# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.4

Exercise 1.4

## Question 1

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

### Solution

**Video 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.

## Question 2

Write down the decimal expansions of those rational numbers in Question \(1\) above which have terminating decimal expansions.

### Solution

**Video Solution**

**Steps:**

(i) \(\begin{align}\;\;\frac{13}{3125}=0.00416\end{align}\)

(ii) \(\begin{align}\,\,\frac{{17}}{8} = 2.125\end{align}\)

(iii) \(\begin{align}\,\frac{{64}}{{455}}\end{align}\) is non - terminating

(iv) \(\begin{align}\,\frac{{15}}{{1600}} = 0.009375\end{align}\)

(v) \(\begin{align}\frac{{29}}{{343}}\end{align}\) is non - terminating

(vi) \(\begin{align}\; \frac{23}{2^{3} \times 5^{2}} &=\frac{23}{200} \\ &=0.115 \end{align}\)

(vii) \(\begin{align}\,\frac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}\end{align}\) it is non-terminating

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

\[\begin{align}\frac{6}{{15}} &= \frac{{2 \times 3}}{{3 \times 5}}\\&= \frac{2}{5}\\ &= 0.4\end{align}\]

(ix) \(\begin{align}\,\frac{{35}}{{50}} = 0.7\end{align}\)

(x) \(\begin{align}\,\frac{77}{210}\end{align}\) it is non-terminating

## Question 3

The following real numbers have decimal expansions as given below.

In each case, decide whether they are rational or not. If they are rational, and of the form \(\begin{align}\frac{p}{q},\end{align}\) what can you say about the prime factor of *\(q\)* ?

(i) \(\,43.123456789\)

(ii) \(\,0.120120012000120000 \dots \dots\)

(iii) \(\,43.\mathop {123456789}\limits^{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\)

### Solution

**Video Solution**

**Reasoning:**

Let *\(x\)* be a rational number whose decimal expansion terminates.

Then *\(x\)* can be expressed in the form \(\begin{align}\frac{p}{q},\end{align}\) where *\(p\)* and *\(q\)* are coprime, and the prime factorisation of *\(q\)* is of the form \({2^n}\, \times \,{5^m}\), where \(n, \;m\) are non-negative integers.

**Steps:**

(i) \(\,43.123456789\)

Since this number has a terminating decimal expansion, it is a rational number of the form \(\begin{align} \frac{p}{q} \end{align}\) and *\(q\)* is of the form \(2^{m} \times \,5^{n}\)

i.e., the prime factors of *\(q\)* will be either \(2\) or \(5 \) or both.

(ii) \(\,0.120120012000120000 \dots \dots\)

The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) \(\,43.\mathop {123456789}\limits^{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\)

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form \(\begin{align} \frac{p}{q} \end{align}\) and *\(q\)* is not of the form \(2^{m}\times 5^{n}\)

i.e., the prime factors of *\(q\)* will also have a factor other than \(2\) or \(5.\)

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school