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**Q.1: Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form.**

**(i) \(\frac{ 23 }{ (2 ^{ 2 } \times 5 ^{ 2 } ) }\) **

**(ii) \(\frac{ 24 }{ 125 }\)**

**(iii) \(\frac{ 171 }{ 800 }\) **

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

**(v) \(\frac{ 17 }{ 320 }\) **

**(vi) \(\frac{ 19 }{ 3125 }\)**

**Sol:**

**i)** \(\frac{ 23 }{ 2 ^{ 3 } \times 5 ^{ 2 } } = \frac{ 23 \times 5 }{ 2 ^{ 3 } \times 5 ^{ 3 } }\) = \(\frac{ 115 }{ 1000 }\) = 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 (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

**ii)** \(\frac{24}{125}= \frac{24}{5^{3}}=\frac{24 X 2^{3}}{5^{3} X 2^{3}}=\frac{192}{1000}\)= 0.21375

We know 5 is not a factor of 23, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

**(iii)** \(\frac{171}{800}= \frac{171}{5^{2} X 2^{5}}=\frac{171 X 5^{3}}{2^{5} X 2^{5}}=\frac{21375}{100000}\)=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 (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

**iv)** \(\frac{15}{1600}= \frac{15}{5^{2} X 2^{6}}=\frac{15 X 5^{4}}{2^{6} X 5^{6}}=\frac{9375}{1000000}\)=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 (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

**v)** \(\frac{17}{320}= \frac{17}{5 X 2^{6}}=\frac{17 X 5^{5}}{2^{6} X 5^{6}}=\frac{53125}{1000000}\)

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 (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

vi) \(\frac{19}{3125}=\frac{19}{5^{5}} = \frac{19 X 2^{5}}{5^{5}X 2^{5}}=\frac{608}{100000}\) =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 (\(2^{m} X 5^{n}\)).

Hence, the given rational is terminating.

**Q.2: Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal.**

**(i) \(\frac{ 11 }{ ( 2 ^{ 3 } \times 3) }\) **

**(ii)** **\(\frac{ 73 }{ ( 2 ^{ 2 } \times 3 ^{ 3 } \times 5) }\)**

**(iii) \(\frac{ 129 }{ ( 2 ^{ 2 } \times 5 ^{ 3 } \times 7 ^{ 2 } ) }\)**

**(iv) \(\frac{ 9 }{ 35 }\) **

**(v) \(\frac{ 77 }{ 210 }\) **

**(vi) \(\frac{ 32 }{ 147 }\) **

**(vii) \(\frac{ 29 }{ 343 }\) **

**(viii) \(\frac{ 64 }{ 455 }\)**

**Sol:**

**(i)** \(\frac{11}{2^{3}X3}\)

We know either 2 or 3 is not a factor of 11, so it is in its simplest form.

Moreover, (2^{3} x 3) \(\neq\) (2^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(ii)**Â We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.

Moreover, (2^{2} x 3^{3} x 5) \(\neq\) (2^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(iii)**Â Â We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.

Moreover, (2^{2} x 5^{7} x 7^{5}) \(\neq\) (2^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(iv)** \(\frac{9}{35}\)= \(\frac{9}{5X7}\)

We know either 5 or 7 is not a factor of 9, so it is in its simplest form. Moreover, (5 x 7) \(\neq\) (2^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(v)** \(\frac{77}{210}\)= \(\frac{77\div 7}{210\div 7}\)= \(\frac{11}{30}\)=

\(\frac{11}{2X3X5}\)

We know 2, 3 or 5 is not a factor of 11, so \(\frac{11}{30}\) is in its simplest form.

Moreover, (2 x 3 x 5) \(\neq\) (2^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(vi)** We know either 3 or 7 is not a factor of 32, so it is in its simplest form.

Moreover, (3 x 7^{2}) \(\neq\) (2^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**(vii)** We know 7 is not a factor of 29, so it is in its simplest form.

Moreover, 7^{3}\(\neq\) (2^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**(viii)** \(\frac{64}{455}\)= \(\frac{64}{5X7X13}\)

We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.

Moreover, (5 x 7 x 13) \(\neq\) (2^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**Q.3: Express each of the following as a fraction in simplest form.**

**(i) \(0.\overline{ 8 }\) **

**(ii) \(2.\overline{ 4 }\)**

**(iii) \(0.\overline{ 24 }\) **

**(iv) \(0.1 \overline{ 2 }\)**

**(v) \(2.2\overline{ 4 }\) **

**(vi) \(0.\overline{ 365 }\)**

**Sol:**

**(i)** Let x = \(\bar{0.8}\)

Therefore, x = 0.888 . . . . . . . . (1)

10x = 8.888 . . . . . . (2)

On subtracting equation (1) from (2), we get

9x = 8\(\Rightarrow\)=x=\(\frac{8}{9}\)

**(ii)** Let x = \(\bar{2.4}\)

Therefore, x =2.444 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â … (1)

10x = 24.444 . . . . . . . . (2)

On subtracting equation (1) from (2), we get

9x = 22 \(\Rightarrow\) x= \(\frac{22}{9}\)

Therefore, 2.4 = \(\bar{\frac{22}{9}}\)

**(iii)** Let x = \(\bar{0.24}\)

Therefore, x = 0.2424 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(1)

100x = 24.2424 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(2)

On subtracting equation (1) from (2), we get

99x = 24 \(\Rightarrow\) x= \(\frac{8}{33}\)

Therefore, 0.24 = \(\bar{\frac{8}{33}}\)

**(iv)** Let x = \(\bar{0.12}\)

Therefore, x = 0.1212 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦â€¦.(1)

100x = 12.1212 Â Â Â Â Â â€¦â€¦.(2)

On subtracting equation (1) from (2), we get

99x=12 \(\Rightarrow\) x=\(\frac{4}{33}\)

Therefore, 0.12 = \(\bar{\frac{4}{33}}\)

**(v)** Let x = \(\bar{2.24}\)

Therefore, Â Â x = 2.2444 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â –(1)

10x = 22.444 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(2)

100x = 224.444 Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€¦(3)

On subtracting equation (2) from (3), we get

90x = 202 \(\Rightarrow\) Â x = \(\frac{202}{90}\) = \(\frac{101}{45}\)

Hence, \(\bar{2.24}\) = Â \(\frac{101}{45}\)

**(vi)** Let x = \(\bar{0.365}\)

Therefore, Â Â Â Â x = 0.3656565 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(1)

10x = 3.656565 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(2)

1000x = 365.656565 Â Â Â Â Â Â Â Â Â â€¦.(3)

On subtracting (2) from (3), we get

990x = 362 = x = Â \(\frac{362}{990}\)= \(\frac{181}{495}\)

Hence, \(\bar{0.365}\) = \(\frac{181}{495}\)

**Q.5: DefineÂ (i) rational numbers (ii) irrational numbers (iii) real numbers.**

**Sol:**

Rational numbers: The numbers of the form \(\frac{p}{q}\) Â where p , q are integers and \(q\neq 0\) are called rational numbers.

Example: \(\frac{2}{3}\)

Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.

Example: \(\sqrt{2}\)

Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.

Example: 2, \(\frac{1}{3}\), \(\sqrt{2}\), -3 etc.

**Q.6: Classify the following numbers as rational and irrational:**

**(i) \(\frac{ 22 }{ 7 }\) **

**(ii) 3.141**

**iii) \(\pi\) **

**(iv) \(3.\overline{ 142857 }\) **

**(v) 5.636363 â€¦ **

**(vi) 2.040040004 â€¦**

**(vii) 1.535335333 â€¦ **

**(viii) 3.121221222 â€¦ **

**(ix) \(\sqrt{ 21 }\)**

**(x) \(\sqrt[ 3 ]{ 3 }\)**

**Sol:**

**(i)** \(\frac{ 22 }{ 7 }\) is a rational number because it is of the form of \(\frac{ p }{ q }\) Â , \(q \neq 0\)

**(ii)** 3.1416 is a rational number because it is a terminating decimal.

**(iii)** \(\pi\) Â is an irrational number because it is a non-repeating and non-terminating decimal.

**(iv)** \(3.\bar{1}\bar{4}\bar{2}\bar{8}\bar{5}\bar{7}\) is a rational number because it is a repeating decimal.

**(v)** 5.636363… is a rational number because it is a non-terminating, repeating decimal.

**(vi)** 2.040040004… is an irrational number because it is a non-terminating and non-repeating decimal.

**(vii)** 1.535335333… is an irrational number because it is a non-terminating and non-repeating decimal.

**(viii)** 3.121221222… is an irrational number because it is a non-terminating and non-repeating decimal.

**(ix)** Â \(\sqrt{21}= \sqrt{3} X \sqrt{7}\) is an irrational number because \(\sqrt{3}\) and \(\sqrt{7}\) Â are irrational and prime numbers.

**(x)** \(\sqrt[3]{3}\) Â is an irrational number because \(\sqrt{3}\) Â is a prime number. So, f is an irrational number.

**Q.7: Prove that each of the following numbers is irrational.**

**(i) \(\sqrt{ 6 }\) **

**(ii) \(\left (2 – \sqrt{ 3 } \right )\)**

**(iii) \(\left (3 + \sqrt{ 2 } \right )\) **

**(iv)** **\(\left (2 + \sqrt{ 5 } \right )\)**

**(v) \(\left (5 + 3 \sqrt{ 2 } \right )\)**

**(vi) \(3 \sqrt{ 7 }\)**

**(vii) \(\frac{ 3 }{ \sqrt{ 5 }}\)**

**(viii) \(\left (2 – 3 \sqrt{ 5 } \right )\)**

**(ix) \(\left ( \sqrt{ 3 } + \sqrt{ 5 } \right )\)**

**Sol:**

**(i)** Let \(\sqrt{6}\) Â =\(\sqrt{2}\) X \(\sqrt{3}\) Â Â be rational.

Hence, \(\sqrt{2}\) , \(\sqrt{3}\) are both rational.

This contradicts the fact that \(\sqrt{2}\) , \(\sqrt{3}\) are irrational.

The contradiction arises by assuming \(\sqrt{6}\) Â is rational.

Hence, \(\sqrt{6}\) Â is irrational.

**(ii)** Let \(2-\sqrt{3}\) Â be rational.

Hence, 2 and \(2-\sqrt{3}\) Â are rational.

(2 – \(2+\sqrt{3}\)) =\( \sqrt{3}\) = rational Â Â Â Â Â Â [Therefore, Difference of two rational is rational]

This contradicts the fact that \( \sqrt{3}\) Â Â Â Â is irrational.

The contradiction arises by assuming \( 2- Â \sqrt{3}\) Â Â is rational.

Hence, \( 2-Â \sqrt{3}\) Â Â Â is irrational.

**(iii)** Let, \( 3+Â \sqrt{2}\) Â Â be rational.

Hence, 3 and \( 3+ \sqrt{2}\) Â Â Â Â are rational.

\( 3+ \sqrt{2}\) – 3 =\( Â Â \sqrt{2}\) = rational Â Â [Therefore, Difference of two rational is rational]

This contradicts the fact that \(\sqrt{2}\) Â is irrational.

The contradiction arises by assuming 3 + \(\sqrt{2}\) Â is rational.

Hence, 3 + \(\sqrt{2}\) Â is irrational.

**(iv)** Let 2 + .VS be rational. Hence, 2Â and .5 are rational. (2 + – 2 = 2 + f – 2 = f = rational [v Difference of two rational is rational] this contradicts the fact that VE is irrational. The contradiction arises by assuming 2 – is rational. Hence, 2 – is irrational.

**(v)** Let, 5 + 3f2 be rational. Hence, 5 and 5 + 3.V2 are rational. (5 + 3f – 5) = 3/2. = rational [v Difference of two rational is rational] x 30 = 1,/2 = rational Product of two rational is rational] This contradicts the fact that V2 is irrational. The contradiction arises by assuming 5 + 3f is rational. Hence, 5 3f is irrational.

**(vi)** Let 30 be rational. x 3f = 1/77 = rational [v Product of two rational is rational] This contradicts the fact that 1/7 is irrational. The contradiction arises by assuming 30 is rational. Hence, 30 are irrational.

**(vii)** Let 3 â€” be rational. 3 1 VS : – ,/5 3 x – = – = rational This contradicts the fact that 1 is irrational.

**(viii)** [Product of two rational is rational]

So, if is rational, then Nig is rational. 5 x 115 = f= rational Hence, 71/- is irrational.

[Product of two rational is rational]3 The contradiction arises by assuming VS â€” is rational.

3 Hence, VS â€” is irrational

**(viii)** Let 2 â€” 3N5 be rational. Hence 2 and 2 â€” 3 f are rational. :. 2 â€” (2 â€” 3A/g = 2 â€” 2 + 3.â€˜,/g = 3.5 = rational [v Difference of two rational is rational] x 3 Vg = f = rational Product of two rational is rational] This contradicts the fact that -N5 is irrational. The contradiction arises by assuming 2 â€” 3, 5 is rational. Hence, 2 â€” 3, 73 is irrational.

**(ix)** Let \(\sqrt{3} + \sqrt{5}\) Â be rational.

Therefore,Â \(\sqrt{3} + \sqrt{5}\) Â = a, where a is rational

Therefore, \(\sqrt{3} \)= Â Â \( a- Â \sqrt{5}\)â€¦..(1)

On squaring both sides of equation (1), we get

3 = \((a-\sqrt{5})^{2}\) = a2 + 5 â€” 2Vga a2 i 2 2a

This is impossible because right-hand side is rational, whereas the left-hand side is irrational.

This is a contradiction.

Hence, \(\sqrt{3} + \sqrt{5}\) Â is irrational.

**Q.8: Prove that \(\frac{ 1 }{ \sqrt{ 3 } }\) is irrational.**

**Sol:**

Let \(\frac{ 1 }{ \sqrt{ 3 } }\) be rational.

Therefore, Â \(\frac{ 1 }{\sqrt{ 3 } }\) Â = \(\frac{ a }{ b }\) , where a, b are positive integers having no common factor other than 1

Therefore, \(\sqrt{ 3 }\) = \(\frac{ b }{ a }\) Â Â Â Â Â Â Â …(1)

Since a, b are non-zero integers, = \(\frac{ b }{ a }\) Â is rational.

Thus, equation (1) shows that \(\sqrt{ 3 }\) Â is rational.

This contradicts the fact that \(\sqrt{ 3 } \) Â Â is rational.

The contradiction arises by assuming \(\sqrt{ 3 }\) Â Â is rational.

Hence, \(\frac{ 1 }{ \sqrt{ 3 }}\) is irrational.

**Q.9: (i) Given an example of two irrationals whose sum is rational.**

**(ii) Give an example of two irrationals whose product is rational.**

**Hint**

**(i) Take ( 2 + \(\sqrt{ 3 }\) ) and 2 – \(\sqrt{ 3 }\) )**

**(ii) Take ( 2 + \(\sqrt{ 2 }\) Â and Â ( 3 – \(\sqrt{ 3 }\))**

**Sol:**

**(i)** Let (2 +\(\sqrt{ 3 }\) ), ( 2 –\(\sqrt{3}\) ) be two irrationals.

Therefore, (2 + \(\sqrt{3}\)) + ( 2 – \(\sqrt{3}\)) = 4 = rational number.

**(ii)** Let 2\(\sqrt{3}\) , 3\(\sqrt{3}\) Â be two irrationals.

Therefore, 2\(\sqrt{3}\) x 3\(\sqrt{3}\) = 18 = rational number.

**Q.10: State whether the given statement is true or false.**

**(i) The sum of two rationalesâ€™ is always rational.**

**(ii) The product of two rationalesâ€™ is always rational. **

**(iii) The sum of two irrationals is always irrational.**

**(iv) The product of two irrationals is always irrational.**

**(v) The sum of a rational and an irrational is irrational.**

**(vi) The product of a rational and an irrational is irrational.**

**Sol:**

**(i)** True

**(ii)** True

**(iii)** False

Counter example: 2 +\(\sqrt{3}\) and 2 –\(\sqrt{3}\) are two irrational numbers. But their sum is 4, which is a rational number.

**(iv)** False

Counter example: 2\(\sqrt{3}\) and 4\(\sqrt{3}\) are two irrational numbers. But their product is 24, which is a rational number.

**(v)** True

**(vi)** True

**Key Features of RS Aggarwal Class 10 Solutions Chapter 1 – Real Numbers Ex 1B**

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