document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1061758>Question 1061758</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Over a specified&#8203; distance, rate varies inversely with time. If a car on a test track goes a certain distance in one dash one-half minute at 150 &#8203;mph, what rate is needed to go the same distance in two dash two-thirds &#8203;minute? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #676634 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=676634\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula for inverse variation is:\r<BR>\n" );
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document.write( "y = k / x\r<BR>\n" );
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document.write( "k is the constant of variation.\r<BR>\n" );
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document.write( "in this problem k will represent the distance, which will be constant.\r<BR>\n" );
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document.write( "in general, you should convert everything to the same units of measure so as to be consistent.\r<BR>\n" );
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document.write( "in this problem, it didn't matter, but i will do the conversion anyway just to be consistent with the general rule, and to find the actual distance accurately.\r<BR>\n" );
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document.write( "first we'll do the conversion.\r<BR>\n" );
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document.write( "1 and 1/2 minutes is equal to 3/2 minutes divided by 60 which is equal to 3/120 hours which is equal to 1/40 hours.\r<BR>\n" );
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document.write( "2 and 2/3 minutes is equal to 8/3 minutes divided by 60 which is equal to 8 / 180 hours which is equal to 2/45 hours.\r<BR>\n" );
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document.write( "if we let k equal the distance and y equal the miles per hour and x equal the time, then the formula of y = k/x becomes rate = distance / time.\r<BR>\n" );
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document.write( "when rate = 150 mph and time = 1/40 hours, we get 150 = distance / (1/40).\r<BR>\n" );
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document.write( "we solve for distance to get distance = 150 / 40 = 3.75 miles.\r<BR>\n" );
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document.write( "now that we know k, we can solve for rate when time is equal to 2/45 hours.\r<BR>\n" );
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document.write( "formula becomes rate = 3.75 / (2/45).\r<BR>\n" );
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document.write( "this is the same as rate = 3.75 * 45 / 2 which results in rate = 84.375 miles per hour.\r<BR>\n" );
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document.write( "if he travels at 150 miles per hour, he covers 3.75 miles in one and a half minutes.\r<BR>\n" );
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document.write( "if he travels at 84.375 miles per hour, he covers 3.75 miles in two and two third minutes.\r<BR>\n" );
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document.write( "note that this problem could have been solved without doing the conversion from minutes to hours, but k would not then have represented the distance.\r<BR>\n" );
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document.write( "it would, however, still have been the constant of variation and could still be used to solve the problem.\r<BR>\n" );
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document.write( "in general, the inverse variation formula is y = k / x.\r<BR>\n" );
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document.write( "note that 1 and 1/2 minutes = 3/2 minutes and 2 and 2/3 minutes = 8/3 minutes.\r<BR>\n" );
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document.write( "if we let y = 150 and x = 3/2, then the formula becomes 150 = k / (3/2).\r<BR>\n" );
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document.write( "solve for k to get k = 225.\r<BR>\n" );
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document.write( "when k = 225 and x = 8/3, the formula becomes y = 225 / (8/3) which is the same as 225 * 3/8 which is equal to 84.375.\r<BR>\n" );
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document.write( "we got the same answer, even though we did not convert minutes to hours.\r<BR>\n" );
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document.write( "in this case, k was just the constant of variation and did not represent distance in miles.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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