RBSE Class 10 Maths Important Questions Chapter 1 Real Numbers

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 1 Real Numbers Important Questions and Answers.

RBSE Class 10 Maths Chapter 1 Important Questions Real Numbers

Objective Type Questions

Question 1.
The learned mathematician Euclid, the explorer of HCF of two numbers, was—
(A) of Greek
(B) of India
(C) of America
(D) of Britain
Answer:
(A) of Greek

Question 2.
A number that has no factor except 1 and itself is called—
(A) Composite number
(B) Prime number
(C) Even number
(D) Odd number
Answer:
(B) Prime number

Question 3.
The smallest prime number is—
(A) 5
(B) 4
(C) 3
(D) 2
Answer:
(D) 2

Question 4.
The highest common factor (HCF) of two or more numbers is—
(A) The smallest common
(B) Only common
(C) The greatest number
(D) The greatest common
Answer:
(D) The greatest common

Question 5.
If in the prime fectorization of the denominator of the rational number written in standard form, there is no other prime factor except 2 or 5 or both digits, then this number is—
(A) non-terminating decimal
(B) terminating decimal
(C) terminating and non-terminating both
(D) None of the above
Answer:
(B) terminating decimal

Question 6.
Real numbers are called—
(A) only rational numbers
(B) only irrational numbers
(C) rational and irrational both
(D) None of the above
Answer:
(C) rational and irrational both

Question 7.
If a number can not be written in the form \(\frac{p}{q}\), where p and q are integers and q ≠
0, than those numbers are called—
(A) whole numbers
(B) rational numbers
(C) irrational numbers
(D) natural numbers
Answer:
(C) irrational numbers

Question 8.
The sum or difference of a rational number and an irrational number is which of the following numbers?
(A) rational number
(B) irrational number
(C) whole number
(D) natural number
Answer:
(B) irrational number

Very Short Answer Type Questions

Question 1.
Which are two important properties of positive integers?
Answer:

1. Euclid’s Division Algorithm
2. The Fundamental Theorem of Arithmetic.

Question 2.
Write Euclid’s Division Algorithm.
Answer:
“Any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.” This is called Euclid’s Division Algorithm.

Question 3.
What is ‘The Fundamental Theorem of Arithmetic’?
Answer:
Every composite number can be expressed uniquely as a product of prime numbers. This fact is called ‘The Fundamental Theorem of Arithmetic’.

Question 4.
Which number in the product or quotient of a non-zero rational and an irrational number?
Answer:
An irrational number.

Question 5.
Give the examples of irrational numbers.
Answer:
\(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) etc.

Question 6.
What is called a composite number?
Answer:
The number that has at least one factor other than 1 and itself is called a composite numebr.

Question 7.
What is the lowest common multiple (L.C.M.)?
Answer:
The lowest common multiple of two or more number is that smallest number which is the multiple of each number.

Question 8.
What is the highest common factor (HCF)?
Answer:
The highest common factor of two or more numbers is that greatest number which completely divides all the given numbers.

Question 9.
If two numbers a and b are given then to which is their product equal?
Answer:
HCF (a, b) × LCM (a. b)

Question 10.
Write the name of the mathematician who explored the technique to compute the HCF of two numbers.
Answer:
Greek Mathematician Euclid.

Question 11.
Write Euclid’s Division Lemma.
Answer:
Given two positive integers a and b. There exist unique whole numbers q and r such that a = bq + r. where 0 ≤ r < b.

Question 12.
Write the number 32760 in the form of product of factors.
Answer:
32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
= 2³ × 3² × 5 × 7 × 13

Question 13.
Define real numebrs.
Answer:
All the rational numbers and irrational numbers taken together give the collection of real numbers.

Question 14.
Write the condition of terminating decimal expansion.
Answer:
Let x = \(\frac{p}{q}\) be a rational number such that the prime factorisation of q is of the fonn 2ⁿ 5^m, where n, m are non-negative integers, then the decimal expansion of x is terminating.

Question 15.
Find the HCF of 48 and 105.
Answer:
48 and 105
Prime factors of 48 = 2 × 2 × 2 × 2 × 3
= 2⁴ × 3
Prime factors of 105 = 3 × 5 × 7
Hence the greatest common number in the two is 3. Hence their HCF will be 3.

Question 16.
Without using Long Division Process write the rational number \(\frac{17}{8}\) in terminating decimal expansion.
Solution:
Suppose that x = \(\frac{17}{8}\)
We can write this as follows :

Question 17.
Find the H.C.F. of numbers 44 and 99.
Solution:
44 = 2 × 2 × 11 = 2² × 11
and 99 = 3 ² 3 ² 11 = 3² × 11
In the factors of 44 and 99, the greatest common number is 11. So, the HCF of these number will be 11.

Question 18.
Is the decimal expansion of the number \(\frac{3}{625}\) terminating or non-terminating recurring? Write it in decimal form.
Solution:
The decimal expansion of the number \(\frac{3}{625}\) is terminating. The decimal form of the number \(\frac{3}{625}\) is 0.0048 625

Question 19.
Find the HCF of 96 and 404 by the Prime Factorisation Method.
Solution:
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
= 2⁵ × 3
Prime factors of 404 = 2 × 2 × 101
= 2² × 101
∴ HCF (96 and 404) = 22 = 4

Short Answer Type Questions

Question 1.
Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2*7 + 1, where q is some integer.
Solution:
Let a be any positive integer and b = 2. Then, by Euclid’s Division Algorithm, for some integer q ≥ 0, a = 2q + r where r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or a = 2q + 1.
If a = 2q then it is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer will be of the form 2q + 1.

Question 2.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is one integer.
Solution:
Let a be a positive odd integer. Now we use Euclid’s Division Algorithm in a and b = 4.
Since 0 ≤ r < 4, therefore the possible remainders are 0, 1, 2 and 3, i.e., a can be of the form of the numbers 4q, 4q + 1, 4q + 2 or 4q + 3, where q is the quotient. Since a is an odd integer, therefore a can not be of the form 4q and 4q + 2 (because both are divisible by 2)
Therefore, any positive odd integer will be of the form 4q + 1 or 4q + 3.

Question 3.
A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
Solution:
This work can be done by trial and error method. But to do it systematically, we find HCF (420, 130). Then this HCF will give the maximum number of barfis that can be placed in each stack so that the number of stacks will be the least and these barfis will cover the least area in the tray.
Now using Euclid’s Division Algorithm
420 = 130 × 3 + 30
130 = 30 × 4 + 10
30 = 10 × 3 + 0
Hence the HCF of 420 and 130 is 10.
Therefore, the sweet seller can make stacks of 10 for both kinds of barfis.

Question 4.
If the HCF of 1261 and 1067 is 97 then find their L.C.M.
Solution:
Given : a = 1261 and b = 1067
H.C.F. = 97, L.C.M. = ?
Formula : L.C.M. = \(\frac{a \times b}{\text { H.C.F. }}\)
= \(\frac{1261 \times 1067}{97}\)
= 1261 × 1067
= 13871

Question 5.
Find the HCF and LCM of the numbers 6 and 20 by prime factorisation method.
Solution:
Here 6 = 2¹ × 3¹
and 20 = 2 × 2 × 5 = 2² × 5¹
Here HCF (6, 20) = 2
and LCM (6, 20) = 2 × 2 × 3 × 5 = 60
Now here note that
HCF (6, 20) = 2¹ = Product of the smallest power of each common prime factor involved in the numbers.
and LCM (6, 20) = 2² × 3¹ × 5¹ = Product of the greatest power of each prime factor involved in the numbers.

Question 6.
Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Solution:
The prime factorisation of 96 and 404 gives
96 = 2⁵ × 3
404 = 2² × 101
Therefore, the HCF of these two integers = 2² = 4
Also, LCM (96, 404) = \(\frac{96 \times 404}{\mathrm{HCF}(96,404)}\)
= \(\frac{96 \times 404}{4}\)
= 9696

Question 7.
Use Euclid’s Division Algorithm to And the HCF of 135 and 225.
Solution:
Step I : Here 225 >135 Now applying Euclid’s Division Lemma on 225 and 135
225 = 135 × 1 + 90

Step II : Since remainder 90 ≠ 0, therefore applying Euclid’s Division Lemma on 135 and 90
135 = 90 × 1 + 45

Step III : Taking the new divisor 90 and new remainder 45 applying Euclid’s Division Lemma
90 = 45 × 2 + 0
Here the remainder has become zero, so procedure stops here. Since the divisor at this stage is 45, therefore the HCF of 135 and 225 is 45

Long Answer Type Questions

Question 1.
Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.
Solution:
Step 1 : Here 12576 > 4052. Now7 applying Euclid’s Division Lemma on 12576 and 4052
12576 = 4052 × 3 + 420

Step 2 : Since the remainder 420 ≠ 0, therefore applying Euclid’s Division Lemma on 4052 and 420
4052 = 420 × 9 + 272

Step 3 : Taking the new divisor 420 and new remainder 272 applying Euclid’s Division Lemma
420 = 272 × 1 + 148
Taking the new divisor 272 and new remainder 148 applying Euclid’s Division Lemma
272 = 148 × 1 + 124
Taking the new divisor 148 and new remainder 124 applying Euclid’s Division Lemma
148 = 124 × 1 + 24
Taking the new divisor 124 and new remainder 24 applying Euclid’s Division Lemma
124 = 24 × 5 + 4
Taking the new divisor 24 and new remainder 4 applying Euclid’s Division Lemma 24 = 4 × 6 + 0
Here the remainder has become zero. So procedure stops here. Since the divisor at this stage is 4, therefore the HCF of 12576 and 4052 is 4.

Question 2.
Prove that \(\sqrt{2}\) is an irrational number.
Solution:
Let us assume, to the contrary, that \(\sqrt{2}\) is a rational number.
So, we can find two integers r and s such that \(\sqrt{2}\) = \(\frac{r}{s}\) and s(≠ 0).
Suppose r and s have no common factor other than 1. Then, we divide r and s by this common factor to get \(\sqrt{2}\) = \(\frac{a}{b}\), where a and b are coprime.
So, b\(\sqrt{2}\) = a
Squaring on both sides and rearranging, we get
2b² = a²
Therefore, 2 divides a².
Therefore by theorem 1.3, 2 will divide a. So, we can write a = 2c, where c is some integer.
Substituting the value of a
2b² = 4c², i.e., b² = 2c²
This means that 2 divides b², and so 2 divides b (again using theorem 1.3 with p = 2) Therefore a and b have at least one common factor 2. But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our in correct assumption that \(\sqrt{2}\) is a rational number.
So, we conclude that \(\sqrt{2}\) is an irrational number. Hence Proved

Question 3.
Prove that \(\sqrt{3}\) is an irrational number.
Solution:
Let us assume, to the contrary, that \(\sqrt{3}\) is a rational number, i.e., we can find two integers a and b (≠ 0) such that\(\sqrt{3}\) = \(\frac{a}{b}\)
If a and b have a common factor other than 1, the we can make a and b coprime by dividing by that common factor.
So, b\(\sqrt{3}\) = a
Squaring on both sides and rearranging.
we get, 3b² = a²
Therefore, a² is divisible by 3. Therefore a in also divisible by 3. So. we can write a = 3c for some- integer c. Substituting main value of a in 3b² = a²,
3b² = 9c²
i.e., b² = 3c²
This means that b² is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least one common factor 3.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{3}\) is a rational number.
So. we conclude that A/3 is an irrational number. Hence Proved

Question 4.
Show that 5 – \(\sqrt{3}\) is an irrational number.
Solution:
Let us assume, to the contrary, that 5 – \(\sqrt{3}\) is a rational number, i.e., we can find coprime a and b (b ≠ 0) such that
5 – \(\sqrt{3}\) = \(\frac{a}{b}\)
Therefore, 5 – \(\frac{a}{b}\) = \(\sqrt{3}\)
Rearranging this equation, we get
\(\sqrt{3}\) = 5 – \(\frac{a}{b}\) = \(\frac{5 b-a}{b}\)
Since a and b are integers, therefore 5 – \(\frac{a}{b}\) is a rational number, i.e., \(\sqrt{3}\) is a rational number.
But this contradicts the fact that \(\sqrt{3}\) is an irrational number.
This contradiction has arisen because of our incorrect assumption that 5 – \(\sqrt{3}\) is a rational number.
So, we conclude that 5 – \(\sqrt{3}\) in an irrational number. Hence Proved

Question 5.
Show that 3\(\sqrt{2}\) is an irrational number.
OR
Prove that 3\(\sqrt{2}\) is an irrational number.
Solution: Let us assume, to the
contrary that 3 A/2 is a rational number, i.e., we can find coprime numbers a and b (b ≠ 0) such that
3\(\sqrt{2}\) = \(\frac{a}{b}\)
Rearranging, we get \(\sqrt{2}\) = \(\frac{a}{3b}\)
Since 3, a and b are integers, therefore \(\frac{a}{3b}\) is a rational number. Therefore \(\sqrt{2}\) wall also be a rational number.
But this contradicts the fact that \(\sqrt{2}\) is an irrational number.
So, we conclude that 3\(\sqrt{2}\) is an irrational number.

Question 6.
Prove that 2 + \(\sqrt{5}\) is an irrational number.
Solution:
Let 2 + \(\sqrt{5}\) be a rational number.
∴ 2 + \(\sqrt{5}\) = \(\frac{p}{q}\)
(Where p and q both are integers and q ≠ 0)
⇒ \(\sqrt{5}\) = \(\frac{p}{q}\) – 2
Since p and q both are integers, therefore \(\frac{p}{q}\) – 2 is a rational number.
So \(\sqrt{5}\) is also a rational number.
But this contradicts the fact that \(\sqrt{5}\) is an irrational number.
Therefore our assumption that 5 + \(\sqrt{5}\) is a rational number, will be incorrect.
Hence 2 + \(\sqrt{5}\) is an irrational number. Hence Proved

Question 7.
What is Euclid’s Division Lemma? Using this find the H.C.F. of the numbers 196, 3820.
Solution:
Euclid’s Division Lemma— Given two positive integers a and b, there exist unique whole numbers q and r satisfying a = bq + r, 0 ≤ r < b Euclid’s Division Algorithm in is based on this lemma. To Find the HCF of 196 and 3820— [Using Euclid’s Division Lemma] Step I : ∵ 3820 > 196 So, applying
Euclid’s Division Lemma on 3820 and 196
3820 = 196 × 19 + 96

Step II : ∵ Remainder 96 ≠ 0, therefore
now applying Euclid’s Division Lemma on 196 and 96
196 = 96 × 2 + 4

Step III : ∵ Remainder 4 ≠ 0, therefore now applying Euclid’s Division Lemma on 96 and 4
96 = 4 × 24 + 0
The remainder has now become zero, so this procedure stops. The division in step III is 4. Hence H.C.F. of 196 and 3820 is 4.

Question 8.
Find the greatest positive integer that divides of 396, 436 and 542 and leaves a remainder 5, 11 and 15 respectively.
Solution:
It is given that on dividing 396 by the required number, there is a remainder of 7. Therefore 396 – 5 = 391 is exactly divisible by the required number, i.e., the required number is a factor of 396. Similarly, 436 – 11 = 425 and 542 – 15 = 527 are also exactly divisible by the required number. Since the required number is that greatest number which divides 391, 425 and 527 exactly. Hence the required number is the H.C.F. of 391, 425 and 527. Using factor tree, the required factors of 391, 425 and 527 are as folllws :
391 = 17 × 23
425 = 5 × 5 × 17 = 5² × 17
527 = 17 × 31
∴ HCF of 391, 425 and 527 is 17.

Question 9.
Find the H.C.F. and L.C.M. of the numbers 180, 72 and 252.
Solution:
Factors of 180 = 2 × 2 × 3 × 3 × 5
= 2² × 3² × 5
Factors of 72 = 2 × 2 × 2 × 3 × 3
= 2³ × 3²
and Factors of 252 = 2 × 2 × 3 × 3 × 7
= 2² × 3² × 7
Therefore H.C.F. of the given numbers
= 2² × 3²
= 4 × 9 = 36
and L.C.M. = 2³ × 3² × 5 × 7
=8 × 9 × 5 × 7
= 2520