chetan2u
naumyuk
If r is a prime number and s is a positive integer, is a \(\frac{r^3}{c}\) terminating decimal?
(1) r \(\geq\) 5
(2) s = 75
Editing your Q as in present state answer is E..
c in \(\frac{r^3}{c}\) must be 's'..
so we have to find if \(\frac{r^3}{s}\) is a terminating decimal...
the value depends on the denominator in its simplified form..
lets see the statements..
(1)
r \(\geq\) 5
it just tell us that r cannot br 2 and 3 nothing else..
insuff
(2)
s = 75
the denominator is 75...
if the numerator is 3 it will be a terminating decimal, otherwise not...
insuff..
combined we know numerator does not have 3, so it will not be a terminating decimal, as we have 3 in denominator..
suff
C
I am sorry I am slightly confused with the explanation. As far as my understanding goes, if the denominator has any other prime number in prime factorisation other than 2 and 5 and if that prime number can be cancelled with the numerator, then the fraction is terminating, otherwise not.
Here, the numerator is >or = to 5 i.e. it could also be 6 or 9 or 12 and those cases the 3 in the denominator will get cancelled. Therefore, there is no way to know what the value of r could be. Hence, my answer was E.
Kindly explain.