Question 1059625
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if a number is 20% more than the other, how much percent is the second number less than the first.
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Let me re-phrase it equivalently in this way:

<pre>
     if a first number is 20% more than the second, how much percent is the second number less than the first.
</pre>

Let n1 be the first number and n2 be the second number. Then


n1 = 1.2*n2, according to the condition. &nbsp;&nbsp;&nbsp;&nbsp;(We measure numbers in terms of &nbsp;n2&nbsp; in this case)
<pre>
                                            (By saying ". . . than n2", we do agree to measure everything in terms of n2)
</pre>It means that &nbsp;n2 = {{{n1/1.2}}} = 0.8(3)*n1.


Hence, &nbsp;n2 &nbsp;is &nbsp;83.(3)% &nbsp;of &nbsp;n1. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(We measure numbers in terms of &nbsp;n1&nbsp; in this case)
<pre>
                                            (By saying ". . . than n1", we do agree to measure everything in terms of n1)
</pre>In other words, &nbsp;n2 &nbsp;is &nbsp;16.(6)% &nbsp;less &nbsp;than &nbsp;n1.


<U>Answer</U>.  &nbsp;If a first number is &nbsp;20%&nbsp; more than the second, &nbsp;then the second number is &nbsp;16.(6)% &nbsp;less than the first.