exercise 1.4 class 10 maths

exercise 1.4 class 10 maths involve complete answers for each question in the exercise 1.4. The solutions provide students a strategic methods to prepare for their exam. exercise 1.4 class 10 maths questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. exercise 1.4 class 10 maths prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way.

 

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)${\displaystyle \frac{13}{3125}}$

(ii)${\displaystyle \frac{17}{8}}$

(iii)${\displaystyle \frac{64}{455}}$

(iv)${\displaystyle \frac{15}{1600}}$

(v)${\displaystyle \frac{29}{343}}$

(vi) ${\displaystyle \frac{23}{2^3{.5}^2}}$

(vii) ${\displaystyle \frac{129}{2^2{.5}^7{.7}^2}}$

(viii)${\displaystyle \frac{6}{15}}$

(ix)${\displaystyle \frac{35}{50}}$

(x)${\displaystyle \frac{77}{210}}$

Solution 1:

(i)${\displaystyle \frac{13}{3125}}$

$3125=5^5$

The denominator is of the form $5^m$.

Hence, the decimal expansion of ${\displaystyle \frac{13}{3125}}$ is terminating.

(ii)${\displaystyle \frac{17}{8}}$

$8=2^3$

The denominator is of the form $2^m$.

Hence, the decimal expansion of ${\displaystyle \frac{17}{8}}$ is terminating.

(iii)${\displaystyle \frac{64}{455}}$

$455=5\times 7\times 13$

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)${\displaystyle \frac{15}{1600}}$

$1600=26\times 52$

The denominator is of the form $2^m\times 5^n$.

Hence, the decimal expansion of ${\displaystyle \frac{15}{1600}}$ is terminating.

(v)${\displaystyle \frac{29}{343}}$

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

(vi) ${\displaystyle \frac{23}{2^3{.5}^2}}$

The denominator is of the form $2^m\times 5^n$.

Hence, the decimal expansion of ${\displaystyle \frac{23}{2^3{.5}^2}}$ is terminating.

(vii) ${\displaystyle \frac{129}{2^2{.5}^7{.7}^2}}$

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 ${\displaystyle \frac{129}{2^2{.5}^7{.7}^2}}$ is non-terminating repeating.

(viii)${\displaystyle \frac{6}{15}}$

The denominator is of the form $5^n$.

Hence, the decimal expansion of ${\displaystyle \frac{6}{15}}$ is terminating.

(ix)${\displaystyle \frac{35}{50}}$

The denominator is of the form $2^m\times 5^n$.

Hence, the decimal expansion of ${\displaystyle \frac{35}{50}}$ is terminating.

(x)${\displaystyle \frac{77}{210}=\frac{11}{30}}$

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 ${\displaystyle \frac{77}{210}}$ is non-terminating repeating.

Question 2.

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

Solution 2:

(i)${\displaystyle \frac{13}{3125}=0.00416}$

(ii)${\displaystyle \frac{17}{8}=2.125}$

(iv)${\displaystyle \frac{15}{1600}=0.009375}$

(vi) ${\displaystyle \frac{23}{2^3{.5}^2=2.125}}$

(viii)${\displaystyle \frac{6}{15}=0.4}$

(ix)${\displaystyle \frac{35}{50}=0.7}$

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

(i) $43.123456789$

(ii) $0.120120012000120000\dots$

(iii) $43.\overline{123456789}$

Solution 3:

(i) $43.123456789$

Since this number has a terminating decimal expansion, it is a rational number of the form ${\displaystyle \frac{p}{q}}$ 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) $0.120120012000120000\dots$

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

(iii) $43.\overline{123456789}$

Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form ${\displaystyle \frac{p}{q}}$ 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 $5$.