Haryana State Board HBSE 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Exercise Questions and Answers.
Haryana Board 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.4
Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or 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) \(\frac{129}{2^2 5^7 5^5}\)
(viii) \(\frac{6}{15}\)
(ix) \(\frac{35}{50}\)
(x) \(\frac{77}{210}\)
Solution :
(i) \(\frac{13}{3125}\)
= \(\frac{13}{5^5}=\frac{13}{2^0 \times 5^5}\)
Since the denominator of \(\frac{13}{3125}\) is of the form 2m5n, where m and n are non-negative integers hence, it has terminating decimal expansion.
(ii) \(\frac{17}{8}\)
\(\frac{17}{8}=\frac{17}{2 \times 2 \times 2}=\frac{17}{2^3}=\frac{17}{2^3 \times 5^0}\)
Since the denominator of \(\frac{17}{8}\) is of the form 2m5n, where m and n are non-negative integers hence, it has terminating decimal expansion.
(iii) \(\frac{64}{455}=\frac{64}{5 \times 7 \times 13}\)
Since the denominator of \(\frac{64}{455}\) is not of the form 2m5n hence, it has non-terminating repeating decimal expansion.
(iv) \(\frac{15}{1600}\)
\(\frac{15}{1600}=\frac{3 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)= \(\frac{3}{2^6 \times 5^2}\)
Since the denominator of \(\frac{15}{1600}\) is of the form of 2m5n, where m and n are non-negative integers hence, it has terminating decimal expansion.
(v) \(\frac{29}{343}\)
\(\frac{29}{343}=\frac{29}{7 \times 7 \times 7}=\frac{29}{7^3}\)Since the denominator of \(\frac{29}{343}\) is not of the form of 2m5n hence, it has non-terminating repeating decimal expansion.
(vi) \(\frac{23}{2^3 5^2}\)
Since the denominator of \(\frac{23}{2^3 5^2}\) is of the form of 2m5n, where m and n are non-negative integers hence, t has terminating decimal expansion.
(vii) \(\frac{129}{2^2 5^7 7^5}\)
Since the denominator of \(\frac{129}{2^2 5^7 7^5}\) is not of the form of 2m5n hence, it has non-terminating repeating decimal expansion.
(viii) \(\frac{6}{15}\)
\(\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}=\frac{2}{2^0 \times 5^1}\)Since the denominator of is of the form of 2m5n where m and n are non-negative integers hence, it has a terminating decimal expansion.
(ix) \(\frac{35}{50}\)
\(\frac{35}{50}=\frac{5 \times 7}{2 \times 5 \times 5}=\frac{7}{2^1 \times 5^2}\)Since the denominator of \(\frac{35}{50}\) is of the form of 2m5n where m and n are non-negative integers hence, it has a terminating decimal expansion.
(x) \(\frac{77}{210}\)
\(\frac{77}{210}=\frac{7 \times 11}{2 \times 3 \times 5 \times 7}=\frac{11}{2 \times 3 \times 5}\) is not of the form of 2m5n hence, it has 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) \(\frac{13}{3125}\)
= \(\frac{13}{5^5}=\frac{13}{2^0 \times 5^5}\)
Since denominator of \(\frac{13}{3125}\) is of the form of 2m5n where m and n are non-negative integers hence, it is a terminating decimal expansion.
Now, \(\frac{13}{5^5}=\frac{13 \times 2^5}{5^5 \times 2^5}\)
= \(\frac{13 \times 32}{(5 \times 2)^5}=\frac{416}{10^5}\)
= \(\frac{416}{100000}\)
= 0.00416.
(ii) \(\frac{17}{8}\)
\(\frac{17}{8}=\frac{17}{2 \times 2 \times 2}=\frac{17}{2^3}=\frac{17}{2^3 \times 5^0}\)Since the denominator of \(\frac{17}{8}\) is of the form of 2m5n, where m and n are non-negative integers hence, it is terminating decimal expansion.
Now, \(\frac{17}{8}=\frac{17}{2^3}=\frac{17 \times 5^3}{2^3 \times 5^3}=\frac{2125}{(2 \times 5)^3}\)
= \(\frac{2125}{10^3}=\frac{2125}{1000}\) = 2.125
(iii) \(\frac{64}{455}\)
\(\frac{64}{455}=\frac{64}{5 \times 7 \times 13}\)Since the denominator of \(\frac{64}{455}\) is not of the form of 2m5n so it has non-terminating repeating decimal expansion.
(iv) \(\frac{15}{1600}\)
\(\frac{15}{1600}=\frac{3 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)= \(\frac{3}{2^6 \times 5^2}\)
Since the denominator of is of the form of 2m5n, where m and n are non-negative integers
hence, it has terminating decimal expansion.
Now \(\frac{3}{2^6 \times 5^2}=\frac{3 \times 5^4}{2^6 \times 5^2 \times 5^4}=\frac{1875}{1000000}\) = 0.001875
(v) \(\frac{29}{343}\)
\(\frac{29}{343}=\frac{29}{7 \times 7 \times 7}=\frac{29}{7^3}\)Since the denominator of \(\frac{29}{343}\) is not of the form of 2m5n so, it is non-terminating repeating decimal expansion.
(vi) \(\frac{23}{2^3 5^2}\)
Since the denominator of \(\frac{23}{2^3 5^2}\), is of the form of 2m5n, where m and n are non-negative integers hence, it has terminating decimal expansion.
Now, \(\frac{23}{2^3 5^2}=\frac{23 \times 5}{2^3 \times 5^2 \times 5}\)
\(\frac{115}{1000}\) = 0.115
(vii) \(\frac{129}{2^2 5^7 7^5}\)
Since the denominator of \(\frac{129}{2^2 5^7 7^5}\) is not of the form of 2m5n hence, it has non-terminating repeating decimal expansion.
(viii) \(\frac{6}{15}\)
\(\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}=\frac{2}{2^0 \times 5^1}\)Since the denominator of is of the form 2m5n where m and n are non-negative integers hence, it has terminating decimal expansion.
Now, \(\frac{2}{5}=\frac{2 \times 2}{2 \times 5}=\frac{4}{10}\) = 0.4
(ix) \(\frac{35}{50}\)
\(\frac{35}{50}=\frac{5 \times 7}{2 \times 5 \times 5}=\frac{7}{2 \times 5}\)Since the denominator of \(\frac{35}{50}\) is of the form of 2m5n, where m and n are non-negative integers hence, it has terminating decimal expansion.
Now, \(\frac{7}{2 \times 5}=\frac{7}{10}\) = 0.7
(x) \(\frac{77}{210}\)
\(\frac{77}{210}=\frac{7 \times 11}{2 \times 3 \times 5 \times 7}=\frac{11}{2 \times 3 \times 5}\)
Since the denominator of \(\frac{77}{210}\) form of 2m5n hence, it has non-terminating decimal expansion.
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 the form of \(\frac{p}{q}\) the prime factors of q ?
(i) 43.123456789,
(ii) 0.120120012000120000…
(iii) \(43 . \overline{123456789}\)
Solution :
(i) 43.123456789
Since it is terminating decimal expansion so, it is a rational number.
43.123456789 = \(\frac{43123456789}{1000000000}\)
q = 1000000000= 109
= (2 × 5)9 = 29 × 59
Prime factors of q is of the form 2m × 5n where, m and n are non-negative integers.
(ii) 0.120120012000120000…..
Since it is non-terminating non-repeating decimal expansion so, it is an irrational number.
(iii) \(43 . \overline{123456789}\)
Since it is non-terminating but repeating decimal expansion so, it is a rational number.
Let x = \(43 . \overline{123456789}\)
= 43.123456789123456789…. ……………..(i)
Multiplying both sides by 1000000000
1000000000 x = \(43123456789 . \overline{123456789}\) ……………..(ii)
Subtract (i) from (ii),
999999999 x= 43123456746
x = \(\frac{43123456746}{999999999}\)
q= 999999999
= 3 × 3 × 3 × 3 × 37 × 333667
Prime factorisation of q is not of the form 2m × 5n, where m and n are non-negative integers.