on certain principal the simple interest and compound interest for 2 years at the same rate of interest are Rs.400 and Rs. 420 respectively.What is the principal?
216819
Let us suppose interest rate is ‘r’ and the principal amount was ‘P’\nThen for two years;\n\nSimple Interest = 2rP\nCompound interest = P(1+r)^2 – P\nSince we know both the amount of Simple interest and the compound interest we get a set of two simultaneous equations\n\n2rP = 400(1)\nP(1+r)^2 –P = 420(2)\n\nFrom first equation we get \nP=400/2r or P = 200/r\n\nSubstituting in Eq (2)\n\n(200/r)(1+2r+r^2)-200/r = 420\n200/r(1+2r+r^2-1) = 420\n1/r(2r+r^2) = 420/200\n2+r = 420/200\nr = 420/200-2\nr = 2.10-2\nr = 0.10\nSolving for P in eq(1)\nP = 400/2r\nP = 200/r\nP = 200/0.10\nP = 2000\n\nSo the principal amount is 2000.
53969
There is in fact a law but it has a high limit. I think its something like 33% over current prime rate. this could mean interest as high as 39%.
496592
Original investment = $6,400\nDecrease % = 8.2\nDecrease = (8.2*6400)/100\n                 = 52480/100\n                 = $524.80\nValue at the end of the year = $6400-524.80\n                                             = $5875.20
578398
Annual compounding yields $2,100.00\nQuarterly compounding yields $2,184.03\nMonthly compounding yields $2,203.91\nDaily compounding yields $2,213.69\n\nNotice that the gains made as the compounding periods increase is not proportional.  That is, going from annual to quarterly gives you an additional $84.03, but going from monthly to daily only gives you $9.78 added earnings.  So the more compounding periods there are, the less you will gain by adding more compounding periods.
22313
You're forgetting that interest is often tax-deductible. If you can take advantage of that, you aren't paying as much. And yes, some houses can double or triple in value in a relatively short amount of time. It all depends on where you buy.
652505
The formula for compound interest calculation is  \nFuture Value = Principal (1 + monthly interest rate*12/100 )^no. of years\nHence $100 would become = 100(1+ 1.67*12/100)^5\n= $ 100(1+20.04/100 )^5\n= $ 100 ( 1.2004)^5\n= $ 100 * 2.4924697\n= $ 249.25 (rounded to two decimals)
311203
$13,455. x 9% = $1210.95 Interest\n$13455.00 + $1210.95 = $14,665.95\nUnless the interest was 9% per annum\nThen you only use 1/2 of the total interest amount and the previous answers are correct (or $605.48 + $13455.).
640152
COMPOUND INTEREST\n---------------------------------\n\n S = P (1 + I) n \n\nwhere:\n\nS = the future value\n\nP = the principal amount\n\nI = the periodic rate of interest*\n\nn = the # of compounding periods for the term of the loan\n--------\n\nFor example,\n\nIf the principal amount is $10,000, the periodic rate of interest is 10% and the # of compounding periods is 6, what is the value at maturity?\n\nS = P (1 + I) n \n\nS = $10000 (1 + .10) 6\n\nS = $ 17715.61\n\nhttp://homepages.cambrianc.on.ca/tutorial/thetutorialcentre/mathscience/math/compound_interest.htm\n\n\nSIMPLE INTEREST \n------------------------------\n\nAt start:\n\nT = P\n\nAfter 1 period:\n\nT = P + Pr = P(1 + 1r)\n\nAfter 2 periods:\n\nT = P + Pr + Pr = P + 2Pr = P(1 + 2r)\n\nAfter 3 periods:\n\nT = P + Pr + Pr + Pr = P + 3Pr = P(1 + 3r)\n\nAfter n periods:\n\nT = P(1 + nr)\n\n--------------------\nWhen money is borrowed, interest is charged for the use of that money for a certain period of time. When the money is paid back, the principal (amount of money that was borrowed) and the interest is paid back. The amount to interest depends on the interest rate, the amount of money borrowed (principal) and the length of time that the money is borrowed. \nThe formula for finding simple interest is: Interest = Principal * Rate * Time. If $100 was borrowed for 2 years at a 10% interest rate, the interest would be $100*10/100*2 = $20. The total amount that would be due would be $100+$20=$120. \n\nSimple interest is generally charged for borrowing money for short periods of time. Compound interest is similar but the total amount due at the end of each period is calculated and further interest is charged against both the original principal but also the interest that was earned during that period. \nhttp://www.learningwave.com/lwonline/percent/interest.html\n\n\n\nhttp://www.aaamath.com/g84-simple-interest.html \na calculater ..
271244
first, start with how much money you have, then find the rate, and the time\n\nfor example:\nI have $20,000, at a rate of 12% for 36 months. The interest would be $20,000 times .12 times 36.
521121
1. 50 x (.35)=17.5  \nExplanation %=((part of)/whole )x100\nso part of=whole (%/100)\n2. 40 (.82)=32.8 (33)\nSame as above\n\n3. (1.75 -1.5)/1.5=.166 16.6%\nExplanation again %=((part of)/whole )x100\nIn this case the original whole is 1.5 \n\n4. (12000-8500)/12000=.291 0r 29.1%\nIn this case the original whole is 12000 \n\n5. 20/1.75=11.43\nHere is a bit tricky x(1+%/100)=the new whole (interesting term ha?) or the original whole + increase = the new whole :)\n\n6.2000(1+.04) (1+.04)=2163.2 (made a mistake earlier)\n\n\nyear 1  principal + % or earned interest (1)\nyear 2  principal + % or earned interest (1) +(principal + % or earned interest (1) %)
103188
$1,067,500 more or less. \n\nThe formula is:\n\n(((1+i)^n)-1)/(i*(1+i)^n)*100,000\n\nWhere i = interest rate,  n = number or years
392184
Assuming you start at the beginning of a month\n\nFirst investment = $350\nAfter the 1st half month amount saved =  $(350(1 + r) + 350)\nAfter 2nd half month amount saved = $((350(1 + r) + 350)(1 + r) + 350)\n=$(350(1 + r)² + 350(1 + r) + 350)\nAfter 3rd half month amount = $350((1 + r)² + (1 + r) + 1)(1 + r) + 350)\n= $350((1 + r)³ + (1 + r)² + (1 + r) + 1)\nThat will mean that, at maturity,\nAmount = $350((1 + r)^20 + ...  + (1 + r)³ + (1 + r)² + (1 + r) + 1)\n= $350 ((1 + r)^21 - 1)/((1 + r) - 1)\n= $100000\nSo [(1 + r)^21 - 1] / r = 100000/350\nSo r/[(1 + r)^21 - 1] = 0.0035\nUsing Excel r = 0.2185\nie 21.85% per half year\nor 43.7% pa ... a rather ambitious interest rate!!\n\nAnd it would need to be high considering you are only investing 20 * $350 ie $7000 to mature at $100000 in 10 years
206524
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
384329
If 0.25% hike in interest requires £13 pm extra, \nthis means 13x12 = 156 pa. \nTherefore the annual interest for 7% is \n\n156 / 0.25 x 7 = £4368
164791
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
472141
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
491385
You are correct; \n\nIf you borrow ₤1500 at the interest of 14%\n\nYou will be paying back ₤1710 as the interest is ₤210 (value of 14% of ₤1,500)\n\nYou will be paying the Principal (₤1,500) plus the interest (₤210)\ngiving you a total of ₤1710.\n\nYou are not stupid, no one knows it all.