## RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQs
- RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself

**Very-Short Answer Questions**

**Question 1.
**

**Solution:**

For any two given positive integers a and b there exist unique whole numbers q and r such that

a = bq + r, where 0 ≤ r < b.

Here, we call ‘a’ as dividend, b as divisor, q is quotient and r as remainder.

Dividend = (Divisor x Quotient) + Remainder

**Question 2.**

**Solution:**

Every composite number can be uniquely expressed as a product of two primes, except for the order in which these prime factors occurs.

**Question 3.**

**Solution:**

360 = 2 x 2 x 2 x 3 x 3 x 5 = 2^{3} x 3² x 5

**Question 4.**

**Solution:**

We know that HCF of two primes is

HCF (a, b) = 1

**Question 5.**

**Solution:**

a and b are two prime numbers then their

LCM = Product of these two numbers

LCM(a, b) = a x b = ab.

**Question 6.**

**Solution:**

We know that product of two numbers is equal to their HCF x LCM

LCM = \(\frac { Product of two numbers }{ HCF }\)

= \(\frac { 1050 }{ 25 }\) = 42

LCM of two numbers = 42

**Question 7.**

**Solution:**

A composite number is a number which is not a prime. In other words, a composite number has more than two factors.

**Question 8.**

**Solution:**

a and b are two primes, then their

HCF will be 1

HCF of a and b = 1

**Question 9.**

**Solution:**

\(\frac { a }{ b }\) is a rational number and it has terminating decimal

b will in the form 2^{m} x 5^{n} where m and n are some non-negative integers.

**Question 10.**

**Solution:**

**Question 11.**

**Solution:**

**Question 12.**

**Solution:**

2^{n} x 5^{n} = (2 x 5)^{n} = (10)^{n}

Which always ends in a zero

There is no value of n for which (2^{n} x 5^{n}) ends in 5

**Question 13.**

**Solution:**

We know that HCF is always a factor is its LCM

But 25 is not a factor of 520

It is not possible to have two numbers having HCF = 25 and LCM = 520

**Question 14.**

**Solution:**

Let two irrational number be (5 + √3) and (5 – √3).

Now their sum = (5 + √3) + (5 – √3) = 5 + √3 + 5 – √3 = 10

Which is a rational number.

**Question 15.**

**Solution:**

Let the two irrational number be (3 + √2) and (3 – √2)

Now, their product = (3 + √2) (3 – √2)

= (3)² – (√2)² {(a + b) (a – b) = a² – b²}

= 9 – 2 = 7

Which is a rational number.

**Question 16.**

**Solution:**

a and b are relative primes

their HCF = 1

**Question 17.**

**Solution:**

LCM of two numbers = 1200

and HCF = 500

But we know that HCF of two numbers divides their LCM.

But 500 does not divide 1200 exactly

Hence, 500 is not their HCF whose LCM is 1200.

**Short-Answer Questions**

**Question 18.**

**Solution:**

Let x = 0.4 = 0.444

Then 10x = 4.444….

Subtracting, we get

9x = 4 => x = \(\frac { 4 }{ 9 }\)

\(\bar { 0.4 }\) = \(\frac { 1 }{ 2 }\) which is in the simplest form.

**Question 19.**

**Solution:**

\(\bar { 0.23 }\)

Let x = \(\bar { 0.23 }\) = 0.232323…….

and 100x = 23.232323……

Subtracting, we get

99x = 23 => x = \(\frac { 23 }{ 99 }\)

\(\bar { 0.23 }\) = \(\frac { 23 }{ 99 }\) which is in the simplest form.

**Question 20.**

**Solution:**

0.15015001500015

It is non-terminating non-repeating decimal.

It is an irrational number.

**Question 21.**

**Solution:**

\(\frac { \surd 2 }{ 3 }\) = \(\frac { 1 }{ 3 }\) √2

Let \(\frac { 1 }{ 3 }\) √2 is a rational number

Product of two rational numbers is a rational

\(\frac { 1 }{ 3 }\) is rational and √2 is rational contradicts

But it contradicts the fact

\(\frac { \surd 2 }{ 3 }\) or \(\frac { 1 }{ 3 }\) √2 is irrational.

**Question 22.**

**Solution:**

√3 and 2.

√3 = 1.732 and 2.000

A rational number between 1.732 and 2.000 can be 1.8 or 1.9

Hence, 1.8 or 1.9 is a required rational.

**Question 23.**

**Solution:**

\(\bar { 3.1416 }\)

It is non-terminating repeating decimal.

It is a rational number.

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E are helpful to complete your math homework.

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