document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=385992>Question 385992</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I hope I got the right section.  I'm having trouble with \"Chemical Word Problems\" (that's what my textbook says).  The original problem is \"One solution was 5% bromide,, and the other was 40% bromide.  How much of each should be used to get 60 milliliters of a solution that is 12% bromide?\"  <BR>\n" );
document.write( "The 5% bromide \"container\" I labelled N.  The 40% I labelled D (per the instructions of my text).<BR>\n" );
document.write( "What I did:  I realized that I have two equations: N+d=100 (according to the textbook)which means that n= 100-d ; So I wrote this: (  ) + (  ) = 100  I plugged \"n\" into the first paranthes, and d into the second.  I turned the percents into decimals (5%= 0.5 and 40%= 0.4).  I plugged it in, to get:<BR>\n" );
document.write( "0.5(N) + 0.4(D) = (100)0.12 (0.12=12%)      Since N=100-D, I replaced N with 100-D to get:<BR>\n" );
document.write( "0.5(100-D) + 0.4D = 12   I solved to get 50 - 0.5D + 0.4D = 12 --> 50 - .1D =12   ------> Subtracted 50 from both sides to get ----->-.1DN= -38  ----> divided both sides by -.1 ----> and I got 380.  Logically, that can't be, because N+D=100.  If D = 380 which is greater than 100... What am I doing wrong?!?  I've done other problems of this nature, using the SAME process, and they have all been wrong!  Please Help! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #272957 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=272957\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Try this!<BR>\n" );
document.write( "Let x = the required number of ml. of the 5% bromide solution. Then (60-x) will be the required number of ml. of the 40% bromide solution and the sum of these two amounts (x+60-x) will equal the 60 ml. of the final 12% bromide solution.<BR>\n" );
document.write( "Here's the equation after changing the percentages to their decimal equivalents:<BR>\n" );
document.write( "0.05x+0.4(60-x) = 0.12(60)<BR>\n" );
document.write( "  Simplify and solve for x.<BR>\n" );
document.write( "0.05x+24-0.4x = 7.2 Combine like-terms.<BR>\n" );
document.write( "0.05x-0.4x+24 = 7.2<BR>\n" );
document.write( "-0.35x+24 = 7.2  Subtract 24 from both sides.<BR>\n" );
document.write( "-0.35x = -16.8 Divide both sides by -0.35<BR>\n" );
document.write( "x = 48 and 60-x = 12<BR>\n" );
document.write( "So you'll need to mix 48 ml. of 5% bromide solution with 12 ml. of 40% bromide solution to obtain 60 ml. of 12% bromide solution. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );