document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140097>Question 1140097</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 7.	The value of a lathe originally valued at K3000 depreciates 15%per annum. Calculate its value after 4 years. The machine is sold when its value is less than K550. After how many years is the lathe sold? (hint : use GP)<BR>\n" );
document.write( "8.	On commencing employment a man is paid a salary of £7200 per annum and receives<BR>\n" );
document.write( "              annual increments of £350. Determine his salary in the 9th year and calculate the total<BR>\n" );
document.write( "              he will have received in the first 12 years.<BR>\n" );
document.write( "9.	A hire tool firm finds that their net return from hiring tools is decreasing by 10% per annum. If their net gain on a certain tool this year is £400, find the possible total of all future profits<BR>\n" );
document.write( "from this tool (assuming the tool lasts for ever). <BR>\n" );
document.write( "10.	K1200 is invested at 12.5% simple interest. How much will have accrued after 3 years?<BR>\n" );
document.write( "11.	K10000 was compounded at 12.5% and amounted to K52015.80. How many years did it take?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760614 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=760614\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 7. 3000*(0.85)^4*3000, the 85% is 100%-15% depreciation.  This is K1566.02<BR>\n" );
document.write( "550=3000(.85)^t<BR>\n" );
document.write( "0.18333=.85^t<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "-1.70=t ln .85, round at end<BR>\n" );
document.write( "divide both sides by ln .85<BR>\n" );
document.write( "t=10.44 years\r<BR>\n" );
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document.write( "8. in the 9th year, he has received 8 increments of 350 pounds or 2800.<BR>\n" );
document.write( "his salary is 10,000 pounds after 9 years and 11,150 after 12 years<BR>\n" );
document.write( "Sum12 is (n/2)(2 a1+(n-1)d)=6(14400+11(350))=109,500 pounds\r<BR>\n" );
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document.write( "9.Geometric series r=0.9, sum is a/1-r=400/.1=4000 pounds\r<BR>\n" );
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document.write( "10. I=prt=1200*1/8*3=$450 simple interest accruing after 3 years.\r<BR>\n" );
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document.write( "11. assuming compounding is annual, 52015.80=10000(1+.125)^n<BR>\n" );
document.write( "5.20128=1.125^n<BR>\n" );
document.write( "ln both sides <BR>\n" );
document.write( "then divide by ln 1.125<BR>\n" );
document.write( "1.6489=t 1n 1.125  round at end<BR>\n" );
document.write( "14.00 years if compounded annually.\r<BR>\n" );
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