document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=758451>Question 758451</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Could you please help me solve this problem:-\r<BR>\n" );
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document.write( "There are fewer than 30 students in a math's class and this year the grades are given out as follows - <BR>\n" );
document.write( "one third of the students got an A<BR>\n" );
document.write( "one quarter of the students got a B<BR>\n" );
document.write( "four students got a C and<BR>\n" );
document.write( "three students got an E\r<BR>\n" );
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document.write( "How many students got a D?\r<BR>\n" );
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document.write( "My understanding is that there must be 29 or less students in the class to begin with. My next step I think I have to do is work out what fraction is 1/3 + 1/4 + the 4 + 5 students which got a grade away from this number of 29 or less and this will give me the answer.<BR>\n" );
document.write( "Now I know the answer is 5 (the sheet tells me) but I am lost in how you work out this equation. Could you please show me how to calculate this formula - I am a mature student trying to re-visit long ago maths and am finding many missing parts lol.\r<BR>\n" );
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document.write( "Many thanks & most appreciated <BR>\n" );
document.write( "Angie </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #461490 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=461490\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> this is what i see from this problem.<BR>\n" );
document.write( "your basic equation is x/3 + x/4 + 4 + D + 3 < 10<BR>\n" );
document.write( "this can be simplified to:<BR>\n" );
document.write( "x/3 + x/4 + 7 + D < 30<BR>\n" );
document.write( "since x has to be an integer, than x has to be divisible by 3 and also by 4.<BR>\n" );
document.write( "this means that x has to be a multiple of both 3 and 4 and has to be smaller than 30.<BR>\n" );
document.write( "multiples of 3 less than 30 are:<BR>\n" );
document.write( "3,6,9,12,15,18,21,24,27<BR>\n" );
document.write( "multiples of 4 less than 30 are:<BR>\n" );
document.write( "4,8,12,16,20,24,28<BR>\n" );
document.write( "the least common multiple is 12.<BR>\n" );
document.write( "the next least common multiple is 24.<BR>\n" );
document.write( "those are the only 2 numbers < 30 that are integers that can represent the total number of students that can possibly satisfy the equation.<BR>\n" );
document.write( "when x = 12, you get the following:<BR>\n" );
document.write( "A = 12/3 = 4<BR>\n" );
document.write( "B = 12/4 = 3<BR>\n" );
document.write( "C = 4<BR>\n" );
document.write( "D = D<BR>\n" );
document.write( "E = 3<BR>\n" );
document.write( "The total of students becomes 14 + D which is greater than 12, so the total number of students can't be 12.<BR>\n" );
document.write( "when x = 24, you get the following:<BR>\n" );
document.write( "A = 24/3 = 8<BR>\n" );
document.write( "B = 24/4 = 6<BR>\n" );
document.write( "C = 4<BR>\n" );
document.write( "D = D<BR>\n" );
document.write( "E = 3<BR>\n" );
document.write( "the total of students becomes 21 + D<BR>\n" );
document.write( "since the total of student is equal to 24, D must be equal to 3.<BR>\n" );
document.write( "D = 3 is your solution.\r<BR>\n" );
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document.write( "i don't know how 5 can be a solution and you can wind up with integer values for everything.<BR>\n" );
document.write( "when D = 5, you get:<BR>\n" );
document.write( "A = x/3<BR>\n" );
document.write( "B = x/4<BR>\n" );
document.write( "C = 4<BR>\n" );
document.write( "D = 5<BR>\n" );
document.write( "E = 3<BR>\n" );
document.write( "the total of students is equal to x/3 + x/4 + 12 which must be less than 30.<BR>\n" );
document.write( "this means that x/3 + x/4 must be less than 18.<BR>\n" );
document.write( "once again, x must be an integer and the only common multiple less than 18 is equal to 12 and it was already shown up above that x can't be equal to 12.\r<BR>\n" );
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document.write( "the key to solving this problem is that x has to be an integer.<BR>\n" );
document.write( "without taking that into consideration, you run into difficulty as i did when i didn't make that assumption.\r<BR>\n" );
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document.write( "this is because the equation is equal to less than 30 which makes 29 possible values of x that have to be solved for independently and, when you do that, you will find that most of them wind up with the value of x or the value of D not being an integer.\r<BR>\n" );
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document.write( "the possible solutions for every value of x < 30 where x is an integer are shown in the following table:\r<BR>\n" );
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document.write( "from the table, you can see that D is negative when x is smaller than or equal to 16.\r<BR>\n" );
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document.write( "from the table, when x is greater than 16 and less than 30, you can also see that A, B, and D are integers only when x is equal to 24.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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