SOLUTION: if a number is 20% more than the other, how much percent is the second number less than the first.

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Question 1059625: if a number is 20% more than the other, how much percent is the second number less than the first.
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
The second number is not less than the first.
The second number is 20% more than the first.
Answer by ikleyn(52851)   (Show Source): You can put this solution on YOUR website!
.
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: 
     if a first number is 20% more than the second, how much percent is the second number less than the first.

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

n1 = 1.2*n2, according to the condition.     (We measure numbers in terms of  n2  in this case)
                                            (By saying ". . . than n2", we do agree to measure everything in terms of n2)
It means that  n2 = = 0.8(3)*n1.

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

Answer.  If a first number is  20%  more than the second,  then the second number is  16.(6)%  less than the first.



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