# RD Sharma Solutions Class 10 Real Numbers Exercise 1.6

### RD Sharma Class 10 Solutions Chapter 1 Ex 1.6 PDF Free Download

#### Exercise 1.6

Q.1: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

(i) $\frac{23}{8}$

(ii) $\frac{125}{441}$

(iii) $\frac{35}{50}$

(iv) $\frac{77}{210}$

(v) $\frac{129}{2^{2} \times 5^{7} \times 7^{17}}$

Sol:

(i) The given number is $\frac{23}{8}$

Here, 8 = 23 and 2 is not a factor of 23.

So, the given number is in its simplest form.

Now, 8 = 23 is of the form 2m x 5n, where m = 3 and n = 0.

So, the given number has a terminating decimal expansion.

(ii) The given number is $\frac{125}{441}$

Here, 441 = 32 x 72 and none of 3 and 7 is a factor of 125.

So, the given number is in its simplest form.

Now, 441 = 32 x 72 is not of the form 2m x 5n

So, the given number has a non-terminating repeating decimal expansion.

(iii) The given number is $\frac{35}{50}$ and HCF(35, 50) = 5.

$∴ \frac{35}{60} = \frac{35/5}{50/5} = \frac{7}{10}$

Here, $\frac{7}{10}$ is in its simplest form.

Now, 10 = 2 x 5 is of the form 2m x 5n, where in = 1 and n = 1.

So, the given number has a terminating decimal expansion.

(iv) The given number is $\frac{77}{210}$ and HCF(77, 210) = 7.

$∴ \frac{77:7}{210:7} =\frac{11}{30}$

Here, $\frac{11}{30}$ is in its simplest form. 30

Now, 30 = 2 x 3 x 5 is not of the form 2m x 5n.

So, the given number has a non-terminating repeating decimal expansion.

(v) The given number is $\frac{129}{2^{2} \times 5^{7} \times 7^{17}}$

Clearly, none of 2, 5 and 7 is a factor of 129.

So, the given number is in its simplest form.

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

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

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

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

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

(v) $\frac{129}{2^{4} \times 5^{7}}$

Sol:

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

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

So, the given number has terminating decimal expansion.

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

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

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

So, the given number has terminating decimal expansion.

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

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

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

So, the given number has terminating decimal expansion.

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

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

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

So, the given number has terminating decimal expansion.

$∴ \frac{14588}{625}$ = $\frac{14588 \times 2^{4}}{2^{4} \times 5^{4}}$ = $\frac{233408}{10^{4}}$ = $\frac{233408}{10000}$ = $23.3408$

(v) The given number is $\frac{129}{2^{4} \times 5^{7}}$

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

So, the given number has terminating decimal expansion.

$∴ \frac{129}{2^{2} \times 5^{7}}$ = $\frac{129 \times 2^{5}}{2^{2} \times 5^{7} \times 2^{5}}$ = $\frac{129 \times 32}{(2 \times 5)^{7}}$ = $\frac{4182}{10^{7}}$ = $\frac{4128}{10000000}$ = $0.0004182$

Q.4: what can you say about the prime factorization of the denominators of the following rational:

(i) 43.123456789

(ii) $43. \overline{123456789}$

(iii) $27. \overline{142857}$

(iv) 0.120120012000120000

Sol:

(i) Since 43.123456789 has terminating decimal expansion. So, its denominator is of the form 2m x 5n, where m, n are non-negative integers.

(ii) Since $43. \overline{123456789}$has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

(iii) Since $27. \overline{142857}$ has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

(iv) Since 0.120120012000120000 … has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.