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.

 

 

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