpected Cash Flow in the first year = Sum(px)= $463\n\nYou should be able to calculate the rest..\n\nhttp://www.futureaccountant.com/probability/\n\nhttp://www.schoolingkids.com/
377198
Formula:\n\n\n\nP is the principal (the initial amount you borrow or deposit) \n\nr is the annual rate of interest (percentage)\n\nn is the number of years the amount is deposited or borrowed for.\n\nA is the amount of money accumulated after n years, including interest. \n\nWhen the interest is compounded once a year: \n\nA = P(1 + r)n \n\n\nHowever, if you borrow for 5 years the formula will look like: \n\nA = P(1 + r)5\n\nThis formula applies to both money invested and money borrowed.\n\nFrequent Compounding of Interest:\n\nWhat if interest is paid more frequently? \nHere are a few examples of the formula: \n\nAnnually = P × (1 + r) = (annual compounding) \n\nQuarterly = P (1 + r/4)4 = (quarterly compounding)\n\nMonthly = P (1 + r/12)12 = (monthly compounding)\n\n\nApprox. $716.43 I believe
890824
I haven't seen the original question, but I'm guessing that $817 must be added to $9999 to find the total amount interest is being charged on. If my guess is correct, then p = 10816.  Use t=3.  You only need to use 36 months if, after working out the interest, you have to calculate a monthly payment.  You would then add the interest to p, and divide by 36 to find the monthly repayment.
439816
set up the diffeq:\n\ndM/dt=1.05*M+alpha //money accrued at rate of 5.5% per year plus some amount alpha that is placed into the accout per year\ndm/dt=1.05*m -45000 //rate spent per year after retirement (assuming interest continues to accrue)..\nM(0)=162000; //intial ammount in retirement fund\nM(15)=m(0) //retirement fund begins expenditure with intial value of total saved + interest, or M(15)...\n\n// Note: 12 is measured from date of retirement...\nm(12)=0; //money only lasts 12 years, assume total amount expended...\nsolve dm/dt and plug in m(12)=0 to figure out what the total saved, m(0), must be present at the beginning of retirement in order to completely expend the money in the 12 yr period\n\nfinally, solve for M(t) and use the known values at M(0) and M(15) to compute the value of alpha.
642311
interest-SIMPLE INTEREST
290145
10.\nYou want to figure out the answer yourself, so I'll give you some other months as a clue:\n\nFebruary - 20\nApril - 10\nMay - 5\nJune - 5