Question & Answer
QUESTION

Write the decimal expansion of $\dfrac{91}{1250}$ without actual division.

ANSWER Verified Verified
Hint: Mathematics includes the study of topics which are related to quantity, structure, space and change. For solving this problem, we must express the denominator in the exponent of 10, so that the fraction can automatically get reduced to decimal without actual division.

Complete step-by-step answer:
Let $x=\dfrac{p}{q}$ be a rational number where p and q are integers such that the prime factorization of ' q ' is of the form ${{2}^{m}}\times {{5}^{n}}$, where n , m are non - negative integers. If the exponent of 2 and 5 are equal i.e. m = n, then we can transform this into an exponent of 10. Then x is said to be terminating with a decimal representation.
According to the problem, we are given a fractional number $\dfrac{91}{1250}$. We have to evaluate the decimal expansion without actually dividing the numbers in the numerator and denominator.
First, expressing the denominator in terms of prime factors, we get
$1250=2\times {{5}^{4}}$
Now, rearranging the denominator so that it could be expressed in the form ${{2}^{m}}\times {{5}^{n}}$,
$\begin{align}
  & \dfrac{91}{1250}=\dfrac{91}{2\times {{5}^{4}}} \\
 & \Rightarrow \dfrac{91\times {{2}^{3}}}{2\times {{2}^{3}}\times {{5}^{4}}} \\
 & \Rightarrow \dfrac{91\times {{2}^{3}}}{{{2}^{4}}\times {{5}^{4}}} \\
 & \Rightarrow \dfrac{91\times {{2}^{3}}}{{{10}^{4}}} \\
\end{align}$
Since the above fraction is not expressed as ${{10}^{4}}$, so we can easily evaluate the decimal expansion without actual division. Now, solving the numerator, we get
$\begin{align}
  & \dfrac{91\times {{2}^{3}}}{{{10}^{4}}}=\dfrac{728}{{{10}^{4}}} \\
 & \Rightarrow 0.0728 \\
\end{align}$
Hence the obtained decimal expansion of $\dfrac{91}{1250}$ is 0.0728.

Note: The key step for solving this problem is the conversion of exponent from 2 and 5 to 10. By doing so we can easily convert the fraction and to desired form and evaluate the value without doing actual division.