A) terminating
B) recurring
C) non-terminating and non-recurring
D) None of the above
Correct Answer: A
Solution :
\[\frac{61}{{{2}^{5}}\times {{5}^{3}}}\]here the denominator have only 2 or 5 prime factor (or in the form of \[{{2}^{n}}\,{{5}^{m}}\]), so \[\frac{61}{{{2}^{5}}\times {{5}^{3}}}\]is a terminating decimal.You need to login to perform this action.
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