Question

Without actual division decide which of the following rational numbers have terminating decimal representation:
$(i) \dfrac{33}{375}$
$(ii) \dfrac{15}{28}$
$(iii) \dfrac{16}{45}$
$(iv) \dfrac{12}{35}$
$(v) \dfrac{80}{27}$
$(vi) \dfrac{123}{1250}$

Answer 1

Hint:
Terminating Decimal: A terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point.
If x is a rational number whose simplest form is $\dfrac{p}{q}$, where p and q are integers and $q \ne 0$. Then,
(i) x is terminating only when q is of the form $(2^m \times 5^n)$ for some non-negative integers m and n.
(ii) x is non-terminating and repeating, if $q \neq (2^m \times 5^n)$.

Complete step by step solution:
$(i) \dfrac{33}{375}$
If we do prime factorization of denominator, we get,
$\dfrac{33}{375}=\dfrac{33}{3 \times 5^3}$
The denominator of the fraction contains factors of 3 and 5. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.
$(ii) \dfrac{15}{28}$
If we do prime factorization of denominator, we get,
$\dfrac{15}{28}=\dfrac{15}{2^2 \times 7}$
The denominator of the fraction contains factors of 2 and 7. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.
$(iii) \dfrac{16}{45}$
If we do prime factorization of denominator, we get,
$\dfrac{16}{45}=\dfrac{16}{3^2 \times 5}$
The denominator of the fraction contains factors of 3 and 5. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.
$(iv) \dfrac{12}{35}$
If we do prime factorization of denominator, we get,
$\dfrac{12}{35}=\dfrac{12}{5 \times 7}$
The denominator of the fraction contains factors of 5 and 7. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.
$(v) \dfrac{80}{27}$
If we do prime factorization of denominator, we get,
$\dfrac{80}{27}=\dfrac{80}{3^3}$
The denominator of the fraction contains factors of 3. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.
$(vi) \dfrac{123}{1250}$
If we do prime factorization of denominator, we get,
$\dfrac{123}{1250}=\dfrac{123}{2 \times 5^4}$
The denominator of the fraction contains factors of 2 and 5. Since the denominator of given fraction has only factors of 2 and factors of 5, i.e. denominator is of the form $(2^m \times 5^n)$, thus, the given fraction have terminating decimal representation.

Hence, only (vi) fractions have terminating decimal representation.

Note:
In these types of questions, we just check whether the prime factorization of the denominator of the fraction has only factors of 2 and 5. If it has only factors of 2 and 5, we can conclude that it is a terminating decimal.