document.write( "Question 1844: HOW MANY GRAMS OF PURE SALT MUST BE ADDED TO 40 GRAMS OF A 20% SALT SOLUTION TO MAKE A SOLUTION THAT IS 36% SALT? \n" ); document.write( "
Algebra.Com's Answer #705 by longjonsilver(2297)\"\" \"About 
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For a 20% salt solution, one fifth (ie 20%) is salt. As we have 40g of the solution, then 8g (ie a fifth) must be salt.\r
\n" ); document.write( "\n" ); document.write( "we would work out the % by saying (8/40)*100 = 20%\r
\n" ); document.write( "\n" ); document.write( "However, we now will add some more salt. Call this unknown amount, x grams. We therefore increase the amount of salt by 8 and the total amount by 8 too. And the new salinity is 36%. So, we have \"%28%288%2Bx%29%2F%2840%2Bx%29%29%2A100+=+36\"\r
\n" ); document.write( "\n" ); document.write( "Re-arranging this gives\r
\n" ); document.write( "\n" ); document.write( "\"%28%288%2Bx%29%2F%2840%2Bx%29%29+=+%280.36%29\"\r
\n" ); document.write( "\n" ); document.write( "\"%288%2Bx%29+=+%280.36%29%2840%2Bx%29\"\r
\n" ); document.write( "\n" ); document.write( "\"%288%2Bx%29+=+%2814.4+%2B+%280.36%29%28x%29%29\"\r
\n" ); document.write( "\n" ); document.write( "Rearranging this again gives\r
\n" ); document.write( "\n" ); document.write( "\"%280.64%29%28x%29+=+%286.4%29\"\r
\n" ); document.write( "\n" ); document.write( "hence x=10g of salt required.\r
\n" ); document.write( "\n" ); document.write( "We can see this in the formula...it is now (18/50)*100 which is 36%\r
\n" ); document.write( "\n" ); document.write( "cheers
\n" ); document.write( "Jon.
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