Question

A real number \[\dfrac{{{2^2} \times {3^2} \times {7^2}}}{{{2^5} \times {5^3} \times {3^2} \times 7}}\] will have
(a) Terminating decimal
(b) Non-terminating decimal
(c) Non-terminating and non-repeating decimal
(d) Terminating repeating decimal

Answer 1

Hint: Before solving this question, we must know about the concept of terminating and non-terminating decimals. We will firstly cancel the common factors in the numerator and denominator and then express the denominator in the powers of prime factors only. If the denominator consists of the powers of two and five only, then the decimal expansion is terminating.

Complete step-by-step answer:
If the number has finite terms after the decimal point, then we will say it is terminating decimal. Else, if the number does not have an end term after the decimal point, we say it is a non-terminating decimal.
Also, if in a number, a digit or a sequence of digits in the decimal part keeps repeating itself, we say it is a repeating decimal otherwise it is non repeating decimal.

We have given a real number i.e.,
\[\dfrac{{{2^2} \times {3^2} \times {7^2}}}{{{2^5} \times {5^3} \times {3^2} \times 7}} - - - \left( 1 \right)\]
Now using the concept of fractions, we know that \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
Therefore, equation \[\left( 1 \right)\] can be written as,
\[\dfrac{{{2^{2 - 5}} \times {3^{2 - 2}} \times {7^{2 - 1}}}}{{{5^3}}}\]
On simplifying, we get
\[\dfrac{{{2^{ - 3}} \times {3^0} \times {7^1}}}{{{5^3}}}{\text{ }} - - - \left( 2 \right)\]
And we know that, \[\left( {{a^0} = 1} \right)\]
\[\therefore {\text{ }}{3^0} = 1\]
So, equation \[\left( 2 \right)\] can be written as,
\[\dfrac{{{2^{ - 3}} \times 1 \times {7^1}}}{{{5^3}}}\]
\[ \Rightarrow \dfrac{{{2^{ - 3}} \times {7^1}}}{{{5^3}}}{\text{ }} - - - \left( 3 \right)\]
Now, we know that \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Therefore, \[{2^{ - 3}} = \dfrac{1}{{{2^3}}}\]
So, equation \[\left( 3 \right)\] becomes,
\[\dfrac{{{7^1}}}{{{5^3} \times {2^3}}}{\text{ }} - - - \left( 4 \right)\]
As we know that, \[{a^n} = a \times a \times a \times .....n{\text{ }}times\]
\[\therefore {5^3} = 5 \times 5 \times 5{\text{ }}\left( {{\text{3 }}times} \right)\]
\[ \Rightarrow {5^3} = 125\]
And similarly, \[{2^3} = 2 \times 2 \times 2{\text{ }}\left( {{\text{3 }}times} \right)\]
\[ \Rightarrow {2^3} = 8\]
So, equation \[\left( 4 \right)\] becomes,
\[\dfrac{7}{{125 \times 8}}{\text{ }}\]
\[ \Rightarrow \dfrac{7}{{1000}}{\text{ = 0}}{\text{.007}}\]
As we can see, a number has finite terms after the decimal point.
Therefore, it is a terminating decimal.
So, the correct answer is “Option a”.

Note: This number can also be called as terminating non-repeating decimal as we can see that the digits are not repeating. But usually, we call it a terminating decimal only. Also, there is a point to note that in the question we had a minimum number of terms in the denominators, so we solve their powers manually. But if the question had maximum number of terms, then we will use the formula, i.e., \[{\left( {a \times b} \right)^m} = {a^m} \times {b^m}\] and find our result.
We can also use this formula in this question to get the result,
Like from equation \[\left( 4 \right)\]
i.e., \[\dfrac{{{7^1}}}{{{5^3} \times {2^3}}}{\text{ }}\]
we can also write it as,
\[\dfrac{{{7^1}}}{{{{\left( {5 \times 2} \right)}^3}}} = \dfrac{7}{{{{10}^3}}} = \dfrac{7}{{1000}} = 0.007\] and hence we get our required result easily.
Also, there is a key point about terminating decimals that is, if we can express the denominator of simplified rational number in the form of \[{2^p}{5^q}\] where \[p,q \in N\] then the number has terminating decimal expansion only. There is no need to solve further.
Like in this question, in the denominator we get \[{5^3} \times {2^3}\] which is of the form \[{2^p}{5^q}\] hence it satisfy the condition for terminating decimal expansion, so if we don’t solve it further, that will be fine too.