document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1208925>Question 1208925</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The price of a pair of sneakers was $100 for the last 10 months of last year. On<BR>\n" );
document.write( "January first, the price increased 30%. After the price increase, an employee<BR>\n" );
document.write( "bought these sneakers with a 10% employee discount. What price did the employee<BR>\n" );
document.write( "pay?\r<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 #847451 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=847451\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Working problems involving successive price increases and decreases using addition for price increases and subtraction for price decreases is slow and inefficient, because you have to do separate calculations to find the price after each increase or decrease.<br><BR>\n" );
document.write( "Working the problems using multiplication, you can account for all the increases and decreases in a single calculation.<br><BR>\n" );
document.write( "In your problem (for example), a price increase of 30% is a multiplication of the price by a factor of 100%+30% = 130%, or 1.3; a price decrease by 10% is a multiplication of the price by a factor of 100%-10% = 90%, or 0.9.  Then the final price is the original price, multiplied by 1.3, and then multiplied by 0.9.<br><BR>\n" );
document.write( "(100)(1.3)(0.9) = 100(1.17) = 117<br><BR>\n" );
document.write( "ANSWER: $117<br><BR>\n" );
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