# 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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