# RS Aggarwal Solutions Class 10 Chapter 1- Real Numbers Ex 1B

## RS Aggarwal Class 10 Ex 1B Chapter 1

The RS Aggarwal Solutions of Class 10 Chapter 1 are prepared by our subject experts to provide accurate and easy solutions for all the problems covered in the RS Aggarwal Maths textbook. These solutions develop a strong conceptual base in students which plays a major role in the later stages for successful competition of competitive exams. We at BYJU’S provide detailed solutions of RS Aggarwal Maths for students who can refer these solutions while solving the exercise questions.

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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, (23 x 3) $\neq$ (2m x 5n)

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, (22 x 33 x 5) $\neq$ (2m x 5n)

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, (22 x 57 x 75) $\neq$ (2m x 5n)

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) $\neq$ (2m x 5n)

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) $\neq$ (2m x 5n)

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 72) $\neq$ (2m 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, 73$\neq$ (2m 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) $\neq$ (2m 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.

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

• The solutions covers all the exercise problems.
• The RS Aggarwal Class 10 solutions is the best reference study material while preparing for your Class 10 boards.
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#### Practise This Question

In theΔABC given in the figure below, if DE divides AB and AC in the same proportion, then: