Ex.1.4 Q1 Real Numbers Solution - NCERT Maths Class 10

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Question

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) \(\begin{align}\frac{13}{3125} \end{align}\)

(ii) \(\begin{align}\frac{{17}}{8}\end{align}\)

(iii) \(\begin{align}\frac{64}{455}\end{align}\)

(iv) \(\begin{align}\frac{15}{1600} \end{align}\)

(v) \(\begin{align}\frac{29}{343}\end{align}\)

(vi) \(\begin{align}\frac{{23}}{{{2^3}{5^2}}}\end{align}\)

(vii) \(\begin{align}\frac{{129}}{{{2^2}{5^7}{7^5}}}\end{align}\)

(viii) \(\begin{align}\frac{6}{{15}}\end{align}\)

(ix) \(\begin{align}\frac{{35}}{{50}} \end{align}\)

(x) \(\begin{align}\frac{{77}}{{210}}\end{align}\)

 

 

 Video Solution
Real Numbers
Ex 1.4 | Question 1

Text Solution

 

Reasoning:

Let \(x = \frac{p}{q}\) be a rational number, such that the prime factorization of \(q\) is of the form \({2^n} \times {5^m}\), where \(n,\; m\) are non-negative integers. Then \(x\) has a decimal expansion which terminates.

Steps:

(i) \(\begin{align}\;\frac{{13}}{{3125}}\end{align}\)

The denominator is of the form \({5^5}.\)

Hence, the decimal expansion of \(\begin{align}\,\frac{{13}}{{3125}}\end{align}\)is terminating.

(ii) \(\begin{align}\;\frac{{17}}{8}\end{align}\)

The denominator is of the form \({2^3}.\)

Hence, the decimal expansion of \(\begin{align}\,\frac{{17}}{8}\end{align}\) is terminating.

(iii) \(\begin{align}\;\frac{{64}}{{455}}\end{align}\)

\[455 = 5 \times 7 \times 13\]

Since the denominator is not in the form \({2^m} \times {5^n},\) and it also contains \(7\) and \(13\) as its factors, its decimal expansion will be non-terminating repeating.

(iv) \(\begin{align}\;\frac{{15}}{{1600}}\end{align}\)

\[1600 = {2^6} \times {5^2}\]

The denominator is of the form \({2^m} \times {5^n}.\)

Hence, the decimal expansion of \(\begin{align}\frac{{15}}{{1600}}\end{align}\)is terminating.

(v) \(\begin{align}\;\frac{{29}}{{343}}\end{align}\)
\[343 = {7^3}\]

Since the denominator is not in the form \(\,{2^m} \times {5^n},\) and it has \(7\) as its factor, the decimal expansion of \(\begin{align}\frac{{29}}{{343}}\end{align}\) is non-terminating repeating.

(vi) \(\begin{align}\;\frac{23}{{{2}^{3}}\times {{5}^{2}}}\end{align}\)

The denominator is of the form \({2^m} \times {5^n}.\)

Hence, the decimal expansion of \(\begin{align}\frac{{23}}{{{2^3} \times {5^2}}}\end{align}\)is terminating.

(vii) \(\begin{align}\;\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align}\)

Since the denominator is not of the form \(2^{m} \times 5^{n},\) and it also has \(7\) as its factor, the decimal expansion of \(\begin{align}\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align}\) is non-terminating repeating.

(viii) \(\begin{align}\frac{6}{{15}}\end{align} \)

\[\begin{align}\frac{6}{{15}} &= \frac{{2 \times 3}}{{3 \times 5}}\\&= \frac{2}{5}\end{align}\]

The denominator is of the form \({5^{n}}.\)

Hence, the decimal expansion of \(\begin{align}\frac{6}{{15}}\end{align}\) is terminating.

(ix) \(\begin{align}\,\frac{{35}}{{50}}\end{align}\)

\[\begin{align}\,\frac{{35}}{{50}} &= \frac{{7 \times 5}}{{10 \times 5}}\\&= \frac{7}{{10}}\end{align}\]

\[10 = 2 \times 5\]

The denominator is of the form \({2^m} \times {5^n}.\)

Hence, the decimal expansion of \(\begin{align}\frac{{35}}{{50}}\end{align}\) is terminating.

(x) \(\begin{align}\,\frac{{77}}{{210}}\end{align}\)

\[\begin{align}\,\,\,\,\frac{{77}}{{210}} &= \frac{{11 \times 7}}{{30 \times 7}}\\ &= \frac{{11}}{{30}}\end{align}\]

\[30 = 2 \times 3 \times 5\]

Since the denominator is not of the form \({2^m} \times {5^n},\) and it also has \(3\) as its factors, the decimal expansion of \(\begin{align}\frac{{77}}{{210}}\end{align}\) is non-terminating repeating.

  
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