document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1137698>Question 1137698</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I posted this question twice earlier with no response. Would appreciate your help.\r<BR>\n" );
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document.write( "Antonio’s bank charges an $8 service fee each month if his account goes below $500. In June, Antonio had $510 in his account. He withdrew $25 and then did not have any transactions in his account for 6 months.\r<BR>\n" );
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document.write( "a. Write as an integer the amount by which Antonio’s bank balance changed during the six months.\r<BR>\n" );
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document.write( "b.What was the balance in the account at the end of January?\r<BR>\n" );
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document.write( "c. How much will Antonio have to deposit for his account to be at least $500?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #755553 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=755553\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> in june, antonio has 510 in his account.\r<BR>\n" );
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document.write( "he withdrew 25 dollars and didn't have any other transactions for 6 months.\r<BR>\n" );
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document.write( "in june, his account went from 510 to 485.\r<BR>\n" );
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document.write( "unless his account was earning some interest, it would remain at that level forever.\r<BR>\n" );
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document.write( "you didn't give an interest rate that the account was earning at.\r<BR>\n" );
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document.write( "for example, if the account was earning 12% per year compounded monthly, then the account would be earning 1% per month and the figures would be as follows:\r<BR>\n" );
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document.write( "f = p * (1 + r) ^ n is the formula to use for this.\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the preswent value<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( "since you are going month by month, you only need the interest rate per month and you would do the following.\r<BR>\n" );
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document.write( "in june, his balance was 485 after the 25 dollar withdrawal.\r<BR>\n" );
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document.write( "at the end of june, he paid 8 dollars because the account was less than 500.\r<BR>\n" );
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document.write( "at the end of july, the account grew to 1.01 * 485 = 489.85 which is still less than 500, so he paid an 8 dollar transaction fee.\r<BR>\n" );
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document.write( "at the end of august, the account grew to 1.01 * 489.85 = 494.7485 which is still less than 500, so he paid an 8 dollar transaction fee.\r<BR>\n" );
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document.write( "at the end of september, the account grew to 1.01 * 494.7485 = 499.695985 which is still less than 500, so he paid an 8 dollar transaction fee.\r<BR>\n" );
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document.write( "at the end of october, the account grew to 1.01 * 499.695985 = 504.6929449 which is more than 500, so he didn't have to pay the 8 dollar transaction fee any more as long as the account balance remained above 500.\r<BR>\n" );
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document.write( "the 1% per month is very generous, since most bank accounts don't give 12% interest rate per year.\r<BR>\n" );
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document.write( "it's more like 1.2 or 2.4 percent per year which would be .1 or .2 percent per month.\r<BR>\n" );
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document.write( "i used those number because they divide evenly, not because those are the actual percentages the bank might give you.\r<BR>\n" );
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document.write( "you can use the formula provided to see when the account would go past 500.\r<BR>\n" );
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document.write( "the formula is f = p * (1 + r) ^ n\r<BR>\n" );
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document.write( "f is the future value, which you would set at 500.\r<BR>\n" );
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document.write( "p is the present value which you would set at 485.\r<BR>\n" );
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document.write( "r is the interest rate per time period which is the annual interest rate / 12.\r<BR>\n" );
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document.write( "to get the monthly rate, you divide the annual percent rate by 1200.\r<BR>\n" );
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document.write( "12 to get it to the month and 100 to get it to a rate rather than a percent rate.\r<BR>\n" );
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document.write( "n is the number of months.\r<BR>\n" );
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document.write( "at 12% percent interest rate per year, the interest rate per month is 12/1200 = .01 and the formula becomes 500 = 485 * (1 + .01) ^ n.\r<BR>\n" );
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document.write( "at 1.2 percent interest rate per year, the interest rate per month is 1.2/1200 = .001 and the formula becomes 500 = 485 * (1 + .001) ^ n.\r<BR>\n" );
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document.write( "at 2.4 percent interest rate per year, the interest rate per month is 2.4/1200 = .0024 and the formula becomes 500 = 485 * (1 + .002) ^ n.\r<BR>\n" );
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document.write( "you would solve for n in these formulas.\r<BR>\n" );
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document.write( "i'll show you how to do one and you can do the others if you want to.\r<BR>\n" );
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document.write( "i'll give you the answers for all.\r<BR>\n" );
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document.write( "at 12%, the formula is 500 = 485 * (1 + .01) ^ n\r<BR>\n" );
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document.write( "divide both sides of this formula by 485 and simplify to get 500/485 = 1.01 ^ n\r<BR>\n" );
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document.write( "take the log of both sides of this formula to get log(500/485) = log(1.01^n)\r<BR>\n" );
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document.write( "since log (1.01^n) is equal to n * log(1.01), the formula becomes log(500/485) = n * log(1.01)\r<BR>\n" );
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document.write( "divide both sides of this equation by log(1.01) to get log(500/485) / log(1.01) = n\r<BR>\n" );
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document.write( "solve for n to get n = 3.061125096.\r<BR>\n" );
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document.write( "the future value will be 500 in 3.061125096 months.\r<BR>\n" );
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document.write( "starting from end of june, that becomes end of july is 1 month from end of june, end of august is 2 months from end of june, end of september is 3 months from end of june.   that means the account balance will become 500 some time in october.\r<BR>\n" );
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document.write( "we already did that above and the answer in both cases agree with each other, as they should.\r<BR>\n" );
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document.write( "that's at 12% per year.\r<BR>\n" );
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document.write( "at 1.2% per year and 2.4% per year, i calculated how long it would take the the account to get to 500.\r<BR>\n" );
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document.write( "at 1.2% per year, it would take  30.47443455 months for the account to grow to 500.\r<BR>\n" );
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document.write( "at 2.4% per year, it would take 15.24482827 months for the account to grow to 500.\r<BR>\n" );
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document.write( "use the formula i provided you and substitute your percent interest rate per year by that rate / 1200 and put it in the formula and solve as i did using logs (assuming you know how to do logs).\r<BR>\n" );
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document.write( "in fact, assuming you know how to do logs, you can just do the following.\r<BR>\n" );
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document.write( "take log(500/485) / log(x) to get the number of months.\r<BR>\n" );
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document.write( "x would be your annual interest rate / 1200 + 1.\r<BR>\n" );
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document.write( "at 12%, x would be equal to 12/1200 + 1 which would become x = 1.01.\r<BR>\n" );
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document.write( "at 1.2%, x would be equal to 1.2/1200 + 1 which would become x = 1.001.\r<BR>\n" );
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document.write( "at 2.4%, x would be equal to 2.4/1200 + 1 which would become x = 1.002\r<BR>\n" );
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document.write( "do the same with whatever percent interest rate per year you have and you'll get the numbe of months it takes for the acount to become equal to 500 again.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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