Question

# Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2m × 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) $\frac{129}{{2}^{2}×{5}^{7}}$ [NCERT]

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Solution

## (i) The given number is $\frac{3}{8}$. Clearly, 8 = 23 is of the form 2m × 5n, where m = 3 and n = 0. So, the given number has terminating decimal expansion. $\therefore \frac{3}{8}=\frac{3×{5}^{3}}{{2}^{3}×{5}^{3}}=\frac{3×125}{{\left(2×5\right)}^{3}}=\frac{375}{{\left(10\right)}^{3}}=\frac{375}{1000}=0.375$ (ii) The given number is $\frac{13}{125}$. Clearly, 125 = 53 is of the form 2m × 5n, where m = 0 and n = 3. So, the given number has terminating decimal expansion. $\therefore \frac{13}{125}=\frac{13×{2}^{3}}{{2}^{3}×{5}^{3}}=\frac{13×8}{{\left(2×5\right)}^{3}}=\frac{104}{{\left(10\right)}^{3}}=\frac{104}{1000}=0.104$ (iii) The given number is $\frac{7}{80}$. Clearly, 80 = 24 × 5 is of the form 2m × 5n, where m = 4 and n = 1. So, the given number has terminating decimal expansion. $\therefore \frac{7}{80}=\frac{7×{5}^{3}}{{2}^{4}×5×{5}^{3}}=\frac{7×125}{{\left(2×5\right)}^{4}}=\frac{875}{{\left(10\right)}^{4}}=\frac{875}{10000}=0.0875$ (iv) The given number is $\frac{14588}{625}$. Clearly, 625 = 54 is of the form 2m × 5n, where m = 0 and n = 4. So, the given number has terminating decimal expansion. $\therefore \frac{14588}{625}=\frac{14588×{2}^{4}}{{2}^{4}×{5}^{4}}=\frac{14588×16}{{\left(2×5\right)}^{4}}=\frac{233408}{{\left(10\right)}^{4}}=\frac{233408}{10000}=23.3408$ (v) The given number is $\frac{129}{{2}^{2}×{5}^{7}}$. Clearly, 22 × 57 is of the form 2m × 5n, where m = 2 and n = 7. So, the given number has terminating decimal expansion. $\therefore \frac{129}{{2}^{2}×{5}^{7}}=\frac{129×{2}^{5}}{{2}^{2}×{5}^{7}×{2}^{5}}=\frac{129×32}{{\left(2×5\right)}^{7}}=\frac{4182}{{\left(10\right)}^{7}}=\frac{4182}{10000000}=0.0004182$

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