Ex.1.3 Q1 Number System Solution - NCERT Maths Class 9

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Question

Write the following in decimal form and say what kind of decimal expansion each has:

i) \(\begin{align}\frac{36}{100}\end{align}\)

ii) \(\begin{align}\frac{1}{11}\end{align}\)

iii)  \(\begin{align}4 \frac{1}{8}\end{align}\)

iv)\(\begin{align}\frac{3}{13}\end{align}\)

v) \(\begin{align}\frac{2}{11}\end{align}\)

vi) \(\begin{align}\frac{329}{400}\end{align}\)

 Video Solution
Number Systems
Ex 1.3 | Question 1

Text Solution

Steps:

(i) \(\begin{align}\frac{{36}}{{100}}=0.36\end{align}\)

Terminating decimal.

(ii) \(\begin{align}\frac{1}{11}\end{align}\)

The remainder \(1\) keeps repeating. \(\begin{align}\frac{1}{{11}} = 0.0909\end{align}\) and can be written as

\(\begin{align}\frac{1}{{11}} = 0.\overline {09} \end{align}\)

Non-terminating recurring decimal.

 (iii)  \(\begin{align}4 \frac{1}{8}=\frac{33}{8}\end{align}\)

\(\begin{align}4\frac{1}{8} = 4.125\end{align}\)

Terminating decimal (\(∵\) The remainder is zero)

(iv) \(\begin{align}\frac{3}{13}=0.23076923\end{align}\)

\(\because\) We find the block of numbers \(230769\) keep repeating.

This is non-terminating recurring decimal and is written as:

\(\begin{align}\,\frac{3}{{13}} = 0.\overline {230769} \end{align}\)

(v) \(\begin{align}\frac{2}{{11}} = 0.1818\end{align}\)

Here we find the block of numbers \(18\) keep repeating. Hence this is a non-terminating recurring decimal and is written as:

\(\begin{align}~\frac{2}{11}=0.\overline{18}\end{align}\)

(vi) \(\begin{align}\frac{{329}}{{400}} = \frac{{329}}{{4 \times 100}}\end{align}\)

\(\begin{align}{\frac{82.25}{100}} \\ {=0.8225}\end{align}\)

Terminating decimal (∵ The remainder is zero)

 Video Solution
Number Systems
Ex 1.3 | Question 1
  
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