NCERT solution class 10 chapter 1 Real numbers exercise 1.4

Exercise 1.4

Q - 1 (i,ii & iv), Ex 1.4 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

Q - 1 (v, vii, x), Ex 1.4 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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) $\dfrac{13}{3125}$
Sol :
$3125=5^5$
The denominator is in the form of $5^m$
Hence, the decimal expansion of $\dfrac{13}{3125}$ is terminating.

(ii) $\dfrac{17}{8}$
Sol :
$8=2^3$
The denominator is in the form of $2^m$
Hence, the decimal expansion of $\dfrac{17}{8}$ is terminating.

(iii) $\dfrac{64}{455}$
Sol :
$455=5\times 7\times 13$
Since,the denominator is not in the form of $2^m\times 5^n$ and it also contains 7 and 13 as its factors , its decimal expansion will be non-terminating repeating .

(iv) $\dfrac{15}{1600}$
Sol :
$1600=2^6\times 5^2$
The denominator is of the form $2^m\times 5^n$ .
Hence , the decimal expansion of $\dfrac{15}{1600}$ is terminating .

(v) $\dfrac{29}{343}$
Sol :
$343=7^3$
Since,the denominator is not in the form of $2^m\times 5^n$ and it also contains 7 as its factors , its decimal expansion will be non-terminating repeating .

(vi) $\dfrac{23}{2^3\times 5^2}$
Sol :
The denominator is of the form $2^m\times 5^n$ .
Hence , the decimal expansion of $\dfrac{23}{2^3\times 5^2}$ is terminating .

(vii) $\dfrac{129}{2^2\times 5^7\times 7^5}$
Sol :
The denominator is of the form $2^m\times 5^n$ but it also has 7⁵ as its factor, the decimal expansion of $\dfrac{129}{2^2\times 5^7\times 7^5}$ is non-terminating repeating .

(viii) $\dfrac{6}{15}$
Sol :
$\dfrac{6}{15}=\dfrac{2\times 3}{3\times 5}=\dfrac{2}{5}$
The denominator is of the form 5ⁿ
Hence ,the decimal expansion of $\dfrac{6}{15}$ is terminating .

(ix) $\dfrac{35}{50}$
Sol :
$\dfrac{35}{50}=\dfrac{7\times 5}{10\times 5}=\dfrac{7}{10}$
$10=2\times 5$
The denominator is of the form $2^m\times 5^n$
Hence , the decimal expansion of $\dfrac{35}{50}$ is terminating .

(x) $\dfrac{77}{210}$
Sol :
$\dfrac{77}{210}=\dfrac{11\times 7}{30\times 7}=\dfrac{11}{30}$
$30=2\times 3\times 5$
The denominator is in the form of $2^m\times 5^n$ but it contains 3 as its factors , the decimal expansion of $\dfrac{77}{210}$ is non-terminating repeating .



Question 2
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Sol :
(i) $\dfrac{13}{3125}=0.00416$
$[+preamble]
\newcommand\showdiv[1]{\overline{\smash{)}#1}}
[/preamble]
\begin{array}{rll}
0.00416\\3125\showdiv{13.00000}\\
\underline{\phantom{0}0\phantom{.00000}}\\
130\phantom{0000}\\
\underline{\phantom{00}0\phantom{0000}}\\
1300\phantom{000}\\
\underline{\phantom{000}0\phantom{000}}\\
13000\phantom{00}\\
\underline{12500\phantom{00}}\\5000\phantom{0}\\
\underline{\phantom{00}3125\phantom{0}}\\
\phantom{00}18750\\
\underline{\phantom{00}18750}\\
\underline{\phantom{000000}0}
\end{array}
$

(ii) $\dfrac{17}{8}=2.125$ (working)
(iii) $\dfrac{64}{455}=$
(iv) $\dfrac{15}{1600}=0.009375$
(v) $\dfrac{29}{343}$
(vi) $\dfrac{23}{2^3\times 5^2}$
(vii) $\dfrac{129}{2^2\times 5^7\times 7^5}$
(viii) $\dfrac{6}{15}$
(ix) $\dfrac{35}{50}$
(x) $\dfrac{77}{210}$

Q - 3 (i & iii), Ex 1.4 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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 $\dfrac{p}{q}$, what can you say about the prime factor of q ?
(i) $43.123456789$
Sol :
Since this number has a terminating decimal expansion, it is a rational number of the form $\dfrac{p}{q}$ 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) $0.120120012000120000\dots$
Sol :
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) $43.\overline{123456789}$
Sol :
Since, the decimal expansion is non-terminating recurring, the given number is a rational number of the form $\dfrac{p}{q}$ 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 .