 # Write down the decimal expansions of the following rational numbers by writing their denominators in the form Question:

Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2m × 5n, where, mn are non-negative integers.

(i) $\frac{3}{8}$

(ii) $\frac{13}{125}$

(iii) $\frac{7}{80}$

(iv) $\frac{14588}{625}$

(v) $\frac{129}{2^{2} \times 5^{7}}$      [NCERT]

Solution:

(i) The given number is $\frac{3}{8}$.

Clearly, $8=2^{3}$ is of the form $2^{m} \times 5^{n}$, where $m=3$ and $n=0$.

So, the given number has terminating decimal expansion.

$\therefore \frac{3}{8}=\frac{3 \times 5^{3}}{2^{3} \times 5^{3}}=\frac{3 \times 125}{(2 \times 5)^{3}}=\frac{375}{(10)^{3}}=\frac{375}{1000}=0.375$

(ii) The given number is $\frac{13}{125}$.

Clearly, $125=5^{3}$ is of the form $2^{m} \times 5^{n}$, where $m=0$ and $n=$

So, the given number has terminating decimal expansion.

$\therefore \frac{13}{125}=\frac{13 \times 2^{3}}{2^{3} \times 5^{3}}=\frac{13 \times 8}{(2 \times 5)^{3}}=\frac{104}{(10)^{3}}=\frac{104}{1000}=0.104$

(iii) The given number is $\frac{7}{80}$.

Clearly, $80=2^{4} \times 5$ is of the form $2^{m} \times 5^{n}$, where $m=4$ and $n=1$.

So, the given number has terminating decimal expansion.

$\therefore \frac{7}{80}=\frac{7 \times 5^{3}}{2^{4} \times 5 \times 5^{3}}=\frac{7 \times 125}{(2 \times 5)^{4}}=\frac{875}{(10)^{4}}=\frac{875}{10000}=0.0875$

(iv) The given number is $\frac{14588}{625}$.

Clearly, $625=5^{4}$ is of the form $2^{m} \times 5^{n}$, where $m=0$ and $n=4$.

So, the given number has terminating decimal expansion.

$\therefore \frac{14588}{625}=\frac{14588 \times 2^{4}}{2^{4} \times 5^{4}}=\frac{14588 \times 16}{(2 \times 5)^{4}}=\frac{233408}{(10)^{4}}=\frac{233408}{10000}=23.3408$

(v) The given number is $\frac{129}{2^{2} \times 5^{7}}$.

Clearly, $2^{2} \times 5^{7}$ is of the form $2^{m} \times 5^{n}$, where $m=2$ and $n=7$.

So, the given number has terminating decimal expansion.

$\therefore \frac{129}{2^{2} \times 5^{7}}=\frac{129 \times 2^{5}}{2^{2} \times 5^{7} \times 2^{5}}=\frac{129 \times 32}{(2 \times 5)^{7}}=\frac{4182}{(10)^{7}}=\frac{4182}{10000000}=0.0004182$  