document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1042015>Question 1042015</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Samantha takes $28000.00 to the bank to invest into a term deposit.the bank gives her three option of interest rate with different compounding.\r<BR>\n" );
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document.write( "option 1 - simple interest type @ 10.2 % annually<BR>\n" );
document.write( "option 2 - compounded interest type @ 10% annually<BR>\n" );
document.write( "option 3 - compounded interest type @ 9.8% daily\r<BR>\n" );
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document.write( "compute her total amount after 30 months for all three option.conclude which option she should take to get maximum return on her investment. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #657159 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=657159\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> her present amount is 28000<BR>\n" );
document.write( "she will be investing for 30 months.\r<BR>\n" );
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document.write( "formula for simple interest is i = p * r * n.<BR>\n" );
document.write( "i is the interest.<BR>\n" );
document.write( "p is the present amount.<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "formula for compound interest is i = f - p.<BR>\n" );
document.write( "first you find the future value and then you find the interest.<BR>\n" );
document.write( "formula for future value is f = p * (1 + r)^n.<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present amount<BR>\n" );
document.write( "r is the interest rate per time period<BR>\n" );
document.write( "n is the number of time periods<BR>\n" );
document.write( "i = interest\r<BR>\n" );
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document.write( "in the formulas you need to use interest rate as a decimal and not interest rate as a percent.<BR>\n" );
document.write( "interest rate as a decimal is equal to interest rate as a percent divided by 100.\r<BR>\n" );
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document.write( "option 1 is simple interest type at 10.2% annually.<BR>\n" );
document.write( "formula is i = p * r * n.<BR>\n" );
document.write( "p = 28000.<BR>\n" );
document.write( "r = 10.2% per year divided by 100 = .102 per year divided by 12 = .0085 per month.<BR>\n" );
document.write( "n = 30 months.<BR>\n" );
document.write( "formula becomes:<BR>\n" );
document.write( "i = 28000 * .0085 * 30 = 7140.<BR>\n" );
document.write( "interest is equal to 7140.\r<BR>\n" );
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document.write( "option 2 is compound interest type at 10% compounded annually.<BR>\n" );
document.write( "p = 28000.<BR>\n" );
document.write( "r = 10% per year divided by 100 = .10 per year.<BR>\n" );
document.write( "n = 30 months divided by 12 = 2.5 years.<BR>\n" );
document.write( "formula of f = p * (1 + r)^n becomes:<BR>\n" );
document.write( "f = 28000 * (1.10)^2.5<BR>\n" );
document.write( "solve for f to get f = 35533.64<BR>\n" );
document.write( "i = f - p<BR>\n" );
document.write( "this says that the interest is equal to the future value minus the present amount.<BR>\n" );
document.write( "formula becomes:<BR>\n" );
document.write( "i = 35533.64 - 28000 = 7533.64<BR>\n" );
document.write( "interest is 7533.64\r<BR>\n" );
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document.write( "option 3 is compound interest type at 9.8% compounded daily.<BR>\n" );
document.write( "here it gets sticky because the number of days in a year is not exactly 365 and the number of days in a month is not exactly 30.<BR>\n" );
document.write( "you have to make some assumptions unless you are told what to assume is the number of days in a year and, consequently, the number of days in a month.\r<BR>\n" );
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document.write( "you can use 365 days in a year or 360 days in a year or 365.25 days in a year, or if you really want to get fancy, some other number that takes into account you have a leap year every 4 years and some other adjustment every 100 years.\r<BR>\n" );
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document.write( "all of them will yield an answer that should be very close to each other and should be consistent.\r<BR>\n" );
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document.write( "consistent means that the number is either the greater of each of the other two or not the greater of each of the other two.\r<BR>\n" );
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document.write( "if we assume 360 days in a year, we would get:\r<BR>\n" );
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document.write( "p = 28000<BR>\n" );
document.write( "r = .098 per year divided by 360 = .000272222 per day<BR>\n" );
document.write( "n = 30 months * 360/12 = 900 days.<BR>\n" );
document.write( "formula of f = p * (1 + r)^n becomes:<BR>\n" );
document.write( "f = 28000 * (1.000272222)^900<BR>\n" );
document.write( "solve for f to get f = 35772.20<BR>\n" );
document.write( "i = f - p<BR>\n" );
document.write( "i = 7772.20\r<BR>\n" );
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document.write( "if we assume 365 days in a year, we would get:\r<BR>\n" );
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document.write( "p = 28000<BR>\n" );
document.write( "r = .098 per year divided by 365 = .000268493 per day.<BR>\n" );
document.write( "n = 30 months * 365/12 = 912.5 days.<BR>\n" );
document.write( "formula of f = p * (1 + r)^n becomes:<BR>\n" );
document.write( "f = 28000 * (1.000268493)^912.5<BR>\n" );
document.write( "solve for f to get f = 35772.2204<BR>\n" );
document.write( "i = f - p<BR>\n" );
document.write( "i = 7772.22\r<BR>\n" );
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document.write( "your interest for the 3 options are:\r<BR>\n" );
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document.write( "7140 for option 1.<BR>\n" );
document.write( "7533.64 for option 2.<BR>\n" );
document.write( "7772.20 for option 3 assuming 360 days in a year.<BR>\n" );
document.write( "7772.22 for option 3 assuming 365 days in a year.\r<BR>\n" );
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document.write( "option 3 is the winner whether or not you assumed 360 days in a year or 365 days in a year.\r<BR>\n" );
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document.write( "option 3 would still be the winner even if you assumed 365.25 days in a year or something even more exotic, so i didn't bother to calculate using those.\r<BR>\n" );
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document.write( "they should really have told you how many days in a year to assume.<BR>\n" );
document.write( "this takes away the ambiguity.<BR>\n" );
document.write( "not telling you makes you work harder that you need to.<BR>\n" );
document.write( "in this case, it didn't matter.<BR>\n" );
document.write( "either assumption would have led to the same conclusion.\r<BR>\n" );
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document.write( "the more times you compound, the greater the effective rate, given that the present amount is the same and the annual interest rate is the same.<BR>\n" );
document.write( "the number of times you compound in a year reaches a limit which culminates into the continuous compounding formula of f = p * e^(r * n).<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present amount<BR>\n" );
document.write( "e is the scientific constant of 2.718281828.....<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.<BR>\n" );
document.write( "in this formula, it doesn't matter whether the time period is in years or months of days or whatever because r*n will always be the same.<BR>\n" );
document.write( "for example:<BR>\n" );
document.write( "if r is 12% per year and n is 1 year, then r*n becomes .12 * 1 = .12<BR>\n" );
document.write( "divide 12% per year by 12 and you get 1% per month.<BR>\n" );
document.write( "multiply 1 year by 12 and you get 12 months.<BR>\n" );
document.write( "r * n becomes .01 * 12 = .12.<BR>\n" );
document.write( "same number.\r<BR>\n" );
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document.write( "if you had used the continuous compounding formula for option 3, your answer would have been f = 28000 * e^(.098 * 2.5) = 35773.39  which would have resulted in i = 7773.39.\r<BR>\n" );
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document.write( "7773.39 is not much greater than 7772.22<BR>\n" );
document.write( "this tells you that assuming daily compounding gives you a close approximation to continuous compounding.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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