Question 1142767
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<pre>
A + 20 = 10*(A-10)


A + 20 = 10A - 100

20 + 100 = 10A - A

120 = 9A


A = {{{120/9}}}.


Do you see here an integer solution ?


I do not see it, too.


Hence, the problem is DEFECTIVE, by the definition.
</pre>


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John pointed me that one third of the year is 4 months.


First, &nbsp;I am very glad that John returned back to the forum (!)


Second, &nbsp;the months are of &nbsp;UNEQUAL&nbsp; time length, &nbsp;so it has no much sense to say that one third of the year is &nbsp;4 &nbsp;months.


It depends on which &nbsp;month / (what months) &nbsp;and which year.



My diagnosis remains the same:  &nbsp;if in &nbsp;"age word problem"  &nbsp;the answer is not an integer, 

then such a problem is &nbsp;DEFECTIVE &nbsp;by the definition.