Terminating decimal is a such decimal that has only a finite number of non-zero digits. That's why if you will write such decimal as a fraction you will have as a denominator 10,100, 1000, etc. Since the fraction can be simplified, you can have in the denominator at the end some product of 2 or 5.
For example:
\(0.4=4/10=2/5\), 5 in the denominator
\(0.25=25/100=1/4\), 4=2*2 in the denominator.
But anyway, if you have terminating decimal you can't have anything other 2 or 5 in prime factorization of denominator. So we have a rule:
The fraction expressed as a decimal will be the terminating decimal if it can be presented as \(\frac{a}{{2^n\cdot 5^m}}\) where \(a\) is an integer, \(m=0,1,2,3..\). , and \(n=0,1,2,3...\) .Or
if a fraction has in denominator any prime factor different from 2 and 5, such fraction will be infinite decimal.Now, according to your problem:
I. Has only 2 as prime factor, since \(32=2^5\).
TerminatingII. Has 7 as prime factor.
InfiniteIII. Has only 5 as prime factor.
TerminatingThe correct answer is B.
Hope this helps!:)