Question

The decimal expansion of the rational number \[\dfrac{{31}}{{{2^2} \times 5}}\] will terminate after
A. one decimal place
B. two decimal place
C. three decimal place
D. more than three decimal place

Answer 1

Hint: In this question, we will proceed by simplifying the terms in the denominator of the given rational number. And then divide the numerator with its denominator to convert the fraction into decimal numbers. Then count how many numbers of digits are thereafter the decimal to get the required answer. So, use this concept to reach the solution to the given problem.

Complete step-by-step solution:
Given rational number is \[\dfrac{{31}}{{{2^2} \times 5}}\]
Simplifying the given rational number by removing the square term, we have
\[ \Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = \dfrac{{31}}{{2 \times 2 \times 5}}\]
Then multiplying the numbers in the denominator, we have
\[ \Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = \dfrac{{31}}{{20}}\]
And simplifying the fraction, we have
\[\Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = \dfrac{{20 + 11}}{{20}} = \dfrac{{20}}{{20}} + \dfrac{{11}}{{20}} \\
\Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = 1 + \dfrac{{11}}{{20}} \\
   \Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = 1 + \dfrac{1}{2}\left( {\dfrac{{11}}{{10}}} \right) \\
   \Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = 1 + \left( {0.5} \right)\left( {1.1} \right) \\
   \Rightarrow \dfrac{{31}}{{{2^2} \times 5}} = 1 + 0.55 \\
  \therefore \dfrac{{31}}{{{2^2} \times 5}} = 1.55 \\
\]
Therefore, the value of \[\dfrac{{31}}{{{2^2} \times 5}}\] is 1.55
Clearly, we have two terminating decimal places in 1.55
Thus, the correct option is B. two decimal places

Note: Terminating decimals are those whose numbers come to an end after few repetitions after the decimal point. For example; 0.5,2.459,147.143 etc. Non-terminating decimals are those which keep on counting after decimal point (i.e., they go on forever). They don’t come to end or if they do it is after along interval. For example: \[\pi \].