RS Aggarwal Solutions Chapter 1 Real Numbers Exercise  1C Class 10 Maths
Chapter Name  RS Aggarwal Chapter 1 Real Numbers Solutions 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 1C Solutions
1. Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form.
(i) 23/(2^{3} × 5^{2})
(ii) 24/125
(iii) 171/800
(iv) 15/1600
(v) 17/320
(vi) 19/3125
Solution
(i) 23/(2^{3} × 5^{2})
= (23 × 5)/(2^{3 }× 5^{3})
= 115/1000
= 0.115
We know either 2 or 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
(ii) 24/125
= 24/5^{3}
= (24 × 2^{3})/(5^{3} × 2^{3})
= 192/1000
= 0.192
We know 5 is not a factor of 23, so it is in its simplest form. Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
(iii) 171/800
= 171/(2^{5} × 5^{2})
= (171 × 5^{3})/(2^{5} × 5^{5})
= 21375/100000
= 0.21375
We know either 2 or 5 is not a factor of 171, so it is in its simplest form.
Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
(iv) 15/1600
= 15/(2^{6} × 5^{2})
= (15 × 5^{4})/(2^{6} × 5^{6})
= 9375/1000000
= 0.009375
We know either 2 or 5 is not a factor of 15, so it is in its simplest form.
Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
(v) 17/320
= 17/(2^{6} × 5)
= (17 × 5^{5})/(2^{6} × 5^{6})
= 53125/1000000
= 0.053125
We know either 2 or 5 is not a factor of 17, so it is in its simplest form.
Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
(vi) 19/3125
= 19/(5^{5})
= (19 × 2^{5})/(2^{5} × 5^{5})
= 608/100000
= 0.00608
We know either 2 or 5 is not a factor of 19, so it is in its simplest form. Moreover, it is in the form of (2^{m }× 5^{n}).
Hence, the given rational is terminating.
2. Without actual division show that each of the following rational numbers is a non terminating repeating decimal.
(i) 11/(2^{3} × 3)
(ii) 73/(2^{3} × 3^{3} × 5)
(iii) 129/(2^{2} × 5^{7} × 7^{5})
(iv) 9/35
(v) 7/210
(vi) 32/147
(vii) 29/343
(viii) 64/455
Solution
(i) 11/(2^{3} × 3)
We know either 2 or 3 is not a factor of 11, so it is in its simplest form.
Moreover, (2^{3 }× 3) ≠ (2^{m }× 5^{n})
Hence, the given rational is non–terminating repeating decimal.
(ii) 73/(2^{3 }× 3^{3} × 5)
We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.
Moreover, (2^{2}× 3^{3 }× 5) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(iii) 129/(2^{2}× 5^{7 }× 7^{5})
We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.
Moreover, (2^{2}× 5^{7}× 7^{5}) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(iv) 9/35 = 9/(5 × 7)
We know either 5 or 7 is not a factor of 9, so it is in its simplest form.
Moreover, (5 × 7) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(v) 77/210
= (77 + 7)/(210 + 7)
= 11/30
= 11/(2 × 3 × 5)
We know 2, 3 or 5 is not a factor of 11, so 11/30 is in its simplest form.
Moreover, (2 × 3 × 7) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(vi) 32/147 = 32/(3 × 7^{2})
We know either 3 or 7 is not a factor of 32, so it is in its simplest form.
Moreover, (3 × 7^{2}) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(vii) 29/343 = 29/7^{3}
We know 7 is not a factor of 29, so it is in its simplest form.
Moreover, 7^{3 }≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
(viii) 64/455 = 64/(5 × 7 × 13)
We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.
Moreover, (5 × 7 × 13) ≠ (2^{m }× 5^{n})
Hence, the given rational is nonterminating repeating decimal.
3. Express each of the following as a rational number in its simplest form:
(i) 0.8
⇒ 10x = 8.888 …(2)
On subtracting equation (1) from (2), we get
9x = 8
⇒ x = 8/9
∴ x = 2.444 …(1)
10x = 24.444 …(2)
On subtracting equation (1) from (2), we get
9x = 22
⇒ x = 22/9
100x = 24.2424 …(2)
On subtracting equation (1) from (2), we get
99 x = 24
⇒ x = 8/33
⇒ 100x = 12.1212 …(2)
On subtracting equation (1) from (2), we get
99x = 12
⇒ x = 4/33