Question

Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2^m × 5ⁿ, where, m, n 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]

Answer 1

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

Clearly, 8 = 2³ is of the form 2^m × 5ⁿ, where m = 3 and n = 0.

So, the given number has terminating decimal expansion.

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

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

Clearly, 125 = 5³ is of the form 2^m × 5ⁿ, where m = 0 and n = 3.

So, the given number has terminating decimal expansion.

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

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

Clearly, 80 = 2⁴ × 5 is of the form 2^m × 5ⁿ, where m = 4 and n = 1.

So, the given number has terminating decimal expansion.

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

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

Clearly, 625 = 5⁴ is of the form 2^m × 5ⁿ, where m = 0 and n = 4.

So, the given number has terminating decimal expansion.

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

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

Clearly, 2² × 5⁷ is of the form 2^m × 5ⁿ, where m = 2 and n = 7.

So, the given number has terminating decimal expansion.

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