document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1188238>Question 1188238</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Justin had some cards. He gave Alan 1/10 of the cards and bought another 45 <BR>\n" );
document.write( "new cards. Then he used 1/3 of the new total number of cards and bought 40 new<BR>\n" );
document.write( "cards. He had 160 cards in the end. How many cards did he have at first? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #821447 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</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=821447\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Yes, this particular problem is probably solved most easily by working backward.<br><BR>\n" );
document.write( "He finished with 160 cards, after buying 40 more, so before buying those 40 he had 120.<BR>\n" );
document.write( "The 120 cards was after he used 1/3 of what he had.  This part is a bit tricky to do mentally.  Since the 120 was after he used 1/3 of what he had before, the 120 is 2/3 of what he had before; therefore the number he had before was (3/2) times 120, which is 180.<br><BR>\n" );
document.write( "That's the first couple of steps; I'll leave it to you to go the last couple.<br><BR>\n" );
document.write( "If you are going to solve it \"forwards\" using formal algebra, take a moment to see what is happening in the problem before deciding how to set the problem up.<br><BR>\n" );
document.write( "The algebraic solution from one of the other tutors starts immediately by letting x be the initial number of cards; this leads quickly to equations involving fractions, which are always a bit harder to work with.<br><BR>\n" );
document.write( "Instead, noticing that the first thing that happens is he give away 1/10 of his cards, start with the initial number of cards being 10x. That gives us....<br><BR>\n" );
document.write( "initial number: 10x<BR>\n" );
document.write( "after giving 1/10 away: 9x<BR>\n" );
document.write( "after buying 45 more: 9x+45<BR>\n" );
document.write( "after using 1/3 of those (2/3 of 9x+45 are left): 6x+30<BR>\n" );
document.write( "after buying 40 more: 6x+70<br><BR>\n" );
document.write( "At that point he had 160 cards:<br><BR>\n" );
document.write( "6x+70=160<BR>\n" );
document.write( "6x=90<BR>\n" );
document.write( "x=90/6=15<br><BR>\n" );
document.write( "ANSWER: The number he started with was 10x=150<br><BR>\n" );
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