Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

Question 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) \(\frac{13}{3125}\)

(ii) \(\frac{17}{8}\)

(iii) \(\frac{64}{455}\)

(iv) \(\frac{15}{1600}\)

(v) \(\frac{29}{343}\)

(vi) \(\frac{23}{2^{3} 5^{2}}\)

(vii) \(\frac{129}{2^{5} 5^{7} 7^{5}}\)

(viii) \(\frac{6}{15}\)

(ix) \(\frac{35}{50}\)

(x) \(\frac{77}{210}\)

Solution:

(i)Let x = \(\frac{13}{3125}\) ………..(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 13 and q = 3125

Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 5^{5} × 2^{0}

which are of the form 2^{n} × 5^{m} here n = 0, m = 5

which are non negative integers.

∴ x = \(\frac{13}{3125}\) have a terminating decimal expansion.

(ii) Let x = \(\frac{17}{8}\) …………………(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 17 and q = 8

Prime factors of q = 8 = 2 × 2 × 2 = 2^{3} = 2^{3} × 5^{0}

which are of the form 2^{n} × 5^{m} here n = 3, m = 0

and these are non negative integers.

∴ x = \(\frac{17}{8}\) have a terminating decimal expansion.

(iii) Let x = \(\frac{64}{455}\) …………(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 64, q = 455

Prime factors of q = 455 = 5 × 7 × 13 which are not of the form 2^{n} × 5^{m}

∴ x = \(\frac{64}{455}\) has a non – terminating decimal expansion.

(iv) Let x = \(\frac{15}{1600}\) ……….(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 15 and q = 1600

Prime factors of q = 1600

= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2^{6} × 5^{2}

which are of the form 2^{n} × 5^{m}, here n = 6, m = 2 and these are non negative integers.

∴ x = having terminating decimal expansion.

(v) Let x = \(\frac{29}{343}\) ………… (1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 29 and q = 343

prime factors of q = 343

= 7 × 7 × 7 = 7^{3}

which are not of the form 2^{n} × 5^{m}, here n = 0, m = 0

∴ x = \(\frac{29}{343}\) will have a non – terminating decimal expansion.

(vi) Let x = \(\frac{23}{2^{3} 5^{2}}\) ……….(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 23, q = 2^{3}5^{2}

Prime factors of q = 2^{3}5^{2}

which are of the form 2^{n} × 5^{m}, here n = 3, m = 2 and these are non negative integers.

∴ x = \(\frac{23}{2^{3} 5^{2}}\) will have a terminating decimal expansion.

(vii) Let x = \(\frac{129}{2^{5} 5^{7} 7^{5}}\) ……………….(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 129 and q = 2^{5} 5^{7} 7^{5}

Prime factors of q = 2^{5} 5^{7} 7^{5}

which are not of the form 2^{n} × 5^{m},

∴ x = \(\frac{129}{2^{5} 5^{7} 7^{5}}\) have a non – terminating decimal expansion.

(ix) Let x = \(\frac{35}{50}=\frac{7}{10}\) …………. (1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 7, q = 10

Prime factors of q = 10 = 2 × 5 = 21 × 51

Which is of the form 2^{n} × 5^{m} here n = 1, m = 1

both n and m are non negative integer.

∴ x = \(\frac{35}{50}\) have a terminating decimal expansion.

(x) Let x = \(\frac{77}{210}=\frac{11}{30}\) ……………..(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 11, q = 30

Prime factors of q = 30 = 2 × 5 × 3

which are not of the form 2^{n} × 5^{m},

∴ x = \(\frac{77}{210}\) have a non – terminating decimal expansion.

Question 2.

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solution:

(i) Let x = \(\frac{13}{3125}\)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 13,q = 3125

Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 5^{5} × 2^{0}

Which are of the form 2^{n} × 5^{m}, where n = 0, m = 5 and these are non negative integers

∴ x = have a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{13}{3125}=\frac{13}{5^{5} \times 2^{0}}\)

x = \(\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}\)

[∵ we are to make 10 in the denominator so multiply and divide by 2^{5}]

x = \(\frac{13 \times 32}{(2 \times 5)^{5}}\)

x = \(\frac{416}{(10)^{5}}=\frac{416}{100000}\)

x = 0.00416

(ii) Let x = \(\frac{17}{8}\) ………………..(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 17, q = 8

Prime factors of q = 8 = 2 × 2 × 2 = 2^{3} × 5^{0}

Which are of the form 2^{n} × 5^{m}, where n = 3, m = 0 and these are non negative integers

∴ x = \(\frac{17}{8}\) can be expressed as a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{17}{8}=\frac{17}{2^{3} \times 5^{0}}\)

x = \(\frac{17 \times 5^{3}}{2^{3} \times 5^{3}}\)

[Multiply and divide with 5^{3} to make the denominator 10]

x = \(\frac{17 \times 125}{(2 \times 5)^{3}}\)

x = \(\frac{2125}{(10)^{3}}=\frac{2125}{1000}\)

x = 2.125

⇒ \(\frac{17}{8}\) = 2.125

(iii) Let x = \(\frac{15}{1600}\) …………….(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 15, q = 1600

Prime factors of q = 1600

= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2^{6} × 5^{2}

Which are of the form 2^{n} × 5^{m}, where n = 6, m = 2 and these are non negative integers

∴ x = \(\frac{15}{1600}\) can be expressed as a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{15}{1600}\)

x = \(\frac{15 \times 5^{4}}{2^{6} \times 5^{2} \times 5^{4}}\)

[To make denominator a power of 10 multiply and divide by 5^{4}]

x = \(\frac{15 \times 625}{2^{6} \times 5^{6}}\)

x = \(\frac{9375}{(2 \times 5)^{6}}\)

x = \(\frac{9375}{(10)^{6}}=\frac{9375}{1000000}\) = 0.009375

In Decimal form, x = \(\frac{15}{1600}\) = 0.009375

(iv) Let x = \(\frac{23}{2^{3} 5^{2}}\) …………….(1)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 15, q = 2^{3} 5^{2}

Prime factors of q = 2^{3} 5^{2}

Which are of the form 2^{n} × 5^{m}, where n = 3, m = 2 and these are non negative integers

∴ x = \(\frac{23}{2^{3} 5^{2}}\) have a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{23}{2^{3} 5^{2}}=\frac{23 \times 5}{2^{3} \times 5^{2} \times 5}=\frac{115}{2^{3} \times 5^{3}}\)

x = \(\frac{115}{(2 \times 5)^{3}}=\frac{115}{1000}\) = 0.115

In Decimal form,

x = \(\frac{23}{2^{3} 5^{2}}\) = 0.115

(v) Let x = \(\frac{6}{15}=\frac{2}{5}\)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 2, q = 5

Prime factors of q = 5 = 2^{0} × 5^{1}

Which are of the form 2^{n} × 5^{m}, where n = 0, m = 1 and these are non negative integers

∴ x = \(\frac{6}{15}\) have a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{6}{15}=\frac{2}{5}\)

x = \(\frac{2 \times 2^{1}}{2^{1} \times 5^{1}}=\frac{4}{10}\) = 0.4

In Decimal form,

x = \(\frac{6}{15}\) = 0.4

(vi) Let x = \(\frac{35}{50}=\frac{7}{10}\)

Compare (1) with x = \(\frac{p}{q}\)

Here p = 7, q = 10

Prime factors of q = 5 = 2^{1} × 5^{1}

Which are of the form 2^{n} × 5^{m}, where n = 1, m = 1 and these are non negative integers

∴ x = \(\frac{7}{10}\) have a terminating decimal expansion.

To Express in Decimal form

x = \(\frac{35}{50}\)

x = \(\frac{7}{10}\)

x = \(\frac{7}{2^{1} \times 5^{1}}\)

x = \(\frac{7}{(2 \times 5)^{1}}=\frac{7}{(10)^{1}}\) = 0.7

Hence in decimal form, x = 0.7

Question 3.

The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form f, what can you say about the prime factors of q?

(i) 43.123456789

(ii) O.120120012000120000……

(iii) 4.3.123456789

Solution:

(i) Let x= 43.123456789 ……….. (1)

It is clear from the number that x is rational number.

Now remove the decimal from the number

∴ x = \(\frac{43123456789}{1000000000}\)

= \(\frac{43123456789}{10^{9}}\) …………….(2)

From (2) x is a rational number and of the \(\frac{p}{q}\).

Where p = 43123456789 and q = 10^{9}

Now, Prime factors of q = 10^{0} = (2 × 5)^{9}

⇒ Prime factors of q are 2^{9} × 5^{9}

(ii) Let x = 0.120120012000120000

It is clear from the number that it is an irrational number.

(iii) Let x = 43.123456789 …. (1)

It is clear that the given number is a rational number because it is non-terminating and repeating decimal.

To show that (i) is of the form \(\frac{p}{q}\)

Multiply (1) with 10^{9} on both sides,

10^{9} x = 43123456789.123456789 …………….(2)

Subtract (1) from (2), we get:

which is rational number of the form \(\frac{p}{q}\)

x = \(\frac{4791495194}{111111111}\)

Here p = 4791495194, q = 111111111

x = \(\frac{4791495194}{3^{2}(12345679)}\)

Hence, prime factors of q are 3^{2} (123456789)