document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=319261>Question 319261</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The intensity of a light source varies inversely as the square of the distance from the source. If the distance from a source increases by 25%, by what percent does the intensity decrease? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #228592 by </B><A HREF=/tutors/aboutme.mpl?userid=devjyotsingh24><B>devjyotsingh24(4)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=devjyotsingh24><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=228592\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the orginal intensity be 100 units.<BR>\n" );
document.write( "Now according to the problem if distance increases by 25% i.e. 1/4 times the intensity should decrease by sqrt(1/4) times=1/2 times.<BR>\n" );
document.write( "Thus decrease in intensity=1/2(100)=50 units.<BR>\n" );
document.write( "Therefore % decrease=(decrease in intensity/orignal intensity)*100=50% </SPAN>\n" );
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