CBSE 10th Maths - Real Numbers - Exercise 1.4

Solution:

(i) $\dfrac{13}{3125}$

3125= 5x5x5x5x5=5 $^{5}$

The denominator is of the form 5$^{m}$. Hence, the decimal expansion of $\dfrac{13}{3125}$ is terminating.

(ii)$\dfrac{17}{8}$

8=2x2x2 = 2$^{3}$

The denominator is of the form 2$^{m}$.

Hence, the decimal expansion of$\dfrac{17}{8}$ is terminating.

(iii)$\dfrac{64}{455}$

455 = 4x7x13

Since the denominator is not in the form 2$^{m}$x 5$^{n}$, and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating.

(iv)$\dfrac{15}{1600}$

1600=2×2×2×2×2×2×5 = $2^{6}5^{2}$

The denominator is of the form 2$^{m}$x 5$^{n}$

Hence, the decimal expansion of $\dfrac{15}{1600}$ is terminating.

(v)$\dfrac{29}{343}$

343= 7x7x7 = $7^{3}$

Since the denominator is not in the form 2$^{m}$x 5$^{n}$, and it has 7 as its factor, the decimal expansion of $\dfrac{29}{343}$ is non-terminating repeating.

(vi) $\dfrac{23}{2^{3}5^{2}}$

Denominator = $2^{3}5^{2}$

The denominator is of the form 2$^{m}$x 5$^{n}$.

Hence, the decimal expansion of $\dfrac{23}{2^{3}5^{2}}$ is terminating.

(vii)$\dfrac{129}{2^{2}5^{7}7^{5}}$

Since the denominator is not of the form 2$^{m}$x 5$^{n}$, and it also has 7 as its factor, the decimal expansion of $\dfrac{129}{2^{2}5^{7}7^{5}}$ is non-terminating repeating.

(viii).$\dfrac{6}{15}$

$\dfrac{6}{15}$=$\dfrac{2 \times 3}{5 \times 3}$=$\dfrac{2}{5}$

The denominator is of the form 5$^{n}$.

Hence, the decimal expansion of $\dfrac{6}{15}$ is terminating.

(ix)$\dfrac{35}{50}$

$\dfrac{35}{50}$=$\dfrac{5 \times 7}{2 \times 5 \times 3}$=$\dfrac{7}{10}$

Denominator = 10=2x5

The denominator is of the form 2$^{m}$x 5$^{n}$.

Hence, the decimal expansion of $\dfrac{35}{50}$ is terminating.

(x)$\dfrac{77}{210}$

$\dfrac{77}{210}$=$\dfrac{7 \times 11}{2 \times 3 \times 5 \times 7}$=$\dfrac{11}{30}$

Denominator = 30 = 2 x 3 x 5

Since the denominator is not of the form 2$^{m}$x 5$^{n}$, and it also has 3 as its factors, the decimal expansion of $\dfrac{77}{210}$ is non-terminating repeating.

Solution:

(i) $\dfrac{13}{3125}$

$\dfrac{13}{3125}$ = $\dfrac{13}{5 \times 5 \times 5 \times 5 \times 5}$ = $\dfrac{13}{5^{5}}$ X $\dfrac{2^{5}}{2^{5}}$ = $\dfrac{13 \times 32}{(5 \times 2)^{5}}$ = $\dfrac{416}{10^{5}}$ = 0.00416

(ii)$\dfrac{17}{8}$

$\dfrac{17}{8}$ = $\dfrac{17}{2 \times 2 \times 2}$X $\dfrac{5^{5}}{5^{5}}$ = $\dfrac{17 \times 125}{(2 \times 5)^{3}}$ = $\dfrac{2125}{10^{3}}$ = 2.125

(iii)$\dfrac{64}{455}$

Decimal expansion is non-terminating repeating.

(iv)$\dfrac{15}{1600}$

$\dfrac{15}{1600}$ = $\dfrac{3 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}$ = $\dfrac{3}{2^{6} \times 5}$X $\dfrac{5^{5}}{5^{5}}$ = $\dfrac{3 \times 3125}{(2 \times 5)^{6}}$ = $\dfrac{9375}{10^{6}}$ = 0.009375

(v)$\dfrac{29}{343}$

Decimal expansion is non-terminating repeating.

(vi) $\dfrac{23}{2^{3}5^{2}}$

$\dfrac{23}{2^{3}5^{2}}$ = $\dfrac{23}{2^{3} \times 5^{2}}$X $\dfrac{5}{5}$ = $\dfrac{23 \times 5}{(2 \times 5)^{3}}$ = $\dfrac{115}{10^{3}}$ = 0.115

(vii)$\dfrac{129}{2^{2}5^{7}7^{5}}$

Decimal expansion is non-terminating repeating.

(viii)$\dfrac{6}{15}$

$\dfrac{6}{15}$ = $\dfrac{2 \times 3}{3 \times 5}$ = $\dfrac{2}{5}$X $\dfrac{2}{2}$ = $\dfrac{2 \times 2}{2 \times 5}$ = $\dfrac{4}{10}$ = 0.4

(ix)$\dfrac{35}{50}$

$\dfrac{35}{50}$ = $\dfrac{7}{2 \times 5}$ = $\dfrac{7}{10}$ = 0.7

(x)$\dfrac{77}{210}$

Decimal expansion is non-terminating repeating.