document.write( "Question 1059625: if a number is 20% more than the other, how much percent is the second number less than the first. \n" ); document.write( "

Algebra.Com's Answer #674792 by ikleyn(52873) $\"\"$ $\"About$

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\n" ); document.write( "if a number is 20% more than the other, how much percent is the second number less than the first.
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\n" ); document.write( "\n" ); document.write( "Let me re-phrase it equivalently in this way:\r
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document.write( "     if a first number is 20% more than the second, how much percent is the second number less than the first.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Let n1 be the first number and n2 be the second number. Then\r
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\n" ); document.write( "\n" ); document.write( "n1 = 1.2*n2, according to the condition. (We measure numbers in terms of n2 in this case)
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document.write( "                                            (By saying \". . . than n2\", we do agree to measure everything in terms of n2)\r\n" );
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It means that n2 = $\"n1%2F1.2\"$ = 0.8(3)*n1.\r
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\n" ); document.write( "\n" ); document.write( "Hence, n2 is 83.(3)% of n1. (We measure numbers in terms of n1 in this case)
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document.write( "                                            (By saying \". . . than n1\", we do agree to measure everything in terms of n1)\r\n" );
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In other words, n2 is 16.(6)% less than n1.\r
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\n" ); document.write( "\n" ); document.write( "Answer. If a first number is 20% more than the second, then the second number is 16.(6)% less than the first.\r
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