A + 20 = 10*(A-10)
A + 20 = 10A - 100
20 + 100 = 10A - A
120 = 9A
A = .
Do you see here an integer solution ?
I do not see it, too.
Hence, the problem is DEFECTIVE, by the definition.
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John pointed me that one third of the year is 4 months.
First, I am very glad that John returned back to the forum (!)
Second, the months are of UNEQUAL time length, so it has no much sense to say that one third of the year is 4 months.
It depends on which month / (what months) and which year.
My diagnosis remains the same: if in "age word problem" the answer is not an integer,
then such a problem is DEFECTIVE by the definition.
