Haryana State Board HBSE 10th Class Maths Important Questions Chapter 1 Real Numbers Important Questions and Answers.

## Haryana Board 10th Class Maths Important Questions Chapter 1 Real Numbers

Short/Long Answer Type Questions

Question 1.

Show that square of any positive integer cannot be of the form (54 + 2) or (5q + 3) for any integer q.

OR

Prove that one of three consecutive positive integer is divisible by 3.

Solution :

Let be any positive integer and applying Euclid’s division Lemma it is of the form 5. For 5P + 1 or 5P + 2 or 5P + 3 or 5P + 4.

So, we have the following cases

Case I : When x = 5P

⇒ x^{2} = 25P^{2} = 5 (5P^{2})

⇒ x^{2} = 5q [Where q = 5P^{2}]

Case II : When

x = 5P + 1

x^{2} = (5P + 1)^{2}

= 25p^{2} + 10P + 1

= 5(5P^{2} + 2P) + 1

= 59 +1

[Where q = 5P^{2} + 2P]

Case III : When x = 5P + 2

x^{2} = (5P + 2)^{2}

= 25P^{2} + 20P + 4

= 5(5P^{2} + 4P) + 4

= 59 + 4

(Where q = 5P^{2} + 4P)

Case IV : When x = 5P + 3

⇒ x^{2} = (5P + 3)^{2}

= 25P^{2} + 30P + 4

= 25P^{2} + 30P + 5 + 4

= 5 (5P^{2} + 6P + 1) + 4

= 5 + 4

(Where q = 5P^{2} + 6P + 1)

Case V : When x = 5P + 4

⇒ x^{2} = (5P + 4)^{2}

= 25P^{2} + 40P + 16

= 25P^{2} + 40P + 15 + 1

= 5(5P^{2} + 8P + 3) + 1

= 5q + 1

(Where q = 5P^{2} + 8P + 3)

So, square of any positive integer cannot be of the form (5q + 2) or (5q + 3).

Or

Solution :

Let x be any positive integer. By Euclid’s division lemma x = 3q + r, where 0 < r ≤ 3

[∴ r = 0, 1, 2]

Putting r = 0, we get

x = 3q + 0 = 3q which is divisible by 3.

Putting r = 1, we get

x = 3q + 1 which is not divisible by 3.

Putting r = 2, we get

x = 3q + 2, which is not divisible by 3.

So, one of every three consecutive positive integers is divisible by 3.

Question 2.

If n is an odd integer, then show that n^{2} – 1 is divisible by 8.

Solution :

We know that any odd positive integer x can be written in form 4q + 1 or 4q + 3

So, according to the question

Case I : When x = 4q + 1

Then, x^{2} – 1 = (4q + 1)^{2} – 1

= 16q^{2} + 8q + 1 – 1

= 89 (2q + 2) …(1)

Which is divisible by 8.

Case II : When x = 4q + 3

Then, x^{2} – 1 = (4q + 3)^{2} – 1

= 16q^{2} + 24q + 9 – 1

= 16q^{2} + 24q + 8

= 8(2q^{2} + 3 + 1) ………(2)

Which is divisible by 8. Therefore, from equations (1) and (2), it is clear that, if x is an odd positive integer. x^{2} – 1 is divisible by 8.

Question 3.

What is the HCF of the smallest prime number and the smallest composite number,

Solution :

Smallest prime number = 2

Smallest composite number = 4

∴ 2 = 2

and 4 = 2 × 2 = 2^{2}

HCF of (2, 4) 2.

Question 4.

Write the smallest number which is divisible by both 306 and 657.

Solution :

The required smallest number is the LCM of 306 and 657.

We have 306 = 2 × 3 × 3 × 17

= 2 × 3^{2} × 17

And 657 = 3 × 3 × 73

= 3^{2} × 73

LCM (306, 657) = 2 × 3^{2} × 17 × 73

= 22938

Hence, the required smallest number = 22338

Question 5.

The length, breadth and height of a room are 8m 50cm, 6m 26cm and 4m 75cm respectively. Find the length of longest rod that can measure the dimensions of the room exactly.

Solution :

Dimensions of a room are:

Length = 8m 50cm = 850 cm.

Breadth = 6m 25cm = 625 cm

And Height = 4m 75cm = 475 cm.

The required length of longest rod is the HCF of 850 cm, 625 cm and 475 cm.

850 = 2 × 5 × 5 × 17

= 2 × 5^{2} × 17

625 = 5 × 5 × 5 × 5 = 5^{4} And

475 = 5 × 5 × 19 = 5^{2} × 19

HCF of (850, 625, 475) = 5^{2} = 25 cm

Hence, required length of longest rod = 25 cm.

Question 6.

State the fundamental theorem of Arithmetic. Find the LCM of numbers 2520 and 10530 by prime factorization method.

Solution :

Fundamental Theorem of Arithmetic : Every composite number can be expressed (or factorized) as a product of primes, and this factorisation is unque, apart from the order in which the prime factors occur. Then factorisation of x can be written as x = P_{1} × P_{2} × P_{3}, × ………. P_{x}, Where P_{1}, P_{2}, ……..P_{x} are primes and written in ascending order.

So, we have

2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7

= 2^{3} × 3^{2} × 5 × 7

And 10530 = 2 × 3 × 3 × 3 × 3 × 5 × 13

= 2 × 3^{4} × 5 × 13

LCM (2520, 10530) = 2^{3} × 3^{4} × 5 × 7 × 13

= 294840

Question 7.

Show that 3\(\sqrt{7}\) is an irrational number.

Solution :

Let us assume that 3\(\sqrt{7}\) is a rational. It can be expressed as form of \(\frac {a}{b}\),where a and b are coprime positive integers and b ≠ 0.

∴ 3\(\sqrt{7}\) = \(\frac {a}{b}\)[HCF of a and b is 1 and b ≠ 0]

⇒ \(\sqrt{7}\) = \(\frac {a}{3b}\) ……………(1)

⇒ \(\frac {a}{3b}\) = rational

[∵ a and b are positive integers]

So, from equ. (1) \(\sqrt{7}\) is rational number. But this contradicts the fact \(\sqrt{7}\) is irrational number. So, our asumption that 3\(\sqrt{7}\) is an irrational number, is wrong.

Hence, that \(\sqrt{6}\) is an irrational numbers.

Question 8.

Prove that \(\sqrt{6}\) is an irrational number.

Solution :

Let us assume that \(\sqrt{6}\) is rational. It can be express in the form of \(\frac {a}{3b}\), where a and b are coprime positive integers and b ≠ 0.

∴ \(\sqrt{6}\) = \(\frac {a}{b}\) (Where a and b are coprime ∴ HCF of a and b = 1 ……..(i)

⇒ 6 = \(\frac{a^2}{b^2}\) (Squaring both sides)

⇒ 6b^{2} = a^{2} ……………(i)

Therefore 6 divides a^{2}. It follows that 6 divides a.

[By theorem 1.3]

Let a = 6c and put this value in equ. (i) we get

6b^{2} = (6c)^{2}

⇒ 6b^{2} = 36c^{2}

⇒ \(\frac{36 c^2}{6}\) = b^{2}

⇒ 6c^{2} = b^{2} ………..(ii)

It means b^{2} is divisible by 6. It follows that b, is divisible by 6.

(By theorem 1.3)

From equations (i) and (ii) we say that 6 is a common factor of both a and b. But this contradicts the fact that a and b are coprime, so we have no common factor. So, our assumption that \(\sqrt{6}\) is a rational number is wrong. Therefore, \(\sqrt{6}\) is an irrational number. Proved.

Question 9.

Given that \(\sqrt{2}\) is an irrational number, then prove that (5 + 3\(\sqrt{2}\)) is an irrational number.

Solution :

Let us assume that 5 + 3\(\sqrt{2}\) is a rational number. It can be express in the form of \(\frac {a}{b}\), where a and b are coprime positive integers and b ≠ 0.

∴ 5 + 3\(\sqrt{2}\) = \(\frac {a}{b}\)

[Where HCF of a and b = 1]

⇒ 5 – \(\frac {a}{b}\) = 3\(\sqrt{2}\)

⇒ \(\frac{5 b-a}{b}\) = 3\(\sqrt{2}\)

⇒ \(\frac{5 b-a}{3}\) = \(\sqrt{2}\)

∵ a and b are positive integers

⇒ \(\frac{5 b-a}{3}\) is rational

Therefore, \(\sqrt{2}\) is rational But given that \(\sqrt{2}\) is irrational. So, our assumption that 5 + 3\(\sqrt{2}\) is rational is wrong.

Hence, 5 + 3\(\sqrt{2}\) is an irrational number.

Proved.

Question 10.

What type of decimal expansion does a rational number has? How can you distinguish it from decimal expansion of irrational numbers?

Solution :

A rational number may has its decimal expansion either terminating decimal expansion or a non-terminating repeating. But an irrational number has its decimal expansion non-repeating and non-terminating.

Question 11.

Write whether rational number \(\frac {7}{75}\) will have terminating decimal expansion or a non-terminating decimal.

Solution :

= \(\frac{7}{3 \times 5^2}\)

Since, denominator of given rational number is not form 2^{m} × 5^{n}.

Hence, it is non-terminating decimal expansion

Question 12.

After how many decimal places will the decimal expansion of \(\frac{23}{2^4 \times 5^3}\) terminate?

Solution :

We have

\(\frac{23}{2^4 \times 5^3}=\frac{23 \times 5}{2^4 \times 5^3 \times 5}=\frac{23 \times 5}{2^4 \times 5^4}\)

= \(\frac{115}{(10)^4}=\frac{115}{10000}\) = 0.0115

Hence, \(\frac{23}{2^4 \times 5^3}\) will terminate after 4 decimal places.

Fill in the Blanks

Question 1.

The sum or difference of a rational and an irrational number is.

Solution :

irrational

Question 2.

Every composite number can be factorized as the product of ………….

Solution :

primes

Question 3.

The product and quotient of a …………..rational and irrational number is irrational.

Solution :

non-zero

Question 4.

………………is the least prime and…………is the least composite number.

Solution :

2, 4

Question 5.

The HCF of two co-prime numbers is always …………..

Solution :

1

Question 6.

If a and b are co-primes, then a^{2} and b^{2} are …………….

Solution :

co-prime.

Multiple Choice Questions

Question 1.

Euclid’s division Lemma states that for two positive integers a and b, there exists unque integers q and r satisfying a = bq + r, and :

(a) o < r < b (b) 0 > r ≤ b

(c) 0 ≤ r < b

(d) 0 ≤ r ≤ b

Solution :

(c) 0 ≤ r < b

Question 2.

The total number of factors of prime numbers is :

(a) 1

(b) 0

(c) 2

(d) 3

Solution :

(c) 2

∵ We know that prime numbers h only two factors 1 and number itself.

Question 3.

Sum of the exponents of prime factors in the prime factorization of 196 is :

(a) 3

(b) 4

(c) 5

(d) 2

Solution :

(b) 4

∵ We have 196 = 2 × 2 × 7 × 7

= 2^{2} × 7^{2}

It’s sum of exponents

= 2 + 2 = 4

Question 4.

The HCF and LCM of 12, 21, 15 respectvely are :

(a) 3, 140

(b) 12, 420

(c) 3, 420

(d) 420, 3

Solution :

(c) 3, 420

We have 12 = 2 × 2 × 3 = 2^{2} × 3

21 = 3 × 7 × 3 × 7 and

15 = 3 × 5 = 3 × 5

LCM (12, 21, 15) = 2^{2} × 3 × 5 × 7 = 420

And HCF (12, 21, 15) = 3

So, HCF and LCM of (12, 21, 15) = 3, 420

Question 5.

Which of the following rational number.

(a) \(\frac {1}{2}\)

(b) \(\frac {1}{2}\)

(c) \(\frac{343}{2^3 \times 5^2 \times 7^3}\)

(d) \(\frac{31}{2^4 \times 3^5}\)

Solution :

(c) \(\frac{343}{2^3 \times 5^2 \times 7^3}\)

Question 6.

The decimal representation of \(\frac{11}{2^3 \times 5}\) will :

(a) terminate after 1 decimal place

(b) terminate after 2 decimal place

(c) terminate after 3 decimal place

(d) not terminate

Solution :

(c) terminate after 3 decimal place

∵ Since, \(\frac{11}{2^3 \times 5}=\frac{11}{8 \times 5}=\frac{11}{40}\) = 0.275

So, \(\frac{11}{2^3 \times 5}\) will terminate after 3 decimal places.