# Real Numbers (Chapter 1, Exercise 1.4) – Class 10 Maths Solutions NCERT

This page contains Exercise 1.4, Chapter 1 of Class 10 Mathematics under NCERT Syllabus. The Chapter 1 of Class 10 Mathematics is about the Real Numbers. This part contains the Solutions of Questions based on Rational Numbers and Their Decimal Expansions which the part of Chapter of Class 10.

The solutions of all the three questions of Exercise 1.4 is given below.

## 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.$\frac{13}{3125}$

Solution:

The prime factorisation of the denominator $3125$ is

$3125=5\times 5\times 5\times 5\times 5=5^5$

$\therefore \frac{13}{3125}=\frac{13}{5^5}=\frac{p}{q}$

Here, the denominator $q$ is in the form $2^n5^m$ (where, $n=0$ and $m=5$).

Therefore, the given rational number has terminating decimal expansion.

### ii.$\frac{17}{8}$

Solution:

The prime factorisation of the denominator $8$ is

$8=2\times 2\times 2=2^3$

$\therefore \frac{17}{8}=\frac{17}{2^3}=\frac{p}{q}$

Here, the denominator $q$ is in the form $2^n5^m$ (where, $n=3$ and $m=0$).

Therefore, the given rational number has terminating decimal expansion.

### iii.$\frac{64}{455}$

Solution:

The prime factorisation of the denominator $455$ is

$455=5\times 7\times 13$

$\therefore \frac {64}{455}= \frac{64}{5\times 7\times 13}= \frac{p}{q}$

Here, the denominator $q$ is not in the form $2^n5^m$.

Therefore, the given rational number has non-terminating repeating decimal expansion.

### iv.$\frac{15}{1600}$

Solution:

The prime factorisation of the denominator $1600$ is

$1600=2^6\times 5^2$

$\therefore \frac{15}{1600} = \frac{15}{2^6\times 5^2}= \frac{p}{q}$

Here, the denominator $q$ is in the form $2^n5^m$ (where, $n=6$ and $m=2$).

Therefore, the given rational number has terminating decimal expansion.

### v.$\frac{29}{343}$

Solution:

The prime factorisation of the denominator $343$ is

$343=7^3$

$\therefore \frac{29}{343} = \frac{29}{7^3}= \frac{p}{q}$

Here, the denominator $q$ is not in the form $2^n5^m$.

Therefore, the given rational number has non-terminating repeating decimal expansion.

## vi.$\frac{23}{2^35^2}$

Solution:

Here, the denominator $q$ is in the form $2^n5^m$ (where, $n=3$ and $m=2$).

Therefore, the given rational number has terminating decimal expansion.

### vii.$\frac{129}{2^2 5^7 7^5}$

Solution:

Here, the denominator $q$ is not in the form $2^n5^m$.

Therefore, the given rational number has non-terminating repeating decimal expansion.

### viii.$\frac{6}{15}$

Solution:

The prime factorisation of the denominator $15$ is

$15=3\times 5$

$\therefore \frac{6}{15} = \frac{2\times 3}{3\times 5}=\frac{2}{5} \frac{p}{q}$

Here, the denominator $q$ is in the form $2^n5^m$, where $n=0$ and $m=1$).

Therefore, the given rational number has terminating decimal expansion.

### ix.$\frac{35}{50}$

Solution:

The prime factorisation of the denominator $50$ is

$50=2\times 5^2$

$\therefore \frac{35}{50} = \frac{5\times 7}{2\times 5^2}= \frac{7}{2\times 5}=\frac{p}{q}$

Here, the denominator $q$ is in the form $2^n5^m$ (where, $n=1$ and $m=1$).

Therefore, the given rational number has terminating decimal expansion.

### x. $\frac{77}{210}$

Solution:

The prime factorisation of the denominator $210$ is

$210=2\times 3\times 5\times 7$

$\therefore \frac{77}{210} = \frac{7\times 11}{2\times 3\times 5\times 7}= \frac{11}{2\times 3\times 5}=\frac{p}{q}$

Here, the denominator $q$ is not in the form $2^n5^m$.

Therefore, the given rational number has non-terminating repeating decimal expansion.

## Question 2: Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solution:

The following rational numbers in the Question 1 have terminating decimal expansion.

Serial No. (i) $\frac{13}{3125}$

$\frac{13}{3125}$

$=\frac{13}{5^5}$

$=\frac{13\times 2^5}{2^5\times 5^5}$

$=\frac{13\times 32}{10^5}$

$=\frac{416}{100000}$

$=0.00416$

Therefore, the decimal expansion of $\frac{13}{3125}$ is $0.00416$.

Serial No. (ii) $\frac{17}{8}$

$\frac{17}{8}$

$=\frac{17}{2^3}$

$=\frac{17\times 5^3}{2^3\times 5^3}$

$=\frac{17\times 125}{10^3}$

$=\frac{2125}{1000}$

$=2.125$

Therefore, the decimal expansion of $\frac{17}{8}$ is $2.125$.

Serial No. (iv) $\frac{15}{1600}$

$\frac{15}{1600}$

$=\frac{3\times 5}{2^6\times 5^2}$

$=\frac{3\times 5\times 5^4}{2^6\times 5^2\times 5^4}$

$=\frac{3\times 5^5}{2^6\times 5^6}$

$=\frac{9375}{10^6}$

$=\frac{9375}{1000000}$

$=0.009375$

Therefore, the decimal expansion of $\frac{15}{1600}$ is $0.009375$.

Serial No. (vi) $\frac{23}{2^3\times 5^2}$

$\frac{23}{2^3\times 5^2}$

$=\frac{23\times 5}{2^3\times 5^2\times 5}$

$=\frac{115}{2^3\times 5^3}$

$=\frac{115}{10^3}$

$=\frac{115}{1000}$

$=0.115$

Therefore, the decimal expansion of $\frac{23}{2^3\times 5^2}$ is $0.115$.

Serial No. (viii) $\frac{6}{15}$

$\frac{6}{15}$

$=\frac{2}{5}$

$=\frac{2\times 2}{5\times 2}$

$=\frac{4}{10}$

$=0.4$

Therefore, the decimal expansion of $\frac{6}{15}$ is $0.4$.

Serial No. (ix) $\frac{35}{50}$

$\frac{35}{50}$

$=\frac{5\times 7}{2\times 5^2}$

$=\frac{7}{2\times 5}$

$=\frac{7}{10}$

$=0.7$

Therefore, the decimal expansion of $\frac{35}{50}$ is $0.7$.

## 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, $\frac{p}{q}$ what can you say about the prime factors of q?

### i. 43.123456789

Solution:

Since the decimal expansion is terminating, the given real number is rational.

Converting the real number into $\frac{p}{q}$ form,

$43.123456789=\frac{43123456789}{1000000000}$

$=\frac{43123456789}{10^9}$

$=\frac{43123456789}{(2\times 5)^9}$

$=\frac{43123456789}{2^9\times 5^9}$

Therefore, the prime factors of $q$ is in the form $2^n5^m$, where $n=9$ and $m=9$.

### ii. 0.120120012000120000…

Solution:

Since the given decimal expansion is non-terminating non-repeating, the number is irrational.

Therefore, the number cannot be expressed in $\frac{p}{q}$ form.

### iii. $43.\overline{123456789}$

Solution:

Since the decimal expansion is non-terminating repeating, the given real number is rational.

Converting the real number into $\frac{p}{q}$ form,

Let, $x=43.\overline{123456789}$

$\therefore x=43.12345678912345…\,\,\,\,\,……(i)$

Multiplying both sides on the equation $(i)$ by $1000000000$

$1000000000x=43123456789.12345…$

$\Rightarrow 1000000000x=43123456746+43.12345…$

$\Rightarrow 1000000000x=43123456746+x$

$\Rightarrow 1000000000x-x=43123456746$

$\Rightarrow 999999999x=43123456746$

$\Rightarrow x=\frac{43123456746}{999999999}$

Here, $q=999999999$ which is not in the form $2^n5^m$.

I hope the solutions of Class 10, Exercise 1.4, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.4, feel free to contact me at [email protected] or [email protected] or fill the form here.

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