Rational numbers and their decimal expansions - class-VI

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.
Here, $5$ divided by $8$ gives $0.625$.

$\displaystyle \frac { 2 }{ 7 }  = 0.285714285.. = 0.\overset { \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _  }{ 285714 } $

The value of $\displaystyle \frac { 3 }{ 11 }$ is $ 0.272727272727.. 0.\overset { \ _ \ _ \ _ \ _  }{ 27 } $

While expressing a fraction in the decimal form, when we perform division we get some remainder. 

A terminating decimal is a decimal that ends. It's a decimal with a finite number of digits.

$2\displaystyle \frac { 1 }{ 6 }  = \frac { 13 }{ 6 }  = 2.166666666.. = 2.1\overset { \ _ \ _  }{ 6 } $

$\frac {65}{455}=0.\overline { 142857 } $
Therefore, the decimal representation will be non-terminating repeating.

$\displaystyle \frac { 5 }{ 6 }= 0.83333333..= 0.8\overset { \ _ \ _  }{ 3 } $ 
This is a pure recurring decimal as $3$ is repeated continuously.

Write the decimal number as a fraction $0.2020020002...... $
It is Non- terminating decimal and cannot be represented as a quotient of two integers.
Therefore, B is the correct answer.

We know $\dfrac{7}{4}=1.75$

A number having non-terminating and non-recurring decimal expansion is an Irrational Number

$2.13113111311113....$ follows a definite pattern but the digits after decimal places are non-terminating and non-recurring. 

$\dfrac{7}{9}=0.777...$, Non terminating decimal number.

Since, $\sqrt { 2 } $ is a irrational number. So, its decimal expansion is non- terminating non recurring. 

Given, $\displaystyle \frac {15}{1600}= \frac {15}{5^{2}2^{6}}$
As it is in the form of ${ 2 }^{ m }\times { 5 }^{ n }$ where ($n=6,m=2$).
So, the rational number $\displaystyle \frac {15}{1600}$ has a terminating decimal expansion

Given, $\displaystyle \frac {29}{343}= \frac {29}{7^{3}}$
As it is not in the form of ${ 2 }^{ m }\times { 5 }^{ n }$.
So, the rational number $\displaystyle \frac {29}{343}$ has a non terminating decimal expansion