## RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Test Yourself

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

**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

**Question 1.**

**Solution:**

**(b)**

Its decimal will be nonterminating repeating decimal.

**Question 2.**

**Solution:**

**(b)** \(\frac { p }{ q }\) is terminating decimal if q = 2^{m} x 5^{n}

Now, 91 = 7 x 13, 45 = 3^{2} x 5

80 = 2^{4} x 5, 42 = 2 x 3 x 7

80 is of the form 2^{m} x 5^{n}

\(\frac { 19 }{ 80 }\) is terminating decimal expansion,

**Question 3.**

**Solution:**

**(b)** Divisor = 9 and remainder = 7

Let b be the divisor, then

n = 9b + 7

Multiplying both sides by 3 and subtracting 1.

3n – 1 = 3(9b + 7) – 1

3n – 1 = 27b + 21 – 1

3n – 1 = 9(3b) + 9 x 2 + 2

3n – 1 = 9(3b + 2) + 2

Remainder = 2

**Question 4.**

**Solution:**

**(b)** \(0.\bar { 68 }\) + \(0.\bar { 73 }\)

0.686868 ……… + 0.737373……

= 1.424241 = \(1.\bar { 42 }\)

**Short-Answer Questions (2 marks)**

**Question 5.**

**Solution:**

4^{n}, n ∈ N

4^{1} = 4

4^{2} = 4 x 4 = 16

4^{3} = 4 x 4 x 4 = 64

4^{4} = 4 x 4 x 4 x 4 = 256

4^{5} = 4 x 4 x 4 x 4 x 4 = 1024

We see that value of 4^{n}, ends with 4 or 6 only.

Hence, the value of 4^{n}, n ∈ N, never ends with 0.

**Question 6.**

**Solution:**

HCF of two numbers = 27 and LCM =162

One number = 81

**Question 7.**

**Solution:**

\(\frac { 17 }{ 30 }\) = \(\frac { 17 }{ 2 x 3 x 5 }\)

Here, q is in the form of 2^{m} x 5^{n}

It is not terminating decimal.

**Question 8.**

**Solution:**

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

Let (2 + √3) is rational and 2 is rational.

Difference of them is also rational.

=> (2 + √3) – 2 = 2 + √3 – 2

= √3 is rational

But it contradicts the fact.

(2 + √3) is irrational.

**Short-Answer Questions (3 marks)**

**Question 11.**

**Solution:**

HCF of 12, 15, 18, 27

12 = 2 x 2 x 3 = 2^{2} x 3

15 = 3 x 5

18 = 2 x 3 x 3 = 2 x 3^{2}

27 = 3 x 3 x 3 = 3^{3}

Now, HCF = 3

and LCM = 2^{2} x 3^{3} x 5 =2 x 2 x 3 x 3 x 3 x 5

= 4 x 27 x 5 = 540

**Question 12.**

**Solution:**

Let 2 + √3 and 2 – √3 are two irrational number.

Sum = 2 + √3 + 2 – √3 = 4 which is a rational.

**Question 13.**

**Solution:**

4620

**Question 14.**

**Solution:**

1008

**Question 15.**

**Solution:**

**Question 16.**

**Solution:**

Give numbers are 546 and 764 and remainders are 6 and 8 respectively.

Remaining number 546 – 6 = 540

and 764 – 8 = 756

Now, required largest number = HCF of 540 and 756 = 108

**Long-Answer Questions (4 marks)**

**Question 17.**

**Solution:**

Let √3 is a rational number.

Let √3 = \(\frac { p }{ q }\) where p and q are integers and have no common factor, other than 1 and q ≠ 0

Squaring both sides.

3 = \(\frac { { p }^{ 2 } }{ { q }^{ 2 } }\) => 3q^{2} – p^{2}

=> 3 divides p^{2}

=> 3 divides p

Let p = 3c for some integer c

3q^{2} = 9c^{2} => q^{2} – 3c^{2}

=> 3 divides q^{2} (3 divides 3c^{2})

=> 3 divides q

3 is common factors of p and q

But it contradicts the fact that p and q have

no common factors and also contradicts that √3 is a rational number.

Hence, √3 is irrational number.

**Question 18.**

**Solution:**

Let n be an arbitrary odd positive integer on dividing n by 4, let m be the quotient and r be the remainder.

By Euclid’s division lemma,

n = 4q + r where 0 ≤ r < 4

n = 4q or (4q + 1) or (4q + 2) or (4q + 3)

Clearly, 4q and (4q + 2) are even number

since n is odd.

n ≠ 4q and n ≠ (4q + 2)

n = (4 q + 1) or (4q + 3) for same integer n

Hence, any positive odd integer of the form 4q + 1 or 4q + 3 for some integer q.

**Question 19.**

**Solution:**

On dividing n by 3, let q be the quotient and r be the remainder, then

n = 3q + r where 0 ≤ r < 3 => n = 3q + r where r = 0, 1 or 2

n = 3q or n = 3q + 1 or n = 3q + 2

(i) Case (I)

If n = 3q then n is divisible by 3

(ii) Case (II)

If n = (3q + 1) then n + 2 = 3q + 3 = 3q (q + 1) which is divisible by 3

In this case, n + 2 is divisible by 3

(iii) Case (III)

If n = (3q + 2) then n + 1 (n + 1) = 3q + 3 = 3(q + 1) which also divisible by 3

In this case, (n + 1) is divisible by 3

Hence, one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

**Question 20.**

**Solution:**

Let (4 + 3√2) is rational number and 4 is also a rational number.

Difference of two rational numbers is also a rational number.

4 + 3√2 – 4 = 3√2 is a rational number

Product of two rational numbers is rational

3 is rational and √2 is rational

But it contradicts the fact

√2 is irrational

Hence, (4 + 3√2 ) is irrational.

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

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