Question

Classify the decimal form of the given rational number into terminating non-terminating recurring type.

• A. $\frac{13}{5}$
• B. $\frac{2}{11}$
• C. $\frac{29}{16}$
• D. $\frac{17}{125}$
• E. $\frac{11}{6}$

Answer 1

The correct option is A $\frac{13}{5}$

We have,

$A)$ $\frac{13}{5}=\frac{13}{5^1\times 2^0}$

So, the denominator $5$ of $\frac{13}{5}$ is of the form $2^m\times 5^n$, where $m,n$ are non-negative integers.

Hence, $\frac{13}{5}$ has terminating decimal expansion.

$B)$ $\frac{2}{11}$

So, the denominator $11$ of $\frac{2}{11}$ is not of the form $2^m\times 5^n$, where $m,n$ are non-negative integers.

Hence, $\frac{2}{11}$ has non-terminating decimal expansion.

$C)$ $\frac{29}{16}=\frac{29}{2^4\times 5^0}$

So, the denominator $16$ of $\frac{29}{16}$ is of the form $2^m\times 5^n$, where $m,n$ are non-negative integers.

Hence, $\frac{29}{16}$ has terminating decimal expansion.

$D)$ $\frac{17}{125}=\frac{17}{5^3\times 2^0}$

So, the denominator $125$ of $\frac{17}{125}$ is of the form $2^m\times 5^n$, where $m,n$ are non-negative integers.

Hence, $\frac{17}{125}$ has terminating decimal expansion.

$E)$ $\frac{11}{6}$

So, the denominator $6$ of $\frac{11}{2\times 3}$ is not of the form $2^m\times 5^n$, where $m,n$ are non-negative integers.

Hence, $\frac{11}{6}$ has non-terminating decimal expansion.