RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1B

  Exercise 1B

PAGE 18

Question 1:

Write actual division, find which of the following rational numbers are terminating decimals.
(i) $\frac{13}{80}$$\frac{13}{80}$

(ii) $\frac{7}{24}$$\frac{7}{24}$

(iii) $\frac{5}{12}$$\frac{5}{12}$

(iv) $\frac{31}{375}$$\frac{31}{375}$

(v) $\frac{16}{125}$$\frac{16}{125}$

Answer 1:

(i) $\frac{13}{80}$$\frac{13}{80}$
Denominator of $\frac{13}{80}$$\frac{13}{80}$ is 80.
And,
80 = 2⁴$\times$$\times$5
Therefore, 80 has no other factors than 2 and 5.
Thus, $\frac{13}{80}$$\frac{13}{80}$ is a terminating decimal.

(ii) $\frac{7}{24}$$\frac{7}{24}$
Denominator of $\frac{7}{24}$$\frac{7}{24}$ is 24.
And,
24 = 2³$\times$$\times$3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{7}{24}$$\frac{7}{24}$ is not a terminating decimal.

(iii) $\frac{5}{12}$$\frac{5}{12}$
Denominator of $\frac{5}{12}$$\frac{5}{12}$ is 12.
And,
12 = 2²$\times$$\times$3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{5}{12}$$\frac{5}{12}$ is not a terminating decimal.
 
(iv) $\frac{31}{375}$$\frac{31}{375}$
Denominator of $\frac{31}{375}$$\frac{31}{375}$ is 375. 
$375=5^3\times 3$$375=5^3\times 3$
So, the prime factors of 375 are 5 and 3.
Thus, $\frac{31}{375}$$\frac{31}{375}$ is not a terminating decimal. 

(v) $\frac{16}{125}$$\frac{16}{125}$
Denominator of $\frac{16}{125}$$\frac{16}{125}$ is 125.
And,
125 = 5³
Therefore, 125 has no other factors than 2 and 5.
Thus, $\frac{16}{125}$$\frac{16}{125}$ is a terminating decimal.

PAGE 19

Question 2:

Write each of the following in decimal form and say what kind of decimal expansion each has.
(i) $\frac{5}{8}$$\frac{5}{8}$

(ii) $\frac{7}{25}$$\frac{7}{25}$

(iii) $\frac{3}{11}$$\frac{3}{11}$

(iv) $\frac{5}{13}$$\frac{5}{13}$

(v) $\frac{11}{24}$$\frac{11}{24}$

(vi) $\frac{261}{400}$$\frac{261}{400}$

(vii) $\frac{231}{625}$$\frac{231}{625}$

(viii) $2\frac{5}{12}$$2\frac{5}{12}$

Answer 2:

(i) $\frac{5}{8}$$\frac{5}{8}$ = 0.625
By actual division, we have:

It is a terminating decimal expansion.

(ii) $\frac{7}{25}$$\frac{7}{25}$
$\frac{7}{25}$$\frac{7}{25}$ = 0.28
By actual division, we have:

It is a terminating decimal expansion.

(iii) $\frac{3}{11}$$\frac{3}{11}$ = $0.\overline{27}$


It is a non-terminating recurring decimal.

(iv) $\frac{5}{13}$$\frac{5}{13}$ = $0.\overline{384615}$


It is a non-terminating recurring decimal.


(v) $\frac{11}{24}$$\frac{11}{24}$
$\frac{11}{24}$$\frac{11}{24}$ = $=0.458\overline{3}$
By actual division, we have:

It is nonterminating recurring decimal expansion.

(vi) $\frac{261}{400}$$\frac{261}{400}$$=0.6525$

It is a terminating decimal expansion.


(vii) $\frac{231}{625}$$\frac{231}{625}$$=0.3696$$=0.3696$

It is a terminating decimal expansion.


(viii) $2\frac{5}{12}$

$=2.41\overline{6}$

2$\frac{5}{12}$$\frac{5}{12}$ = $\frac{29}{12}$$\frac{29}{12}$ = 

By actual division, we have:


It is non-terminating decimal expansion.

Question 3:

Express each of the following decimals in the form $\frac{p}{q}$$\frac{p}{q}$, where p, q are integers and q ≠ 0.
(i) $0.\overline{2}$$0.\overline{2}$
(ii) $0.\overline{53}$$0.\overline{53}$
(iii) $2.\overline{93}$$2.\overline{93}$
(iv) $18.\overline{48}$$18.\overline{48}$
(v) $0.\overline{235}$$0.\overline{235}$
(vi) $0.00\overline{32}$$0.00\overline{32}$
(vii) $1.3\overline{23}$$1.3\overline{23}$
(viii) $0.3\overline{178}$$0.3\overline{178}$
(ix) $32.12\overline{35}$$32.12\overline{35}$
(x) $0.40\overline{7}$$0.40\overline{7}$

Answer 3:

(i) $0.\overline{2}$$0.\overline{2}$
Let x = 0.222...                               .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222...                                 .....(ii) 
Subtracting (i) from (ii) we get
$9x=2\Rightarrow x=\frac{2}{9}$$$\begin{gathered} 9x=2 \\ \Rightarrow x=\frac{2}{9} \end{gathered}$$

(ii) $0.\overline{53}$$0.\overline{53}$
Let x = 0.5353...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353...                                 .....(ii) 
Subtracting (i) from (ii) we get
$99x=53\Rightarrow x=\frac{53}{99}$$$\begin{gathered} 99x=53 \\ \Rightarrow x=\frac{53}{99} \end{gathered}$$

(iii) $2.\overline{93}$$2.\overline{93}$
Let x = 2.9393...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393...                                 .....(ii) 
Subtracting (i) from (ii) we get
$99x=291\Rightarrow x=\frac{291}{99}=\frac{97}{33}$$$\begin{gathered} 99x=291 \\ \Rightarrow x=\frac{291}{99}=\frac{97}{33} \end{gathered}$$

(iv) $18.\overline{48}$$18.\overline{48}$
Let x = 18.4848...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848...                                 .....(ii) 
Subtracting (i) from (ii) we get
$99x=1830\Rightarrow x=\frac{1830}{99}=\frac{610}{33}$$$\begin{gathered} 99x=1830 \\ \Rightarrow x=\frac{1830}{99}=\frac{610}{33} \end{gathered}$$

(v) $0.\overline{235}$$0.\overline{235}$
Let x = 0.235235...                               .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235...                                 .....(ii) 
Subtracting (i) from (ii) we get
$999x=235\Rightarrow x=\frac{235}{999}$$$\begin{gathered} 999x=235 \\ \Rightarrow x=\frac{235}{999} \end{gathered}$$

(vi) $0.00\overline{32}$$0.00\overline{32}$
Let x = 0.003232...                               .....(i)
we multiply x by 100.
100x = 0.3232...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=32\Rightarrow x=\frac{32}{9900}=\frac{8}{2475}$$$\begin{gathered} 9900x=32 \\ \Rightarrow x=\frac{32}{9900}=\frac{8}{2475} \end{gathered}$$

(vii) $1.3\overline{23}$$1.3\overline{23}$
Let x = 1.32323...                               .....(i)
we multiply x by 10.
10x = 13.2323...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323...                           .....(iii)
Subtracting (ii) from (iii) we get
$990x=1310\Rightarrow x=\frac{131}{99}$$$\begin{gathered} 990x=1310 \\ \Rightarrow x=\frac{131}{99} \end{gathered}$$

(viii) $0.3\overline{178}$$0.3\overline{178}$
Let x = 0.3178178...                               .....(i)
we multiply x by 10.
10x = 3.178178...                                 .....(ii) 
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178...                           .....(iii)
Subtracting (ii) from (iii) we get
$9990x=3175\Rightarrow x=\frac{3175}{9990}=\frac{635}{1998}$$$\begin{gathered} 9990x=3175 \\ \Rightarrow x=\frac{3175}{9990}=\frac{635}{1998} \end{gathered}$$

(ix) $32.12\overline{35}$$32.12\overline{35}$
Let x = 32.123535...                               .....(i)
we multiply x by 100.
100x = 3212.3535...                                 .....(ii) 
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=318023\Rightarrow x=\frac{318023}{9900}$$$\begin{gathered} 9900x=318023 \\ \Rightarrow x=\frac{318023}{9900} \end{gathered}$$

(x) $0.40\overline{7}$$0.40\overline{7}$
Let x = 0.40777...                               .....(i)
we multiply x by 100.
100x = 40.7777...                                 .....(ii) 
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777...                           .....(iii)
Subtracting (ii) from (iii) we get
$900x=367\Rightarrow x=\frac{367}{900}$$$\begin{gathered} 900x=367 \\ \Rightarrow x=\frac{367}{900} \end{gathered}$$
 

Question 4:

Express $2.\overline{36}+0.\overline{23}$$2.\overline{36}+0.\overline{23}$ as a fraction in simplest form.

Answer 4:

Given: $2.\overline{36}+0.\overline{23}$$2.\overline{36}+0.\overline{23}$
Let 
$x=2.\overline{36}...\left(i\right)y=0.\overline{23}...\left(\mathrm{ii}\right)$$$\begin{gathered} x=2.\overline{36} ...\left(\mathrm{i}\right) \\ y=0.\overline{23} ...\left(\mathrm{ii}\right) \end{gathered}$$
First we take x and convert it into $\frac{p}{q}$$\frac{p}{q}$
100x = 236.3636...        ...(iii)
Subtracting (i) from (iii) we get
$99x=234\Rightarrow x=\frac{234}{99}$$$\begin{gathered} 99x=234 \\ \Rightarrow x=\frac{234}{99} \end{gathered}$$
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves. 
$100y=23.2323...$$100y=23.2323...$               ...(iv)
Subtracting (ii) from (iv) we get
$99y=23\Rightarrow y=\frac{23}{99}$$$\begin{gathered} 99y=23 \\ \Rightarrow y=\frac{23}{99} \end{gathered}$$
Adding x and y we get
$2.\overline{36}+0.\overline{23}$$2.\overline{36}+0.\overline{23}$ = $x+y=\frac{234}{99}+\frac{23}{99}=\frac{257}{99}$$x+y=\frac{234}{99}+\frac{23}{99}=\frac{257}{99}$

Question 5:

Express in the form of $\frac{p}{q}:0.\overline{38}+1.\overline{27}$$\frac{p}{q}: 0.\overline{38}+1.\overline{27}$.

Answer 5:

$\mathrm{Let}0.\overline{38}=x1.\overline{27}=y$$$\begin{gathered} \mathrm{Let} 0.\overline{38}=x \\ 1.\overline{27}=y \end{gathered}$$
x = 0.3838...                                  ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838...                          ...(ii)
Subtracting (i) from (ii) we get
99x = 38
$\Rightarrow x=\frac{38}{99}$$\Rightarrow x=\frac{38}{99}$
Similarly, we take 
y = 1.2727...                                  ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727...                       ...(iv)
Subtract (iii) from (iv) we get
99y = 126
$\Rightarrow y=\frac{126}{99}\mathrm{Now},x+y=\frac{38}{99}+\frac{126}{99}=\frac{164}{99}$$$\begin{gathered} \Rightarrow y=\frac{126}{99} \\ \mathrm{Now}, x+y=\frac{38}{99}+\frac{126}{99}=\frac{164}{99} \end{gathered}$$