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

## RS Aggarwal Class 10 Chapter 1 - Real Numbers Ex 1C Solutions Free PDF

For subjects like Maths practice is an important task to score well and these RS Aggarwal maths solution will help you in doing so. The RS Aggarwal Class 10 Solutions provide complete information and knowledge of each and every concept that prepares students to face all kind of questions irrespective of their difficulty. The more you practice and solve the exercise questions, faster you can solve the questions asked during the exam. The students are required to go through solutions of RS Aggarwal Class 10 thoroughly before the final exams to score well and enhance their problem-solving abilities.

These solutions are designed specially by our experts to provide complete and accurate solutions for each and every question mentioned in the RS Aggarwal Book. Along with their board exams, these solutions will also help students to score well in highly competitive exams. The RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers act as a guide for the students while preparing effectively for their exam.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 1 – Real Numbers Ex 1C

Q.1: Prove that $2 \sqrt{ 3 } – 1$ is an irrational number.

Sol:

Let x = 2 $\sqrt{ 3 }$ -1 be a rational number.

x =2 $\sqrt{ 3 }$-1

$\Rightarrow$ $x ^{ 2 }$ = $( 2 \sqrt{ 3 } – 1) ^{ 2 }$

$\Rightarrow$ $x ^{ 2 }$  = $( 2\sqrt{ 3 } ) ^{ 2 }$ +$1 ^{ 2 }$$2 ( 2\sqrt{ 3 } ) ( 1 )$

$\Rightarrow$ $x ^{ 2 }$ = 12 + 1 – $4 ( \sqrt{ 3 } )$

$\Rightarrow$ $x ^{ 2 }$ – 13 =- $4 ( \sqrt{ 3 } )$

$\Rightarrow$ $\frac{ 13 – x ^{ 2 } } { 4 }$= $\sqrt{ 3 }$

Since x is a rational number, $x ^{ 2 }$  is also a rational number.

$\Rightarrow$ 13 — $x ^{ 2 }$  is a rational number

$\Rightarrow$ $\frac{ 13 – x ^{ 2 } } { 4 }$ is a rational number

$\Rightarrow$  $\sqrt{ 3 }$  is a rational number

But $\sqrt{ 3 }$  is an irrational number, which is a contradiction.

Hence, our assumption is wrong.

Thus, ($2 \sqrt{ 3 }$ – 1) is an irrational number.

Q.2: Prove that $\left ( 4 – 5 \sqrt{ 2 } \right )$ is an irrational number.

Sol:

Let x = 4-$5 \sqrt{ 2 }$ be a rational number.

x = 4 – $5 \sqrt{ 2 }$

$\Rightarrow$ $x ^{ 2 }$= $( 4 – 5 \sqrt{ 2 } ) ^{ 2 }$

$\Rightarrow$ $x ^{ 2 }$ = $4 ^{ 2 }$ + $( 5 \sqrt{ 2 } )^{ 2 }$ – 2 ( 4 ) ( $( 5 \sqrt{ 2 } )$)

$\Rightarrow$ $x ^{ 2 }$ = 16 + 50 – $40 \sqrt{ 2 }$

$\Rightarrow$ $x^{2}$ – 66 = – $40 \sqrt{ 2 }$

$\frac{ 66 – x ^{ 2 } } { 40 }$ = $\sqrt{ 2 }$

Since x is a rational number, $x ^{ 2 }$ is also a rational number.

$\Rightarrow$  66 — $x ^{ 2 }$  is a rational number

$\Rightarrow$  $\frac{66 – x ^{ 2 } } { 40 }$is a rational number

$\sqrt{ 2 }$ is a rational number

But $\sqrt{ 2 }$  is an irrational number, which is a contradiction.

Hence, our assumption is wrong.

Thus, (4-$5 \sqrt{ 2 }$) is an irrational number.

Q.3: Prove that $5 \sqrt{ 2 }$ is irrational.

Sol:

Let $5 \sqrt{ 2 }$  is a rational number.

Therefore, $5 \sqrt{ 2 }$ = $\frac{ p } { q }$ ,  where p and q are some integers and HCF(p, q) = 1 ….(1)

$\Rightarrow$  $5 \sqrt{ 2 }$ q = p

$\Rightarrow$ $( 5 \sqrt{ 2 } q ) ^{ 2 }$= $p ^{ 2 }$

$\Rightarrow$ $2 ( 25 q ^{ 2 } )$ = $p ^{ 2 }$

$\Rightarrow$ $p ^{ 2 }$  is divisible by 2

$\Rightarrow$ p is divisible by 2      ………(2)

Let p = 2m, where m is some integer.

Therefore, $5 \sqrt{ 2 }$  = 2 m

$\Rightarrow$ $( 5 \sqrt{ 2 } q ) ^{ 2 }$= $2m ^{ 2 }$

$\Rightarrow$ $2 ( 25 q ^{ 2 } )$ = $4m ^{ 2 }$

$\Rightarrow$ 25 $q ^{ 2 }$ = 2 $m ^{ 2 }$

$\Rightarrow$ $q ^{ 2 }$  is divisible by 2

$\Rightarrow$ q is divisible by 2   ………..(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong.

Thus, $5\sqrt{ 2 }$  is irrational.

Q.4: Prove that $\frac{ 2 }{ \sqrt{ 7 } }$ is irrational.

Sol:

$\frac{ 2 } { \sqrt{ 7 } }$ = $\frac{ 2 }{ \sqrt{ 7 } }$ x $\frac{ \sqrt{ 7 } }{\sqrt{ 7 } }$ = $\frac{ 2 }{ 7 } \sqrt{ 7 }$

Let $\frac{ 2 }{ 7 } \sqrt{ 7 }$ is a rational number.

Therefore, $\frac{ 2 }{ 7 } \sqrt{ 7 }$ = $\frac{ p }{ q }$  , where p and q are some integers and HCF(p, q) = 1 ….(1)

$\Rightarrow$ $2\sqrt{ 7 } q =7 p$

$\Rightarrow$ ($2\sqrt{ 7 } q ) ^{ 2 } = 7 q ^{ 2 }$

$\Rightarrow$ $7 ( 4 q ^{ 2 } )$ = $49 p ^{ 2 }$

$\Rightarrow$ $4 q ^{ 2 }$ = $7 p ^{2}$

$\Rightarrow$ $q ^{2}$  is divisible by 7

$\Rightarrow$ q is divisible by 7   …… (2)

Let q = 7m, where in is some integer.

$2 \sqrt{ 7 } q$ = 7p

$[2 \sqrt{ 7 }( 7m ) ] ^{2} )$ = $7 p ^{2}$

343($4 m ^{ 2 }$ ) = 49 $p ^{ 2 }$

7 ( $4 m ^{ 2 }$) = $p ^{ 2 }$

$p ^{ 2 }$ is divisible by 7

p is divisible by 7   ……. (3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong.

Thus, $\frac{ 2 }{ \sqrt{ 7 } }$ is irrational.

Q.5: State Euclid’s division lemma.

Sol:

Euclid’s division lemma, states that for any two positive integers  a and b, there exist unique whole numbers q and r, such that a=b x q + r  where $0\leq r< b$

Q.6: State fundamental theorem of arithmetic.

Sol:

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

Q.7: Express 360 as the product of its prime factors.

Sol:

Prime factorization:

360 = 23 x 32 x 5

Q.8: If a and b are two prime numbers then find HCF (a , b).

Sol:

Prime factorization:

a = a

b = b

HCF = product of smallest power of each common prime factor in the numbers = 1

Thus, HCF (a, b) = 1

Q.9: If a and b are two prime numbers then LCM (a, b).

Sol:

Prime factorization:

a = a

b=b

LCM = product of greatest power of each prime factor involved in the numbers = a x b

Thus, LCM (a, b) = ab.

Q.10: If the product of two numbers is 1050 and their HCF is 25, find their LCM.

Sol:

HCF of two numbers = 25

Product of two numbers = 1050

Let their LCM be x.

Using the formula, Product of two numbers = HCF x LCM

We conclude that,

1050 = 25.X

X = $\frac{ 1050 }{ 25 }$ = 42

Hence, their LCM is 42.

Q.11: What is a composite number?

Sol:

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

Q.12: If a and b are relatively prime then what is their HCF?

Sol:

If two numbers are relatively prime then their greatest common factor will be 1.

Thus, HCF (a, b) = 1.

Q.13: If the rational number $\frac{ a }{ b }$ has a terminating decimal expansion, what is the condition to be satisfied by b?

Sol:

Let x be a rational number whose decimal expansion terminates.

Then, we can express x in the form $\frac{a}{b}$, where a and b are co prime, and prime factorization of b is of the form ($2^{m}$  x$5^{n}$ ), wherein and n are non negative integers.

Q.14: Simplify : $\frac{ (2 \sqrt{ 45 } + 3 \sqrt{ 20 } ) }{ 2 \sqrt{ 5 } }$

Sol:

$\frac{ 2\sqrt{ 45 }+ 3 \sqrt{ 20 } } { 2 \sqrt{ 5 } }$ = $\frac{ 2 \sqrt {3 X 3 X 5} + 3\sqrt{2 X 2 X 5} } { 2 \sqrt{ 5 } }$

$=$  $\frac{ 2 X 3 \sqrt{ 5 } + 3 X 2\sqrt{ 5} } {2 \sqrt{ 5 } }$

$=$ $\frac{ 6 \sqrt{ 5 } + 6 \sqrt{ 5 } } { 2 \sqrt{ 5 } }$

$=$ $\frac{ 12 \sqrt{ 5 } }{ 2 \sqrt{ 5 } }$ = 6

Thus, simplified form of   $\frac{ 2 \sqrt{ 45 } + 3 \sqrt{ 20 } }{ 2 \sqrt{ 5 } }$ is 6.

Q.15: Write the decimal expansion of $\frac{ 73 }{ ( 2 ^{ 4 } \times 5 ^{ 3 } ) }$

Sol:

Decimal expansion:

$\frac{ 73 }{ 2 ^{ 4 } \times 5^{ 3 } }$ = $\frac{ 73 X 5 }{ 2^{ 4} \times 5 ^{ 4 } }$

= $\frac{ 365 }{ ( 2 X 5 )^{ 4 } }$ = $\frac{ 365 }{ ( 10 ) ^{ 4 } }$

= $\frac{ 365 }{ 10000 }$ = 0.0365

Thus, the decimal expansion of   $\frac{ 73 }{ 2^{ 4 } X 5^{ 3 } }$ is 0.0365.

Q.16: Show that there is no value of n for which $\left ( 2 ^{ n } \times 5 ^{ n } \right )$ ends in 5.

Sol:

We can write:

(2″ x 5″) = (2 x 5)n

= 10n

For any value of n, we get 0 in the end.

Thus, there is no value of n for which (2″ x 5″) ends in 5.

Q.17: Is it possible to have two numbers whose HCF is 25 and LCM is 520?

Sol:

No, it is not possible to have two numbers whose I-ICF is 25 and LCM is 520.

Since HCF must be a factor of LCM, but 25 is not a factor of 520.

Q.18: Give an example of two irrationals whose sum is rational.

Sol:

Let the two irrationals be 4 – $\sqrt{ 5 }$ and 4 + $\sqrt{ 5 }$.

(4 – $\sqrt{ 5 }$) + (4 + $\sqrt{ 5 }$) = 8

Thus, sum (i.e., 8) is a rational number.

Q.19: Give an example of two irrationals whose product is rational.

Sol:

Let the two irrationals be 4 $\sqrt{ 5 }$ and 3 $\sqrt{ 5 }$.

(4 $\sqrt{ 5 }$) X (3 $\sqrt{ 5 }$ ) = 60

Thus, sum (i.e., 60) is a rational number.

Q.20: If a and b are relatively prime, what is their LCM?

Sol:

If two numbers are relatively prime then their greatest common factor will be 1.

HCF (a, b) = 1

Using the formula, Product of two numbers = HCF x LCM we conclude that,

a x b = 1 x LCM

LCM = ab

Thus, LCM (a, b) is ab.

Q.21: The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?

Sol:

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500.

Since HCF must be a factor of LCM, but 500 is not a factor of 1200.

Q.22: Express $0.\bar{4}$ as a rational number in simplest form.

Sol:

Let x be $0.\bar{4}$.

x = $0.\bar{4}$  ………  (1)

Multiplying both sides by 10, we get

10x = $4.\bar{4}$       ………(2)

Subtracting (1) from (2), we get

10x – x = $4.\bar{4}$– 0.4

$\Rightarrow$ 9x = 4

$\Rightarrow$x=$\frac{ 4 }{ 9 }$

Thus, simplest form of $0.\bar{4}$ — as a rational number is $\frac{4}{9}$.

Q.23: Express $0.\bar{ 23 }$ as a rational number in simplest form.

Sol:

Let x be $0.\bar{23}$.

x = $0.\bar{4}$.   ……. (1)

Multiplying both sides by 100, we get

100x = $23.\bar{23}$  ……   (2)

Subtracting (1) from (2), we get

100x – x = $23.\bar{23}$$0.\bar{23}$.

$\Rightarrow$ 99 x = 23

$\Rightarrow$x=$\frac{23}{99}$

Thus, simplest form of $0.\bar{23}$ as a rational number is $\frac{ 23 }{ 99 }$

Q.24: Explain why 0.15015001500015 … is an irrational number.

Sol:

Irrational numbers are non-terminating non-recurring decimals.

Thus, 0.15015001500015 … is an irrational number.

Q.25: Show that $\frac{ \sqrt{ 2 } }{ 3 }$ is irrational.

Sol:

Let  $\frac{ \sqrt{ 2 } }{ 3 }$ is a rational number.

Therefore, $\frac{ \sqrt{ 2 } }{ 3 }$ = $\frac{ p } { q }$,

where p and q are some integers and HCF(p, q) = 1        … ….(1)

$\Rightarrow$  $\sqrt{ 2 }$ q = 3p

$\Rightarrow$ $( \sqrt{ 2 } q ) ^{ 2 }$ = $( 3 p ) ^{ 2 }$

$\Rightarrow$  $2 ( q ) ^{ 2 }$ = $9 ( p ) ^{ 2 }$

$\Rightarrow$  $( p ) ^{ 2 }$ is divisible by 2

$\Rightarrow$ p is divisible by 2         …….(2)

Let p = 2m, where in is some integer.

Therefore, $\sqrt{ 2 }$ q = 3p

$\Rightarrow$  $\sqrt{ 2 }$ q = 3 ( 2m )

$\Rightarrow$  $(\sqrt{ 2 } q ) ^ {2 }$ = $( 3 ( 2 m ) ) ^{ 2 }$

$\Rightarrow$ $2 q ^{ 2 } = 4 ( 9 p ^{ 2 } )$

$\Rightarrow$ $q ^{ 2 } = 2 ( 9 p ^{ 2 } )$

$\Rightarrow$ $q ^{ 2 }$  is divisible by 2

$\Rightarrow$ q is divisible by 2 …….(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).

Hence, our assumption is wrong.

Thus, $\frac{ \sqrt{ 2 } }{ 3 }$  is irrational.

Q.26: Write a rational number between $\sqrt{ 3 }$ and 2.

Sol:

Since, $\sqrt{ 3 }$  = 1.732….

So, we may take 1.8 as the required rational number between $\sqrt{ 3 }$  and 2.

Thus, the required rational number is 1.8

Q.27: Explain why $3.\overline{1416}$ is a rational number.

Sol:

Since, $3.\bar{1}\bar{4}\bar{1}\bar{6}$  is a non-terminating repeating decimal.

Hence, is a rational number.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 1 – Real Numbers Ex 1C

• The solutions are prepared by our subject experts in a step by step approach for better and easy understanding.
• It is prepared as per the latest CBSE syllabus of Class 10 Maths.